Pesquisavel Aircraft Structures - D.J.peery

f J H f l U

AIRCRAFT STRUCTURES Second Edition

David J. Pecry, PI . .D.

' l.ate Professor of Aeronautical Engineering The Pennsylvania Stale University

J. J. Azar, Ph.D. Professor of Petroleum ami Mechanical Engineering '

University of Tulsa

McGraw-Hill Book Company New York Si. Louis San Francisco Auckland Bogota Hamburg

Johannesburg London Madr id Mexico Montreal New Delhi Panama Paris Sfto Paulo Singapore Sydney Tokyo Toronto

T O THE MEMORY O F

Dr. David Peery

This book was set in Times Roman by Santypc-Byrd. The editors were Diane D. Heibcrg and J. W. Maisel; the production supervisor was Charles Hess. New drawings were done by J & R Services, Inc. The covcr was designed by Kao & Kuo Associates. Ilalliday Lithograph Corporation was printer and binder.

AIRCRAFT STRUCTURES 2 O S O " J

Copyright ifr> 1982 by McGraw-Hill. Inc. AH rights reserved. Copyright 1950 by McGraw-Hill. Inc. All rights reserved. Printed in the United States of America. Except as permitted under the United States Copyright Act of 19.76, no part of this publication may be reproduced or distributed in any form or by any means, or stored in a data base or retrieval system, without the prior written permission of the publisher.

2 3 4 5 6 7 8 9 0 H D I I D 8 9 8 7 6 5 4 3

I S B N D - 0 7 - D 4 t J l e l t i - f l

Library of Congress Cataloging in Publication Data

Peery. David J. Aircraft structures.

Bibliography: p. Includes index. 1. Airframes. I- Azar, Jamal J., date

II. Title TLG7I.6.P4 1982 629.1 MM I 81-17196 ISBN 0-07-049196-8 AACR2

CONTENTS

Preface ix

Chapter I Static Analysis of Structures l 1.1 Introduction 1 1.2 Structural System 1 1.3 -- Load Classification 2

. /1 .4 Supports and Reactions 3 1.5 Equations of Static Equilibrium , 5 1.6 Statically Determinate and Indeterminate Structures 6 1.7 Applications 7

Problems 21

Chapter 2 Flight-Vehicle Imposed Loads 26 2.1 Introduction 26 2.2 General Considerations 26

/i.3 Basic Flight Loading Conditions 27 ^2.4 Flight-Vehiclc Aerodynamic Loads 31

•' 2.5 Flight-Vehicle Inertia Loads 34 ' 2 . 6 Load Factors for Translational Acceleration 37

2.7 Velocity-Load Factor Diagram 41 2.8 Gust Load Factors 43 2.9 Examples 46

Problems 58

Chapter 3 Elasticity of'Structures 62 3.1 Introduction 62 3.2 Stresses 62 3.3 Stress Equilibrium Equations in a Nonuniform Stress Field 65 3.4 Strains and Strain-Displacement Relationships 66 3.5 Compatibility Equations for Plane-Stress and

Plane-Strain Problems 69 3.6 Boundary Conditions 70

r i CONTENTS

3.7 Stress-Strain Relationships 71 3.8 Transformation of Stresses and Strains 73

Problems 75

Chapter 4 Behavior and Evaluation of Vehicle Material 78 4.1 Introduction 78 4.2 Mechanical Properties of Materials 79 4.3 Equations for Stress-Strain Curve Idealization 82 4.4 Fatigue 84

• 4.5 Strength-Weight Comparisons of Materials 89 4.6 Sandwich Construction 92 4.7 Typical Design Data for Materials 94

Problems 95

Chapter 5 Stress Analysis V 5.1 Introduction 97 5.2 Force-Stress Relationships 97 5.3 Normal Stresses in Beams 99 5.4 Shear Stresses in Beams 101 5.5 Shear Flow in Thin Webs 107 5.6 Shear Center 109 5.7 Torsion of Closed-Section Box Beams 112 5.8 Shear Flow in Closed-Section Box Beams 113 5.9 Spanwise Taper Effect 118

5.10 Beams with Variable Stringer Areas 122 5.11 Airy Stress Function 126

Problems 130

Chapter 6 Deflection Analysis of Structural Systems 139 6.1 Introduction 139 6.2 Work and Complementary Work: Strain and Complementary

Strain Energies 140 6.3 Principle of Virtual Displacements and Related Theorems 140 6.4 Principle of Virtual Forces and Related Theorems 144 6.5 Linear F.laslic Structural Systems 147 6.6 Castigliano's Second Theorem in Deflection Analysis of Structures 150 6.7 Rayleigh-Ritz Method in Deflection Analysis of Structures 156 6.8 Finite Difference Method in Deflection Analysis of Structures 158 6.9 Redundant Structures and the Unit-Load Method 165

6.10 Structures with Single Redundancy 167 6.11 Structures with Multiple Redundancy 173 6.12 Shear Lag 181

Problems 186

Chapter 7 Finite Element Stiffness Method in Structural Analysis 190

7.1 Introduction 190 7.2 Mathematical Model o f the Structure 191

CONTENTS vii

7.3 Element Discretization 191 7.4 Applications of Finite Element Matrix Method 192 7.5 Coordinate System 192 7.6 Forces, Displacements, and Their Sign Convention , 194 7.7 Stiffness Method Concept 197 7.8 Formulation Procedures for Element Structural Relationships 202 7.9 Element Shape Functions 207

7.10 F.lenient Stillness Matrices 209 7.11 From Element to System Formulations 211 7.12 Global Stiffness Matrices Tor Special Beam Elements 214

Problems 226

Chapter 8 Analysis of Typical Members of Semimonocoque Structures 233

8.1 Introduction 233 8.2 Distribution or Concentrated Loads to Thin Webs 233 8.3 Loads on Fuselage Bulkheads 238 8.4 Analysis of Wing Ribs 243 8.5 Shear Flow in Tapered Webs 246 8.6 Cutouts in Semimonocoque Structures 2-51 s

8.7 Shearing Deformations 260 8.8 Torsion of Box Beams 261 8.9 Elastic Axis or Shear Center t 263

8.10 Warping of Beam Cross Sections 266 8.11 Redundancy of Box Beams 269 8.12 Torsion ofMulticell Box Beams 270 8.13 Beam Shear in Multicell Structures 272

Problems 275

Chapter 9 Thermal Stresses 279 9.1 Introduction 279 9.2 Thermal Stress Problem: Philosophy 279 9.3 Formulation of Equations for Thermal Stress Analysis 281 9.4 Solution Methods for Thermoelastic Problems 284 9.5 Thermal Stresses in Unrestrained Beams with Temperature

Variation through the Depth Only 289 9.6 Thermal Stresses in Built-up Structures 291

Problems 295

Chapter 10 Design of Members in Tension, Bending, or Torsion 299

10.1 Tension Members 299 10.2 Plastic Bending 300 10.3 Constant Bending Stress 302

. 10.4 Trapezoidal Distribution of Bending Stress 304 10.5 Curved Beams 307 10.6 Torsion of Circular Shafts 310 10.7 Torsion of a Noncircular Shaft 311

¥111 CONTENTS

10.8 10.9

10.10 10.11

Chapter 11 11.1 11.2 11.3 11.4 11.5 11.6 11.7 11.8 11.9

11.10 11.11 11.12 11.13

11.14 11.15 11.16 11.17 11.18 11.19

Chapter 12 12.1 12.2 12.3 12.4 12.5

End Restraint of Torsion Members Torsional Stresses above the Elastic Limit Combined Stresses and Stress Ratios Failure Theories in Structural Design Problems

Buckling Design of Structural Members Beam-Deflection Equations Long Columns Eccentrically Loaded Columns Short Columns Column End Fixity Empirical Formulas for Short Columns Dimensionless Form of Tangent Modulus Curves Buckling of Isotropic Flat Plates in Compression Ultimate Compressive Strength of Isotropic Flat Sheet Plastic Buckling of Flat Sheet Nondimensional Buckling Curves Columns Subject to Local Crippling Failure Needham and Gerard Methods for Determining Crippling Stresses Curved Sheet in Compression Elastic Shear Buckling of Flat Plates Elastic Buckling of Curved Rectangular Plates Pure Tension Field Beams Angle of Diagonal Tension in Web Semitension Field Beams Problems

Joints and Fittings Introduction Bolted or Riveted Joints Accuracy of Fitting Analysis Eccentrically Loaded Connections Welded Joints Problems

316 317 320 325 327

328

328 330 331 332 334 339 342 344 349 353 357 358

362 364 368 370 373 377 380 387

389

389 390

.395 401 404 408

Appendixes A Moments of Inertia and Mohr's Circle B Matrix Algebra

409 434

References Index

446

450

PREFACE

The purpose of this book is twofold: (1) to provide the reader with fundamental concepts in the analysis and design of flight structures, and (2) to develop unified analytical tools for the prediction and assessment of structural behavior re-gardless of field application.

The text is written in a manner which allows the reader to develop a work-ing knowledge in both the classical and the modern computer techniques of structural analysis. In addition, it helps the reader develop a thorough under-standing of the important factors which must be considered in the design of structural components.

The scope covers areas that the authors feel are essential fundamentals from which the reader may progress to the analysis and design of more complex and larger structural systems. The definitions of structural systems and its con-stituents, loads, supports and reactions, (he concepts in statics, and the funda-mental principles of mechanics are covered in the first portion of the book. The basic elasticity relations (stress-strain, strain displacements, etc.), and material behavior and selection' are discussed. Load analysis of flight vehicles and the analysis and design of specific flight-vehicle structural components are presented. Fatigue analysis, thermal stress analysis, and instability analysis of structures are also included. Energy methods, finite-difference methods, and the stiffness matrix methods in the deflection analysis of structures arc provided. Numerous examples are solved in every section of the book to close the gap between the theoretical developments and its application in solving practical problems.

The b o o k is written for a first course in structural analysis and design. It is intended for senior-level students. The essential prerequisites are strength of ma-terials and a basic course in calculus. The book also serves as an excellent reference for the practicing engineer.

J. J. Azar

i x

C H A P T E R

ONE STATIC ANALYSIS OF

STRUCTURES

1.1 INTRODUCTION

The full understanding of both the terminology in statics and the fundamental principles of mechanics is an essential prerequisite to the analysis and design of structures. Therefore, this chaptcr is devoted to the presentation and the appli-cation of these fundamentals.

1.2 STRUCTURAL SYSTEM

Any deformable solid body which is capable of carrying loads and transmitting these loads to other parts or the body is referred to as a structural system. The constituents of such systems are beams, plates, shells, or a combinat ion of the three.

Bar elements, such as shown in Fig. 1.1, are one-dimensional structural mem-bers which are capable of carrying and transmitting bending, shearing, torsional, and axial loads or a combination of all four.

Bars which arc capablc of carrying only axial loads are referred to as axial rods or two-force members. Structural systems constructed entirely out of axial rods arc called trusses and frequently are used in many atmospheric, sea, and land-based structures, since simple tension or compression members are usually the lightest for transmitting forces.

Plate elements, such as shown in Fig. 1.2, are two-dimensional extensions of bar elements. Plates made to carry only in-plane axial loads are called mem-

1

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viga

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chapa

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casca

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casca

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treliça - só tem carga axial (definição)

2 AIRCRAFT STRUCTURES

Figure 1.1 Bar elements, (a) Gen-eral ba r ; (/)) axial rod ; (c) tor-sional rod.

braues. T h o s e which are capable of carrying only in-plane shearing loads are referred to as shear panels; frequently these are found in missile fins, aircraft wings, and tail surfaces.

Shells are curved plate elements which occupy a space. Fuselages, building domes, pressure vessels, etc., are typical examples of shells.

1.3 LOAD CLASSIFICATION

Loads which act on a structural system may be generally classified in accordance with their causes. Those which are produced by surface contact are called surface loads. Dynamic and/or static pressures are examples of surface loads. If the area of contact is very small, then the load is said to be concentratcd; otherwise, it is called a distributed load. (Sec Fig. 1.3.)

Loads which depend on body volume are called body loads. Inertial, mag-netic, and gravitational forces are typical examples. Generally, these loads are assumed to be distributed over the entire volume of the body.

Loads a l so m a y be categorized as dynamic, static, or thermal. Dynamic loads arc time-dependent, whereas static loads are independent of time. Thermal loads arc created on a restrained structure by a uniform and/or nonuniform temper-ature change .

Regardless of the classification of the externally imposed loads, a structural member, in general , resists these loads internally in the form of bending, axial, shear, and torsional actions or a combinat ion of the four.

In order to present definitions for internal loads, pass a hypothetical plane so

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carga dinâmica - depende do tempo; estática - independe

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STATIC ANALYSIS Oh' STRUCTURES Z3

Figure 1.2 Plate elements, (a) General plate element; (fc) mem-brane element; (r) shear panel.

that it cuts the face of a structural member perpendicular to the member axis, as shown in Fig. 1.4. Thus, a bending moment may be defined as a force whose vector representation lies in and parallel to the plane of the cut, while a torque is a force whose vector representation is normal to that cut. On the other hand, shear load is a force which lies in and is parallel to the plane of the cut, while axial load is a force which acts normal to the plane of the cut.

1.4 SUPPORTS AND REACTIONS

The primary function of supports is to provide, at some points of a structural system, physical restraints that limit the freedom of movement to only that

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definição de momento de flexão e de torsão, e força cisalhante e axial

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limitações, condições de contorno

I w

\

"'i C o n c e n t r u l e d i

Di.slr ibutcd

l('2A;

r m

Figure 1 3 Concentrated and distributed loads. (<i) Actual loads: (IT) idealised loads; (r) wing pressure load.

intended in the design. The types of supports that occur in ordinary practice are s h o w n in Fig. 1.5.

The forces induccd at points of support are callcd reactions. For example, a hinge support is designed to al low only rotation at the point of connect ion, and

l ' — shear loads, lb.


-G

YEZ:

Pin

(ft)

- Roller

Knife edge

M

\

\ \

I K .

id)

Figure 1.5 Support types, (a) Hinge suppor t : {/>) hinge-roller support ; (r) fixed suppor t ; (dj fixed-roller support .

thus reactive forces (reactions) are developed in the other directions where move-ments are not allowed.

Likewise, a hinge-roller support allows rotation and a translation in only the .v direction, and hcnce there exists one reactivc force (reaction) in the y direction. A fixed support normally is designed to provide restraints against rotation and alKranslations; therefore, reactive forces and moments (reactions) are developed along the directions where movements are not permitted.

1.5 EQUATIONS O F STATIC EQUILIBRIUM

One of the first steps in the design of a structural system is (he determination of internal loads acting on each system member. Any solid body in space or any part cut out of the body is said to be in a state of stable static equilibrium if it simultaneously satisfies:

I F , = 0 I M ; = 0

( U )

where I F , = 0 implies that the vector sum of all forces acting on the system or on part of it must add to zero in any one direction along a chosen set of system axes x, y, and z; likewise, E Mt = 0 means that the vector sum of all moments at

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permite rotação

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permite rotação e translação em uma direção

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any preselected point must add to zero around any one of the chosen set of axes x, y, and z.

While F.q. (1.1) applies for general space structures, for the case of planar-type structures it reduces to

Z F i = 0 £ M, = 0 0" = J') (1.2)

N o t e that only six independent equat ions exist for any free body in space and three independent equations exist for a coplanar free body. If, for example, an attempt is made to find four unknown forces in a coplanar free body by using the two force equat ions and moment equat ions at two selected points, the four equat ions cannot be solved because they are not independent (i.e., one of the equat ions can be derived from the other three). The fol lowing equat ions cannot be solved for the numerical values of the three unknowns because they are not independent:

f i + F , + F 3 = 3

F , + F 2 + 2 F y = 4

2 F , + 2 F 2 + 3 F 3 7

In matrix form these are as fol lows.

" i l l Fi = ~ 3 " 1 1 2 F2 = 4 2 2 3 F}_ = _ 7 _

The third equat ion may be obtained by adding the first two equations , and consequent ly it does not represent an independent condition. The dependence of these equat ions is more readily established if an attempt is made to find the inverse of the 3 x 3 matrix of coefficients. It can be shown easily that the deter-minant of the matrix is zero; hence equat ion dependence does, in fact, exist. A matrix w h o s e determinant is zero is said to be singular and therefore cannot be inverted.

1.6 STATICALLY DETERMINATE AND INDETERMINATE STRUCTURES

A structure is said to be determinate if all its external reactions and the internal loads on its members can be obtained by utilizing only the static equa-tions of equilibrium. Otherwise the structure is said to be statically indeterminate. In the latter, or what is c o m m o n l y referred to as a redundant structure, there are more u n k n o w n forces than the number of independent equations of statics which can be utilized. The additional equat ions required for the analysis of redundant structures can be obtained by considering the deformations (displacements) in the

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Figure 1.6 Indeterminate structures.

structure. This is studied in detail in later chapters. External reactions, internal loads, or a combination of both may cause a structural system to be statically redundant. The number of redundancies in a structure is governed by the number of external reactions and/or the number of members that may be taken out without the stability of the structure being affected. For example, in Fig. 1.6a if member 1-3 and/or reaction R l x is removed, then the structure becomes stati-cally deterniinate and maintains its stability. Likewise, in Fig. 1.6b if members 2 -4 and 2-6 and either reaction R2y. or R3y are removed, then the structure bccomes determinate and stable. If, on the other hand, additional members such as 3-5 and 1-5 are removed, then the structure becomes a mechanism or un-stable. Mechanisms cannot resist loads and therefore are not used as structural systems.

1.7 APPLICATIONS

While the set of [Eqs. (1.1) and (1.2)] is simple and well known, it is very important for a student to acquire proficiency in the application of these equa-tions to various types of structural systems. Several typical structural systems are analyzed as illustrative examples.

Example 1.1 Find the internal loads acting on each member of the structure shown in Fig. 1.7.

SOLUTION First, disassemble the structure as shown in Fig. 1.8 and make a free-body diagram for each member. Since members 1-2 and 4—6 are two-force members (axial rods), the forces acting on them are along the line joining the pin joints of these members. All directions of forces are chosen arbitrarily and must be reversed if a negative value is obtained; that is, F3x is assumed to act to the right on the horizontal member and therefore must act to the left since its magnitude came to be negative.


Figure 1.7

F o r t h e p u l l e y

I M , = o ( + r v )

1000 x 2 — IT = 0

T = 10001b

XFy = 0{±*)

s \ - / \ \ W \ \ W W W \ W '-.V 1', \ W W W W W V

A , A ,

r imi rc 1.8 Disassembled structure.


Flx- 1000 = 0

Flx = 1000 lb

ZF, = 0(+T) Fly- 1000 = 0

Fly = 1000 lb

For member 3-6-7,

£Af3 = 0 ( +

, 1000 x 7 - 2.4F4 = 0

F 4 = 2915 lb

S F , = 0 ( ±>)

F3x + 2915 cos 36.9° - 1000 = 0

Fix = - 1 3 3 5 lb

IFy = 0 (+ f )

Fiy + 2915 sin 36.9° - 1000 = 0

Fis.= - 7 5 0 lb

Since the magnitudes of F3x and F3j. came to be negative, the assumed direc-tion must be reversed. A common practice is to cross out {rather than erase) the original arrows.

For member 2-3-4-5,

2 M s = 0 ( + r v )

1335 x 5 - 2915 x 2 cos 36.9° - 4F, = 0

F, = 500 lb

I Fx = 0 ( ±* )

Fix - 2915 cos 36.9" + 1335 - 500 cos 60° = 0

FSx - 1250 lb

ZFy = 0 I + t)

FSy - 2915 sin 36.9" + 750 - 500 sin 60" = 0

F s , = 1433 lb

N o w all internal loads have been obtained without the use of the entire structure as a free body. The so lut ion is checked by applying the three


equations of equilibrium to the entire structure:

1250 - 1000 - 500 cos 60° = 0

SF, = 0 ( + t)

1433 - 1000 - 500 sin 60° = 0

r w 5 = o ( + r > )

1000 x 9 - 1000 x 7 - 500 x 4 = 0

This equilibrium check should be made wherever possible to detect any errors that might have occurred during the analysis.

Example 1.2 Find the internal load in member 5 of the coplanar truss struc-ture shown in Fig. 1.9.

SOLUTION Several methods are available for analyzing truss structures; two are discussed and applied in solving this example. (a) Method of joints. In the analysis of a truss by the method of joints, the

two equations of static equilibrium, TFX = 0 and ~LFy = 0, are applied for each joint as a free body. Two unknown forces may be obtained for each joint. Since each member is an axial rod (two-force member), it exerts equal and opposite forces on the joints at its ends. The joints of a truss must be analyzed in sequence by starting at a joint which has only two members meeting with unknown forces. Then the joints are analyzed in the proper sequence until all joints have been considered, if necessary.

To find the unknown reactions, consider the entire structure as a free body.

E M 4 = 0 ( + r \ )

2000 x 10 + 4000 x 10 +• 1000 x 30 - 20R6v = 0 6 >•

R6V = 4500 lb

4000 lb

2000 lb [ ( ? ) ' 2 ® 3

1000 lb

Figure 1.9 Truss structure.

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STATIC ANAI.YSIS Ol STRUCTUR1S I 1

Z F ^ 0 ( i > )

2000 ~ /? 4 j t .

R4i. = 2000 lb

IF,. = 0 ( + t)

K4v - 4000 - 1000 + 4500 = 0

R ^ = 500 lb

The directions of unknown forces in each member are assumed, as in the previous example, and vectors are changed on the skctch when they are found to be negative. (Sec Fig. 1.10.)

Isolating joint 4, we have

IFX = 0 ( + )

/•', - 2000 = 0

F2 = 2000 lb

ZFy = 0 ( + T)

500 - F, = 0

F, = 500 lb

Isolating joint 1 gives

IF,. = 0 ( + T)

500 - F 4 sin 45° = 0

F 4 = 707 lb

1FX = 0(4-2000 + 707 cos 45°

F3 = 2500 lb

2000

4 ,

500

2000 —

500

F , = 0

1000

Figure 1.18


4000 lb

20001b 1 2 3

1000 Ih 2000 lb

500 lb 4500 lb

Figure L I (

Finally, isolating joint 5 to obtain the load in member 5, we get

ZF, = 0( + T) 707 sin 45° - Fs = 0

Fs = 500 lb 200C

707

Arrows acting toward a joint indicate that a member is in com-pression, and arrows acting away from a joint indicate tension.

(b) Method of sections. As in (his example, often it is desirable to find the internal loads in certain selected members of a truss without analyzing the entire truss. Usually the method of joints is cumbersome in this case, since the loads in all the members to the left or right of any member must be obtained before the force is round in that particular member. An analysis by the method of sections will yield the internal load in any preselected member by a single operation, without the necessity of finding loads in the other members. Instead of considering the joints as free bodies, a cross section is taken through the truss and the part of the truss on either side is considered as a free body. The cross section is chosen so that it cuts the members for which the forces are desired and preferably only three members.

In our example, the interna! load in member 5 is desired; the free body is as shown in Fig. 1.11. The load in member 5 may be found by summing forces in the y direction on either part of the cut truss. Con-sidering the left part as a free body, we get

Example 1.3 Analyze the structural system shown in Fig. 1.12.

SOLUTION Quite often, structural systems are made up largely of axial rods but contain some members which are loaded laterally, as shown in Fig. 1.12.

= 0 ( + T)

500 - Fs = 0

Fs = 500 lb


20 in of* 20. in-

These structures usually are classified as trusses, since the analysis is similar to that used for trusses. As shown in Fig. 1.12, members 1-2 and 2-3 are not axial rods, and separate free-body diagrams for these members, as shown in Fig. 1.13a and b, are required. Since each of these members has four un-known reactions, the equations of static equilibrium are not sufficient to find all four unknowns. It is possible, though, to find the vertical reactions

= Rzy = R2= RIY = 100 lb and to obtain the relations RLX = R2X and R2,X = RIX from the equations of equilibrium.

When the unknown reactions obtained from members 1-2 and 2-3 are applied to the remaining part of the structure as a free body (Fig. 1.136), it is apparent that the rest of the analysis is similar to that of the previous examples. Ail members except 1-2 and 2-3 may now be designed as simple tension or compression members. The horizontal members (1-2 and 2-3) must be designed for bending combined with axial and shear loads.

Example 1—4 Find the reactions at supports A, B, and C, for the landing gear of Fig. 1.14. Members OB and OC are two-force members. Member OA resists bending and torsion, but point A is hinged by a universal joint so that the member can carry torsion but no bending in any direction at this point.

SOLUTION First consider the components of the torsional reaction at point A. The resultant torsional vcctor 'l'A, shown in Fig. 1.15b, must be along the member, and it has components TAX and TAY.

Any vector force Q acting in space may be resolved into components along the chosen set of axes by the method of direction cosines.

Qx = Q cos a

Q, = Q cos fi

QZ = Q cos Y (1.3)

10 lb/ in 10 Ill/in

— — * - (

«2,-

!)-<

* .V (a)

Figure 1.13

Here a, /?, and y are the angles between the .\\ y, and z axes and the line of action of the force vector Q, respectively.

Thus, from Fig. 1.15b, utilizing Eq. (1.3), we have

TAY=TA COS ft = T a (F§) = 0 . 8 7 ^

TAX = Ta cos CL = TA (f§) = 0 . 6 ^

TAZ = TA cos y = TA (&) = 0

In the free-body diagram for the entire structure, shown in Fig. 1 . 1 5 a ,

there are six unknown reaction forces. The six equations of static equilibrium are just sufficient to determine these unknown forces. Taking moments about an axis through points A and B gives

'EMJIN = 4000 x 36 - 0.8 F(- x 30 = 0

FC = 6000 lb

UM'-J I J

4 0 in

4000 lb

Figure 1.14

T h e torsional m o m e n t TA m a y be found by taking m o m e n t s about line 0A. N o t e that all u n k n o w n forces act through this line.

XMa0 = 1000 x 4.8 - T a = 0

TA = 4800 in • lb

Figure 1.15

r


The other forces are obtained from the following equations, which are chosen so that only one unknown appears in each equation.

ZM 0 X = 2880 - 40F a z = 0

FAX = 7 2 l b

The subscript "0X" designates an axis through point 0 in the .v direction.

I M , = 1000 + 12 - 6000 x 0.6 + 0.371 F'„ = 0

FB = 6820 lb

Y.FX = FAX - 6820 x 0.557 = 0

FAX = 3800 lb

ZFV = 4000 + 6820 x 0.743 - 6000 x 0.8 - F A Y = 0

FAR = 4270 lb

Check: ZMAY = - 1 0 0 0 x 36 + 6000 x 0.6 x 30 - 6820 x 0.557

x 20 + 3840 = 0

Example 1.5 Find the internal loads on all members of the landing gear shown in Fig. 1.16.

SOLUTION For convenience, the reference axes V, D, and S are taken as shown in Fig. 1.16 with the V axis parallel to the oleo strut. Free-body diagrams for the oleo strut and the the horizontal member are shown in Fig. 1.17. Forces perpendicular to the plan of the paper are shown by a circled dot for forces toward the observer and a circled cross for forces away from

Figure 1.16

i ; s = (.300 lb

TF: = 41.720 in-lb _ ) 0 8 3 2 c r , , q,44o u, 0.707611 - .

= 9.030 lb

0.707811 11

y.'Jju ib'' rL

/•',.= 19,790 lb

/ '„ = 4740 lb

1-y = 1.9,790 Ib

r !-:s = (.300 |h -!•:„ = 4740 Ib

c j _ j ) r , , = 41.720 in-ih 9930 II.

1 0.555 CC, -<i.1IKIII.

5190 lb A

10,300 Ib

94401b (.300 lb

L

I v v U 111

tyZlmi=,4-n

5100 lb

M

19,300 Ib

(f) Figure 1.17

the observer. The V component of the 20,000-lb forcc is

20,000 cos 15° = 19,300 lb

The D component is

20,000 sin 15" = 5190 lb

The angle of the side-brace member CG with the V axis is

tan"1 H = 33.7°


The V and S components of the force in member CG are

CG cos 33.7° = 0.832CG CG sin 33.7" = 0.555CG

The drag-brace member BH is at an angle of 45° with the V axis, and the components of the force in this member along the V and D axes are

BH cos 45° = 0.707BM BH sin 45° = 0.707BH

The six unknown forces acting on the olco strul are now obtained from the following equations:

ZM £ V = 51?0 x 8 - 7e = 0

Te = 41,720 in • lb

I M £ j = 5190 x 44 - 0.707BH x 20 - Q.1Q1BH x 3 = 0

BH = 14,050 lb

0.707BH = 9930 lb

ZMED = 0.555CC x 20 + 0.832CG x 3 - 19,300 x 8 = 0

C G = 11,350 lb

0.555CG = 6300 lb

0.832CG = 9440 lb

ZF„ = 19,300 + 9930 - 9440 - E, = 0

Ev = 19,700 lb

ZF, = £ , - 6300 = 0

Es = 6300 lb

ZFd = —5190 H- 9930 ~ Ed = 0 i

E, = 4740 lb / The horizontal member IJ is now considered as a free body. The forces

obtained above are applied to this member, as shown in Fig. 1.17c and d, and the five unknown reactions are obtained as follows:

EF, = / , = 0

ZMI0 = 19,790 x 3 + 9440 x 18 + 6300 2 - 20J r = 0

Jv= 12,100 lb

ZF,. = 19,790 + 9440 - 12,100 - /„ = 0

IB= 17,130 lb

STATIC ANAI.YSIS Ol STRUCTUR1S I 1

ZMIV = 41,720 - 4740 x 3 + 20JD = 0

JD= - 1 3 7 5 lb

E F j = 4740 + 1375 - I d = 0

/«, = 6115 lb

The reactions are now checked by considering the entire structure as a Tree body, as shown in Fig. 1.17e.

EF„ = 19,300 - 17,130 - 12,100 + 9930 = 0

- 5 1 9 0 + 1375 - 6115 + 9930 = 0

ZFS = 0

Y.M lv = 5190 x 11 - 1375 x 20 - 9930 x 3 = 0

IJV/ lo = 19,300 x 11 - 12,100 x 20 + 9930 x 3 = 0

Z.W„ = 5190 x 44 - 9930 x 23 = 0

Example 1.6 Find t h e loads on the lifl and drag-truss members of the e x t e r -

nally braced monoplane wing shown in Fig. 1.18. The air load is assumed to be uniformly distributed along the span of the wing. The diagonal drag-truss members are wires, with the tension diagonal effective and the other diagonal carrying no load.

SOLUTION The vertical load of 20 lb/in is distributed to the spars in inverse proportion to the distance between the center of pressure and the spars. The

.4. c

30 in ' JU in -40 in ->-j"*-40 in-»j-*-40 in

.100 in_ - 1 8 0 i n - 20 lli/i»

.'I. II

Figure 1.18

20 lb/in

• H U M n m m n 5 lb / in JO in

20 AIRCRAFT STRUCTLIRRS

load on the front spar, is therefore 16 lb/in, and that on the rear spar is 4 lb/in. If the front spar is considered as a free body, as shown in Fig. 1.19a, the vertical forces at A and G may be obtained.

•£M a = - 16 x 180 x 90 + 100G. = 0

G, = 2590 lb

Gj_ 100

2590 60

G/= 4320 lb

2FZ = 16 x 180 - 2590 - AR = 0

A, = 290 lb

Force AY cannot be found at this point in the analysis, since the drag-truss members exert forces on the front spar which are not shown in Fig. 1.19a.

If the rear spar is considered as a free body, as shown in Fig. 1.196, the

-180 in-

16 lb/ill - 1 0 0 i n -

(2880 lb)

• M • • M M •

290 lb

(a)

U, = 2590 lb

G> = 4320 lb

II. = 12 I f Ey -

4 lb/in (720 lb)

* • I I * H I I * M

1080 lb N !•:, = 324 lb N h~ = 648 lb

75 1b 150 1b 10801b

175 lb \ 200 1b 200 lb 2700 lb

1001b

1225 lb I f ) Figure 1.19


vertical forces at B and E may be obtained:

XMB = - 4 x 180 x 90 -I- 100£j = 0

E. = 648 lb

EX ER 648 30 ~ 100 ~ 60

EX = 324 lb = 1080 lb

EF. = 4 x 180 - 648 - B. = 0

Br = 72 lb

The loads in the plane of the drag truss can be Obtained now. The forward load of 5 lb/in is applied as concentrated loads at the panel points, as shown in Fig. 1.19c. The components of the forces at G and E which lie in the plane of the truss also must be considered. The remaining reactions at A and B and the forces in all drag-truss members now can be obtained by the methods of analysis for coplanar trusses, shown in Fig. 1.19c.

1.1 A 5000-lb airplane is in a steady glide wilh Ihc (light palh at an angle (I below Ihc horizontal (see i 'ig. I'1.1). The drag for te in (he direction of the llighl palh is 750 lb. Kind the lift force L normal lo the flight palh and Ihe angle 0.

1.2 A jet-propelled airplane in steady (light has forces acling as shown in l-'ig. PI.2. l ind the jet thrust T, lift L, and the tail load P.

PROBLEMS

Flight path Figure PI . I

L . P

150 in — -6 in-

8000 lb Figure PI.2


1.3 A wind-tunnel model of an airplane wing is suspended as shown in l'ig. I'1.1 and P1.4. I-ind the loads in members B, C, and £ tf the forces at A are L = 43.8 i b , l) -- 3.42 lb, and M = - 2 0 . 6 in • Ih 1.4 For (he load of Prob. 1.3, find ihe forces I.. D, and M at a point A if the measured forces are fl = 40.2, C = 4.16, and £ = 3.74 lb.

B C

1.5 Find the forces at points A and B of I he landing gear shown iri Fig. PI.5.

1.6 Find the forces at points A, II, and C of (he structure of (he hraccd-wing monoplane shoH?n in F i t PI.6.

Figure PI .6

1.7 Find (he forces I ' and M at the cut cross section of the beam shown in Fig. PI.7.


4000 ( in-lb I

— 3 0 in-iOifa/in

±LU

200 lb 20 in-1

^ \ M

300 1b ' Figure PI.7

1.8 Find the internal loads in all members of the truss slructure shown in Fig. PI.8.

1 0 0 0 l b

(a)

1 0 i n 10 in

Ch)

1(100 111 2000 lit 6 0 0 I t )

800 Ih

1000 lb

Figure P l .8

1.9 Find the internal loads on all members of the fuselage truss structure shown in Fig. PI.9.

JD -O JZ> J3

15 ill

15 in

Figure P1.9

1.10 All members of the structure shown in Fig. PI .10 are two-force members, except member ABC. Find the reactions on member ABC and the loads in other members of the structure.

1.11 The bending moments about the x and z axes in a plane perpendicular to the spanwise axis of a wing are 400,000 and 100,000 in-lb, as shown in Fig. l ' l .11. Find the bending moments about the .V, and 2, axes which are in the same plane but rotated 10° counterclockwise.

1.12 The main beam of the wing shown in Fig. PI.12 has a swcepback angle of 30°. First the moments of 300,000 and 180,000 in • lb are computed about the x and y axes which are parallel and perpendicular to the centerline of the airplane. Find the moments about the x' and y' axes.

1.13 Find the forces acting on all members of the nose-wheel structure shown in Fig. P1.13. Assume the V axis is parallel to the oleo strpt.

Figure PI .12 Figure PI .13


1.14 Analyze (he landing gear structure of Example 1.5 for a 15,000-lb load up parallel to the V axis and a 5000-lb load aft parallel to the D axis. The loads are applied at the same point of the axle as the ioad in Example 1.5. 1.15 Write a computer program to calculate the reactions of the beam structure shown in Fig. PI.15.

i 1 1 1

- i r

Figure P1.15

1.16 Write a computer program to calculate the internal shear and bending moments al every station of the cantilevered wing shown in Fig. PI.16. Assume the center of pressure is at 25 percent of the chord length measured from the x axis.

Figure P I . 16

C H A P T E R

TWO FLIGHT-VEHICLE IMPOSED LOADS

2.1 INTRODUCTION

Before the final selection of member sizes on flight vehicles can be made, all load conditions imposed on the structure must be known. The load conditions are those which are encountered both in flight and on the ground. Since it is impos-sible to investigate every loading condition which a flight vehicle might encounter in its service lifetime, it is normal practice to select only those conditions that will be critical for every structural member of the vchicle. These conditions usually are determined from past investigation and experience and are definitely specified by the licensing or procuring agencies.

Although the calculations of loads imposed on flight-vehicle structures are the prime responsibility of a special group in an engineering organization called the loads group, a basic general overall knowledge of the loads on vehicles is essential to stress analysts. Therefore, in I his chapter we present the fundamentals and terminology pertaining to flight-vehicle loads.

2.2 GENERAL CONSIDERATIONS

Every flight vchicle is designed to safely carry out specific missions. This results in a wide variety of vehicles relative to size, configuration, and performance. Com-mercial transport aircraft are specifically designed to transport passengers from one airport to another. These types of aircraft arc never subjected to violent intentional maneuvers. Military aircraft, however, used in fighter or dive-bomber operations, are designed to resist violent maneuvers. The design conditions u-

2 6

FLIGHT-VF.H1CLF. IMPOSED LOADS 2 7

sually are determined from the maximum acceleration which the human body can withstand, and the pilot will lose consciousness before reaching the load factor (load factor is related to acceleration) which would cause structural failure of the aircraft.

To ensure safety, structural integrity, and reliability of flight vehicles along with the optimality of design, government agencics, both civil and military, have established definite specifications and requirements in'regard to the magnitude of loads to be used in structural design of the various flight vehicles. Terms are defined below which arc generally used in the specification of loads on flight vehicles.

The limit loads used by civil agencies or applied loads used by military agencies are the maximum anticipated loads in the entire service life-span of the vehicle. The ultimate loads, commonly referred to as design loads, are the limit loads multiplied by a factor of safety (FS):

ultimate load FS =

limit load

Generally, a factor of safety which varies from 1.25 for missile structures to 1.5 for aircraft structures is used in practically every design because of the uncertainties involving

1. The simplifying assumption used in the theoretical analyses 2. The variations in material properties and in the standards of quality control 3. The emergency actions which might have to be taken by the pilot, resulting in

loads on the vehicle larger than the specified limit loads.

The limit loads and ultimate loads quite often are prescribed by specifying certain load factors. The limit-load factor is a factor by which basic loads on a vehicle are multiplied to obtain the limit loads. Likewise, the ultimate load factor is a factor by which basic vehicle loads are multiplied to obtain the ultimate loads; in other words, it is the product of the limit load factor and the factor of safety.

2.3 BASIC FLIGHT LOADING CONDITIONS

One_QF four .basic conditions will probably produce the highest load in any part of the airplane for.aay flight condition. Usually these conditions arc railed posi-tive high angle of attack, positive low angle of attack, negative high angle of attack, and negative low angle of attack. All these conditions represent symmetri-cal flight maneuvers; i.e., there is no motion normal to the plane of symmetry of the airplane.

The positive high angle of attack (PHAA) condition is obtained in a puliout at the highest possible angle of attack on the wing. The lift and drag forces are

perpendicular and parallel respectively, to the relative wind, which is shown as horizontal in Fig. 2.1a. The resultant R of these forces always has an aft com-ponent with respect to the relative wind, but will usually have a forward com-ponent C with respect to the wing chord line, because of the high angle of attack a. The maximum forward component C will be obtained when A has a maximum value. In order to account for uncertainties in obtaining the stalling angle of attack under unsteady flow conditions, most specifications arbitrarily require that a value or a be used which is higher than the wing stalling angle under steady How conditions. An angle of attack corresponding to a coefficient of lift of 1.25 times the maximum coefficient of lift for steady flow conditions is often .used,_and_

"acroclynamic data are extrapolated from data measured for steady flow condi-tions. Experiments show that these high angles of attack and high lift coefficients may be obtained momentarily in a sudden pull-up before the airflow reaches a steady condition, but it is difficult to obtain accurate lift measurements during the unsteady conditions.

In the PHAA condition, the bending moments from the normal forces N, shown in Fig. 2.1 a, produce compressive stresses on the upper side of the wing, and the moments from the chordwise forces C produce compressive stresses on the leading edge of the wing. These compressive stresses will be additive in the upper flange of the front spar and the stringers adjacent to ii. The I*HAX condi-tion, therefore, will be critical for compressive stresses in the upper forward

(a) PHAA (/)) 1'I.AA

(t) NIIAA

Figure 2.1


region of the wing cross section and for tensile stresses in the lower aft region of the wing cross section. For normal wings, in which the aerodynamic pitching-moment coefficient is negative, the line of action of the resultant force R is farther forward on the wing in the PHAA condition than in any other possible flight attitude producing an upload on the wing. The upload on the horizontal tail in this condition usually will be larger than for any other positive flight attitude, since pitching accelerations are normally neglected and the load on the hor-izontal tail must balance the moments of other aerodynamic forces about the center of gravity of the airplane.

In the positive low angle of attack (PLAA) condition, the wing has the smallest possible angle of attack at which the lift corresponding to the limit-load factor may be developed. For a given lift on the wing, the angle of attack decreases as the indicated airspeed increases, and consequently the PLAA condi-tion corresponds to the maximum indicated airspeed at which the airplane will dive. This limit on the permissible diving spe'ed depends on the type of aircraft, but usually is specified as 1.2 to 1.5 times the maximum indicated speed in level flight, according to the function of the aircraft. Some specifications require that the terminal velocity of the aircraft—the velocity obtained in a vertical dive sustained until the drag equals the airplane weight—be calculated and the limit on the diving speed be determined as a function of the terminal velocity. Even fighter aircraft are seldom designed for a diving speed equal to the terminal velocity, since the terminal velocity of such airplanes is so great that difficult aerodynamic and structural problems are encountered. Aircraft are placarded so that the pilot will not exceed the diving speed limit.

In the PLAA condition, shown in Fig. 2.1 b, the chordwise force C is the largest force acting aft on the wing for any positive flight a t t i t u d e j h e wing bending moments in this condition produce the roaximuiiLiXtniPxessivf... stresses on the upper rear spar flange and adjacent, stringers...and maximum tensile, stresses on the lower front spar flange and adjacent stringers. In this condition, thelmFoTaction of theTesultant wing force R is farther aft than for any other positive flight condition. The moment of this force about the centerofjravity of the airplane has the maximum negative (pitching) value; consequently, the down-load on the horizontal tail required to balance the moments of other aerody-namic forces will be larger than for any positive flight condition.

The negative high angle of attack (NHAA) condition, shown in Fig. 2.1c, occurs in intentional flight maneuvers in which the air loads on the wing are down or when the airplane strikes sudden downdrafts while in level flight. The load factors for intentional negative flight attitudes are considerably smaller than for positive flight attitudes, because conventional aircraft engines cannot be oper-ated under a negative load factor for very long and because the pilot is in the uncomfortable position of being suspended from the safety belt or harness. Gust load factors are also smaller for negative flight attitudes, since in level flight the weight of the airplane adds to the inertia forces for positive gusts but subtracts from the inertia forces for negative gusts.

In the NHAA condition, usually the wing is^assumed to be at the negative

3 0 AIRCRAFT STRUCTURES

stalling angle of attack for steady flow conditions. The assumption used in the PHAA condition—the maximum lift coefficient momentarily exceeds that for steady (low—is seldom used because it is improbable that negative maneuvers will be entered suddenly. The wing bending moments in the negative high angle of attack condition produce the highest compressive stresses in the lower forward region of the wing cross section and the highest tensile stresses in the upper aft region of the wing cross section. The line of action of the resultant force R is farther aft than for any other negative flight attitude, and it will probably pro-duce (he greatest balancing upload on (he horizontal tail for any negative flight attitude. -

The negative low angle of attack (NLAA) condition, shown in Fig. 2Ad, occurs at the diving-speed limit of the airplane. This condition may occur in an intentional maneuver producing a negative load factor or in a negative gust condition. The aft ioad C is a maximum for any negative flight attitude, the compressive bending stresses have a maximum value in the lower aft region of the wing cross section, and the tensile bending stresses have a maxim.um value in the upper forward region of the wing cross section. The resultant force R is farther forward than in any other flight attitude, and the download on the hor-izontal tail will probably be larger than in any other negative flight attitude.

In summary, one of the four basic symmetrical flight conditions is critical for the design of almost every part of the airplane structure. Xn the stress analysis of a. S-OmmUaflaLadi ng, it is necessary to investigate each cross section "for each of the fgui^ggfflditk&aSi^Then each stringer or spar flange"is designed for the maximum* tension and the maximum compression obtained in any of the conditions. The probable critical conditions for each region of the cross section are shown in Fig. 2.2.

Some specifications require the investigation of additional conditions of medium-high angle of attack and medium-low angle of attack which may be critical for stringers midway between the spars, but usually these conditions are not considered of sufficient importance to justify the additional work required for the analysis. The wing, of course, must be strong enoujgh to resist loads at medium angles of attack, but normally the wing will have adequate strength if it meets the requirements for the four limiting conditions.

Figure 2.16

FLIGHT-VF.H1CLF. IMPOSED LOADS 31

For aircrafts such as transport or cargo aircrafts, in which the load may be placed in various positions in the gross-weight condition, it is necessary to deter-

m i n e the balancingjâil loads for the most forward and rn^êarwaja center-ot-gravity positions at vThic|i't te^ jaVrEj^ l ^â^ê^f wna I. Each of t'he"'fo"ur m'gfit concfitions must be investigaledforeacrexfreme*po^tTon of the" center_of gravity.|For smaller aircraft, in which the,.u3sMlQadcannqt be shjQed appreciably, there may be only one position of the center of gravity at the gross-weight condition. *To account lor^greafcf'"'Ealancing 'TairToIia-i "wliich may occiTr To?"anoTH'er location ot' the center of gravity, it may bis possible to makej5gme_ g m s e r v a ^ only one location.

The gust load factors on an aircraft are greater when it is flying at the minimum flying weight than they are at the gross-weight condition. While this is seldom critical for the wings, since they have less weight to carry, it is critical for a structure such as the engine mount which carries the same weight at a higher load factor. It is therefore necessary to calculate gust load factors at the minimum weight at which the aircraft will be flown.

For aircraft equipped with wing flaps, other high lift dcvices, or dive brakes, additional flight loading conditions must be investigated for the flaps extended. These conditions usually are not critical for wing bending stresses, since the specified load factors are not large, but may be critical for wing torsion,"shear in the rear spar, or down tail loads, since the negative pitching moments may be quite high. The aft portion of the wing, which forms the flap supporting structure, will be critical for the condition with flaps extended.

Unsymmetrical loading conditions and pitching-acceleration conditions for commercial aircraft are seldom of sufficient importance to justify extensive analy-sis. Conservative simplifying assumptions usually are specified by the licensing agency for use in the structural design of members which will be critical for these conditions. The additional structural weight required to meet conservative design assumptions is not sufficient to justify a more accurate analysis. Some military aircraft must perform violent evasive.maneuvers such as snap rolls, abrupt rolling pullouts, and abrupt pitching motions. The purchasing agcncy for such airplanes specifies the conditions which should be investigated. Such investigations require the calculation of the mass moment of inertia of the airplane about the pitching, rolling, and yawing axes. The aerodynamic forces on the airplane are calculatcd and set in equilibrium by inertia! forces on the airplane.

2.4 FLIGHT-VEHICLE AERODYNAMIC LOADS

Extensive aerodynamic information is required to investigate the per-formance, control, and stability of a proposed aircraft. Only the information which is required for the structural analysis is considered here, although normally this would be obtained as part of a much more extensive program. The first aerodynamic data required for the structural analysis are the lift, drag, and


pitching-moment force distributions for the complete aircraft with the horizontal tail removed, through the range of angles of attack from the negative stalling angle to the positive stalling angle. While these data can be calculated accurately for a wing with a conventional airfoil section, similar data for the combination of the wing and fuselage or the wing, fuselage, and nacelles are more difficult to calculate accurately from published information because of the uncertain effects of the aerodynamic interference of various components. It is therefore desirable to obtain wind tunnel data on a model of the complete airplane less horizontal tail. It is often necessary, of course, to calculate these data from published infor-mation in order to obtain approximate air loads for preliminary design purposes.

Wind tunnel tests of a model of the complete airplane with the horizontal tail removed provide values of the lift, drag, and pitching moment for all angles of attack. Then components of the lift and drag forces with respect to airplane reference axes are determined. The aircraft reference axes may be chosen as shown in Fig. 2.3. The force components are CzqS and CxqS along these axes, where q = pV2/2 is the dynamic pressure and S is the surface wing area. The nondimensional force coefficients C. and Cx are obtained by projecting the lift and drag coefficients, respectively, for the airplane less horizontal tail along the reference axes by the following equations:

The angle 0 is measured from the flight path to the x axis, as shown in Fig. 2.3, and is equal to the difference between the angle or attack a. and the angle of wing incidence i.

The pitching moment about the airplane's center of gravity is obtained from wind tunnel data and is C^^cqS, where Cma_l is the dimensionless pitching-moment coefficient of the airplane less tail and c is the mean aerodynamic chord of the wing. The mean aerodynamic chord (MAC) is a wing reference chord which usually is calculated from the wing planform. If every, airfoil section along

C- = C c o s 0 + CD sin 0

Cx = C„ cos 0 - CL sin 0

(2.1) (2.2)

•—— t

.V

Figure 1 3


the wing span has the same pitching-moment coefficient cm , the MAC is deter-mined so that the total wing pitching moment is cmcqS. For a rectangular wing planform the value of c (the MAC) is equal to the wing chord; for a trapezoidal planform of the semiwing, the value of c is equal to the chord at the centroid of the trapezoid. The MAC is actually an arbitrary length, and any reference length would be satisfactory if it were used consistently in all wind tunnel tests and calculations. For irregular shapes of planforms, some procuring or licensing agencies require that the mean chord (wing area divided by wing span) be used as the reference chord.

The balancing air load on the horizontal tail, C,qS, is obtained from the assumption that there is no angular acceleration of the airplane. The moments of the forces shown in Fig. 2.3 about the center of gravity are therefore in equilib-rium : , .

C.qSL, = C„^tcqS

or C, = j - C m _ (2.3)

where C, is a dimensionless tail force coefficient expressed in terms of the wing area and L, is the distance from the airplane's center of gravity to the resultant air load on the horizontal tail, as shown in Fig. 2.3. Since the pressure dis-tribution on the horizontal tail varies according to the attitude of the airplane, L, theoretically varies for different loading conditions. This variation is not great, and it is customary to assume L, contant, by using a conservative forward pos-ition of the center of pressure on the horizontal tail. The total aerodynamic force on the airplane in the z direction, CZa qS, is equal to the sum of the force C2 qS on the airplane less tail and the balancing tail load C,qS:

Cx,qS = C.qS + C,qS

or CZa = C. + C, (2.4)

For power-on flight conditions, the moment of the propeller or jet thrust about the center of gravity of the airplane should also be considered. This adds another term to the Eq. (2.3).

Now the aerodynamic coefficients can be plotted against the angle of attack a, as shown in Fig. 2.4. If more than one position of the center of gravity is considered in the analysis, it is necessary to calculate the curves forC„_ r, C,, and

for each ccntcr-of-gravity position. The right-hand portions of the solid curves shown in Fig. 2.4 represent the aerodynamic characteristics after stallihg of the wing. Since stalling reduces the air loads on the wing, these portions of the curves are not used. Instead, the curves are extrapolated, as shown by the dotted lines, in order to approximate the conditions of a sudden pull-up, in which high lift coefficients may exist for a short time. For the PHAA condition, the angle of attack corresponding to the force coefficient of 1.25 times the maximum value of C.o is used, and the curves are extrapolated to this value, as shown in Fig. 2.4.


2.5 FLIGHT-VEHICLE INERTIA LOADS

The maximum load on any part of a flight-vehicle structure occurs when the vehicle is being accelerated. The loads produced by landing impact, maneuvering, gusts, boost and staging operations, launches, and dockings are always greater than the loads occurring when all the forces on the vehicle are in equilibrium. Before any structural component can be designed, it is necessary to determine the inertia loads acting on the vehicle.

In many of the loading conditions, a flight vehicle may be considered as being in pure translation or pure rotation. The inertia force on any element of mass is equal to the product of the mass and the acceleration and acts in the direction opposite to the acceleration. If the applied loads and inertia forces act on an element as a free body, these forces are in equilibrium. For example, a body of mass m under the action of a force vector F moves so as to satisfy the equation

F = ma (2.5)

where m is the mass and a is the acceleration relative to a newtonian frStme of reference. If a cartesian system of x, y, and z axes is chosen in this frame, then Eq. (2.5) gives, upon resolving into components,

Fx = m.v, Fy = my, F. = m'z (2.6) a

where Fx, Fy, and F. are the components of F along the x, y, and z axes, respectively, and y, and z are the components of acceleration along the x, y, and z axes. In the preceding discussion, all parts of the rigid body were moving in straight, parallel lines and had equal velocities and accelerations. In many engin-eering problems, it is necessary to consider the inertia forces acting on a rigid body which has other types of motion. In many cases where the elements of a rigid body are moving in curved paths, they are moving in such a way that each


element moves in only one plane and all elements move in parallel planes. This type of motion is called plane motion, and it occurs, for example, when a vehicle is pitching and yet has no rolling or yawing motion. All elements of the vehicle move in planes parallel to the plane of symmetry. Any type of plane motion can be considered as a rotation about some instantaneous axis perpendicular to the planes of motion, and the following equations for inertia forces are derived on the assumption that the rigid body is rotating about an instantaneous axis perpen-dicular to a plane of symmetry of the body. The inertia forces obtained may be used for the pitching motion of a vehicle, but when they are utilized for rolling or yawing motions, it is necessary to first obtain the principal axes and moments of inertia of the vehicle.

The rigid mass shown in Fig. 2.5 is rotating about point O with a constant angular velocity o . The acceleration of any point a distance r from the center of rotation is wir and is directed toward the center of rotation. The inertia force acting on an element of mass dM is the product of the mass and the acceleration, or co2r dM, and is directed away from the axis of rotation. This inertia force has components u>2x dM parallel to the x axis and a>2y dM parallel to the y axis. If the x axis is chosen through the center of gravity C, the forces are simplified. The resultant inertia force in the y direction for the entire body is found as follows:

The angular velocity w is constant for all elements of the body, and the integral is zero because the .* axis was chosen through the center of gravity. The inertia force in the x direction is found in the same manner:

The term x is the distance from the axis of rotation O to the center of gravity C, as shown in Fig. 2.5.

If the body has an angular accderation a, the element of mass dM has an

Fy = o)2y dM = co2 y dM = 0

r

oj- r (h\I , .2.. .iM

Z Figure 2 3


additional inertia force ar dM acting perpendicular to r and opposite to the direction of acceleration. This force has components ctx dM in the y direction and ay dM in the x direction, as shown in Fig. 2.6. The resultant inertia force on the entire body in the x direction is

Fx = Jay dM = a j y dM = 0

The resultant inertia force in the y direction is

Fy = J ax dM = aj" dM = axM (2.8)

The resultant inertia torque about the axis of rotation is found by integrating the terms representing the product of the tangential force on each element ar dM and its moment arm r:

T0 ar2 dM = a r1 dM = a I0 (2.9)

The term / 0 represents the moment of inertia of the mass about the axis of rotation. It can be shown that this moment of inertia can be transferred to a parallel axis through the center of gravity by use of the following relationship:

l0 = Mx2 +JC (2.10)

where lc is the moment of inertia of the mass about an axis through the center of gravity, obtained as the sum of the products of mass elements dM and the square of their distances rc from the center of gravity:

r* dM

By substituting the value of IQ from F.q. (2.10) in F.q. (2.9), the following ex-pression for the inertia torque is obtained:

T 0 = Mxza + lca (2.11)

/

Figure 2.16


The inertia forces obtained in Eqs. (2.7), (2.8), and (2.11) may be represented as forces acting at the center of gravity and the couple Ic a, as shown in Fig. 2.7. The force OLXM and the couple /r a must both produce moments about point 0 which arc opposite to the direction of A. The force CU2XM must act away from point O.

It is seen from Fig. 2.7 that the forces at the centroid represent the product of the mass of the body and the components of acceleration of the center of gravity. In many cases, the axis of rotation is not known, but the components of acceler-ation of the center of gravity can be obtained. In other cases, the acceleration of one point of the body and the angular velocity and angular acceleration are known. If the point"0 in Fig. 2.8 has an acceleration a 0 , an inertia force at the center of gravity of Ma0, opposite to the direction of a0, must be considered in addition to those previously taken into account.

2.6 LOAD FACTORS FOR TRANSLATIONAL ACCELERATION

For flight or landing conditions in which the vehicle has only translational accel-eration, every part of the vehicle is acted on by parallel inertia forces which are proportional to the weight of the part. For purposes of analysis, it is convenient to combine these inertia forces with the forces of gravity, by multiplying the weight of each part by a load factor n, and thus to consider the combined weight and inertia forces. When the vehicle is being accelerated upward, the weight and inertia forces add directly. The weight of w of any part and the inertia force wa/g have a sum nw:

a nw = w + vv —

9 "—-

or n = 1 + - (2.12) 9

The combined inertia and gravity forces are considered in the analysis in the same manner as weights which are multiplied by the load factor n.

In the case of an airplane in flight with no horizontal acceleration, as shown in Fig. 2.9, the engine thrust is equal to the airplane drag, and the horizontal


components of the inertia and gravity forces arc zero. The weight and the inertia force on the airplane act down and will be equal to the lift. The airplane lift L is the resultant of the wing and tail lift forces. The load factor is defined as follows:

I ' j r l i f t Load factor = • — — -

weight

or n = £ (2.13)

This value for the load factor can be shown to be the same as that given by Eq. (2.12) by equating the lift nW to the sum of the weight and inertia forces:

L = nW = W + W -(J

a or n = 1 4- -

9

which corresponds to Eq. (2.12). Flight vehicles frequently have horizontal acceleration as well as vertical

acceleration. The airplane shown in Fig. 2.10 is being accelerated forward, since the engine thrust T is greater than the airplane drag D. Every element of mass in the airplane is thus under the action of a horizontal inertia force equal to the product of its mass and the horizontal acceleration. It is also convejiifent to consider the horizontal inertia loads as equal to the product of a load factor nx

I

1 — t g a Figure Z 9

FLIGHT-VEHICLE IMPOSED LOADS 3 9

L

and the weights. This horizontal load factor, often called the thrust had factor, is obtained from the equilibrium of the horizontal forces shown in Fig. 2.10:

„ W = — W T - D (I

T D W

(2.14)

A more general case of translational acceleration is shown in Fig. 2.11, in which the airplane thrust line is not horizontal. It is usually convenient to obtain components of forces along x and z axes which are parallel and perpendicular to the airplane thrust line. The combined weight and inertia load on any element has a component along the 2 axis of the following magnitude:

nw = vv cos 0 + w —

a. or n — cos 0 + —

0

From summat ion of all forces a long the z axis,

L = W-'^cos 0 + ~

(2.15)

(2.16)

Figure 2.24


By combining Eqs. (2.15) and (2.16),

or

L = Wn

W which corresponds with the value used in Eq. (2.13) for a level attitude of the airplane.

The thrust load factor for the condition shown in Fig. 2.11 is also similar to that obtained for the airplane in level attitude. Since the thrust and drag forces must be in equilibrium with the components of weight and inertia forces along the x axes, the thrust load factor is obtained as follows:

W „xW = — ax-W sin 0 = 7" - D

or = ' T-D

W

This value is the same as that obtained in Eq. (2.14) for a level attitude of the airplane.

In the case of the airplane landing as shown in Fig. 2.12, the landing load factor is defined as the vertical ground reaction divided by the airplane weight. The load factor in the horizontal direction is similarly defined as the horizontal ground reaction divided by the airplane weight:

n. = 777 (2.17)

and n x =

W

R* w (2.18)

In the airplane analysis, it is necessary to obtain the components of the load factor along axes parallel and perpendicular to the propeller thrust line. However, aerodynamic forces are usually obtained first as lift and drag forces perpendicular and parallel to the direction of flight. If load factors are obtained first along lift and drag axes, they may be resolved into components along other axes, in,the same manner as forces are resolved into components. The force acting on any

H' — a, st '

I t

Figure 2.16


weight vv is wn, and the component of this force along any axis at an angle 0 to the force is wn cos 6. The component of the load factor is then n cos 9.

As a general definition, the load factor n along any axis i is such that the product of the load factor and the weight of an element is equal to the sum of the components of the weight and inertia forces along that axis. The weight and inertia forces are always in equilibrium with the external forces acting on the airplane, and the sum of the components of the weight and inertia forces along any axis must be equal and opposite to the sum of the components of the external forces along the axis 2F ; . The load factor is therefore defined as

v r .

• n ' = W

where ZF,- includes all forces except weight and inertia forces.

2.7 VELOCITY-LOAD-FACTOR DIAGRAM

The various loading conditions for an airplane usually are represented on a graph of the limit-load factor n plotted against the indicated airspeed V. This diagram is often called a V-n diagram, since the load factor n is related to the acceleration of gravity g. In all such diagrams, the indicated airspeed is used, since all air loads are proportional to q o r p V 1 / ! . The value of q is the same for . the air density p and the actual airspeed at altitude as it is for the standard sea-level density p0 and the indicated airspeed, since the indicated airspeed is defined by this relationship. The V-n diagram is therefore the same for all alti-tudes if indicated airspeeds are used. Where compressibility effects are considered, they depend on actual airspeed rather than indicated airspeed and consequently are more pronounced at altitude. Compressibility effects are not considered here.

The aerodynamic forces on an airplane are in equilibrium with the forces of gravity and inertia. If the airplane has no angular acceleration, both the inertia and gravity forces will be distributed in the same manner as the weight of various items of the aircraft and will have resultants acting through the center of gravity of the aircraft. It is convenient to combine the inertia and gravity forces as the product of a load factor n and the weight W, as described previously. The z component of the resultant gravity and inertia force is the force nW acting at the center of gravity of the airplane, as shown in Fig. 2.13. The load factor n is obtained from a summation of forces along the z axis:

C.a qS — nw

C-..PSV2

or n = 2 W (2.20)

The maximum value of the normal force coefficient Crii may be obtained at various airplane speeds. For level flight at a unit-load factor, the value of V corresponding to C-, mai would be the stalling speed of the airplane. In acceler-


(NLAA) Figure 2.13

atcd flight, the maximum coefficient might be obtained at higher speeds. For t 0 t>e obtained at twice the stalling speed, a load factor n = 4 would be

developed as shown by Eq. (2.20). For a force coefficient of 1 . 2 5 r e p r e s e n t -ing the highest angle of attack for which the wing is analyzed, the value of the load factor n is obtained from Eq. (2.20) and may be plotted against the airplane velocity V, as shown by line OA in Fig. 2.13..This line OA represents a limiting condition, since it is possible to maneuver the airplane at speeds and load factors corresponding to points below or to the right of line OA, but it is impossible to maneuver at speeds and load factors corresponding to points above or to the left of line OA because this would represent angles of attack much higher than the stalling angle.

The line AC in Fig. 2.13 represents the limit on the maximum maneuvering load factor for which the airplane is designed. This load factor is determined from the specifications for which the airplane is designed, and the pilot must restrict maneuvers so as not to exceed this load factor. At speeds below that correspond-ing to point A, it is impossible for the pilot to exceed the limit load factor in any symmetrical maneuver, because the wing will stall at a lower load factor.J-or airspeeds between those corresponding to points A and C, it is not practical to design jhe airplane structure so that it could not be overstressed by violent maneuvers. Some types of airplanes may be designed so that the pilot would have to exert large forces on the controls in order to exceed the limit-load factor.

Line CD in Fig. 2.13 represents the limit on the permissible diving speed for the airplane. This value is usually specified as 1.2 to 1.5 times the maximum indicated airspeed in level flight. Line OB corresponds to line OA, except that the wing is at the negative stalling angle of attack, and the air load is down on the wing. The equation for line OB is obtained by substituting the maximum negative value of C.a into Eq. (2.20). Similarly, line BD corresponds to line AC, except that the limit-load factor specified for negative maneuvers is considerably less than for positive maneuvers.

FLIGHT-VEHICLE IMPOSED LOADS 4 3

The aircraft may therefore be maneuvered in such a manner that velocities and load factors corresponding to the coordinates of points within the area OACDB may be obtained. The most severe structural load ng conditions will be represented by the corners of the diagram, points A, B, C, and D. Points A and B represent PHAA and NHAA conditions. Point C represents the PLAA condition in most cases, although the positive gust load condition, represented by point F, may occasionally be more severe. The NLAA condition is represented by point D or by the negative gust condition, point E, depending on which condition pro-duces the greatest negative load factor. The method of obtaining the gust load factors, represented by points E and F, is explained in the following section.

2.8 GUST LOAD FACTORS

When an airplane is in level flight in calm air, the angle of attack a is measured from the wing chord line to the horizontal. If the airplane suddenly strikes an ascending air current which has a vertical velocity KV, the angle of attack is increased by the angle Aa, as shown in Fig. 2.14. The angle Ax is small, and the angle in radians may be considered as equal to its langent:

Aa = : KU (2.21)

The change in the airplane normal force coefficient C.u. resulting from a change in angle of attack Act, may be obtained from the curve of C2a versus a of Fig. 2.4. This curve is approximately a straight line, and it has a slope /J which may be considered constant:

P = AC. -a Aa •

(2.22)

After striking the gust, the airplane normal force coefficient increases by an amount determined from Eqs. (2.21) and (2.22):

AC. flKU (2.23)

The increase in the airplane load factor An may be obtained by substituting the value of AC., from Eq. (2.23) into Eq. (2.20):

AC. pSV2

An = • 2 W

Chord line

F igure 2.24


pSPKUV 2 W or A n = — (2.24)

where p = standard sea-level air density, 0.002378 slug/ft3

S = wing area, ft2

P = slope of the curve of C.a versus a, rad KU = effective gust velocity, ft/s

V = indicated airspeed, ft/s W = gross weight of airplane, lb

For purposes of calculation, it is more convenient to determine the slope /? per degree and the airspeed V in miles per hour. Introducing the necessary constants in Eq. (2.24) yields

where /? is the slope of CIm versus a per degree, V is the indicated airspeed in miles per hour, and other terms correspond to those in Eq. (2.24).

When the airplane is in level flight, the load factor is unity before the plane strikes the gust. The change in load factor A/i from Eq. (2.25) must be combined with the unit-load factor in order to obtain the total gust load factor:

Equation (2.26) may be plotted on the V-n diagram, as shown by the inclined straight lines through points F and H of Fig. 2.13. These lines represent load factors obtained when the airplane is in a horizontal attitude and strikes positive or negative gusts. Equation (2.25) is similarly plotted, as shown by the inclined lines through points G and E of Fig. 2.13. These lines represent load factors obtained when the airplane is in a vertical attitude and strikes positive or nega-tive gusts in directions normal to the thrust line.

The gust load factor represented by point F of Fig. 2.13 may be more severe than the maneuvering load factor represented by point C. In the case shpwn, however, the maneuvering load factor is obviously greater and will represent the PLAA condition. The negative gust load factor represented by point E is greater than the negative maneuvering load factor represented by point D and will determine the NLAA condition. It might seem that the gust load factors should be added to the maneuvering load factors, in order to provide for the possibility of the airplane's striking a severe gust during a violent maneuver. While this condition is possible, it is improbable because the maneuvering load factors are under the pilot's control, and the pilot will restrict maneuvers in gusty weather. Both the maneuvering and gust load factors correspond to the most severe condi-tions expected during the life of the airplane, and there is little probability of a combined gust and maneuver producing a condition which would exceed the limit-load factor for the design condition.


The "effective, sharp-edged gust" velocity KU is the velocity of a theoretical gust which, if encountered instantaneously, would give the same load factor as the actual gust. Actually, it is impossible for the upward air velocity to change suddenly from zero to its maximum value. There is always a finite distance in which the air velocity changes gradually from zero to the maximum gust velocity, and a short time is required for the airplane to move through this transition region. Most specifications require that the airplane be designed for a gust ve-locity U of 30 ft/s with the gust effectiveness factor K of 0.8 to 1.2, depending on the wing loading W/S. Airplanes with higher wing loadings usually are faster and pass through the transition region from calm air to air with the maximum gust velocity in a shorter time, and hence they must be designed for larger values of K. The design values of KU are obtained from accelerometer readings for airplanes flying in turbulent air and represent the maximum effective gust velocities which will ever be encountered during the service life of the airplane. Some specifi-cations require gust velocities of 50 ft/s, with corresponding gust reduction factors K of about 0.6. Since the values of K U in this case are also about 30 ft/s, the net effect is equivalent to a gust velocity U of 30 ft/s with a K of 1.0. The actual maximum vertical air velocities probably exceed 50 ft/s, but the transition is gradual, corresponding to the values of K = 0.6. High gust load factors exist for only a fraction of a second, and the airplane cannot move far in this time.

In order to understand the effect of gusts, it is necessary to study the motion of the airplane after it encounters a gust. If the gust is encountered instanta-neously, the factor K is 1.0, and the effective gust velocity is U. The airplane is. accelerated upward with an initial acceleration a0 and attains a variable vertical velocity v. The gust angle of attack (Aa of Fig. 2.14) has a maximum value of U/V at the time the gust is encountered (f = 0), but this angle of attack is decreased to (U — v)/V after the airplane attains an upward velocity. When the upward ve-locity v is equal to U, the relative wind is again horizontal, and the airplane is no longer accelerated. The variable vertical acceleration a is therefore

d v ILzl n ™ (2.27)

Separating the variables and integrating, we have

dv a0

log

o-U-v U

V — v a0l

dl o

U U

By using the exponential form and substituting the value of a from Eq. (2.27), the following expression for the acceleration a at lime t is obtained:

— = e - ° 0 « u ( 2 . 2 8 )

As a numerical example, consider a gust velocity U of 30 ft/s and an initial


5 !i

0 0.1 0.2 0 . 3 0 . 4 0 . 5

Time, s

Figure 2.15

acceleration a0 of 5g, corresponding to a gust load factor of 6.0. By substituting these values into Eq. (2.28) and plotting a versus t, the curve of Fig. 2.15 is obtained. The gust acceleration is seen to approach zero asymptotically in an infinite time, but it decreases greatly in the first 0.1 s. Thus, if the airplane had a forward speed of 500 ft/s (340 mi/h) it would move forward only 50 ft in 0.1 s. It seems logical to expect that atmospheric conditions are such that it is more than 50 ft from any region of calm air to a regiotrin which the gust velocity is 30 ft/s. The actual gust acceleration probably is represented more accurately by the dotted line of Fig. 2.15, which would indicate an effectiveness factor K of about 0.6. However, since airplane accelerometer readings have shown effective gust velocities KU of 30 ft/s, the true conditions probably are represented by gust velocities U of more than 50 ft/s with effectiveness factors K less than 0.6.

2.9 EXAMPLES

Example 2.1 When landing on a carrier, a 10,000-ib airplane is given a deceleration of 1g (96.6 ft/s2) by means of a cable engaged by an arpe'sting hook as shown in Fig. 2.16.

A It'

N

K „ " - " i 20 in Figure 2.16


(a) Find the tension in the cable, the wheel reaction R, and the distance e from the center of gravity to the line of action of the cable.

(b) Find the tension in the fuselage at vertical sections AA and BB if the portion of the airplane forward section AA weighs 3000 lb and the por-tion aft of section BB weighs 1000 lb.

(c) Find the landing run if the landing speed is 80 ft/s.

S O L U T I O N

(a) First consider the entire airplane as a free body.

W 10,000 „ Ma = — a — — 3g = 30,000 lb

s a

T.FX = T cos 10° - 30,000 = 0

T = 30,500 lb

ZFj, — R — 10,000 - 30,500 sin 10" = 0

R = 15,3001b

HMCS = 20 x 15,300 - 30,500c = 0

e = 10 in

(b) Consider the aft section of the fuselage as a free body, as shown in Fig. 2.17. It is acted on by an inertia force of

Ma = ^ ^ 3g = 3000 lb 9

The tension on section BB is found as follows:

IFX = 30,000 - 3000 - 71 = 0

Ti = 27,000 lb

Since there is no vertical acceleration, there is no vertical inertia force. Section BB has a shear force Vt of 6300 lb, which is equal to the sum of the weight and the vertical component of the cable force.

Consider the portion of the airplane forward of section A A as a free

Ala = 3000 lb

t o A i K l . K A r I M K U C I U K h S

Ma = 9000 Ib V2 = 3000 Ib

n = 9000 Ib

A

3000 lb Figure 2.18

body, as shown in Fig. 2.18. The inertia force is the following.

3000 Ma = 3g = 9000 lb

9

TFX = T2 - 9000 = 0

T2 = 9000 lb

The section AA must also resist a shearing force V2 of 3000 lb and a bending moment obtained by taking moments of the forces shown in Fig. 2.18.

The forces Tu Tz, V„ and V2 may be chccked by considering the equilibrium of the center portion of the airplane, as shown in Fig. 2.19.

6000 Ma = 3g = 18,000 lb

9

XFx = 27,000 - 18,000 - 9000 = 0

2 F , = 15,300 - 3000 - 6000 - 6300 = 0

(c) From elementary dynamics, the landing run s is obtained as follows:

v2 — i'o = 2 as

0 - (802) = 2(—96.6)s

s = 33 ft

Example 2.2 A 30,000-lb airplane is shown in Fig. 2.20a at the time, of landing impact, when the ground reaction on each main wheel is 45,000 lb. («) If one wheel and tire weighs 500 lb, find the compression C and bending

6000 lb

r , =9000 lb

K, = 3000 lb

15.300 lb i = T 18,0001b Figure 2.19


| Ma = 60,000 lb I

H' = 30,000 lb

A

F T 45.000 lb 45,000 lb

1000 lb

W, = 500 lb ilD • a

| - . 2 0 m - H ^ a = 3 0 0 0 l h

^ I If = 1500 lb

T M

45,000 lb

(h)

Figure 2.20

moment m in the oleo strut if the strut is vertical and is 6 in from the centerline of the wheel, as shown in Fig. 2.20b.

(b) Find the shear and bending moment at section A A of the wing if the wing outboard of this section weighs 1500 lb and has its center of gravity 120 in outboard of section AA.

(c) Find the required shock strut deflection if the airplane strikes the ground with a vertical velocity of 12 ft/s and has a constant vertical deceleration until the vertical velocity is zero. This neglects the energy absorbed by the tire deflection, which may be large in some cases.

(r/J Find the time required for the vertical velocity to become zero.

S O L U T I O N

(a) Considering the entire airplane as a free body and taking a summation of vertical forces yield

Y.FS. = 45,000 + 45,000 - 30,000 - Ma = 0

Ma = 60,000 lb

60,000 60,000g " ~ - M ~ 30,000 ~ g

Consider the landing gear as a free body, as shown in Fig. 2.20b. The

inertia force is

w, 500 M,a = - 1 a = 2 g = 1000 lb

9 9

The compression load C in the oleo strut is found from a summation of vertical forces:

T.Fy = 45,000 - 500 - 1000 - C = 0

C = 43,500 lb

The bending moment m is found as follows:

m = 45,000 x 6 - 1000 x 6 - 500 x 6 = 261,000 in • lb

(/>) The inertia force acting on the portion of the wing shown in Fig. 2.20c is

w, 1500 M 2 a = — a — 2g = 3000 lb

9 9

The wing shear at section AA is found from a summation of vertical forces.

1Fy = V - 3000 - 1500 = 0

V = 4500 lb The wing bending moment is found by taking moments about section A A mw = 3000 x 120 + 1500 x 120 = 540,000 in • lb

(<•) The shock strut deflection is found by assuming a constant vertical accel-eration of —2g, or —64.4 ft/s2, from an initial vertical velocity of 12 ft/s to a final zero vertical velocity.

v2 — Vq = las

0 - (122) = 2(—64.4).?

s = 1.12 ft

((/) The time required to absorb the landing shock is found based on el-ementary dynamics. /

v — L'0 — at

0 — 12 = —64.4f

. f = 0.186 s

Since the landing shock occurs for such a short time, it may be less injurious to the structure and less disagreeable to the passengers than would a sustained load.

Example 2-3 A 60,000-lb airplane with a tricycle landing gear makes a hard two-wheel landing in soft ground so that the vertical ground reaction is

rl.Kot! 1 -Vt ' .nn. 1.1:. IM I'U.M.U LUAUS J I

270,000 Ib and the horizontal ground reaction is 90,000 Ib. The moment of inertia about the cenlcr of gravity is 5,000,000 lb • s2 • in, and the dimensions are shown in Fig. 2.21. (n) Find the inertia forccs on the airplane. (/;) Find the inertia forces on a 400-lb gun turret in the tail which is 500 in aft

of the center of gravity. Neglect the moment of inertia of the turret about its own cenlcr of gravity,

(c) If the nose wheel is 40 in from the ground when the main wheels touch the ground, find the angular velocity or the airplane and the vertical velocity of the nose wheel when the nose wheel reaches the ground, assuming no appreciable change in the moment arms. The airplane's center of gravity has a vertical velocity of 12 ft/s at the moment of impact, and the ground reactions are assumed constant until the vertical velocity reaches zero, at which time the vertical ground reaction bccomes 60,000 lb and the horizontal ground reaction becomes 20,000 lb.

(a) The inertia forccs on the entire airplane may be considered as horizontal and vertical forces Max and May, respectively, at the ccntcr of gravity and a couplc /,. a,, as shown in Fig. 2.21. These correspond to the inertia forces shown on the mass of Fig. 2.7, sincc the forccs at the center of gravity represent the product of the mass and the acceleration com-ponents of the ccntcr of gravity.

XFx = 90,000 - Max = 0

Max = 90,000 lb

ZFy = 270,000 - 60,000 - May = 0

May = 210,000 lb

TM,:il = -270,000 x 40 - 90,000 x 100 + I c a = 0

[ c z = 19,800,000 in • Ib

S O L U T I O N

Ma y I

500 j„-t

40 in

i TO.000 lb • • 270,000 Ib

200 in— <- - -•—40 in F igure 2.21

90,000 90,000 M 60,000 9 ~ ' 3

210,000 210,000 M 60,000 (I = 3.5®

/ c « 19,800,000 a = 1 7 = l ^ o a o o o = 3 - 9 r a d / s

(h) The acceleration of the center of gravity of the airplane is now known, and the acceleration and inertia forces for the turret can be obtained by the method shown in Fig. 2.8, where the center of gravity of the airplane corresponds to point 0 of Fig. 2.8 and the center of gravity of the turret corresponds to point C. These forces are shown in Fig. 2.22 and have the following values:

400 Max = 1,5g = 600 lb

9

400 Ma, = 3.5g = 1400 lb

9

400 axM = 3.96 x 500 x — = 2050 lb

386

In calculating the term axM, x is in inches and g is used as 386in/s2. Ifjc is in feet, g will be 32.2 ft/s2. The total force on the turret is 600 lb forward and 3850 lb down. This total force is seen to be almost 10 times the weight of the turret.

(<) The center of gravity of the airplane is decelerated vertically at 3.5g, or 112.7 ft/s2. The time of deceleration from an initial velocity of 12 ft/s to a zero vertical velocity is found from the following.

v = v0 = at

0 - 12 = 112.7f

/ = 0.106 s

I f May = 1400 lb I 1 } axM = 2050 lb Figure 2.22

During this time, the center of gravity moves through a distance found from

s = v0t + %at2

= 12 x 0.10(5 - i x 112.7 x 0.1062

= 0.636 ft, or 7.64 in

The angular velocity of the airplane at the end of 0.106 s after the landing is found from

(D — <Oq — cut

w - 0 = 3.96 x 0.106

co = 0.42 rad/s

The angle of rotation during this time is found from

0, = « 0 f + i « t 2

0, = 0 + |(3.96X0.1062) = 0.0222 rad

The vertical motion of the nose wheel resulting from this rotation, shown in Fig. 2.23, is

s, = 0[X = 0.0222(200) = 4.44 in

The distance of the nose wheel from the ground, after the vertical velocity of the center of gravity of the airplane has become zero, is

s2 = 40 - 7.64 - 4.44 = 27.92 in

The remaining angle of rotation 0 2 , shown in Fig. 2.23, is

2 x 200

Since the ground reaction decreases by the ratio of after the vertical acceleration of the airplane becomes zero, the angular acceler-ation decreases in the same proportion, as found by equating moments

63 a i r O r a i - t s t r u c t u r e s

about the center of gravity:

60,000 270,000

3.96 = 0.88 rad/s2

The angular velocity of (he airplane at the time the nose wheel strikes the ground is found from the following equation.

co2 — a>o = 2a2 02

w2 - (0.422) = 2 x 0.88 x 0.1396

co = 0.65 rad/s

Since at this time the motion is rotation, with no vertical motion of the center of gravity, the vertical velocity of the nose wheel is found as follows:

v = 0.65 x ^ = 10.8 ft/s

This velocity is smaller than the initial sinking velocity of the airplane. Consequently, Ihe nose wheel would strike the ground with a higher velocity in a three-wheel level landing.

It is of interest to find the centrifugal force on the turret UJ2XM at the time the nose wheel strikes the ground. This force was zero when the main wheels hit because the angular velocity a> was zero. For the final value of co, the following value is obtained:

CJ2XM = (0.652) x 500 x = 219 lb

This force is much smaller than other forces acting on the turret, and usually it is neglected. In part c, certain simplifying assumptions are made which do not quite correspond with actual landing conditions. Aerody-namic forces are neglected, and the ground reactions on the landing gear are assumed constant while the landing gear is a combination of the tire deflection, in which the load is approximately proportional to the defor-mation, and the oleo strut deflection, in which the load is almost constant during the entire deformation, as assumed. The tire deflection may be as much as one-third to one-half the total deflection. The aerodynamic forces, which have been neglected, would probably reduce the maximum angular velocity of the airplane, since the horizontal tail moves upward as the airplane pitches, and the combination of upward and forward motions would give a downward aerodynamic force on the tail, tending to reduce the pitching acceleration.

The aerodynamic effects of the lift on the wing and tail surfaces are not shown in Fig. 2.21, but they will not affect the pitching acceleration appreciably if the ground reactions remain the same. Just before the airplane strikes the ground, the lift forces on the wing and tail are in

f l i g h t - v e h i c l e i m p o s e d l o a d s 5 5

equilibrium with the gravity force of 60,000 lb. Since the horizontal ve-locity of the airplane and the angle of attack are not appreciably changed W, = 0.0222 rad = 1.27°), the lift forces continue to balance the weight of the airplane when the center of gravity is being decelerated. Instead of the weight of 60,000 lb shown in Fig. 2.21, there should be an additional inertia force of 60,000 lb down at the center of gravity. The moments about the center of gravity and the pitching accclcralion are not changed, but the vertical deceleration is increased. At the end of the deceleration of the center of gravity, the ground reactions are almost zero, since most of the airplane weight is carried by the lift on the wings. The airplane then pilches forward through the angle 0 2 , which appreciably changes the angle of attack (02 = 0.1396 rad = 8°). The wing lift is then decreased, and most of the weight is supported by the ground reactions on the wheels. For the structural design of the airplane, usually only the loads during the initial impact arc significant.

Example 2-4 Construct the V-n diagram and determine the wing internal load resulting from aerodynamic forces for the airplane (Fig. 2.24) whose wing planform is shown.in Fig. 2.25. The following conditions are specified:

W = airplane gross weight = 8000 lb

S = airplane wing area = 266 ft2

KL' = effective gust velocity = ?4 ft-'s

I 'j — design diving speed = 400 mi 'h

n = limit-load factor = +6.0 and —3.0

The aerodynamic characteristics of the airplane with the horizontal tail removed have been obtained from corrected wind tunnel data and are given in Table 2.1. The moment coefficient Cflf is about the center of gravity of the airplane and is expressed in terms of the wing area and the mean aerody-namic chord for the wing, c = 86 in. The stalling angle of the wing is 20°, corresponding to a maximum lift coefficient of 1.67. The aerodynamic data

•V

Figure 2.26

5 6 a i r O r a i - t s t r u c t u r e s

2 Q-JQ 2.049 2 OG 2.087 2 0 Q S 2 . 0 8 5 0 7 7 2 .0532 035 1 .972 j g o , 1 . 8 5 9 Oi (total normal ' | ~ | | | " | | " [ | " | * | ' | * | | force coefficient I

0 20 4 0 60 80 tOO 120 140 160 180 -200 220 240

H 240 in

Figure 2.25

are extrapolated to the angle of attack of 26°. The negative stalling angle is - 1 7 ° .

The force coefficients acting normal to the thrust line are calculated in Table 2.2. The components of C,_ and CD are calculated in columns 2 and 3. The tail load coefficient C, is calculated in column 4 by means of Eq. (2.3). The final values of C.a, the normal force coefficient for the entire airplane, are obtained in column 5 as the sum of values from columns 2, 3, and 4.

The V-n diagram is constructed from the calculated data for C.o. For the OA portion of the curve of Fig. 2.13, the value of C-.a is assumed to be 1.25 times the value at the stalling angle for the wing, or

CZa = 1.25(1.656) = 2.070

This corresponds with the angle of attack of 26", within the accuracy of the data, and this angle is assumed. The equation for the curvo OA of Fig. 2.13 is

Table 2.1

. a — 0, deg Q G> c*

26 2.132 0.324 0.0400 20 1.670 0.207 0.0350

15 1.285 0.131 0.0280

10 0.900 0.076 0.0185

5 0.515 0.040 0.0070

0 0.130 0.023 - 0 . 0 1 0 5

- 5 - 0 . 2 5 5 0.026 - 0 . 0 3 1 6

- 1 0 - 0 . 6 4 0 0.049 - 0 . 0 5 2 5

- 1 5 - -1 .025 0.092 - 0 . 0 7 7 0

- 1 7 - 1 . 1 8 0 0.115 - 0 . 0 8 6 0


Table 2.2

0, deg C „ sin 0 C , cos 0 C, C.

(I) (2) (31 (4) (5)

26 0.14 J 1,918 0.017 2.078 20 0.071 1.570 0.01S 1.656 15 0.034 1.240 0.012 1.286 10 0.013 O.S87 0.008 0.908 5 0 .004 0.512 0.003 0.519 0 0 0.130 - 0 . 0 0 4 0.126

- 5 - 0 . 0 0 2 - 0 . 2 5 4 - 0 . 0 1 3 - 0 . 2 6 9 - 1 0 - 0 . 0 0 8 - 0 . 6 3 0 - 0 . 0 2 2 - 0 . 6 6 0 - 1 5 - 0 . 0 2 4 —0.990 - 0 . 0 3 2 - 1 . 0 4 6 - 1 7 - 0 . 0 3 4 - 1 . 1 3 0 - 0 . 0 3 6 - 1 . 2 0 0

found as follows:

n = 2.078 = 2.078 x 0.00256 2 W

= 0.0001.772 V2

•256 ( \8000 J

For point A, n = 6 and V = 184 mi/h. The equation for the curve OB of Fig. 2.13 is found as follows:

oSV2

n = - 1.200 = — 0.0001024 K2

For point B, n = — 3 and V = 172 mi/h. Points C and D are plotted with coordinates (400, 6), and (400, - 3). The diagram is shown in Fig. 2.26.

The gust load factors are now obtained from Eqs. (2.25) and (2.26). The slope ft may be obtained from the extreme coordinates of the curve for C.a if we assume a straight-line variation:

„ 2.078 + 1.200 " = — — = 0.0763 per degree

Z D + 1 /

« 1 I OflO^lL-—"

300 _ J

we

H >1=3

VD = 4 0 0

- ) ' . mi/h

« = -0.000)02-11 Figure 2.26

Table 2.3

C h o r d Sta t ion length Force coefficient Shea r Bending momen t

>; C, ('„, Vi A V I 000

0 240 0 0 0 1 220 50 1.859 56(1 5 2 200 66 1.903 1.880 29 3 ISO 73.2 1.972 3.510 83 4 160 76.4 2.035 5,310 171 5 140 79.6 2.053 7.230 296 6 120 82.8 2.077 9.250 461 7 100 86.0 2.085 11.370 667 8 80 89.2 2.095 13.580 917 9 60 9 2 4 2.087 15,870 1.212

10 4 0 95.6 2.06 18,220 1,553 11 20 98.8 2.049 20,630 1.942 12 0 102 2.02 23,090 2,379

From Eq. (2.26),

A _ n | P K U V _ '(0.0763X34) " ' W/S 30

= 0.00865 V

For V — 400 mi/h, An — 3.46. Points F and E represent gust load factors of 4.46 and — 3.46, respectively.

The wing internal bending and shear loads are now calculated for the P1IAA condition, which is represented by point A on the V-n diagram. The wing has an angle of attack of 26° at an indicated airspeed of 184 mi/h. The total force coefficients normal to the wing chord are given in Fig. 2,25.

The shear and bending loads are calculated based on

s

and M, = M, , + 1 {Y, , - X)

where i indicates the station number and </ is the dynamic pressure in pounds per square foot. All results are summarized in Table 2.3.

PROBLEMS

2.1 An a i rp l ane weighing 5000 lb strikes an u p w a r d gust of air which p roduces a wing lift of 25,000 lb (see l ig. I'2.1). W h a t tail load P is required to prevent a pi tching accelera t ion if the d imensions are as

s h o w n ? W h a t will be the vertical accelerat ion of the a i rp l ane? If this lift force acts until the a i rp lane ob ta ins a vertical velocity of 20 ft/s, how much lime is r equ i red?

25,000 ib

Figure P2.1

2.2 An a i rp lane weighing 8000 lb has an upward accelerat ion of 3</ when landing. If the d imens ions are as shown in Fig. P2.2, what are the wheel react ions R , and R2 ' - W h a t t ime is required lo decelerate ihe a i rp lane f rom a vertical velocity of 12 ft/s'.' W h a t is the vertical compression of the landing gear du r ing this dece le ra t ion? What is the shea r and bend ing m o m e n t on a vertical section A A if the weight fo rward of this section is 2(X)0 !h and has a center of gravi ty 40 in from Ihis cross sect ion?

Ma = 2 4 , 0 0 0 lb 1

Figure P2.2

2.3 T h e a i rp lane s h o w n in Fig. P2..1 is mak ing an ar res ted l and ing on a car r ie r deck. Find the load factors « a n d n x , pe rpend icu l a r and parallel to Ihe deck. Tor a poin t a t t he center of gravity, a poin t 200 in afl of the center of gravi ty , and a point 100 in fo rward of Ihe center of gravi ty. Fitid Ihe relative

«' = 10,000 Ib

6 0 a i r c r a f t s t r u c t u r e s

vertical velocity wi th which t h e nose wheel s tr ikes the deck if the vert ical velocity of the center of gravity is 12 f t /s a n d the a n g u l a r velocity is 0.5 rad / s counterc lockwise f o r the posi t ion shown . T h e radius of gyra t ion for the m a s s of the a i rp lane a b o u t Ihe center of gravi ty is 60 in. Assume no change in the d imens ions o r loads s h o w n .

2.4 An a i rp lane is flying at 550 mi/h in level flight when it is suddenly pulled upward in to a curved pa th of 2000-fl rad ius . (See Fig . P2.4.) F ind Ihe load fac tor of the a i rp lane .

Figure P2.4

2.5 If ihe a i rp lane in P rob . 2.4 is given a pi tching accelerat ion of 2 r a d / s 2 . find its load factor , assuming tha t the c h a n g e in lift d u e to pi tching may be neglected.

2.6 A large t r anspor t a i rcraf t is m a k i n g a level landing, as shown in Fig. P2.6. The gross weight of the aircraft is 150,000 lb, and its p i tching mass m o m e n t of inert ia is 50 x 10" lb • in • s 2 a b o u t the center of gravity. T h e l and ing rear-wheel react ion is 350,000 lb at an angle of 15" with the vertical. Deter -mine whe ther passenger A o r B will receive the mos t load. Assume tha t each passenger weights 170 lb and neglect the a i rp l ane lift.

Figure P2.6

f l i g h t - v e h i c l e i m p o s e d l o a d s 61

2.7 Assume thai the center of gravi ty of the a i rp l ane in E x a m p l e 2.4 is m o v e d forward 8 in w i t h o u t chang ing the external a e r o d y n a m i c conf igura t ion . T h e d i s t ance L, is n o w 208 in, and the values of the a e r o d y n a m i c pi tching m o m e n t s a b o u t the center of gravi ty a r e C M — 8C . /86 , where values of CM a r e given in T a b l e 2.1.

(a) Ca lcu la te curves for C, and C._. (/>) C o n s t r u c t a I'-II d i a g r a m , us ing the cond i t ions specified in Sec. 2.7. (c) Ca lcu la te the wing bend ing -momcn t d i ag ram Tor a i r loads n o r m a l to the wing chord for the

PMAA cond i t ion . {</) Ca lcu la te the wing hcud ing-momcul d i ag ram Tor clKirdwi.sc air loads . (<•) Ca lcu la te the a i r - load tors ional m o m e n t s a b o u t the wing's leading edge if the leading edge is

s traight and perpendicu la r t o the p lane of symmet ry of the a i rp lane . A s s u m e the airfoil a t any sect ion to have an a e r o d y n a m i c ccnter a t the quar t e r -chord poin t and t o h a v e a negligible pitching m o m e n t a b o u t this po in t .

2.8 Ca lcu la te t h e wing n o r m a l ani l chordwise b e n d i n g - m o m c n t d i a g r a m s for the PLAA condi t ion for the a i rp l ane analyzed in Sec. 2.7.

2.9 If the a i rp lane wing of F.xamplc 2.4 weighs 4 .0 lb / f t 2 , which is a s sumed dis t r ibuted uniformly o v e r the area , ca lcula te the wing bend ing m o m e n t s resul t ing f rom gravity and iner t ia forces normal t o the wing chord for the four p r imary load ing condi t ions .

C H A P T E R

THREE ELASTICITY OF STRUCTURES

3.1 I N T R O D U C T I O N

This chapter defines stresses and strains and their fundamental relationships. The stress behavior of structures undergoing elastic deformation that is due to the action of external applied loads is also discussed. The term elasticity or elastic behavior is used here to imply a recovery properly of an original size and shape.

3.2 STRESSES

Consider the solid body shown in Fig. 3.1 which is acted on by a set of external forces Qi, as indicated. If we assume that rigid-body motion is prevented, the solid will deform in accordance with the external applied forces; as a result, internal loads between all parts of the body will be produced.

If the solid is separated into two parts by passing a hypothetical plane, as shown in Fig. 3.1 h, then there exist internal forces whose resultants are indicated by Qt and Q„ acting on parts I and II, respectively. y

Thc forces which hold together the two parts of the body are nofmally distributed over the entire surface of the cut plane. If we consider only an infi-nitesimal area SA acted on by a resultant force <5Q, then an average force per unit may be expressed as

i>/t (3.1)

In the limit as HA approaches zero, F.q. (3.1) becomes

dQ clA (3-2)

62

e l a s t i c i t y o f s t r u c t u r e s 6 3

Figure 3.1

where a now is the limiting value of the average force per unit area and, by definition, the stress at that point. A stress is completely defined if its magnitude and direction and the plane on which it acts are all known. For instance, it is not appropriate to ask about the stress at point 0 of the solid shown in Fig. 3.2 unless the plane on which the stress is acting is specified. An infinite number of planes may be passed through point 0, thus resulting in an infinite number of different stresses.

In the most general three-dimensional state of stress, nine stress components may exist:

<TV

ff.. (3.3)

Figure 3.2

where <r;f (i = x, z) are the normal stresses and = try (i ^ / - .x, 3% r) are the shearing stresses. The first subscript on nis (/, j = x, y, z) denotes the plane at a constant i on which the stress is acting, and the second subscript denotes the positive direction of the stress. Figure 3.3 illustrates the stress notation.

In the case of the two-dimensional state of stress, or what is commonly referred to as the plane stress problem (aZ2 = 0 : x = CT.v = 0), F.q. (3.3) becomes

M = 17 xx ff-vjl

= <TyJ (3.4)

Z

Figure P2.6

ELASTICITY OF STRUCTURES 6 5

3.3 STRESS EQUILIBRIUM EQUATIONS IN A NONUNIFORM STRESS FIELD

In general, a solid which is acta! on by a set of external applied loads experiences a state of stress that is not uniform throughout the body. This condition gives rise to a set of equations which are referred to as the equations of equilibrium. Consider the three-dimensional solid shown in Fig. 3.4.

Using the equilibrium equations of statics yields the following:

ZFx = 0

X dx dy dz + (rr tA + axx A dx)dy dz — <rxx dy dz + {<jxy + <rXTiT dy)dx dz

— axy tlx dz + (a.x + a,x . dz)dx dy — a.x dx dy = 0

Or Tjc.Tr. .\ + ,, + <T:X, z + X = 0 (3.5)

Similarly,

2 Fr = 0 < r x v , + ffvv, y + a.y . + Y = 0 (3.6)

Z Fz = 0 <J.X x + <t=v. + (7„ : + Z = 0 (3.7)

M

Figure 3.4

where X, Y, and Z are body forces. The comma denotes partial differentiation with rcspcct to the following subscript:

d<rzx do

In a cylindrical coordinate set of axes, the equilibrium equations may be easily derived from Fig. 3.5:

<r„,f + ^ „ + °fr ~ + <rr!_z + R = 0 (3.8) j

- <fm.« + ffrfl.r + — + z + © = 0 (3.9) r r

' <rr-<*::.:+- <rth.e + Vri.r+-y + Z = 0 (3.10)

where again R, ©, and 7. are body forces. For plane stress problems, the equilibrium equations simplify to the follow-

ing:

<?xx.x + Vxy.y + -Y = 0 (3.1

axy. x + y + Y= 0

or, in cylindrical coordinates.

„ , ' , Irr ~ <T„ , „ „ rr, r + ~ rO, 0 + + R = 0 r r

, , (3->2) 1 2<rr0 - Cm. 0 + OrO r + ~ 1 " + © = 0 r ' r

3.4 S T R A I N S A N D S T R A I N - D I S P L A C E M E N T R E L A T I O N S H I P S

Strains are nondimensional quantities associated with the deformations (displace-ments) of an element in a solid body under the action of external applied loads.

Figure 3.5

ELASTICITY 01- STRUCTURES 67

To arrive at a mathematical definition of the strain components, take the solid body shown in Fig. 3.6 and consider only the infinitesimal elements OA, OB, and OC.

If, after deformations take place, the displacements of point 0 are denoted by qx, qy, and q. in the .x, j\ and z directions, respectively, then the displacements of points A, B, and C which are dx, dy, and dz away from point O will be qx -f qxxdx, qy + qrr dy, and r/_ + <y_ . rfz, respectively. Figure 3.7 shows all the displacements resulting from the application of external applied loads. When these relative displacements occur in a solid body, the body is said to be in a state of strain. The strains associated with the relative change in length are referred to as normal strains, and those related to relative change in angles are called shear-ing strains.

The normal strain is defined as

where AL is the change in length of an clement whose original length was L before deformation took place. On the basis of Eq. (3.13), the normal strain in the .v direction, for instance, is obtained as follows:

(3.13)

A [OA) O'A' - OA ~ OA" OA

OA = dx

O'A' = i(dx + qx,x dx)2 + {qy.x dx)2 + (q:<x dx)2-}

The normal strain in the .v direction is then

= [(1 + <L..v)2 + (ly.xf + (<?,,)']1/2 - 1

,2-, 1/2

-V

Figure 3.6

6 8 AIRORAi-T STRUCTURES

which reduces (o the following by using the binomial expansion technique:

= <?.*., + i[(g.t.,)2 + Uly, J 2 + (f/z.,)2] + • • •

For small displacements, the terms involving the squares of derivatives may be neglected in comparison with the derivative in the first term. Therefore, the .x-direction linear normal strain becomes

In similar manner, the linearized normal strains in the y and z directions may be derived and are given by

e,, = <7,..v ( 3 . l 4 )

and ~ €., = q.. (3.14r)

The shearing strains may be derived by finding the relative change in the angle between a given pair of the three line segments shown in Fig. 3.7.

€.vr = <J.t. y + <?,•. JC

ex. = qx . + q. x (3.15)

= </,.; + q:.y The derivation of F.q. (3.15) may be found in Rcfs. II, 12, and 13.

Thus, in a three-dimensional state of strain, the strain field components may

be compacted in a matrix form:

^xx w = eyx eyy (3.16)

where e;j = e;i for / f j is a shearing strain and for i = j is the normal strain. In the case of two-dimensional state of strain (plane-strain problem),

e. . = e,x = e.y = 0, Eq. (3.16) reduces to

w •C ^yx ew>-I

In cylindrical coordinates, Eqs. (3.14) and (3.15) may be written as

(3.17)

= llr.r <7z.O ,

r

In, a , 1r , , o, 600 = + — Z;r = Qz,r + <lr,z (3-18)

Ir.O . <10 *zz = 'lz.z er0 = + q„,r r r

where qr, q„, and q. are the displacements in the r, 0, and z directions, respec-tively.

3.5 C O M P A T I B I L I T Y E Q U A T I O N S F O R P L A N E - S T R E S S A N D P L A N E - S T R A I N P R O B L E M S

The strain-displacement relationships for plane-strain problems are given by the following.

= Ix. x

= (3-19)

= Ix, y + Qy.x

By examining Eq. (3.19), it is apparent that there exist three components of strain which arc expressed in terms of only two components of displacements. Thus, it may be concluded that not all these strain components are independent of one another. This may be easily verified by differentiating twice the first and the second equations of (3.19) with respect to y and respectively, and the last equation with respcct to x only.

yv ~ elx, xyy

^yy, xx tfy, yxx

^xy, xy 'Ix. xyy "1" xxy

Substituting the first two equations into the third yields

^A V. .Ry ^xx. vy ^yy. xx (3 .20)

which is the compatibility equation of deformation. Thus, the choice of the three strain components cannot be arbitrary, but

must be such that the compatibility equation is satisfied. This compatibility equation ensures the existence of single-valued displacement functions qx and qy

for a solid. In fact, when Fq. (3.20) is expressed in terms of stresses, it ensures the existence of a unique solution for a stress problem, as is illustrated in later chapters.

For three-dimensional state of stress, the compatibility equations may be derived in similar manner to that of Fq. (3.20) and are given by the following.

€xy. XV ^XX, •>••>• e.V}'. XX

- + e . z . xx

y- - -- 1 f .

.v..,-.- = e „ + e xx 26, y. xz •• yz ~ €xz, yy + Zyz.xy 2e. r. .vv ^ vi :zz+exz.yz

3.6 B O U N D A R Y C O N D I T I O N S

Boundary conditions arc those conditions for which the displacements and/or the surface forces are prescribed at the boundary of a given solid. For instance, if the displacements qx, qy, and qz are prescribed at the boundary, then conditions are referred to as displacement boundary conditions and may be written as

q.v ='/.,('/} <iy ='/,-('/) q-. = q=('i) (3.22)

where </A. cjy, and q. arc known functions of displacements at the boundary. On the other hand, if erAV, <Jrr, and <r are prescribed at the boundary. j=Hen

Ihe conditions are called force boundary conditions and are normally expressed as

x r "xx "xy "zx N„ = "xy "yy (frr

_NZ. _ «zx "yz (T--(3.23)

where the Nn (i = v, v, z) are the surface boundary forces, the HtJ (/', /' = x, y, z) are the prescribed stresses at the boundary, and //v, >jy. //. are the direction cosines. For the derivation or F.q. (3.23) see Ref 14.

It is appropriate to note at this point that in order to obtain the exact stress field in any given solid under the action of external loads, the equations of equilibrium, compatibility equations, and the boundary conditions must all be satisfied.

ELASTICITY 01- STRUCTURES 7 1

3.7 STR ESS-STR AI IN RELATIONSHIPS

Structural behavior may be classified into three basic categories based on the functional relationship between stresses and strains:

1. Inelastic nonlinear 2. Elastic nonlinear 3. Linear elastic

Consider a steadily loaded bar as shown in Fig. 3.8. Tf upon loading the func-tional relationship between the stress and its corresponding strain takes on a curved path, and upon unloading it takes a different curved path, as in Fig. 3.9a, then the structural behavior is referred to as inelastic nonlinear behavior. If such relationship follows the same curved path upon both loading and unloading the bar as in Fig. 2.9b, then the behavior is said to be elastic nonlinear behavior. Last, if such relationship lakes on the same straight path as in Fig. 2.9c, then ihc behavior is termed linear elastic behavior.

In the most general ease for a linear clastic anisotropic solid, Hookc's law, which relates stresses to strains, may be written in a matrix form as follows, in which e = strain, a = stress, and ((', j - x, y, z) are the material elastic con-stants (a{j = fij,).

" l l "l2 " l3 «14 "15 " I 6 err "21 "22 "23 a 24 "25 "26 ay.y e:: "31 «32 "33 "34 "35 "36 e.vr a 4.1 «42 "43 "44 "45 "46

e> ; «5I "5 2 "5 3 "54 "55 "56

J'hi "6 2 "63 "64 "65 " 66_ Jxy.

If there exist three orthogonal planes of elastic symmetry throuj

of the solid body, then Eq. (3.24) becomes

«U "l2 " l 3 0 0 0

e.v.r «21 "22 "23 0 0 0

"31 "3 2 "33 0 0 0 a.. 0 0 0 "44 0 0

C~\7 0 0 0 0 "5 5 0

0 0 0 0 0 "(,(._ J*.

(3.24)

(3.25)

The elastic constants a„ in Eq. (3.25) may be defined in terms of the' engineering-material constants as follows:

Figure 3.6


Figure 3.9 (a) Inelastic nonlinear: (h) elastic nonlinear; (r) linear elastic

Eyy "33 =

al2 = a2i =

0 3 2 = - P = liyy

Ey,

. 2 k

2 k

«55 1

G „ "66 :

(3.26)

where E y . G;j, and vy (/, / = x, y, z) are the modulus of elasticity, shear modulus, and Poisson's ratio, respectively.

A body which obeys Eq. (3.25)—i.e., at each point there exist three mutually perpendicular planes of elastic symmetry—is commonly referred to as an ortho-tropic body. Plywood, for instance, and most reinforced plastics may be des-ignated as orlhotropic materials.

If all directions in a solid are elastically equivalent and any plane which passes through any point of the body is a plane of elastic symmetry, then it is called an isotropic body. In this case, the elastic constants in Eq. (3.25) simplify to

1 fl|l = "22 — tf.13 = 7 E

a 44 = o 5 5 = « f i6 = —

(3.27)

a 12 = «21 = "23

E

«32 = al3 = «3l =

G = 2(1 + v)

where E = Young's modulus, G = shear modulus, and v = Poisson's ratio. Most metals, such as aluminum, steel, and titanium, are isotropic materials. It is important to note here that in an anistropic body, normal strains will

induce not only normal stresses but also shearing stresses; likewise, shearing strains will produce normal stresses, as may easily be seen from Eq. (3.24). How-ever, normal strains in an isotropic or orthotropic body will cause only normal stresses, while shearing strains will cause shearing stresses. This may be verified by examining Eq. (3.25).

For plane-stress problems where a . . , a y : , and a x z arc zero, if we assume isotropic material, Eq. (3.25) becomes

1 - V

1 - v 1 = - v 1 .1) E - v 1

0 0

0

1 2(1 + >•)

(Tyy (3.28)

By substituting Eq. (3.28) into Eq. (3.20) and utilizing Eq. (3.11), the compatibility equation in lenns of stresses alone becomes

ff.™. x, + 2oX), xy + <J„, = 0 (3.29)

The bodv forces are assumed lo be zero.

3.8 T R A N S F O R M A T I O N O F STRESSES A N D S T R A I N S

A stress field tr . rrvr), such as shown in Fig. 3.10, which is known in one set of systems axes may be transformed to any other arbitrary set of axes such as q(l

x

"J Figure 3.10

For instance, assume that at a given point in a solid, the stresses (axx, < T V J , ,

and <rxy) are known in reference to the x and y axes, as shown in Fig. 3.11. The object is to lind the set of stresses in reference to a new set of axes, >ilh which are rotated through an angle 0 as shown. The stress <rm may be found by considering the free-body diagram which is cut by a plane along the P axis at an angle 6 from the vertical, as shown in Fig. 3.116. If side OB is assumed to have area A, then sides OC and CB will each have an area of A cos 0 and A sin 0, respectively. Since equilibrium conditions prevail, the summation of forces along each of the >/ and /? directions must be zero, or

<7n A — <rxx A cos 0 cos 6 — ayy A sin 0 sin 6 — <rx>, A cos 0 sin 9

— rryx A sin 0 cos 0 = 0

By simplifying and noting that ax y = tryx, the following is obtained:

ZF„ = 0 + .

a,n = COS + Sm " — axy S,n cos2 0 + (T . sin2 0 — <T..,. sin 20

(3.30«)

or

o K X s i n 20 IT,.,, s i n 20

^~2 + 2 + (I - 2 cos2 0)a: (3.30b)

In a similar manner, (rpp may be obtained and is given by

dpj! = <rxx sin2 0 + (Tyy cos2 0 + <rxy sin 20 (3.30c)

3 0

Figure P2.6

e l a s t i c i t y o f s t r u c t u r e s 7 5

Equations (3.30a). (3.306), and (3.30c) may be assembled in a matrix form as

(3.31 o) 'Tin —

J V _

cos20 sin20 sin2fl cos20 sin 20 sin 20

- s i n 20 sin 20

2 cos 0 "ft

where 0 is (he angle of rotation and is positive in the clockwise direction. In compact matrix form, Eq. (3.28) becomes

where [ T ] is referred to as the transformation matrix. In a similar manner, the strains may be transformed as follows:

e/'/l = [ T ] e„. 1.

- -'if-

(3.316)

(3.32)

where y,s (tensor shearing strain) = 2ey (engineering shearing strain) and the matrix [ T ] is the same as that in Eq. (3.31).

P R O B L E M S

3.1 Der ive tlie compat ib i l i ty equat ion in cylindrical coord ina te s for a two-dimensional s tate of stress.

3.2 Find the di rec t ion a long which a hole may be drilled in the solid s h o w n in Fig. P3.2 such thai n o shear ing stresses exist a l ong the hole direction.

°vr

_ L

n,

Figure P3.2

3.3 F ind the m a x i m u m antl min imum normal stresses and the planes on which they act for the block shown in Fig. I'3.2. Also, de termine the m a x i m u m shea r stress and the p lane on which it ac t s in Prob . 3.2.

3.4 Der ive Fq. (3.30e).

3.5 F o r a two-d imens iona l s tate of stress, show tha t the stress-strain re la t ionship is

V,, s,2 0 " h " tTvr = Si 2 S22 0 u .

fTxt 0 0 u = where S,. a r e the s t i f fness cons tan ts , defined a s fol lows.

S44 - a h t , A = a , , a 2 2 - a f 2

3.6 Find the ma t r ix of the s t i f fness cons tan t s if a new. a r b i t r a r y set of chosen axes is taken as s h o w n in Fig. P3.6. Assume tha t the st iffness cons tan t s with respect to the x a n d y axes are as given in Prob . 3.5.

/ /

/ /

0 / 7

/ /

\ \

\ \

\

11 F igure P3.6

3.7 A n a r r o w canti lever beam h a s unit width and is loaded with a force Q, a s shown in Fig. P3.7. T h e de fo rmat ions in the x and v d i rec t ions are

1, - *£/(-Qx2y - bQf h Q'-y) + ic/iiGr' - Qczy) <h = {EUvQxy2 + iQx3 - QL2s + §Q/j>)

(«) If 17n = 0, find the stress fields for the beam u n d e r cons idera t ion a t a n y point (JC, y). (/>) Show tha t the stress fields in par t (n) are the s t ress fields for the given beam. (r) If Q = 1000 lb, c = 2 in, £ . = 100 in. El - 10", a n d » = 0.25 a n d by utilizing the slress fields

in part («), find the principal n o r m a l stresses and the m a x i m u m shear s t ress a t a point on the beam given by x — I.J2. y - — r /2 .

(</) Cou ld the fol lowing stress fields be possible stress fields for the given beam'.'

, - Q[y2 + »tv2 •

, = e i y + > - < r -

r = -2 Qvxy

I'2)J

•v2)]

y

Figure P3.7

3.8 You are given a rec tangu la r p la te with positive appl ied stresses (<t„ and<r„). What must be the magni tude o r <rv in o r d e r for the cont rac t ion in the x d i rect ion to be p reven ted? If this plate is subjected to the positive stresses (T„ . ff„. and a x y , in wha t direct ion s h o u l d the plate be d r a w n in order to preserve its angles d u r i n g s t re tching?

ELASTICITY OF STRUCTURES 77

3.9 A plate of unit thickness is subjected to a set of loads Pn and P e uniformly distributed over the sides a and />. respectively. (See Fig. P3.9.) What must be the ratio of magni tudes o f P , and Pf in order for the contract ion of the plate in the * direction to be prevented?

Figure P3.9

CHAITER

FOUR BEHAVIOR A N D EVALUATION OF VEHICLE MATERIAL


It is of utmost importance for the structural analyst to have a full understanding of the behavior of vehicle materials and be able to intelligently evaluate and sclcct the material best suited to the constraints and operational requirements of the design.

The materials used in various parts of vehicle structures generally arc selec-ted by different criteria. The criteria are predicated on the constraints and oper-ational requirements of the vehicle and its various structural members. Some of these more important requirements involve

1. Environment S 2. Fatigue 3. Temperature 4. Corrosion 5. Creep 6. Strength and stiffness 7. Weight limitation 8. Cost 9. Human factor

This chapter familiarizes the reader with the roles that some of these require-ments play in the final selection and evaluation of materials to be used in vehicle struclures.

78

b e h a v i o r a n d e v a l u a t i o n o f v e h i c l e m a t e r i a l 7 9

4.2 M E C H A N I C A L P R O P E R T I E S O F M A T E R I A L S

Materials, in general, may be classified according to their constituent com-position as single-phase or multiphase. All metals, such as aluminum, steel, and titanium, are referred to as single-phase materials. All composites, which are made out of filaments (fibers) embedded in a matrix (binder), are called multiphase materials. Plywood and reinforced fiber glass arc examples of multiphase materials.

Almost all important structural properties of single- or multiphase materials are obtained by three basic tests: tensile test, compression test, and shear test. The American Society for Testing Materials (ASTM) sets all the specifications and test procedures for materials testing.

Tensile Test

Figure 4.1 shows the basic configuration of a tensile test specimen. The load P is applied gradually through the use of a tensile testing machine. The normal strain €„ usually is measured cither by utilizing electrical strain gage techniques or by measuring the total elongation <5 in an effective gage length L for various values of the tension load P. For small loads, the elongation is assumed to be uniform over the entire gage length /, and therefore the normal strain may be mathemat-ically expressed in the form

<5 c„ = z (4.1,

where 6 and L are both measured in the same units of length. The corresponding normal stress a„ is also assumed to be uniformly distributed over the cross-sectional area A of the test specimen and is obtained as follows:

^ (4-2)

For common engineering units, the load P is in pounds, the area A is in square inches, and the stress <r is in pounds per square inch. The stress-strain diagram for a material is obtained by plotting values of the stress a against corresponding values of the strain e, as shown in Fig. 4.2. For small values of the

ta)

D C Figure 4.1

Stress __ ^ Ultimate strength

C / * \ Cm) * Jr V II Yield point (o , r )

Proportional limit (o ) L D Strain, e

GO

Figure 4.2

stress, the stress-strain curve is a straight line, as shown by line OA of Fig. 4.2. The constant ratio of stress to strain for this portion of the curve is called the modulus of elasticity E, as defined in the following equation:

E = — (4.3)

where E has units of pounds per square inch. A material such as plain low-carbon steel, which is commonly used for bridge

and building structures, has a stress-strain diagram such as that shown in Fig. 4.2«. At point B, the elongation increases with no increase in load. This stress at this point is called the yield point, or yield stress, <rly and is very easy to detect when such materials are tested.

The stress at point A, where the stress-strain curve first deviates from a straight line, is called the proportional limit a,p and is much more difficult to measure while a test is being conducted. Specifications for structural steel usually are based on the yield stress rather than the proportional limit, because of the ease in obtaining this value.

Flight-vehicle structures arc made of materials such as aluminum alloys, high-carbon steels, and composites which do not have a definite yield point, but which do exhibit stress-strain behavior similar to that shown in Fig. 4.2b. It is convenient to specify arbitrarily the yield stress for such materials as the stress at which a permanent strain of 0.002 in/in is obtained. Point B of Fig. 4.2b rep-resents this yield stress and is obtained by drawing line BD parallel to OA

b K h a v i o K a n d e v a l u a t i o n o f " v e h i c l e m a t e r i a l 8 1

through point D, representing zero stress and 0.002-in/in strain, as shown. When the load is removed from a test specimen which has passed the proportional limit, the specimen does not return to its original length, but retains a permanent strain. For the material represented by Fig. 4.2a, the load might be removed gradually at point C. The stress-strain curve would then follow line CD, parallel to OA, until at point D a permanent strain equal to OD were obtained for no stress. Upon a subsequent application of load, the stress-strain curvc would follow lines DC and CG. Similarly, if the specimen represented by Fig. 4.2b were unloaded at point B, the stress-strain curve would follow line BD until a per-manent strain of 0.002 in/in were obtained for no stress.

It is customary to use the initial area A of the tension test specimen rather than the actual area of the necked-down specimen in computing the unit stress a. While the true stress, calculated from the reduced area, continues to increase until failure occurs, the apparent stress, calculated from the initial area, decreases, as shown by the dotted lines GH of Fig. 4.2. The actual failure occurs at point H, but the maximum apparent stress, represented by point G, is the more important stress to use in design calculations. This value is defined as the ullimute strength o lu . In the design of tension members for vehicle structures, it is accurate to employ the initial area of the member and the apparent ultimate tensile strength rrlu. In using the stress-strain curve to calculate the ultimate bending strength of beams, as shown in a later chapter, the results are slightly conservative because the beams do not neck down in the same manner as tension members.

Compression Test

The compressive strength of materials is more difficult to identify from stress-strain curves than the corresponding tensile strength. Compressive failures for most structural designs in engineering applications are associated with instabil-ities which are related to yield stress rather than ultimate stress. Yield-stress values which are obtained based on the 0.2 percent offset method (0.002-in/in permanent strain) have been proved to be relatively successful for correlating instabilities in most metals; however, the correlation is less satisfactory for non-metals, specifically composites. Specimen geometry and means of supports seem to have considerable elfect on compression test results. The compressive stress-strain diagrams for most materials arc similar to the tensile stress-strain curves.

Shear Test In-plane shear properties are the most difficult to obtain and have the least standardized testing procedures of all major mechanical properties. The design of a shear-test specimen having a test section subjected to a uniform shearing stress is impossible. The closest practical approach is probably a thin-walled circular cylinder loaded in torsion, as shown in Fig. 4.3. For small displacements, the shear strain is expressed as

82 AIRORAi-T STRUCTURES

Figure -1.3

where e, = shear strain ' AS = change in arc length

L = effective gage length

The shear stress may be calculated from the well-known strength-of-material torsional equation as

Tr (4.5)

where cr, = shear stress r = cylinder radius

J = cross-sectional polar moment of inertia

For very thin-wailed cylindrical test specimens, Hq. (4.5) may be written as

TRm (4.6)

where Rm — mean radius = (R„ + /?,)/2 and Jm = InR^t. The shear modulus of elasticity may be obtained from the shear stress-shear

strain diagram as

e. (4.7)

4.3 E Q U A T I O N S F O R STRESS-STRAIN C U R V E IDEALIZATION'

In the design of vehicle structural members, it is necessary to consider the proper-ties of the stress-strain curve at stresses higher than the elastic limit. In other types of structural and machine designs, it is customary to consider only stresses below the elastic limit; but weight considerations are so important in flight vehicle design that it is necessary to calculate the ultimate strength of each member and to provide the same factor of safety against failure for each part of the entire structure. The ultimate bending or compressive strengths of many members are difficult to calculate, and it is necessary to obtain information from destruction tests of complete members. In order to apply the results on tests of members of one material to similar members of another material, it is desirable to

r k i i a v i o r a n d e v a l u a t i o n o f v e h i c l e m a t e r i a l 8 3

obtain an idealized analytical expression for the stress-strain diagrams of various materials.

Romberg and Osgood'" have developed a method of expressing any stress-strain curve in terms of the modulus of elasticity E, a stress a, (which is approxi-mately equal to the yield stress), and a material shape factor n. The equation for the stress-strain diagram is

€ = B + (4.8)

where e and a are dimcnsionlcss terms defined as follows:

i = — (4.9) <7i

and <T = ~ (4.10) ffi

The curves expressed by Eq. (4.8) are plotted in Fig. 4.4 for various values of n. A material such as mild steel, in which the stress remains almost constant above the yield point, is represented by the curve for n = oo. Other materials, with various types of stress-strain diagrams, may be represented by the curves for other values of n. In order to represent the stress-strain diagrams for all materials by the single equation, it is necessary to use the reference stress value of a, rather than the yield stress. The value of IT, is obtained as shown in Fig. 4.5 by drawing the line a = 0.7Ee. from the origin to the stress-strain curve and obtaining the stress coordinate cr, of this point of intersection. The stress <rl is approximately equal to the yield stress for typical flight vehicle materials. The value of n may be determined so that Eq. (4.8) fits the experimental stress-strain curve in the desired region. Romberg "and Osgood show that for most materials the value ofn may be accuratcly determined from the stress ov and a similar stress tr2 on the line a = 0.85£e.

_ i :c t = --"i

Figure 4.4


It is appropriate to note here that similar equations for idealizing shear stress-shear strain curves may be formulated and fitted into various materials' shear test data.

4.4 F A T I G U E

Fatigue is a dynamic phenomenon which may be defined as the initiation and propagation of microcracks into macrocracks as a result of repeated applications of stresses. It is a process of localized progressive structural fracture in material under the action of dynamic stresses. A structure which may not ever fail under a single application of load may very easily fail under the same load if it is applied repeatedly. This failure under repeated application of loads is termed fatigue

failure. In spite of the many studies and vast amount of experimental data accumu-

lated over the years, fatigue is still the most common cause of failure in machin-ery and various structures as well as the least understood of all other structural behavior. This lack of understanding is attributed to the fact that the initiation and the propagation or microscopic cracking arc inherently statistical in nature. In fact, the analyst often is confronted with the wide variations in the statistics-'of what may be estimated as (1) type of service and environment, (2) magnitude of service loads and frequency of occurrence, (3) the quality control during the fabrication operations, (4) the extent and accuracy of the analyses in determining stresses, and (5) the applicability or the material strength data.

The main objective of all fatigue analyses and testing is the prediction of fatigue life of a given structure or machine part subjected to repeated loading. Such loads may have constant amplitude, as indicated in Fig. 4.6; however, in fiight vehicle structures, the load history is usually random in nature, as shown in Fig. 4.7.

Fatigue-life prediction Most of the fatigue-life prediction methods used in the design of structures have been based on fatigue allowable data generated by sine

b e h a v i o r a n d e v a l u a t i o n o f v e h i c l e m a t e r i a l 8 5

I

Figure 4.6 Sine loading.

wave excitation. Only recently closed-loop, servo-controlled hydraulic machines have become available for true random loading testing. Fatigue test data usually are presented graphically (Fig. 4.8) and the curves are referred to as the allowable S-N curves. The curves in conjunction with the "cumulative damage" concept form the basis for most of the methods used in the prediction of fatigue

Among the several theories proposed for fatigue-life prediction, the Palmgrcn-Mincr theory,26 because of its simplicity, seems to be the most widely used. The method hypothesizes that the useful life expended may be expressed as the ratio of the number of applied cycles rj; to the number of cycles /V,- to failure at a given constant stress level a-,. When the sum of all the fractions reaches 1, failure should occur. Mathematically this failure criterion is written as

It is important to note that in Palmgren-Miner theory no provisions are made to take into account the various effects on fatigue life, such as notch sensitivity effect, loading sequence effect (high-low or low-high), and the consequences of

life.

or

Figure 4.7 R a n d o m stress loading.

'I iinc

Figure 4.8 S-N fatigue curves.

different levels of mean stress. In fact, the theory has been shown to yield un-conscrvative results in some test case studies conducted by Gassner,20

Kowalewski,21 and Corten and Dolan.22

To illustrate the use of Eq. {4.II), consider a bracket which supports an electronic box in the aircraft cockpit. In a typical mission, the bracket encounters a stress history spectrum idealized as shown in Fig. 4.9. The fatigue allowable of the bracket material is given in Fig. 4.10. The problem is to find the number of missions the aircraft may accomplish before the bracket fails.

From Figs. 4.9 and 4.10, Table 4.1 may be easily constructed. Therefore, in one mission 0.433 percent of the useful life of the bracket is expended. This means

24

12

a ft <

56.5 llz

65 llz

70111

100 l l z

Time, s

5 0 l l z

120llz

L_ 1U 12 14

Figure 4.9 Bracket stress history spectrum per mission.

B K H A V I O K AND EVALUATION OF "VEHICLE MATERIAL 8 7

Af cycles t o fa i lure

Figure 4.10 .V-<Y curves.

that tiic aircraft might accomplish about 200 missions without bracket failure, based on Eq. (4.6).

Fatigue test data: S-IS' curves Fatigue tests arc conducted for a wide variety of reasons, one of which is to establish materials' fatigue allowables, or what is commonly referred to as the S-N curves. Contrary to results of static tests, it has been observed that the scatter in fatigue test results can be quite large. This inherent scatter characteristic leaves no choice but to treat the results statistically. One of the more widely used statistical distribution functions is the log-normal distribution, whose-mean value is taken as

M = - i log Nt (4.12) " i = i

where n — total number of specimens tested at the same stress level and AT, = number of cycles to failure for specimen /.

Table 4.1

ij,, cycles n,. kips'in '' ;V(. cycles Hi N,

200 6 y. 0 141 20 50.00(1 0.00282 105 10 X 0 260 14 10" 0.00026

25 22 2 x I0 J 0.00125 240 4 j. 0

0.00433


The unbiased standard deviation of M is defined by

rt (log AT.--M)*

' - r - i - J ( 4 - i 3 )

In order to calculate the number of cycles to failure, based on some confidence level, the standard variable ( is taken as

or log N = M + £5 (4.15)

The probability of surviving log N cycles is

1 ProbabiIity(log N) = P(log N) = - 7 = e " W 2 d£ (4.16)

V 2 ? t J{

Equation (4.16) may be used to tabulate the probabilities of survival for various values of the standard variable f, as shown in Table 4.2.

To illustrate the procedure, consider the following actual fatigue test results for seven tested specimens:

Specimen no., / Cycles to failure Nt

1 61,318 2 39,695 3 62,803 4 51,039 5 83,910 6 35,631 7 96,500

From Eq. (4.12) the mean value is

M = - t log £ log N,-n i = l ' i = l

= 4(log 61,318 + log 39,695 + • • • + log 96,500)

= 4.76463

Table 4.2

c Probability of survival, %

-1 .280 90.0 -1.645 95.0 -2.330 99.0

BKHAVIOK AND EVALUATION OF "VEHICLE MATERIAL 8 9

From Eq. (4.13) the unbiased standard deviation is 11/2

5 = ^ (log ,V, - M)2 J

( l o g 6 1 , 3 1 8 - 4 . 7 6 4 6 3 ) 2 + • • • + ~ j (log 96,500 - 4.76463)2~1'/2

6

= 0.0503

From Eq. (4.15), if we assume a probability of survival of 95 percent,

log N = M + £<•> = 4.76463 + (-1.645X0.0503) = 4.68189

Therefore the number of cyclcs to failure for a 95 percent probability of survival is

N = antilog 4.68189 = 48,100 cycles

4.5 S T R E N G T H - W E I G H T C O M P A R I S O N S O F M A T E R I A L S

The criterion commonly used i.n the sclcction of structural materials for aero-space vehicle application is that which yields minimum weight. This involves selecting the proper combination of material and structural proportions with the weight as the objective function to be minimized to yield an optimum design. Although weight comparisons of materials may be based on several factors, such as resistance to corrosion, fatigue behavior, creep characteristics, strength, and so on, the only treatment given here is with respect to strength.

As an illustration, let us consider the three loaded members shown in Fig. 4.11; for simplicity, it is assumed that there exists only one free variable (the thickness /, in this ease) to be chosen in the design. The criteria which govern the design of members in Fig. 4.1 la, b, and c are ultimate uniaxial tension, ultimate

(M U)

Figure 4.11 I.onded mcmbci.s. («) I 'ure Icnsion; ih) pure compression; (r) pure bending.

uniaxial compression or buckling, and ultimate bending. The expressions relating the applied external loads to the induced actual stresses are

P Tension: a, = — (A = bi) (4.17)

Compression (buckling): a = — — l / = — - J (4.18)

„ Aff Bending: __ a" = J f ( 4 J 9 )

where <r,, a c , ah = ultimate tensile, compression, and flcxurai stresses, respectively A = cross-sectional area I = moment of inertia of member

The weight of the member may be expressed in terms of the material density (f> pounds per cubic inch) and the fixed and free geometric dimensions as

IV -= Lbtp (4.20)

Solving for the free variable t from Eqs. (4.17) through (4.19) and substituting into Eq. (4.20) yields the material weight required to meet each specified design cri-terion. Thus

(tension) (4.21)

l3bp f 12^1/2

W = — ( — J (compression) (4.22)

6 M f e V ' 2

Ob W = Lp ( J (bending) (4.23)

Willi Eqs. (4.21) to (4.23) available, weight comparisons of different materials may be conducted. Thus the weights of two different materials required to carry the axial load P may be readily obtained from Eq. (4.21) as

f a a a

where the subscripts 1 and 2 refer to materials 1 and 2, respectively, and a,, and a,2 arc the ultimate tensile stresses of materials I and 2, respectively.

I1! ' = I'l h i ( 4 . 2 4 )

Pi

Similarly, the ratio of weights of two members of two different materials resisting the same bending moment may be easily obtained by utilizing Eq. (4.23):

BKHAVIOK AND EVALUATION OF "VEHICLE MATERIAL 91

Likewise, the ratio of weights for two members of two different materials resisting the same compressive (buckling) load may be written immediately by using Eq. (4.22):

Typical aerospace vehicle sheet materials arc compared in Table 4.3 by means of liqs. (4.24) through (4.26). The weights of the various materials are compared with the aluminum alloy 2024-T3. The weight ratios for tension mem-bers, shown in column 5. do not vary greatly for the different materials. For members in bending, however, the lower-density materials have a distinct advan-tage, as shown in column 6. Similarly, the lower-density materials have an even greater advantage in compression buckling, as indicated in column 7. Values of a vary with sheet thickness, and those shown are used only for comparison.

The computations of Table 4.3 indicate that the last three materials, having lower densities, are superior to the aluminum alloys. However, it is important to note that magnesium alloys arc more subject to corrosion than aluminum alloys, while wood and plastic materials are less ductile. Brittle materials are undesirable for structures with numerous bolted connections and cutouts which produce local high-stress concentrations. Ductile materials, which have a large unit elongation at the ultimate tensile strength, will yield slightly at points of high local stress and will thus relieve the stress, whereas brittle materials may fail under the same conditions. Fiber-reinforced plastics have been used successfully for acrospace vehicle structures as long ago as the late 1940s and early 1950s. In those days, the main reinforcement was glass fiber in fabric form with polyester resin as the bonding agent. Since then and primarily in the last few years, development of new high-modulus fibers (such as boron, silicon carbide, graphite, and beryllium) in combination with high-modulus, high-temperature-resistant resins (such as cycloaliphatic epoxies, polymeric and polybenzimidazole resins) has added a new dimension to materials Tor applications in aerospace and marine- and land-based structures. These new fibers and resins arc being combined in a unidirectional,

Table 4.3 Strength-weight comparisons of materials

(4.26)

Rat io of weight to weight of 2024-T3 a l u m i n u m alloy

T e n s i o n : Bending: Buckl ing:

Sheet material {I)

a. kip\ in" average />. lb in"

(21 (3)

/•;(in1). r, "j kips./inJ p2 a ,

(4) (5)

Stainless s tcd Aluminum alloy 2024-T* Aluminum alloy 7075-Tf» Magnesium alloy Laminated plastic Sprucc wtunl

1X5 0.2X6 Wi n.100 77 0.101 4 0 0.065 30 0.050

9.4 (I.OI56

26 10.5 10.4

.5 2.5 1.3

1.23 1.72 1.00 1.00 0.87 0 9 3 1.07 0.83 1.10 0.74 1.09 0.42

2.12

1.00

1.01 0.77 0.83 0.31


preimpregnated form which gives the analyst complete freedom to tailor the composite (a composite is made of a certain number of unidirectional plies) to meet the imposed load requirements in both magnitude and direction. Studies have indicated that through the use of composite materials, the total weight of an aerospace vehicle could be reduced by more than 35 percent.

4.6 S A N D W I C H C O N S T R U C T I O N

The problem of increasing weight which accompanies increasing material thick-ness is being met frequently by the use of sandwich construction in application to aerospace vehicles. This type of construction consists of thin, outer- and inner-facing layers of high-density material separated by a low-density, thick core material.

In aerospace applications, depending on the specific mission requirements of the vehicle, the material of the sandwich facings may be reinforced composilcs, titanium, aluminum, steel, ctc. Several types of core shapes and core materials may be utilized in the construction of the sandwich. The most popular core has been the "honeycomb" core, which consists of very thin foils in the form of hexagonal cells perpendicular to the facings.

Although the concept of sandwich construction is not new, only in the last decade has it gained great prominence in the construction of practically all aerospace vehicles, including missiles, boosters, and spacecraft. This is primarily a result of the high structural efficiency that can be developed with sandwich construction. Other advantages offered by sandwich construction are its excellent vibration and flutter characteristics, superior insulating qualities, and design ver-satility.

An element of a sandwich beam is shown in Fig. 4.12. For simplicity, the facings are assumed to have equal thickness t f , and the core thickness is tc. It is also assumed that the core carries no longitudinal normal stress a. Let us con-sider the case where it is desired to find the optimum facing thickness which results in a minimum weight of the sandwich beam carrying a bending moment M. Without any loss in generality, H and L may be taken as unit values, and thus

' / C o r e Fac ing

Figure 4.12 Sandwich beam clement.

BKHAVIOK AND EVALUATION OF "VEHICLE MATERIAL 9 3

the resisting bending moment can be expressed as

M = otj (tf + tc)

If we assume that t, is small compared to te, which is normally the case, then the above equation reduces to

M = atf tc

or

M = apt\ (4.27)

where t j is expressed in terms of tc by the equation tj- = The weight of a unit element of the beam is approximately

W = Pc tc + 2pfptc (4.28)

where pc and pf equal the core and facing densities, respectively. Eliminating the variable tc from Eqs. (4.27) and (4.28) yields

W = (pr + 2 p f P ) J ^ (4.29)

The value of /? for the minimum weight may be obtained by differentiating Eq. (4.29) wilh respect lo [I and equaling the derivative to zero. Performing the differentiation and solving for p yield

P = (4-30) 2 pf

Equation (4.30) indicates that for a sandwich material resisting bending moment, the minimum weight is obtained when the two layers of the face material have approximately the same total weight as the core. Note that this condition does not yield minimum weight if the beam element is under the action of compressive (buckling) load.

It is now possible to compare the weight of a sandwich-construction beam element with that of a solid element corresponding to the sandwich face material, if we assume that they both resist the same loads. A sandwich element designed to resist bending moments will have a total weight equal to twice the total weight of the facings, if the face and core materials have equal weights. Thus, from Eq. (4.28)

W = 4pfit (4.31)

By solving for t from Eq. (4.27) and substituting into Eq. (4.31), the following weight for the sandwich element is obtained:

W = 4pP / P (432) V P a


The weight Ws of a solid beam element from Eq. (4.23) is

(4.33)

Hencc. the ratio of the weight of a sandwich beam element to that of a solid element of the corresponding sandwich face material is

It is important to note that Eq. (4.34) is valid only for a sandwich in which the total weight of the facings is equal to the weight of the core. In order to compare the weights of a sandwich with solid elements studied in Table 4.3, consider a sandwich whose facings arc made of 2024-T3 aluminum alloy and a core material whose density is 0.01 lb/in3. From Eq. (4.30)

From Eq. (4.34)

^ = ^ 0 . 0 5 2 pf 2(0.1)

^77= 1.63 y a 0 5 = 0.37 w ;

Thus, the sandwich has only 37 percent of the weight of a solid element resisting the same bending moment. Also note that the value of 0.37 is less than any of the other values in column 6 of Table 4.3.

In the preceding discussion, it was assumed that the proportions for the sandwich clement were limited only by theoretical considerations. In actual struc-tures, practical considerations are much more important. The thickness of the face material, for example, usually is greater than the theoretical value, because it might not be feasible to manufacture and form very thin sheets. Likewise, the core was assumed to support the facings sufficiently to develop the same unit stress as in a solid element, whereas the actual low-density materials might not provide such support. /

4.7 T Y P I C A L D E S I G N D A T A F O R M A T E R I A L S

In the manufacture of materials, it is not possible to obtain cxactly the same structural properties for all specimens of a material. In a large number of tested specimens of the same material, the ultimate strength may vary as much as 10 percent. In the design of an aerospace vehicle structure, therefore, it is necessary to use stresses which are the minimum values that may be obtained in any specimen of the material. These values are termed the minimum guaranteed values of the manufacturer. The licensing and procuring agencies spccify the minimum values to be used in the design of aerospace vehicles. These values are contained

m-IIAVlOR AND EVALUATION OF VF.H1CI.K MATERIAL 9 5

Table 4.4 Typical mechanical data for materials

Tension Ultimate stress Yield stress Proportional limit

E Modulus of elasticity c Elongation

Compression Ultimate (block) stress Yield stress

ffCP Proportional limit Column yield stress Modulus of elasticity

Shear Ultimate stress Torsional modulus of rupture

a'c Proportional limit (torsion) a Modulus of rigidity (torsion)

Bcarimf Ultimate stress Yield stress

in Military Handbooks, such as M1L-HDBK-5A, MIL-HDBK-5, M1L-HDBK.-17, MIL-HDBK-2B. Tabic 4.4 shows the typical mechanical data required in the design of aerospace structures. Normally these data are presented in accordance with one of the following bases:

A hash: At least 99 percent of all mechanical property values are expected to fall above the specified property values with a confidence of 95 percent.

B basis: At least 90 percent of all mechanical property values are expected to fall above the specified property values with a confidence of 95 percent.

S basis: Minimum mechanical properly values as specified by various agencies.

I 'ROBLKMS

4.1 The buckling load for a sandwich column is approximately given by P ~ x2 El/I.'. Kind ihe thickness ratio of facing lo the core which results in an op t imum design (minimum weight) for the column. Sec Fig. P4.1.

<f I

~ 7 ~ l m m n r 1 1 1 1 1 1 1 1 1 \ 1 1 1 1 1 1 1 1 1 M 1 1 1 1 1

l f Figure P4.1

4.2 Find the weight ra t io or a sandwich column to that of a solid column whose material is the same as thai of the sandwich facings.

4 J Work Prob. 4.2 for a core density of 0.015 lh/in'1 and the following specific cases of facing materials:

(а) 2024-T3 aluminum alloy (density = O.I lb/in3) (б) 6A1-4V t i tanium (density = 0.16 lb/in3) (c) 321 stainless steel (density = 0.286 lb/ in ' ) (J) Inconel (density = 0.3 Ih/in ') (<•) Beryllium (density = 0.069 lb/in3) IJ) Reinforced composite (unidirection)

(!) Glass fiber (density = 0.09 lb/in ') • (2) Boron fiber (density = 0.095 lb/ in J)

(3) Graphi te (density = 1.053 lb/in3) 4.4 A missile-holding fixture on an aircraft is subject dur ing each High! to Ihc stress-load history shown in Fig. P4.4. After how many flights will the fixture fail if the material fatigue allowable is that shown in Fig. 4 . I0?

60 —

S 5 0 ~ . 1 40 -

8 30 -

20 -

10 -

Figure P4.4

6 S 10 T i m e , S

14 16

4S Ten par ts were futiguc-teslcd al the same stress level, and the following failure cycles were repor ted :

Specimen no. Cycles to failure

1 300,000 2 190,000 3 225,000 4 - 350,000 5 260,000 6 280,000 7 490,000 8 310.000 9 360,000

10 390.000

After how many cycles should the part be replaced so that only the following percentage of the parts in service fails before replacement'.'

(a) 1 i (ft) 5 ( d 10

C H A P T E R

FIVE STRESS ANALYSIS


In order to select sizes of structural members to meet the design load require-ments on a specific aerospace vehicle, it is necessary to find the unit stresses acting on the cross section of each structural element. The unit stress referred to here is the force intensity at any point, and it has units of force per unit area, or pounds per square inch in common engineering units.

It was shown in Chap. 3 that there exist two distinct components of stress, normal and shear stress. A normal stress is a unit stress which acts normal to the cross section of the structural element, while the shear stress is parallel and in the plane of the cross section. A normal stress is induced by bending moments and axial forces. A shear stress, however, is caused by torsional moments and shear forces. This chapter discusses the theory and the application of these two funda-mental slrcss components.

5.2 F O R C E - S T R E S S R E L A T I O N S H I P S

The stress field at any choscn point in a solid beam may be entirely defined by the components of force resultants or stresses acting along the directions of some "gaussian" coordinate system, as shown in Fig. 5.1. The forces and stresses are taken to he positive if they act in the positive direction of the corresponding coordinate axis.

97

1 1

/ 7 /

(b)

Figure 5.1 Stresses and force resultants.

From Fig. 5.1 the force resultants may be related to the stresses as follows:

P = o\ dA M

(TX). dA M

V. = axz dA JA

where P = axial force K., V. = shear forces j *

M_, Mr = bending moments T - torque

<Txx = normal stress (Ttv, <Jx. = shearing stresses

= = - I3"7" = £ <IA dA

(5.

z<rxf) dA

STR1-SS ANALYSIS 9 9

5.3 N O R M A L S T R E S S E S IN B E A M S

The normal stresses in beam elements are induced by bending and/or extensional actions. Two approaches may be used to determine stresses; the first is based on the theory of elasticity, and the second is based on strength-of-materials theory. The latter, which is used here, assumes that plane sections remain plane after extensional-bending deformation takes place. This assumption implies that the deformations due to transverse shear forces (K and Vy) are very small and there-fore may be neglected. In addition, this assumption allows the displacements (deflections) of any point in the beam to be expressed in terms of the displace-ments of points located on the beam axis.

Assume that the displacement in the x direction of any point in the beam is represented by qx{\, y. z). If we take ujs) to be the extensional displacement of any point on the beam axis [y = z = 0) and tj/. and ipy to be the rotational displacements of the beam cross section, then

</,(.v. y, z) = „ , ( .v) - y xj,z(x) + z,j,y(x) ( 5 . 2 )

The axial strain from Fq. (3.14A) is defined by

«.v.v = qx, x Hence, from Eq. (4.2)

e.vv = !<>. v - V'A;. + zi/\.., (5.3)

At any given cross section .v = .x0,

dux(xQ) dx

dUx o)

dx

#y(-v o) dx

— const = Bl

= const = B2

= const = B3

where Eq. (5.3) becomes

<Exx= Bt + B2y + B3z (5.4)

In order to determine the stresses which correspond to the strains in Eq. (5.4), the stress-strain relationship in Chap. 3 is utilized. By assuming that the stresses a . , and ay r are negligible compared to a x x , the following relationship for an isotropic material may be obtained easily from Eq. (3.25):

= Eexx (5.5)

where F = modulus of elasticity of the material. Substituting Eq. (5.4) into Eq. (5.5) yields

=/•;(«, + B2y+ B3Z) (5.6)


Constants S, , B», and S , may now be determined through the use of Eq. (5.1), or or

P = E(B, + B 2 y + B_,z) J.4

M. = - | EY{B, + B2 y + B3 2) J / t

M, = | £r(B» + B2 y + B3 z)

Carrying out the integrations yields

- = B , / i + B 2 y + B 3 z td

•M ^ = B , y + L B2 + FY. B3

MR

where ,4 = cross-sectional area

J>

/v . = I yz (M = product moment of inertia of cross section JA

z CLA

(5.7)

/ . = y'2 (IA = moment of inertia of cross section about x axis (5.8«)

dA — moment of inertia of cross section about y axis (5.8h)

(5.8c)

45-9)

If the z and y axes are taken through the geometric ccntroid of the cross section, then y and z become identically zero. Hence Eq. (5.7) reduces to

-M. = /__ B, + / , , (5.10)

STR1-SS ANALYSIS 101

Solving fiqs. (5.10) for the unknown constants yields

> - 7 i

_LMt + !rzM, 3 E(Iy / . — Iyz)

Substituting Eqs. (5.11) into Eq. (5.6) yields the general expression of the normal stress:

P /„ M. + 7R= M.. . IZMV + IVZMZ

When Eq. (5.12) is used, it is important to observe the sign convention used in the derivation. See Fig. 5.1. In cases where y and z axes are principal axes of the cross-sectional area, the product of the moment of inertia Iyz about these axes is zero. For this condition, Eq. (5.12) reduces to

P M- My

If there is no axial force acting on the beam and bending occurs about the z axis only, then Eq. (5.13) reduces to the familiar strength-of-material pure bending equation

ax x = ^ (5.14) 1-y

5.4 S H E A R STRESSES IN B E A M S

The shear stresses in beams are induced by pure shear force action and/or tor-sional action. In this section, only shear stresses due to shear forces are con-sidered, while the latter are dealt with in a separate section.

Consider a small section of a beam, as shown in Fig. 5.2. For simplicity, assume that the beam cross section is symmetrical and the theory of strength of materials holds. The shear force Vy parallel to the beam cross section produces shear slresses irxy of varying intensity over the cross-sectional area. Correspond-ing to the vertical shear stress axy there exists a shear stress ayx in the xz plane which is equal to a x t at the points of intersection of the two planes. Thus, the expression of the vertical shear stress ax y at any point in the cross section is obtained by determining (he shear stress ayx on a horizontal plane through the point.

The bending stresses on the left and right sections of the beam element (Fig.

102 AIRCRAFT s t r u c t u r e s

y

/ M: + I-"v 5.v

x

A Figure 5.2 lie;mi element.

5.2) arc shown in Fig. 5.3. At any point a distance >' from the neutral axis, the bending stress will be M.y/L on the left Face and M.y/Iz + Vy>iy/L on the right facc. In order to obtain the shear stress at a distance y = yl above the neutral axis, the portion of the beam above that point is considered as a free body, as shown in Fig. 5.3c. For equilibrium of the horizontal forces, the force produced by the shear stress ayx on the horizontal area of width t and length 5x must be equal to the difference in the normal forces on the two cross sections. Summing forccs in the horizontal direction yields

where the integral represents the moment of the area of the cross section above the point where the shear stress is being determined, with the moment arms measured from the neutral axis. The cross-sectional area considered is shown by the shaded portion in Fig. 5.3a It is important to note that Eq. (5.16) is appli-

(5.15)

Equation (5.15) may be written in standard form as

V Cc

( 5 . 1 6 )

,1Ly + I V u '

'1

Neutral ;ixis

(«) (/•) (<'}

Figure 5.3

s t r i 5 s a n a l y s i s 1 0 3

cubic only to beams of uniform, symmetrical cross sections. Tapered beams and beams of unsymmetrical cross section are considered later.

FA a in pic 5.1 Find the maximum normal stress in the beam in Fig. 5.4 and the shear stress distribution over the cross section.

S O L U T I O N The maximum normal stress due to bending will occur at the point of maximum bending moment, or at the fixed end of the beam. Since the shear force is constant throughout the beam span, the shear stress dis-tribution will be the same at any cross section. The moment of inertia for the cross is obtained as follows:

/ . = 2 3 x + 3 x 2.51 + 1 1 x - 1 = 4 3 . 3 i n 4

= P = My = 0

The maximum normal stress is

M.y <TA(

40 x 20( + 3) 2 ^ = +55.4 kips/in

For a point 1 in below the top of the beam, the integral of Eq. (5.16) is equal to the moment of the area of the upper rectangle about the neutral axis:

y tIA = 2.5(3) = 7.5 in3 .

The average shear stress just above this point, where f = 3 in is

Vr " x * = 7 . >' d A = . t ^ 0 0 0 , 7.5 = 2310 lb/in2

43.3 x 3

The average shear stress just below this point, where t = 1 in is

K

I. ydA = | ^ - 7 . 5 = 6930 lb/in2

For a point 2 in below the top of the beam, the integral of Eq. (5.16) is

y dA = 2.5 x 3 + 1.5 x 1 = 9.0 I' NJ

\

\

\

\

T 3 in

J - -3 ill -—J 1

" T 1 in

40 kips

Figure 5.4

1 0 4 a i r c r a f t s t r u c t u r e s

This shear stress at this point is

K = 9.0 = 8320 lb/in2

At a point on the neutral axis of Ihe beam, the shear stess is

cy, = d A = < 1 5 x 3 f 1 x 2) = 8780 lb/in2

The distribution of shear stress over the cross section is shown in Fig. 5.5. The stress distribution over the lower half of the beam is similar to the distribution over the upper half because of the symmetry of the cross section about the neutral axis.

Alternative Solutions for Shear Stresses

Fn some problems it is more convenient to find shear stresses by obtaining the forces resulting from the changc in bending stresses between two cross sections than it is to use Eq. (5.16). Portions of the beam between two cross sections a unit distance apart are shown in Fig. 5.6. The bending moment increases by VY in this unit distance, and the bending stresses on the left face of the beam are larger than those on the right face by an amount Vy>]jJl:, where IJ = 1. At the top of the beam, this difference is

K.y 40(3) r 2

f=^n=2-77kips/in

The differences in bending stresses at other points of the cross scction are obtained by substituting various values of y and are shown in Fig. 5.6ft. Cutting sections and utilizing the equations of static equilibrium in each case (Fig. 5.6c, d, and <?) yield the shearing stresses at these various points:

(yyx = 6930/3 = 2910 lb/in2 at 1 in below top of beam

cy x = 8320/1 = 8320 lb/in2 at 2 in below top or beam /

<jyx = 8780/1 = 8780 lb/in2 at neutral axis

Note that these shear stress values are the same as shown in Fig. 5.5.

R

d

2 3 1 0 lb / in 2

/ 6 9 3 0 lb/ in 2

8 3 2 0 lb / in 2

^ t — S 7 S 0 lb / in 2

E ^ r S320 lb / in 2

ft''30 lb/ in 2

2 3 1 0 lb/ in 2 F i g u r e 5 .5

s t r 1 - s s a n a l y s i s 1 0 5

Example 5.2 In ihe beam cross section shown in Fig. 5.7, the webs are considered to be ineffective in resisting normal stresses but capable of trans-mitting shear. Fach stringer area of 0.5 in2 is assumed to be lumped at a point. Find the shear stress distribution in the webs.

S O L U T I O N If we neglect the moments of inertia of the webs and of the string-ers about their own ccntroids, the cross-sectional moment of inertia about the neutral axis is

' Iz = 2(0.5)(62) + 2(0.5)(22) = 40 in4

If two cross sections 1 in apart are considered, the difference in bending stresses l^y/l- on Ihe two cross sections will be = 1.2 kips/in2 on the outside stringers and 8(^5) = 0.4 kips/in2 on the inside stringers. The differ-ences in axial, loads on the stringers at the two cross sections are found as the product of these stresses and the stringer areas and are shown in Fig. 5.7c.

The shear stress in the web at a point between the upper two stringers is found from the equilibrium of Spanwise forces on the upper stringer.

<tva(0.04)(1) = 600 or

<ryx = 15,000 lb/in2

v."" 'M

« . . . . = 1 5 . 0 0 0 I h / i n

:o ,ooo I win-

15.000 lb/in-

Wl

1 0 6 a i r c r a f t s t r u c t u r h s

If the webs resist no bending stress, the shear stress will be constant along each web, as shown in Fig. 5.7d. If the webs resist bending stresses, the shear stress in each web will vary along the length of the web and will be greater at the end nearer the neutral axis. The shear stress in the web between the two middle stringers is found by considering spanwise forces on the two upper stringers:

<t^(0.04XI) = 600 + 200

or

aJX = 20,000 lb/in2

In problems involving shear stresses in thin webs, the shear force per inch length of web often is obtained rather than the shear stress. The shear per inch, or shear flow, is equal to the product of the shear stress and the web thickness. The shear flow for each web, shown in Fig. 5.7c, is equal to the sum of the longitudinal loads above the web.

The shear stresses may also be obtained by using Eq. (5.16). For a point belwcen the two upper stringers,

K 8000

y l y d A = 4 0 7 0 0 4 0 ( ° " 5 * 6 ) = 1 5 , 0 0 0 l b / ' n 2

For a point between the two middle stringers,

d A = 4 0 x T O 4 Q ( ° - 5 X 6 + ° ' 5 * 2 ) = 2 ° m ' b / i n 2

Example 5.3 Find expressions for the normal stress for all beams whose unsymmetrical cross sections are given in Fig. 5.8a and b.

I \ .1/ - 10 kips-iu

2 in |

M. = 100 kip-in

/2 = 693.3 in 4 . f,. = 173.3 in4

/,,, = 240 in''

(a)

Figure 5.8

,\lr = 42.5 kip-in

15' 14 13 I -

/ , = 71 .23 in 4 , / , . = 913.71 in

= 5.3 in4

(M

s t r e s s a n a l y s e 107

S O L U T I O N From Fq. ( 5 . 1 2 ) with P set to zero, the normal stress for the beam in Fig. 5.8a is

' / , . A / , f / A/, , / . M . + / . , . M , ff-" = - T~, — + ^ T ^ 1,1;- 1U ' ' lylZ ~ lyz '

73.3( 100,000) 4- 240( 10,000) 693.3(10,000) + 240(100,000) : ~2 ^ + T^O Z

(693.3)( 173.3) — 240 " (693.3X173.3 - 240

= --315_r f 494:

Similarly, for the beam in Fig. 5.8b, the normal stress expression is

( j „ = — 6457jj + 9.06z

5.5 S H E A R F L O W IN TIIIN W E B S

Shear How is defined as the product of the shear stress and the thickness of the web. For all practical purposes, it is sufficiently accurate to assume that shear stresses in thin webs arc always parallel to the surfaces for the entire thickness of the web. In Fig. 5.9. a curved web representing the leading edge of a wing is shown, and the shear stresses are parallel to the surfaces of the web at all points. Air loads normal to the surface must, of course, be resisted by shear stresses perpendicular to the web, but these stresses usually are negligible and are not considered here. It might appear that a thin, curved web is not an efficient structure for resisting shearing stresses, but this is not the case. The diagonal tensile and compressive stresses <r, and ac are shown in Fig. 5.9 on principal planes at 45" from the planes of maximum shear a, . From Mohr's circle for a condition of pure shear, it may be shown that the diagonal compressive stress ae

and the diagonal tensile stress a, are both equal to the maximum shear stress crs. If the diagonal compression alone were acting on the curved web, it would bend the web to an increased curvature. The diagonal tensile stress, however, tends to decrease the curvature, and the two effects countcract each other. Consequently, the curved web will resist high shear stress without deforming from its original curvature.

The shear flow q, which is the product of the shear stress <rv and the web thickness f, usually is more convenient to use than the shear stress. The shear llow may be obtained before the web thickness is determined, but the shear stress depends on the web thickness. Often it is necessary to obtain the resultant force on a curved web in which the shear flow q is constant for the length of the web. The element of the web shown in Fig. 5.10 has length ds, and the horizontal and vertical components of this length are dz and dy, respectively. The force on this element of length is q ds, and the components of the force are q dz horizontally and q dy vertically. The total horizontal force is

F. = q dz = qz Jo

(5.17)

where z is the horizontal distance between the ends of the web. The total vertical force on the web is

q dy = qy (5.18)

where v is the vertical distance between the ends of the web. The resultant force is qL, where L is the length of the straight line joining the ends of the web, and the resultant force is parallel to this line. Equations (5.17) and (5.18) are independent of the shape of the web, but depend on the components of the distance between the ends of the web. The induccd torsional moment of the resultant force depends on the shape of the web. The torque induced at any point such as 0 , shown in Fig. 5.1 la, is equal to rq ds. The area dA of the triangle formed by joining point 0 and the extremities of the element of length ds is r ds/2. Then the torque induced by the shear flow along the entire web may be obtained as follows:

T = qr ds = j" 2q dA — 2q J dA

(5.19)

T = 2Aq

<ly 'I I. /

Figure P2.6

I s t r i W a n a l y s i s 109

Figure 5.11

where A is the area enclosed by the web and the lines joining the ends of the web with point 0 , as shown in Fig. 5.1 lb. The distance e, shown in Fig. 5.11 b, of the resultant force from point 0 may be obtained by dividing the torque by the force:

2 Aq 2 A qL~ L

(5.20)

It is important to note that the shear flow q is assumed to be constant in the derivation of Eqs. (5.19) and (5.20).

5.6 S H E A R C E N T E R

Open-section thin web beams, such as in Fig. 5.12, are unstable in carrying -t&tSlQMUfiSjjg^.^ 1ms it a beam cross section is symmetrical about a vertical axiT* then the vertical loads must be applied in the plane of symmetry in order to produce no torsion on the cross section. However, if the beam cross section is not symmetrical, then the loads must be applied at a point such that they produce no torsion. This point is callcd the shear center and may be obtained by finding the position of the resultant of the shear stresses on any cross section. The simplest type of beam for which the shear center may be calculated is made of two concentrated flange areas joined by a curved shear web, as shown in Fig. 5.12.

(m Figure 5.12

110 a i r c r a f t s t r u c t u r e s

The two flanges must lie in the same vertical plane if the beam carries a vertical load. If the web resists no bending, the shear flow in the web will have a constant value (/. The resultant of the shear How will be qL = V, and the position of this resultant from Eq. (5.20) will be a distance e = 2A/L to the left of the flanges, as shown in Fig. 5.12a. Therefore, all loads must be applied in a vertical plane which is a distance e from the plane of the flanges.

A beam with only two flanges that are in a vertical plane is not stable for horizontal loads. The vertical location of the shear center would have no signifi-cance for this beam. For beams which resist horizontal loads as well as vertical loads, it is necessary to determine the vertical location of the shear center. If the cross section is symmetrical about a horizontal axis, the shear center must lie on the axis of symmetry. If the cross section is not symmetrical about a horizontal axis, the vertical position of the shear center may be calculated by taking mo-ments of the shear forces produced by horizontal loads. The method of calcu-lating the shear center of a beam can be illustrated best by numerical examples.

Example 5.4 Find the shear flows in the webs of the beam shown in Fig. 5.13a. Each of the four flange members has an area of 0.5 in2. The webs are assumed to carry no bending stress. Find the shear center for the area.

S O L U T I O N T W O cross sections 1 in apart are shown in Fig. 5 .13 / J , The in-crease in bending moment in the 1-in length is equal to the shear V. The increase of bending stress on the flanges in the 1-in length is

Vyv 10,000 x 5 „ r 2

/ . 50 = 1000 lb/in2

The load on each 0.5-in2 area resulting from this stress is 500 lb and is shown in Fig. 5.13/). The actual magnitude of the bending stress is not needed in the shear-flow analysis, since the shear flow depends on only the change in bending moment or the shear. If each web is cut in the spanwise direction, as shown, the shear forces on the cut webs must balance the loads on the

s •T

10 in

UL V "

1000 lb/in

(<i)

Figure 5.13

s t r e s s a n a l y s i s i i i

flanges. The force in web ab must balance the 500-lb force on flange a, and since this span wise force acts on a 1 -in length, the shear flow in the web will be 500 lb/in in the direction shown. The shear [low in web be must balance the 500-lb force on flange b as well as the 500-lb spanwise force in web ab, and consequently the shear flow has a value of 1000 lb/in. The shear flow in web cd must balance the 1000-lb spanwise force in web be as well as the 500-lb force on llangc c, which is in the opposite direction. The shear flow in web cd is therefore 500 lb/in and is checked by the equilibrium of flange d.

The directions of the shear flow on the vertical beam cross section are obtained from the directions of the spanwise forces. Since each web has a constant thickness, the shear (low, like shear stresses, must be equal on perpendicular planes. The shear flow on a rectangular element .must, form two equal and opposite, couples..The directions of all shear flows are shown in Fig. 5.136 and the back section in Fig. 5.13a. The shear center is found by taking moments about point c:

The shear centcr will be on a horizontal axis of symmetry, since a horizontal force along this axis will produce no twisting of the beam.

Example 5.5 Find the shear flows in the webs of the beam shown in Fig. 5.14a. Each of the four flanges has an area of 1.0 in2. Find the shear center for the area.

S O L U T I O N The moment of inertia of the area about the horizontal centroidal axis is

I Tc = oO

or

- 10,000e + 500(4)(10) = 0

e = 2 in

I. = 4(1 x 42) = 64 in' ,4

1000 lb/in

(b)

Figure 5.14

112 a i r O r a i - t s t r u c t u r e s

The change in axial load in each flange between the two cross sections 1 in apart is

£ M - i S 3 » x 4 x t - l 0001b i 2

The axial loads and shear flows are shown in Fig. 5.14ft. The shear flows in the webs are obtained by a summation of the spanwise forces on the el-ements, as in Example 5.4.

The distance e to shear center is found by taking moments about a point below c, on the juncture of the webs. The shear flow in the nose skin pro-duces a moment equal to the product of the shear flow and twice the area enclosed by the semicircle. The shear flow-in the upper horizontal web has a resultant force of 6000 lb and a moment arm of 10 in. The short vertical webs at a and d each resist forces of 1000 lb with a moment arm of 6 in. The resultant forces on the other webs pass through the centers of moment:

I Tc = 0+)

- 16,000e + 2(39.27)(2000) + 6000(10) + 2(1000X6) = 0

or e = 14.32 in

5.7 T O R S I O N O F C L O S E D - S E C T I O N B O X B E A M S

The thin-web, open-section box beams previously considered are capable of re-sisting loads which are applied at the shear center but become unstable under torsional loads. In many structures, especially in aerospace vehicles, the resultant load takes on different positions for different loading conditions and conse-quently may produce torsion. On an aircraft wing, Tor example, the resultant aerodynamic load is farther forward on the wing at high angles of attack than at low angles of attack. The position of this load also changes when the ailerons or wing flaps are deflected. Thus a closed-section box beam, which is capable of resisting torsion, is used for aircraft wings and similar structures. Typical type§. of wing construction are shown in Fig. 5.15. The wing section of Fig. 5.15a has*6nly

Spar

r r

(a)

Stringers (longitudinals)

(h) Figure 5.15 Typical wing cons t ruc t ion .

STR1-SS ANALYSIS 1 1 3

Figure 5.16 Box b e a m loaded in torsion.

one spar, and the skin forward of this spar forms a closed section which is designed to resist the wing torsion, whereas the portion aft of the spar is lighter and is designed not to resist any loads on the wing but to act as an aerodynamic surface. The wing seclion shown in Fig. 5.15/? has two spars which form a closed-section box beam. In some wings, two or more closed boxes may act together in resisting torsion, but such sections are statically indeterminate and are considered in a later chapter.

The box section shown in Fig. 5.16 is loaded only by a torsional moment T. Since the axial loads in the stringers are produced by wing bending, they are zero for the condition of pure lorsion. If the upper stringers arc considered as a free body, as shown in Fig. 5.16/\ the spanwise forces must be in equilibrium; that is, qa = f/la or q = qt. If similar sections containing other flanges are considered, it becomes obvious that the shear flow at any point must be equal to q. The constant shear flow q around the circumference has no resultant horizontal or vertical force, since in the application of Eqs. (5.17) and (5.18) the horizontal and vertical distances between the endpoints of the closed web are zero. The resultant of the shear flow is thus a torque equal to the applied external torque T, taken about any axis perpendicular to the cross section. If we take point 0 in Fig. 5.16c as a reference, the following may be immediately written from Eq. (5.19):

T = X 2(&A)q = 2Aq (5.21)

where A is the sum of the triangular areas A A and is equal to the total area enclosed by the box section. The area A is the same regardless of the position of point 0 , since the moment of a couple is the same about any point. If point O is chosen outside the section, some of the triangular areas AA will be negative, corresponding to the direction of the moment of the shear flow about point O, but the algcbraic sum of all areas S.A will be equal to the enclosed area A.

5.8 SHEAR F L O W IN C L O S E D - S E C T I O N B O X BEAMS

Consider a box beam containing only two stringers, as shown in Fig. 5.17. Since this seclion is stable under the action of torsional loads, the vertical shear force V


Figure 5.17 One-ce l l - two-s t r ingers box beam.

may be applied at any point in the cross section. Note that this beam is unstable under the action of a horizontal load, since the two stringers in the same vertical plane cannot resist a bending moment about a vertical axis. If two cross sections 1 in apart are considered, as shown in Fig. 5Mb, the difference in axial load on the stringers, AP, between the two cross sections may be found from the differ-ence in the bending stress axv = —M.y/Iz = Vy{\)y/1., or | AP| = axxAr = VyAshJIz = VyA2h2/Iz, where Ax and A2 are stringer areas. These loads must be balanced by the shear flow shown in Fig. 5.17/J. If we consider equilibrium, the summation of forces on the upper stringer in the spanwise direction must be zero:

^(1 in) + g0(l i n ) - i ^ i = 0 * 2

q • = " ^ - 4 0 (5.22)

The shear flow q0 may be found by summing torsional moments for the back section about a perpendicular axis through the lower stringer:

VyC-2Aq0 = 0

V.C " - 2 7

where A is the total area enclosed by the box. Substituting this value in Hq. (5.22) yields

K M . YJSL

— t r ( 5 2 3 }

The shear flows in box beams with several slringc..; may be obtained by a method similar to that previously used. From a summation of spanwise loads on various stringers the shear flows may all be expressed in terms of one unknown shear flow. Then this shear flow may be obtained by equating the moments of the shear flows to the external torsional moment about a spanwise axis. For the box beam shown in Fig. 5.18, all the shear flows q,q2,..., q„ may be expressed in

STR1-SS ANALYSIS 1 1 5

terms of the shear flow q0 by considering the spanwise equilibrium of the string-ers between web 0 and the web under consideration:

I/, = i/o + A P,

<12 = </o + A Pt + AP2

(5.24)

or <•/„ = Qo + Z o

where APn represents the summation of loads AP between 0 and any web n. After all the shear flows are expressed in terms of the unknown q0, the value of q0 may be obtained from the summation of the torsional moments. Note that the shear flow in any other web could have been considered as the unknown q0. For the case of general bending, the difference in axial load on the stringers AP between two sections 1 in apart may be found from Eq. (5.12). Making the substitutions M. = VJ. 1 in), My = V.( \ in), and P = 0 yields

I V + / V I V + ! V r v )'- z , 2 r z ' yz r v

I I - I2 } I I - I2 * V ' - ' f? * V * T * V A/', - ( - y + ^ ^ ^ Uf (5-25)

where v and z are the coordinates of the stringer area Aj.

Example 5.6 Find the shear flow in all webs of the box beam shown in Fig. 5.19a.

S O L U T I O N The moment of inertia of the beam cross section about the neutral axis is / = (4 x 0.5 | 2 x IX52) = 100 in4. The difference in bending stress between the two cross sections 1 in apart is K(l)y/1 = 10,000 (l)(5)/!00 = 500 lb/in2. This produces compressive loads AP of 500 lb on the 1-in2 upper stringer areas and 250 lb on the 0.5 in2 stringer areas, as shown in Fig. 5. The shear flow in the leading-edge skin is considered as the unknown <y0 > although the shear How in any other web could have been considered as the unknown. Now the shear llow in all other webs may be obtained in terms of c/0 bv considering the equilibrium of the spanwise forces on the stringers, as shown in Fig. 5.19h.

The value of q„ is obtained by considering the equilibrium of torsional


250 Ib 5001b j- 10 - 1 0 i n —

1.0 in2

0.5 in2 0.5 in2 t 10 in

0.5 ill2 0.5 in2 1 -—8 in—*

1.0 in2

10 kips

q n 250 V o 500

(/') qa 250 ? „ • - 500

Figure 5.19

m o m e n t s d u e t o t h e s h e a r f l o w a n d t h e e x t e r n a l a p p l i e d l o a d s i n r e f e r e n c e t o ; i ny a x i s n o r m a l l o ( h e b a c k c r o s s s c c t i . . n o f t h e b e a m . T a k i n g I h e a x i s t h r o u g h p o i n t 0, f o r e x a m p l e , a s s h o w n i n F i g . 5 . 1 9 c , a n d s u m m i n g t o r q u e s t o z e r o y i e l d

I T 0 = 0 +

- 1 0 , 0 0 0 ( 8 ) + (q0 - p O X 100) + (q0 - 5 0 0 ) ( 1 0 0 ) + c / 0 (278 .5 ) = 0

o r q0 = 3 2 4 l b / i n

T h e s h e a r f l o w in t h e r e s t o f t h e w e b s m a y b e c o m p u t e d e a s i l y f r o m F i g .

f 5 . 1 9 c .

E x a m p l e 5 . 7 F i n d t h e s h e a r f l o w in t h e w e b s o f I h e b o x b e a m s h o w n i n F i g .

5 .20 .

SOLUTION T h e c h a n g e in b e n d i n g s t r e s s b e t w e e n t w o c r o s s s e c t i o n s is o b -

t a i n e d f r o m E q . (5 .25) . T h e t e r m s t o b e u s e d i n t h i s e q u a t i o n a r e o b t a i n e d a s

f o l l o w s :

I z = (2 x 3 + 2 x l ) ( 5 2 ) = 2 0 0 i n 4

/ , = ( 2 x 3 + 2 x 1) (10 2 ) = 8 0 0 i n 4

g ) = 1 ( 5 ) ( - 10) + 3( - 5 ) ( - 10) + 1(10)( — 5 ) + 3 ( 5 X 1 0 ) = 2 0 0 i n 4

\ X

V. — A k i p s

V y = 10 k i p s

STRESS ANALYSIS 1 1 7

V \ s s

- V

/ V S

Figure 5.20 Box beam.

3 in2

10 in

1 in J

K, = 4 kips * <

-10 i n -

1 in2

> 3 in2

-10 ill—

VY = 10 kips

Section C-C-

T h c s u b s t i t u t i o n i n l ; . q . ( 5 . 2 5 ) y i e l d s

AP = (23.33z - 7333y)Af

W i t h t h e a b o v e e q u a t i o n , t h e A P o n e a c h s t r i n g e r m a y b e o b t a i n e d e a s i l y b y

m a k i n g t h e p r o p e r s u b s t i t u t i o n f o r t h e f l a n g e a r e a a n d i t s c o r r e s p o n d i n g

c o o r d i n a t e s . T h e r e s u l t s a r c s h o w n i n F i g . 5 . 2 1 a . N o w t h e s h e a r f l o w i n e a c h

w e b c a n b e o b t a i n e d f r o m t h e i n c r e m e n t s o f t h e flange l o a d s , a s w a s d o n e i n

E x a m p l e 5 . 6 f o r t h e s y m m e t r i c a l b o x b e a m . T h e s h e a r flow i n t h e l e f t - h a n d

w e b i s d e s i g n a t e d q 0 . T h e s h e a r flows i n t h e r e s t o f t h e w e b s a r e o b t a i n e d b y

c o n s i d e r i n g t h e e q u i l i b r i u m o f f o r c e s i n t h e s p a n w i s e d i r e c t i o n a n d a r e g i v e n

i n F i g . 5 . 2 1 a . N o w t h e u n k n o w n s h e a r flow q0 i s o b t a i n e d f r o m t h e e q u i l i b -

r i u m o f t o r s i o n a l m o m e n t s . T a k i n g p o i n t O a s a r e f e r e n c e p o i n t a n d s u m -

n i i n g m o m e n t s a b o u t t h e .x a x i s t h r o u g h O y i e l d

(</„ - 4 0 0 ) ( 1 0 0 ) + f /„ ( 1 0 0 ) + (<70 - 6 0 0 ) ( 1 0 0 ) + (q0 - 1 0 0 0 ) ( 1 0 0 ) = 0

o r 4 0 0 r / ( ) - 2 0 0 , 0 0 0 = 0

(/„ = 5 0 0 l b / i n

f In -f/u - 1000 = - 5 0 0

Ui)

Figure P5.28


T h e f i n a l s h e a r - f l o w r e s u l t s a r e i n d i c a t e d o n F i g . 5 . 2 1 / ; . T h e m i n u s s i g n

i m p l i e s t h a t t h e w r o n g d i r e c t i o n o f s h e a r flow h a s b e e n a s s u m e d .

I n t h e p r e c e d i n g a n a l y s i s o f s h e a r s t r e s s e s i n b e a m s , w c a s s u m e d t h a t t h e c r o s s

s e c t i o n o f t h e b e a m r e m a i n e d c o n s t a n t . S i n c e i n a e r o s p a c e v e h i c l e s t r u c t u r e s a

m i n i m u m w e i g h t i s a l w a y s s o u g h t , u s u a l l y t h e b e a m s a r e t a p e r e d i n o r d e r t o

a c h i e v e m a x i m u m s t r u c t u r a l e f f i c i e n c y . W h i l e t h i s v a r i a t i o n i n c r o s s s e c t i o n m a y

n o t c a u s e a p p r e c i a b l e e r r o r s in t h e a p p l i c a t i o n o f t h e f W t u r e f o r m u l a f o r b e n d i n g

s t r e s s e s , o f t e n i t c a u s e s l a r g e e r r o r s i n t h e s h e a r s t r e s s e s d e t e r m i n e d f r o m E q .

A s a n i l l u s t r a t i o n , c o n s i d e r t h e b e a m s h o w n i n F i g . 5 . 2 2 w h i c h , f o r s i m p l i c i t y ,

is a s s u m e d t o c o n s i s t o f t w o s t r i n g e r s j o i n e d b y a v e r t i c a l w e b t h a t r e s i s t s n o

b e n d i n g . T h e r e s u l t a n t a x i a l l o a d s i n t h e s t r i n g e r s m u s t b e i n t h e d i r e c t i o n o f t h e

s t r i n g e r s a n d m u s t h a v e h o r i z o n t a l c o m p o n e n t s Px = M J h . T h e v e r t i c a l c o m -

p o n e n t s o f t h i s l o a d w h i c h a c t o n t h e s t r i n g e r s , P t a n a i a n d P t a n a 2 , a s s h o w n

in F i g . 5.22b, r e s i s t p a r t o f t h e e x t e r n a l a p p l i e d s h e a r Vy. B y d e s i g n a t i n g t h e s h e a r

5.9 SPANWISE TAPER EFFECT

( 5 . 1 6 ) .

c \ b- K \ \ \ \ \ \ \ \ \ \ \ K.

\ r t a n a .

U

I' tan « 2

(ft)

Figure 5.22

STRILSS ANALYSIS 1 1 9

r e s i s t e d h y t h e s t r i n g e r s u s Vr a n d t h a t r e s i s t e d b y t h e w e b s a s V„,

K = + K ( 5 . 2 6 a )

V s = P ( t a n a , + t a n <x2) ( 5 . 2 6 f t )

F r o m t h e g e o m e t r y (if t h e b e a m , t a n a , = hJC, t a n a2 = / i 2 / C , a n d t a n a , +

t a n a , = ( / i , t- h2), ( ' = It/C. S u b s t i t u t i n g t h i s v a l u e i n t o E q . ( 5 . 2 6 6 ) y i e l d s

1 / = l ' ~ ( 5 . 2 7 )

E q u a t i o n ( 5 . 2 7 ) w i l l a p p l y f o r a b e a m w i t h a n y s y s t e m o f v e r t i c a l l o a d s . F o r t h e

p r e s e n l l o a d i n g , t h e v a l u e o f P is Vyhfh. S u b s t i t u t i n g t h i s v a l u e f o r P i n t o E q .

( 5 . 2 7 ) y i e l d s

Vf ( 5 . 2 8 )

F r o m E q s . ( 5 . 2 6 « ) a n d ( 5 . 2 S ) a n d f r o m t h e g e o m e t r y ,

K = Vj £ ( 5 - 2 9 )

E q u a t i o n s ( 5 . 2 8 ) a n d ( 5 . 2 9 ) c a n b e e x p r e s s e d i n t e r m s o f t h e d e p t h s / i 0 a n d h o f t h e

b e a m b y m a k i n g u s e o f t h e p r o p o r t i o n a/C = h0/h:

K / ! o Vw = ( 5 . 3 0 a )

a n d Vr = V '' ~ ( 5 . 3 0 6 ) h

T h e s h e a r f l o w i n t h e w e b s n o w c a n b e f o u n d b y u s i n g E q . ( 5 . 1 6 ) i n c o n j u n c -

t i o n w i t h t h e s h e a r l '„ i n E q . (5.30<i) . F o r i n s t a n c e , if w e a s s u m e t h a t t h e a r e a s o f

b o t h s t r i n g e r s i n F i g . 5 . 2 2 a r e t h e s a m e , t h e s h e a r H o w a t a s e c t i o n w h e r e t h e

d i s t a n c e b e t w e e n t h e s t r i n g e r s is h m a y b e c a l c u l a t e d a s f o l l o w s :

I ' </ = T y d A = m i T = T (5"31)

" W h e n t h e b e a m h a s s e v e r a l s t r i n g e r s , t h e s h e a r f l o w m a y b e o b t a i n e d i n a m a n n e r

s i m i l a r t o t h a t f o r t h e t w o - s t r i n g e r b e a m a s l o n g a s t h e s t r i n g e r a r e a s r e m a i n

c o n s t a n t a l o n g t h e s p a n . I f t h e s t r i n g e r a r e a s v a r y a l o n g t h e s p a n a n d n o t a l l v a r y

i n t h e s a m e p r o p o r t i o n , E q . ( 5 . 1 6 ) c a n n o t b e a p p l i e d .

E x a m p l e 5 . 8 F i n d t h e s h e a r H o w s i n t h e w e b o f t h e b e a m s h o w n i n F i g . 5 . 2 3

a t 2 0 - i n i n t e r v a l s a l o n g I h e s p a n .

S o L t r n o N T h e s h e a r f l o w s a r e o b t a i n e d b y t h e u s e o f E q s . ( 5 . 3 0 a ) a n d ( 5 . 3 1 ) .

T h e s o l u t i o n o f t h e s e e q u a t i o n s i s s h o w n i n T a b l e 5 . 1 .

120 AIRC'RAIT STRUC TURES

100 in-

Figure 5.23

W h i l e s l i d e - r u l e a c c u r a c y i s s u f f i c i e n t f o r s h e a r - f l o w c a l c u l a t i o n s , t h e v a l u e s

i n T a b l e 5 .1 a r e c o m p u t e d t o f o u r s i g n i f i c a n t figures f o r c o m p a r i s o n w i t h a

m e t h o d t o b e d e v e l o p e d l a t e r .

E x a m p l e 5 . 9 F i n d t h e s h e a r flows a t s e c t i o n A A o f t h e b o x b e a m s h o w n i n

F i g . 5 . 2 4 .

SOLUTION T h e m o m e n t o f i n e r t i a o f t h e c r o s s s e c t i o n a t A A a b o u t t h e n e u -

t r a l a x i s i s

I = 2 ( 2 + 1 + 1 X 5 2 ) = 2 0 0 i n 4

T h e b e n d i n g s t r e s s e s a t s e c t i o n A A a r e

M . y 4 0 0 , 0 0 0 x 5 , „ , = = ^ = 1 0 , 0 0 0 l b / i n

T h e h o r i z o n t a l c o m p o n e n t s o f t h e f o r c e s a c t i n g o n t h e 2 - i n 2 s t r i n g e r s a r e

2 0 , 0 0 0 l b , a n d t h e f o r c e s o n t h e 1 - i n 2 s t r i n g e r s a r e 1 0 , 0 0 0 l b , a s s h o w n i n F i g .

5 . 2 5 a . T h e v e r t i c a l c o m p o n e n t s a r e o b t a i n e d b y m u l t i p l y i n g t h e f o r c e s a n d

t h e t a n g e n t s o f t h e a n g l e s b e t w e e n s t r i n g e r s a n d t h e h o r i z o n t a l . T h e s u m o f

t h e v e r t i c a l c o m p o n e n t s o f f o r c e s o n a l l s t r i n g e r s V f i s 4 0 0 0 l b , a n d t h e

r e m a i n i n g s h e a r V „ o f 4 0 0 0 l b is r e s i s t e d b y t h e s h e a r flows in t h e w e b s . I f o n e

o f t h e u p p e r w e b s i s c u t , a s s h o w n i n F i g . 5 . 2 5 6 , t h e s h e a r flows i n t h e w e b s

m a y b e o b t a i n e d f r o m v f y

q = f y dA

T a b l e 5 . 1

h 1 =•

0 10 1 10.000 1,000 20 12 0.8333 8.333 694.4 40 14 0.7143 7,143 510.2 60 16 0.6250 6,250 390.6 80 18 0.5555 5,555 308.6

100 20 0.5 5,000 250.0

STRESS ANALYSIS 121

Figure 5.24

w h e r e t h e i n t e g r a l r e p r e s e n t s t h e m o m e n t o f t h e a r e a s b e t w e e n t h e c u t w e b

a n d t h e w e b u n d e r c o n s i d e r a t i o n . T h e c h a n g e i n b e n d i n g s t r e s s o n a s t r i n g e r

b e t w e e n t h e t w o c r o s s s e c t i o n s 1 i n a p a r t i s V w y / l w h e n t h e e f f e c t o f t a p e r i s

c o n s i d e r e d , a n d t h e c h a n g e i n a x i a l l o a d o n a s t r i n g e r o f a r e a A j i s

A P = y M /

T h e s e a x i a l l o a d s a r c s h o w n i n F i g . 5 . 2 5 b i n t h e s a m e w a y a s t h e y w e r e

s h o w n p r e v i o u s l y f o r b e a m s w i t h n o t a p e r . T h e e q u i l i b r i u m o f f o r c e s i n t h e

s p a n w i s e d i r e c t i o n y i e l d s t h e s h e a r flow i n t e r m s o f q0 i n a l l t h e w e b s , a s

s h o w n i n F i g . 5 . 2 5 b . N o w t h e s h e a r f l o w q0 c a n b e f o u n d b y s u m m i n g

t o r s i o n a l m o m e n t s a b o u t t h e z a x i s t h r o u g h p o i n t 0 f o r t h e b a c k s e c t i o n , a s

s h o w n i n F i g . 5.26a. T h e f i n a l s h e a r flow i n e a c h w e b i s s h o w n i n F i g . 5 . 2 6 b .

I T 0 = 0 +

8 0 0 0 ( ' 2 ) + (qu - 1 0 0 ) ( 2 0 0 ) + <J o (100) + (q0 + 1 0 0 X 1 0 0 )

- 2 ( 5 0 0 X 1 0 ) - 2 ( 5 0 0 ) ( 2 0 ) = 0

qg ~ 6 0 l b / i n

(") (6)

Figure 5.25


10001b 5001b 500 1b

• 8 0 0 0 lb

J J Figure 5.26

5.10 BEAMS WITH VARIABLE STRINGER AREAS

I n S e e . 5 . 9 , b e a m s w e r e c o n s i d e r e d w h i c h v a r i e d i n d e p t h b u t h a d s t r i n g e r s w h o s e

c r o s s s e c t i o n s w e r e c o n s t a n t . I n m a n y a e r o s p a c e s t r u c t u r a l b e a m s , t h e c r o s s -

s e c t i o n a l a r e a o f t h e s t r i n g e r m e m b e r s v a r i e s a s w e l l a s t h e d e p t h o f t h e b e a m . I f

t h e a r e a s o f a l l t h e s t r i n g e r m e m b e r s a r e i n c r e a s e d b y a c o n s t a n t r a t i o , t h e

m e t h o d o f S e c . 5 . 9 c a n b e u s e d ; if t h e a r e a s a t o n e c r o s s s e c t i o n a r e n o t p r o -

p o r t i o n a l t o t h e a r e a s a t a n o t h e r c r o s s s e c t i o n , t h e m e t h o d w o u l d b e c o n s i d e r a b l y

i n e r r o r . T h e a i r p l a n e w i n g s e c t i o n i n F i g . 5 . 2 7 r e p r e s e n t s a s t r u c t u r e i n w h i c h t h e

v a r i a t i o n i n s t r i n g e r a r e a s m u s t b e c o n s i d e r e d . T h e s t r i n g e r a r e a s i n t h i s w i n g a r e

d e s i g n e d i n s u c h a m a n n e r t h a t t h e b e n d i n g s t r e s s e s a r e c o n s t a n t a l o n g t h e s p a n .

I n o r d e r t o r e s i s t t h e l a r g e r b e n d i n g m o m e n t s n e a r t h e r o o t o f t h e w i n g , t h e

b e n d i n g s t r e n g t h i s a u g m e n t e d b y i n c r e a s i n g t h e d e p t h o f t h e w i n g a n d t h e a r e a

o f s p a r c a p s A a n d B. T h e s t r i n g e r s w h i c h r e s i s t t h e p a r t o f t h e b e n d i n g m o m e n t

n o t r e s i s t e d b y t h e s p a r c a p s h a v e t h e s a m e a r e a f o r t h e e n t i r e s p a n . S i n c e t h e

a x i a l s t r e s s e s o n t h e s e s t r i n g e r s a r e t h e s a m e a t e v e r y p o i n t a l o n g t h e s p a n , t h e

i n c r e m e n t s o f l o a d i n c r e a s e A I' w i l l b e z e r o e x c e p t o n s p a r c a p s A a n d B. I t m a y

b e s e e n f r o m F . q . ( 5 . 2 4 ) t h a t t h e s h e a r flow m u s t b e c o n s t a n t a r o u n d t h e e n t i r e

l e a d i n g e d g e o f t h e w i n g a n d c h a n g e s o n l y a t t h e s p a r c a p s . C o n s e q u e n t l y , t h e

m e t h o d s o f a n a l y s i s p r e v i o u s l y u s e d a r e n o t a p p l i c a b l e t o t h i s p r o b l e m .

- T h e b e n d i n g s t r e s s e s a n d t o t a l s t r i n g e r l o a d s m a y b e c a l c u l a t e d f o r t w o c r o s s

s e c t i o n s o f t h e b e a m . T h e a c t u a l d i m e n s i o n s a n d s t r i n g e r a r e a s f o r e a c h c r o s s

s e c t i o n a r c u s e d , s o t h a t a n y c h a n g e s b e t w e e n t h e c r o s s s e c t i o n s a r e t a k e n i n t o

Figure 5.22


Figure 5.28

c o n s i d e r a t i o n . T h e s t r i n g e r l o a d s Ptt a n d Ph a r e s h o w n i n F i g . 5 . 2 8 f o r t w o

s e c t i o n s a d i s t a n c e a p a r t . T h e i n c r e a s e i n l o a d i n a n y s t r i n g e r i s a s s u m e d t o b e

u n i f o r m i n l e n g t h a . T h e i n c r e a s e i n s t r i n g e r l o a d p e r u n i t l e n g t h a l o n g t h e s p a n

i s

' P — P A P = * " ( 5 . 3 2 )

a

T h i s t y p i c a l f o r c e is s h o w n i n F i g . 5 . 2 8 b . N o w t h e s h e a r flow c a n b e o b t a i n e d

f r o m t h e s e v a l u e s , o f A P , a s i n t h e p r e v i o u s a n a l y s i s .

I t i s s e e n t h a t t h e s h e a r f o r c e i s n o t u s e d i n finding t h e v a l u e s o f A P ;

c o n s e q u e n t l y , i t i s n o t n e c e s s a r y t o c a l c u l a t e t h e v e r t i c a l c o m p o n e n t s o f t h e

s t r i n g e r l o a d s . T h e c f f e c t o f b e a m t a p e r a n d c h a n g e s i n s t r i n g e r a r e a a r e i m -

p l e m e n t e d a u t o m a t i c a l l y w h e n t h e m o m e n t s o f i n e r t i a a n d b e n d i n g s t r e s s e s a r e

c a l c u l a t e d . S i n c e it i s n e c e s s a r y t o d e t e r m i n e t h e w i n g b e n d i n g s t r e s s e s a t f r e q u e n t

s t a t i o n s a l o n g t h e w i n g s p a n i n o r d e r t o d e s i g n t h e s t r i n g e r s , t h e t e r m s P„ a n d Pb

c a n b e o b t a i n e d w i t h o u t l o o m a n y a d d i t i o n a l c a l c u l a t i o n s . T h u s t h i s m e t h o d o f

a n a l y s i s i s o f t e n s i m p l e r a n d m o r e a c c u r a t e t h a n t h e m e t h o d w h i c h c o n s i d e r e d

v a r i a t i o n s i n d e p t h b u t n o t v a r i a t i o n s i n s t r i n g e r a r e a .

T h e d i s t a n c e a b e t w e e n t w o c r o s s s e c t i o n s m a y b e a n y c o n v e n i e n t v a l u e . I t i s

c o m m o n p r a c t i c e t o c a l c u l a t e w i n g b e n d i n g s t r e s s e s a t i n t e r v a l s o f 15 t o 3 0 i n

a l o n g t h e s p a n . T h e i n t e r v a l s a r e q u i t e s a t i s f a c t o r y f o r s h e a r - f l o w c a l c u l a t i o n s .

N o t e t h a t f o r v e r y s m a l l v a l u e s o f u , s m a l l p e r c e n t a g e e r r o r s i n Pa a n d Ph r e s u l t i n

l a r g e p e r c e n t a g e e r r o r s i n A P. H o w e v e r , if a i s t o o l a r g e , t h e a v e r a g e s h e a r f l o w

o b t a i n e d b e t w e e n t w o s e c t i o n s m a y n o t b e q u i t e t h e s a m e a s t h e s h e a r f l o w

m i d w a y b e t w e e n t h e s e c t i o n s .

E x a m p l e 5 . 1 0 F i n d t h e s h e a r f l o w s i n t h e b e a m o f F i g . 5 . 2 3 b y t h e m e t h o d o f

u s i n g d i f f e r e n c e s i n b e n d i n g s t r e s s e s .

1 2 4 AIRCRAFT STRUCTURES

Table 5.2

.X

(II

M (2)

h (3) (4)

P--P. (5)

10 100,(XX) 11 9,091 20 (3,986 30 300.000 13 23.077 40 10,256 50 500,000 15 33,333 60 7,843 70 700,000 17 41.176 80 6,192 90 900,000 19 47,368

M Ph- P. <? =

20 Percentage error (6) (7)

699.3 0.7

512.8 0.5

392.1 0.4

309.6 0.3

S O L U T I O N F o r t h i s t w o - f l a n g e b e a m , t h e a x i a l l o a d i n t h e flanges h a s a

h o r i z o n t a l c o m p o n e n t P = M/h. T h e v a l u e s o f P f o r v a r i o u s s e c t i o n s a r e

c a l c u l a t c d i n c o l u m n 4 o f T a b l e 5 . 2 . I n c o m p u t i n g t h e s h e a r a t a n y c r o s s

s e c t i o n , v a l u e s o f t h e a x i a l l o a d s a t c r o s s s e c t i o n s 1 0 i n o n e i t h e r s i d e a r e

f o u n d . T h e f r e e - b o d y d i a g r a m s a r e s h o w n i n F i g . 5 . 2 9 . T h e c i r c l e d n u m b e r s

r e p r e s e n t s t a t i o n s , o r t h e d i s t a n c e f r o m t h e c r o s s s e c t i o n t o t h e l e f t e n d o f t h e

b e a m . T h e d i f f e r e n c e i n h o r i z o n t a l l o a d s o n t h e u p p e r p a r t o f t h e b e a m

b e t w e e n t h e c r o s s s e c t i o n s 2 0 i n a p a r t m u s t b e b a l a n c e d b y t h e r e s u l t a n t o f

t h e h o r i z o n t a l s h e a r flow, 2 O q . T h e d i f f e r e n c e s i n a x i a l l o a d s a r e t a b u l a t e d i n

c o l u m n 5 , a n d t h e s h e a r flows q = (Pb — PJ/20 a r e s h o w n i n c o l u m n 6 . T h e

v a l u e o f t h e s h e a r f l o w a t s t a t i o n 2 0 t h u s i s a s s u m e d t o b e e q u a f t o t h e

a v e r a g e h o r i z o n t a l s h e a r b e t w e e n s t a t i o n s 1 0 a n d 3 0 . E v e n t h o u g h t h e s h e a r

2 0 9 1 l b 2 3 . 0 7 7 l b

3 3 . 3 3 3 l b

( 3 3 . 3 3 3 l b

4 7 . 3 6 H l b

[47.-S6X I

Figure 5.22

STRESS ANALYSIS 1 2 5

10 in - 4 - 10 in »

2 in2 1 I in2 1 in2 10 in—»J« 10 in — •

» a i l i n 2

I »

' I 2 in2 l m 2

' t 9 in

I 11 in

1

4 s 1 , 1 . 1 Station <1(1

(«t

figure 5 J 0

Station 60

(b)

d o e s n o t v a r y l i n e a r l y a l o n g t h e s p a n , t h e e r r o r i n t h i s a s s u m p t i o n i s o n l y 0 . 7

p e r c e n t , a s f o u n d b y c o m p a r i s o n w i t h t h e e x a c t v a l u e o b t a i n e d i n T a b l e 5 . 1 .

T h i s e r r o r i s e v e n s m a l l e r a t t h e o t h e r s t a t i o n s .

E x a m p l e 5 . 1 1 F i n d t h e s h e a r f l o w s a t c r o s s s e c t i o n AA o f t h e b o x b e a m

s h o w n i n F i g . 5 . 2 4 b y c o n s i d e r i n g t h e d i f f e r e n c e i n b e n d i n g s t r e s s e s a t c r o s s

s e c t i o n s 1 0 i n o n e i t h e r s i d e o f A A.

S O L U T I O N T h e m o m e n t o f i n e r t i a a t s t a t i o n 4 0 ( 4 0 i n f r o m t h e l e f t e n d ) i s

f o u n d f r o m t h e d i m e n s i o n s s h o w n i n F i g . 5 . 3 0 a . T h e b e n d i n g s t r e s s e s a t

s t a t i o n 4 0 , r e s u l t i n g f r o m t h e b e n d i n g m o m e n t o f 3 2 0 , 0 0 0 i n • l b , a r e

My 3 2 0 , 0 0 0 ( 4 . 5 )

1 6 2 = 8 8 8 8 l b / i n 2

T h e l o a d s o n t h e J - i n 2 a r e a s a r e 8 8 8 8 l b , a n d t h e l o a d s o n t h e 2 - i n 2 a r e a s a r e

1 7 , 7 7 7 l b , a s s h o w n i n F i g . 5 . 3 1 a . T h e m o m e n t o f i n e r t i a a t s t a t i o n 6 0 i s

f o u n d f r o m t h e d i m e n s i o n s s h o w n i n F i g . 5 . 3 1 6 :

L = 8 ( 5 . 5 2 ) = 2 4 2 i n 4

T h e b e n d i n g s t r e s s e s r e s u l t i n g f r o m t h e b e n d i n g m o m e n t o f 4 8 0 , 0 0 0 i n • l b

(b)

Figure P5.28

a r e

M . y 4 8 0 , 0 0 0 x 5 . 5 ° x x = ~ = = 1 0 , 9 0 9 l b / i n 2

T h e l o a d s o n t h e s t r i n g e r s a r e 1 0 , 9 0 9 a n d 2 1 , 8 1 8 l b , a s s h o w n in F i g . 5 . 3 h i .

T h e i n c r e m e n t s o f f l a n g e l o a d s A P i n a 1 - i n l e n g t h a r e f o u n d f r o m E q . ( 5 . 3 2 ) .

F o r t h e a r e a o f 2 i n 2 ,

A „ 2 1 , 8 1 8 - 1 7 , 7 7 7 A P = — — : = 2 0 2 b 20

F o r t h e a r e a o f 1 .0 i n 2

0 , 9 0 9 - 8 8 8 8 A P = — — — = 1 0 1 l b

20

T h e v a l u e s o f AP a r e s h o w n i n F i g . 5 . 3 1 f t . T h e r e m a i n i n g s o l u t i o n i s i d e n t i c a l

t o t h a t o f E x a m p l e 5 . 9 . T h e v a l u e s o f A P a r e 1 p e r c e n t h i g h e r t h a n t h e e x a c t

v a l u e s s h o w n i n F i g . 5 . 2 5 f t . T h e r e a s o n f o r t h i s s m a l l d i s c r e p a n c y i s t h a t t h e

a v e r a g e s h e a r ( l o w b e t w e e n s t a t i o n s 4 0 a n d 6 0 i s 1 p e r c e n t h i g h e r t h a n t h e

s h e a r f l o w a t s t a t i o n 5 0 . T h e o t h e r a s s u m p t i o n s u s e d i n t h e t w o s o l u t i o n s a r e

i d e n t i c a l . T h e m e t h o d o f u s i n g d i f f e r e n c e s i n b e n d i n g s t r e s s e s a u t o m a t i c a l l y

c o n s i d e r s t h e e f f e c t s o f t h e s h e a r c a r r i e d b y t h e s t r i n g e r s , a n d it is n o t n e c e s -

s a r y t o c a l c u l a t e t h e a n g l e s o f i n c l i n a t i o n o f t h e s t r i n g e r s . I t i s , h o w e v e r ,

n e c e s s a r y t o f i n d t h e t o r s i o n a l m o m e n t s a b o u t t h e p r o p e r a x i s if t h e s t r i n g e r

f o r c e s a r e o m i t t e d i n t h e m o m e n t e q u a t i o n .

5.11 AIRY STRESS FUNCTION

I t is s h o w n i n C h a p . 3 t h a t a s t r e s s field d e s c r i b e s t h e e x a c t s t a t e o f s t r e s s i n a

s o l i d if a n d o n l y if i t s a t i s f i e s t h e c o n d i t i o n s o f e q u i l i b r i u m , c o m p a t i b i l i t y , a n d

p r e s c r i b e d b o u n d a r y s t r e s s e s . F o r t w o - d i m e n s i o n a l s t r e s s p r o b l e m s i n t h e a b s e n c e

o f b o d y f o r c e s , t h e e q u i l i b r i u m a n d c o m p a t i b i l i t y e q u a t i o n s a r e y

* + ff"- = ° ( e q u i l i b r i u m ) ( 3 . 1 1 ) °xy.x + <?yy.y = 0

a X X t + 2(7V). x y + rxvv r j = 0 ( c o m p a t i b i l i t y ) ( 3 . 2 9 )

I f a s t r e s s f u n c t i o n <t> (.v. y ) i s a s s u m e d s u c h t h a t t h e s t r e s s e s i n a s o l i d a r c

d e f i n e d b y

= 

= .AX

= -

STRILSS ANALYSIS 127

t h e n F q s . ( 3 . 1 1 } a r c i d e n t i c a l l y s a t i s f i e d . U p o n s u b s t i t u t i n g E q s . ( 5 . 3 3 ) i n t o E q .

( 3 . 2 9 ) , t h e f o l l o w i n g i s o b t a i n e d :

+ 2 0 x x y r + ® y y y y = 0 ( 5 . 3 4 )

T h e s o l u t i o n o f E q . ( 5 . 3 4 ) s a t i s f i e s b o t h t h e e q u i l i b r i u m a n d t h e c o m p a t i b i l i t y

e q u a t i o n s a n d t h e r e f o r e g i v e s a p o s s i b l e s t r e s s field i n a n e l a s t i c s o l i d . I n o r d e r f o r

t h e o b t a i n e d s t r e s s field t o d e s c r i b e t h e t r u e s t a t e o f s t r e s s f o r a s p e c i f i c p r o b l e m ,

t h e p r o s c r i b e d b o u n d a r y c o n d i t i o n s m u s t b e s a t i s f i e d a l s o .

T h e s o l u t i o n o f E q . ( 5 . 3 4 ) m a y b e o b t a i n e d b y t w o m e t h o d s : t h e p o l y n o m i a l

s o l u t i o n a n d t h e F o u r i e r s e r i e s s o l u t i o n .

P o l y n o m i a l s o l u t i o n If t h e s t r e s s f u n c t i o n ^ ( J ^ y ) is a s s u m e d t o h a v e a s o l u t i o n o f

t h e f o r m

N 0 ( . x , y) = .r) + <t>2(.r, }>) + ••• + d>„(x, y) = £ $„(*, y) ( 5 . 3 5 )

n= 1

w h e r e

<!>„(a\ y) = £ Ain X"'' Y' ( 5 . 3 6 ) i = 0

t h e n , b y c o n s i d e r i n g v a r i o u s d e g r e e s o f p o l y n o m i a l s a n d s u i t a b l y a d j u s t i n g t h e i r

c o e f f i c i e n t s Ain, a n u m b e r o f p r a c t i c a l p r o b l e m s m a y b e s o l v e d . F o r i n s t a n c e ,

t a k i n g

<I)(.v, y ) = I <D„(A% y ) = £ ( f A ^ X ' - ' Y 1

n = 1 n = 1 \ i = 0

a n d a s s u m i n g a l l c o e f f i c i e n t s t o b e z e r o A22 a n d A01 y i e l d

= <!>.„. = 2 / 1 2 2

^ - = 2A02 ( 5 . 3 8 )

= 0

B y e x a m i n i n g l-!q. ( 5 . 3 7 ) , i t m a y b e s h o w n e a s i l y t h a t t h i s i s t h e s o l u t i o n o f a

t w o - d i m e n s i o n a l s o l i d l o a d e d a s s h o w n i n F i g . 5 . 3 2 a . I f , i n a d d i t i o n , A01 i s t a k e n

a s z e r o , t h e n t h e s o l u t i o n c o r r e s p o n d s t o t h a t o f F i g . 5 . 3 2 f c . F i g u r e 5 . 3 2 c a n d h

r e p r e s e n t s t h e c o n d i t i o n s w h e r e A12 a n d A 3 i , r e s p e c t i v e l y , a r e t h e o n l y n o n z e r o

s e t o f c o e f f i c i e n t s i n E q . ( 5 . 3 6 ) .

T o i l l u s t r a t e t h e u s e o f t h e A i r y s t r e s s f u n c t i o n i n t h e s o l u t i o n o f p r a c t i c a l

p r o b l e m s , l e t u s c o n s i d e r t h e b e a m p r o b l e m s h o w n i n F i g . 5 . 3 3 a . I t i s a s s u m e d

t h a t t h e e x t e r n a l s h e a r l o a d V is d i s t r i b u t e d a t t h e s u r f a c e jc = x 0 a c c o r d i n g t o

t h e s t r e n g t h - o f - m a t e r i a l p a r a b o l i c s h e a r - s t r e s s d i s t r i b u t i o n , a n d t h e a x i a l f o r c e S

i s a p p l i e d u n i f o r m l y o v e r t h a t s u r f a c e .

A l s o , i t i s a s s u m e d t h a t t h e b e n d i n g s t r e s s e s v a r y l i n e a r l y , a s s h o w n i n F i g .

( 5 . 3 7 )

5 . 3 3 / ) . T h r o u g h c o m p a r i s o n o f F i g s . 5 . 3 2 / ) , c , a n d d a n d 5 .33/> , it c a n b e s e e n t h a t

t h e l o a d i n g o f F i g . 5 . 3 3 6 m a y b e o b t a i n e d b y s u p e r p o s i n g t h e l o a d s s h o w n i n F i g .

5 . 3 2 6 , c, a n d d. T h u s , t h e s t r e s s field f o r t h e b e a m i n F i g . 5 . 3 3 « i s

axx ~ 2 / 1 2 2 + 6 - 4 3 4 X V

<x,, = 0

<Txy = As2-3A3Ay2

( 5 . 3 9 )

T h e b o u n d a r y c o n d i t i o n s w h i c h m u s t b e s a t i s f i e d f o r t h e g i v e n b e a m i n F i g . 5 . 3 3 a

a r e

<T,J. = 0

<x j y dy = V

"xx <b' = s

a t

a t

a t

y = ±h

x = 0

-XT = 0

U t i l i z i n g F .qs . ( 5 . 3 9 ) i n E q s . ( 5 . 4 0 ) y i e l d s

- A 1 2 - 3/134. /J2 = 0

(-A12-3AJ4y2)dy =

4 5 . 4 0 )

( 5 . 4 1 )

2A21 dy = S

S o l v i n g E q . ( 5 . 4 1 ) s i m u l t a n e o u s l y g i v e s

3 V = 7 7

4 h V

A s

H c n c c t h e t r u e b e a m s t r e s s e s a r e

S 3 V S V rT™ = v , - 2 l s x y : = T h - 7 x y

< r „ = 0 ( 5 . 4 2 )

3 v 3 V i V m 2^

F o u r i e r s e r i e s s o l u t i o n T h e F o u r i e r s e r i e s t e c h n i q u e i s u s e d w h e n e v e r t h e l o a d

d i s t r i b u t i o n i s d i s c o n t i n u o u s o v e r a p o r t i o n o f t h e s o l i d , a s s h o w n i n F i g . 5 . 3 4 , f o r

e x a m p l e . T h i s m e t h o d a s s u m e s t h a t t h e s o l u t i o n o f E q . ( 5 . 3 4 ) m a y b e e x p r e s s e d a s

fl>(.v, y) = <l\(x)<t>y(y) (5.43)

w h e r e ' I ^ U " ) m a y b e t a k e n , in t h e f o r m o f F o u r i e r s e r i e s , a s

. mitx (I>Xx) = s i n —

o r ( 5 . 4 4 )

, , , m 7 C X

<!>,(.v) = c o s — j —

, / T

M Figure 5.34

U t i l i z i n g E q . ( 5 . 4 3 « ) o r (5 .43 /? ) in E q . ( 5 . 4 3 ) a n d t h e n s u b s t i t u t i n g i n t o E q .

( 5 . 3 4 ) y i e l d

dv * - 2 [I , Mr 2 + [f<X>r = 0 ( 5 . 4 5 )

w h e r e / ! = m n / L .

E q u a t i o n ( 5 . 4 5 ) i s a f o u r t h - o r d e r h o m o g e n e o u s d i f f e r e n t i a l e q u a t i o n w i t h

c o n s t a n t c o e f f i c i e n t s . I t s s o l u t i o n c a n b e o b t a i n e d e a s i l y i n t e r m s o f h y p e r b o l i c

f u n c t i o n s a s

( p y = C , s i n h py + C2 c o s h Pv + C , y s i n h fly + C 4 y c o s h f l y ( 5 . 4 6 )

C o n s t a n t s C ( , ( \ . • • • a r e d e t e r m i n e d f r o m t h e b o u n d a r y c o n d i t i o n s o f t h e s o l i d

u n d e r c o n s i d e r a t i o n . T h e t o t a l s o l u t i o n o f E q . ( 5 . 3 4 ) t h u s b e c o m e s

<l>(.v, y ) = ( C , s i n h Py + C , c o s h Py + C . , y s i n h Py

, , . mux + C 4 y c o s h Py) s i n — —

L

<[)(.\, y ) = ( C , s i n h py + C2 c o s h f l y + C , y s i n h Py

max + C 4 y c o s h Py c o s — —

( 5 . 4 7 a ) .

( 5 . 4 7 b )

T h e c h o i c c o f t h e t r i g o n o m e t r i c f u n c t i o n d e p e n d s o n t h e s y m m e t r y o f t h e

l o a d i n g . F o r i n s t a n c e , if t h e l o a d i n g is s y m m e t r i c a l a b o u t t h e c h o s e n y a x i s o f t h e

b e a m , a s s h o w n i n F i g . 5 . 3 4 a , t h e n t h e c o s i n e f u n c t i o n m u s t b e u s e d . F o r t h e

b e a m o f F i g . 534b. t h e s i n e f u n c t i o n m u s t b e a s s u m e d b e c a u s e o f t h e a n t i s y m -

m e t r y o f t h e l o a d i n g a b o u t t h e c h o s e n y a x i s .

STRILSS ANALYSIS 131

PROBLEMS

5.1 Find Ihe m a x i m u m tensile and max imum compressive stresses resul t ing f r o m bending of the beam shown in Fig. 1*5.1. Kim! the dis t r ibut ion of shear stresses over the cross sect ion at Ihe seclion where the shear is a m a x i m u m , consider ing points in the cross scction at vert ical intervals of I in.

2 0 l h / i n

, 2 0 0 0 l b

f t t t t

, 2 0 0 0 l b

! f i — 3 0 i n - * - { — * — ' 4 0 i n — - - - - 3 0 i n - -L i - * - 3 0 i n — j

1 in

C I

3 i n

r 3^in

~ T 3 i n

Figure P5.I

5.2 Find Ihe max imum *hcar and bending stresses in the b e a m cross sect ion s h o w n in Fig. P5.2 if the shear V is 10,090 lb and the bend ing momen t M is 400,000 in • lb. Bo th angles h a v e the same cross section. Assume the web to be effective in resisting bending stresses.

1 0 i n

r i

0 . 5 i n

T

- 0 . 0 5 1 i n

/ - 0 . 1 5 i n 4

A = 0 . S i n 2

Figure P5.2

5.3 Find the shear stress and the shear-f low dis t r ibut ion over the cross section of the beam shown in Fig. P5.3. Assume the web to be ineffective in resisting bend ing a n d the s t r inger areas to be con-centrated at points .

f \

\

\ 0 . 0 4 0 i n

\

1 . 0 i n 2

r

10 in2

Sin

N 2.0 in2

1 0 , 0 0 0 l b Figure P5.3

1 3 2 A I R C R A F T S T R U C T U R E S

5.4 Each o r Ihe live u p p e r s t r ingers has an area o r 0.4 in 2 , and each or the five lower str ingers has an area ofO.X in i . F i n d Ihe shear f lows in all the webs if the vertical shear ing force is 12,000 Ib.

— r 2 in

" 6 in

J. m t Figure P5.4

S>-5j E:\ch of the six s t r ingers of the cross section s h o w n in Fig. P5.5 a n d P5.6 has an area of 0.5 in 2 . Find Ihe shea r f lows in all webs and the location of the shea r ccntcr for a vertical shear ing force of

^10,000 Ib.

' S j Find the shear flows in all wchs in Fig. P5.5 and P5.fi for a hor izonta l shear ing force of 3000 Ib. Each s l r inger has a n area of 0.5 in2 .

Figure P5.5 and P5.6

fe.7\l?ind a general expression for the shear-Row dis t r ibut ion a r o u n d Ihe circular lube shown in Fig. PV7. A . Assume the wall thickness t to be small c o m p a r e d with the rad ius R.

Figure P5.7


5.8 Use Hqs. (5.17) and (5. IS) to find the shear flow in the webs of the two-str inger beams shown in Fig. P5.8 under the action of a vertical shear Vr. ,

5.9 Find the shear-flow distr ibution for the section shown in Fig. PS.9.

1 in

8 in

-12 i

- 8 in-

1 in

10,0001b Figure P5.9

5.10 Find the shear (lows in the webs of the box beam shown in Fig. P5.10 if the area is symmetrical abou t a horizontal centerline. •

10.000 lb Figure P5.10

5.11 Find the shear Hows in the webs of the beam shown in Fig. P 5 . l l and P5.12. All stringers have areas of 1.0 in 2 . 5.12 Assume that the two right-hand stringers in Fig. P 5 . l l and P5.12 have areas of 3.0in J and the other stringers have areas of 1.0 in2 . Kind the shear flows in the webs by two methods.

3 4 A I R C ' R A I T S T R U C T U R E S

j - 10 in 10 in ~

in i t 10 in

1 -*-<. in

8000 Ib - Figure 5.1! and P5.12

5.13 Find the shear flows in all webs if Ihe two right-hand stringers shown in Fig. P5.13 have areas of 1.5 in2 and the other stringers have areas of 0.5 in2 .

5.14 Find the shear-now distribution in all webs shown in Fig. P5.14. Ml parts of the cross section resist bending stresses.

1 ' 1 * t

0.1 in **~0.l in (> in

I

"**4 in-*

12,000 Ib Figure P5.I4

5.15 Solve Example 5.8 if Ihe beam depth varies from 5 in at the free end to 15 in at the suppor t . (Sec Fig. P5.15.)

STRILSS A N A L Y S I S 1 3 5

Figure P5.15

5.16 Find Ihc shear flows for Ihe cross section at x — SO in. Consider only this one cross section, but calculate the torsional moments by two methods.

(a) Select the torsional axis arbitrarily, and calculate the in-plane components of the Range loads.

(ft) Take moments about a torsional axis joining the centroids of the various cross sections.

5.17 Repeal Prob. 5.16 if there is an additional chordwise load of 6000 lb act ing to the left at the center of the lip cross section. 5.18 A cantilever beam 30 in long carries a vertical load or 1000 lb at the free end. The cross section is rectangular and is 6 by 1 in. Find the maximum bending stress and the location of the neutral axis if (a) Ihe 6-in side is vertical, (ft) the 6-in side is tilted 5° from the vertical, and (c) the 6-in side is tilted 10" from the vertical.

5.19 A horizontal beam with a square cross section resists vertical loads. Find the angle of the neutral axis with the horizontal if one side of the beam makes an angle 0 with the horizontal. At what angle should the beam be placed for the bending stress to have a minimum value?

5.20 Find the bending stresses and stringer loads for the box beam whose cross section is shown in Fig. P5.20 if M . = 100,000 and M y = - 4 0 , 0 0 0 in - lb. Assume the areas of the stringer members are as follows:

(d) a = b~c = il = 2 in2

(ft) a = b = 3 in2 . c = <1 1 in' (<1 a = d - 3 in2 h = r - i in ;

W a = c = 3 in2 . h - </ = 1 in'

(c) a = c = 1 in2 . /, = </ = 3 in

10 in

- 2 0 i n - Figurc P5.20

5,21 A beam with the cross section shown in Fig. P5.21 resists a bending moment M . - 100 in • lb. Calculate the bending stresses a t points A, I), and C.

1 3 6 A I R C ' R A I T S T R U C T U R E S

i u

t x t „

0.1 in •a-o

C 0.1 ->- Figure P5.21

5.22 The box beam shown in Fig. P5.22 resists bending moments of Af. = 1,000,000 and M = 120.000 in • lb. I ind the bending stress in each stringer member . Assume that the webs are ineffective in bending and the areas and coordinates of the stringers are as follows:

No. Area, in 2 z, in y, in

1 1.8 2.62 8.3 2 0.4 - 10.81 9.12 3 0.8 - 2 4 . 7 0 9.75 4 2.3 - 2 4 . 7 0 - 1 . 3 5 1.0 2.62 - 1 . 2

5.23 Find the shear flows at the cross section shown in Fig. P5.23 and P5.24 for x = 50 in. Consider only Hie one cross seclion, and calculate the in-plane components of the stringer loads.


STRESS ANALYSIS 137

5.24 Repeal Prob. 5.23. using the differences in stringer loads shown in Fig. P5.23 and P5.24 at the cross section for x — 40 and — 60 in. 5.25 Calculate the shear flows in the webs of the cross section shown in Fig. P5.25 and P5.26 at .v = 50 in. Assume the flange areas as follows:

(«) a •- b = 3 in2, r <1 - I in2

(/>) a " c - I in ' . h - <1 3 in" (<•) u — i = 3 in2, h -11 - I in2

Consider only the one cross section, and calculate the in-plane components of the llange loads.

5.26 Repeat Prob. 5.25, using the dillercnces in llange loads at the cross section shown in Fig. P5.25 and P5.26 for A* = 40 and 60 in. Use a torsional axis joining the cenlroids or the cross sections.

a h

d r Figure P5.I5 and P5.26

5.27 Kind the shear Hows at station 100 of ihe fuselage shown in Fig. P5.27. Assume all stringer areas to be I in2.

Station 100

O ' i

10.0011 Ih

Figure P5.27

5.28 Using Ihe Airy stress function, find Ihe stresses in the beams shown in Fig. P5.2X.

-100 in—

Uli

20 Ili/in

r B

4 in i n \

-\ . \

\

4 in -100 in-

tb)

Figure P5.28


5.29 A circular , thin p la te is under the act ion of uni formly d is t r ibu ted pressure applied a r o u n d the ouler edge. Kind the stresses, using the Airy stress funct ion. Assume the pla te thickness is equal to I. (See Fig. P5.29.)

C H A P T E R

SIX DEFL12CTJON ANALYSIS OF

STRUCTURAL SYSTEMS

6.1 INTRODUCTION

T h e m o s t i m p o r t a n t a p p l i c a t i o n s o f t h e m e t h o d s f o r c a l c u l a t i n g d i s p l a c e m e n t s

( d e f l e c t i o n s ) a r e i n t h e a n a l y s i s o f r e d u n d a n t ( i n d e t e r m i n a t e ) s t r u c t u r a l s y s t e m s , a s

is d e m o n s t r a t e d h e r e a n d in l a t e r c h a p t e r s .

T h e d e f l e c t i o n s o f m o s t e n g i n e e r i n g s t r u c t u r e s a r e s m a l l a n d v e r y s e l d o m a r e

u s e d a s a n i m p o r t a n t d e s i g n c r i t e r i o n . H o w e v e r , t h e r e l a t i v e r i g i d i t y o f v a r i o u s

e l e m e n t s in r e d u n d a n t s t r u c t u r a l s y s t e m s a f f e c t s t h e s t r e s s d i s t r i b u t i o n i n t h e

s t r u c t u r e ; t h e r e f o r e , it is n c c c s s a r y t o c o n s i d e r t h e d e f o r m a t i o n s i n t h e a n a l y s i s o f

s u c h s t r u c t u r a l s y s t e m s .

T h e m e t h o d s o f d c f l c c t i o n a n a l y s e s p r e s e n t e d i n t h i s c h a p t e r a r e C a s t i g l i a n o ' s

m e t h o d , t h e R a y l e i g h - R i t z m e t h o d , a n d t h e finite d i f f e r e n c e m e t h o d . T h e finite-

e l e m e n t m e t h o d is p r e s e n t e d in a s e p a r a t e c h a p t e r . A l s o p r e s e n t e d h e r e i s t h e

u n i t - l o a d m e t h o d f o r t h e a n a l y s i s o f s i m p l e r e d u n d a n t s t r u c t u r e s . T h e e n e r g y

m e t h o d s t r e a t e d a r c d e r i v e d in a c c o r d a n c e w i t h t h e p r i n c i p l e s o f v i r t u a l a n d

c o m p l e m e n t a r y v i r t u a l w o r k a s s o c i a t e d w i t h v i r t u a l d i s p l a c e m e n t s a n d v i r t u a l

f o r c e s , respectively. W e a s s u m e s m a l l s t r a i n s a n d c o r r e s p o n d i n g s m a l l d i s p l a c e -

m e n t s i n a l l t h e d e v e l o p m e n t s o f t h i s c h a p t e r . I n a d d i t i o n , t h e m a t e r i a l i s a s s u m e d

p c r f c c t l y c l a s t i c .

139


6.2 WORK AND COMPLEMENTARY WORK: STRAIN AND COMPLEMENTARY STRAIN ENERGIES

C o n s i d e r a s t r u c t u r e t o b e a c t e d o n b y a s e t o f g e n e r a l i z e d f o r c e s Q,{i = 1, 2 , . . . , n) w h i c h r e s u l t i n a c o r r e s p o n d i n g s e t o f g e n e r a l i z e d d i s p l a c e m e n t s q,{i = 1, 2 , . . . , n). T h e f o r c e - d i s p l a c e m e n t r e l a t i o n s h i p f o r a t y p i c a l f o r c e Q„ a n d i t s c o r r e -

s p o n d i n g d i s p l a c e m e n t q„ a r e s h o w n g r a p h i c a l l y in F i g . 6 . 1 a . T h e a r e a u n d e r t h e

c u r v e r e p r e s e n t s t h e w o r k W d o n e b y f o r c e Q„ i n m o v i n g t h r o u g h t h e c o r r e -

s p o n d i n g d i s p l a c e m e n t qH. T h e a r e a a b o v e t h e c u r v e i s d e f i n e d a s t h e comp-lementary work W. F o r a s y s t e m w h i c h i s i n a s t a t e o f s t a t i c e q u i l i b r i u m , a n d if

h e a t d i s s i p a t i o n is n e g l e c t e d , t h e n b a s e d o n t h e c o n s e r v a t i o n o f e n e r g y , i t c a n b e

s t a t e d t h a t t h e w o r k d o n e o n t h e s y s t e m is e q u a l t o t h e s t r a i n e n e r g y s t o r e d i n t h e

s y s t e m , o r

W = U ( 6 . 1 ) .

a n d W = U (6 .2 )

w h e r e U a n d U a r e t h e s t r a i n e n e r g y a n d c o m p l e m e n t a r y s t r a i n e n e r g y , r e s p e c t -

i v e l y . F i g u r e 6 .1 / ) s h o w s a g r a p h i c a l r e p r e s e n t a t i o n o f t h e s e q u a n t i t i e s .

6.3 PRINCIPLE O F VIRTUAL DISPLACEMENTS AND RELATED THEOREMS

I n v a r i a t i o n a l m e c h a n i c s , a s t r u c t u r a l s y s t e m i s i m a g i n e d t o h a v e g o n e t h r o u g h a

s e t o f i n f i n i t e s i m a l d i s p l a c e m e n t s c o n s i s t e n t w i t h t h e c o n s t r a i n t s , w h e n , i n r e a l i t y ,

n o s u c h d i s p l a c e m e n t s e x i s t . T h e s e fictitious m o v e m e n t s o f t h e s t r u c t u r e a r e c o m -

m o n l y r e f e r r e d t o a s t h e virtual displacements, a n d t h e c o r r e s p o n d i n g w o r k i s

c a l l e d t h e virtual work. C o n s i d e r t h a t a s t r u c t u r e i s g i v e n a s m a l l v a r i a t i o n i n v i r t u a l d i s p l a c e m e n t

5q„, a s s h o w n in F i g . 6 . 1 . T h i s i n d u c e s a v a r i a t i o n i n v i r t u a l w o r k 5W a n d a

Figure 6.1 («) work and complementary work: (h) strain and complementary strain energies.

l5l:i'Li:C T I O N A N A L Y S I S O F S T R U C T U R A L SYSTt iMS 1 4 1

c o r r e s p o n d i n g v a r i a t i o n in s t r a i n e n e r g y 3U, a s i n d i c a t e d b y t h e v e r t i c a l s t r i p s i n F i g . 6 . 1 . I f w e a s s u m e t h a t t h e e x t e r n a l a p p l i e d f o r c e s { Q } a n d t h e i n d u c e d i n t e r n a l s t r e s s e s { £ } r e m a i n c o n s t a n t d u r i n g t h e v i r t u a l d i s p l a c e m e n t s , t h e n t h e c h a n g e s i n v i r t u a l w o r k a n d v i r t u a l s t r a i n e n e r g y c a n b e o b t a i n e d r e a d i l y f r o m F i g . 6.1 a n d a r e g i v e n b y

AW = (5VK = Q„ Sqn

A U* = dU* = ade

(6 .3 )

(6 .4 )

w h e r e <5 d e n o t e s t h e first v a r i a t i o n , a n d i t o p e r a t e s i n t h e s a m e m a n n e r a s t h e d i f f e r e n t i a l o p e r a t o r </, a n d 6 U * is t h e v a r i a t i o n i n s t r a i n e n e r g y p e r u n i t v o l u m e ( e n e r g y d e n s i t y ) , i .e. .

<•51/ = t>U* dV ( v o l u m e i n t e g r a l ) (6 .5 )

If t h e v a r i a t i o n s in <>W a n d <>U a r e c o n s i d e r e d f o r t h e v a r i a t i o n o f a l l d i s p l a c e -m e n t s (i = 1 , 2 n) c o n s i s t e n t w i t h t h e c o n s t r a i n t s , t h e n

<W= I QlSql i— 1

(6.6)

o r , i n m a t r i x c o m p a c t f o r m , F q . (6 .6) b e c o m e s

SW = L^Jtfi} (6-7)

w h e r e t h e s y m b o l s | J a n d [ j i n d i c a t e r o w a n d c o l u m n m a t r i c e s , r e s p e c t i v e l y .

L i k e w i s e , t h e c o r r e s p o n d i n g v a r i a t i o n i n s t r a i n e n e r g y d e n s i t y c a n b e e x p r e s s e d

a s

W * = L<5£J{£} (6 .8 )

w h e r e t h e s t r a i n a n d s t r e s s l i e lds , in g e n e r a l , a r e g i v e n b y

L.SFC'L = <5[E„ 6 „ . E , J (6 .9 )

a n d

( i - j — (6.10)

T h e p r i n c i p l e o f v i r t u a l d i s p l a c e m e n t s s t a t e s t h a t a n e l a s t i c d c f o r m a b l e s t r u c -

t u r a l s y s t e m is in a s l a t e o f e q u i l i b r i u m if t h e v i r t u a l w o r k SW d o n e b y f o r c e s { Q }

is e q u a l t o t h e v i r t u a l s t r a i n e n e r g y <51/ f o r e v e r y a r b i t r a r y v i r t u a l d i s p l a c e m e n t

c o n s i s t e n t w i t h t h e c o n s t r a i n t s o f t h e s t r u c t u r e . M a t h e m a t i c a l l y , t h i s p r i n c i p l e is

e x p r e s s e d a s

$W = 5U

L<5<?J{G} = L<5EJ{2} dV

(6.11)

(6.12)

w h e r e t h e e x t e r n a l a p p l i e d f o r c e s { Q } a n d t h e i n t e r n a l s t r e s s e s { S } a r e a s s u m e d

t o b e i n e q u i l i b r i u m a n d t h e c o o r d i n a t e d i s p l a c e m e n t s {<5c/} a n d t h e s t r a i n s { i 5 £ }

s a t i s f y t h e c o m p a t i b i l i t y c o n d i t i o n

{•5 £ } = W { 5 q } ( 6 . 1 3 )

S u b s t i t u t i n g E q . ( 6 . 1 3 ) i n t o E q . ( 6 . 1 2 ) y i e l d s

M Q } = M A ] T { £ } d v ( 6 . 1 4 )

o r , in e x p a n d e d f o r m ,

[<5<j, S q 2 - - - S q r - "

<2,

Qi

Q:

L Q,.

Sq2 • • • Sqt • • • <5<j„]

A , , A , 2 A>3 ' • X

A 2 I A 2 2 ^ 2 3 • A

A . „ A 3 2 A 3 3 • •

A«2 A . 3 ' • A

Cl

."•ft J

dV ( 6 . 1 5 )

w h e r e t h e s u p e r s c r i p t T d e n o t e s m a t r i x t r a n s p o s e ( i .e . , r o w s a n d c o r r e s p o n d i n g

c o l u m n s a r e e x c h a n g e d ) .

If e v e r y v i r t u a l d i s p l a c e m e n t <5qr [ ( r = 1, 2 , . . . n ) , r # i ] i s s e t e q u a l t o z e r o

w i t h t h e / t h d i s p l a c e m e n t <5qt a l l o w e d t o b e a u n i t d i s p l a c e m e n t , t h e n i t c a n b e

s e e n e a s i l y f r o m E q . ( 6 . 1 5 ) t h a t E q . ( 6 . 1 4 ) r e d u c e s t o

1 • Qi = 1 • 1 A ! { E } dv

l5l:i'Li:C TION ANALYSIS OF STRUCTURAL SYSTtiMS 1 4 3

o r

Qt = IAI W dV ( 6 . 1 6 ) J i'

w h e r e t ! i e r o w m a t r i x [ / . J is d e l i n e d b y E q . ( 6 . 1 3 ) , w h i c h u n d e r t h e c o n d i t i o n

f O f o r r # i d'lr = 1 , ,

( 1 f o r r = i

b e c o m c s

L;„.j = ( 6 . 1 7 )

E q u a t i o n ( 6 . 1 6 ) d e s c r i b e s w h a t is c o m m o n l y r e f e r r e d t o a s t h e unit-displacement met hud. S e e A r g y r i s a n d K . c i s e y 2 - ' f o r m o r e d e t a i l e d i n f o r m a t i o n o n t h i s t h e o r e m .

F o r c o n s e r v a t i v e s t r u c t u r a l s y s t e m s , E q . ( 6 . 1 1 ) m a y b e e x p r e s s e d in t h e f o r m

S(U + P) = dV = 0 ( 6 . 1 8 )

w h e r e 6 V i s t h e f i r s t v a r i a t i o n o f I h e t o t a l p o t e n t i a l e n e r g y a n d

&\V = £ Q , <•)£/, i= I

E q u a t i o n ( 6 . 1 8 ) is t h e p r i n c i p l e o f s t a t i o n a r y t o t a l p o t e n t i a l a n d m a y b e s h o w n

r e a d i l y t o b e e q u i v a l e n t t o s t a t i n g t h a t

- ' - = 0 (f- = 1 , 2 , . . . , n ) ( 6 . 1 9 )

w h i c h i s t h e b a s i s o f t h e R a y l c i g h - R i t z m e t h o d i n s t r u c t u r a l a n a l y s i s .

A n o t h e r t h e o r e m t h a t i s b a s e d o n t h e p r i n c i p l e o f v i r t u a l d i s p l a c e m e n t s i s

C a s t i g l i a n o ' s f i r s t t h e o r e m .

F r o m E q s . ( 6 . 3 ) a n d ( 6 . 1 1 ) .

&U = M 6 } (6-2°)

B y n o t i n g t h a t til' i s a f u n c t i o n o f t h e c o o r d i n a t e g e n e r a l i z e d " d i s p l a c e m e n t s

[<s ( / . ( i = 1 , 2 » ) ] a n d u s i n g t h e T a y l o r s e r i e s e x p a n s i o n , E q . ( 6 . 2 0 ) b e c o m e s

" ru . 1 " " 82U . t d U = ^ + £ £ -r-T- 5qt 6qj + • • •

i i " / . ' 2 i - i dqi dqj

o r . i n m a t r i x f o r m ,

W = L' '/J + iL 'WJl^} + • • • (6.21)

w h e r e t h e s u m b o l P i l e n o t e s p a r t i a l d i f f e r e n t i a t i o n a n d t h e m a t r i x [ S ] i s a s t i f f n e s s

m a t r i x w h o s e e l e m e n t s a r e d e l i n e d b y

= ( / , ; = 1 , 2 , . . . , « ) ( 6 . 2 2 ) t(Jj

If o n l y first-order v a r i a t i o n i n 6q i s r e t a i n e d i n E q . (6 .21) , t h e n f r o m E q . (6 .20 ) ,

o r

L < 5 d ( j e } ~ j — H = o

S i n c e [<•%] a r e i n d e p e n d e n t d i s p l a c e m e n t s ,

r l 0

o r

- { S F o r / = r , E q . ( 6 . 2 3 ) s t a t e s t h a t

{<2} = M (6.23)

dU Qr = i r ( 6 - 2 4 ) -dqr

E q u a t i o n ( 6 . 2 4 ) is C a s t i g l i a n o ' s first t h e o r e m

6.4 PRINCIPLE OF VIRTUAL FORCES AND RELATED THEOREMS

T h e t r e a t m e n t o f v i r t u a l f o r c e s in v a r i a t i o n a l m e c h a n i c s i s a n a l o g o u s t o t h a t o f t h e v i r t u a l d i s p l a c e m e n t s p r e s e n t e d in S e c . 6 .3 . T h u s , f r o m F i g . 6 .1 i t c a n b e s e e n e a s i l y t h a t f o r a g i v e n s m a l l v a r i a t i o n i n v i r t u a l f o r c e SQ„, c o n s i s t e n t w i t h t h e s t a t i c e q u i l i b r i u m c o n d i t i o n s , t h e c o r r e s p o n d i n g v a r i a t i o n s i n c o m p l e m e n t a r y v i r t u a l w o r k a n d c o m p l e m e n t a r y v i r t u a l s t r a i n e n e r g y d e n s i t y a r e

AW = 5W = q„5Q„ (6,25)

a n d

AO* = dU* = e da (6.26)

w h e r e

50 = 5U* dV ( v o l u m e i n t e g r a l ) ( 6 . 2 7 ) Jv

If w e c o n s i d e r t h e v a r i a t i o n s in 5W a n d SU f o r t h e v a r i a t i o n o f a l l f o r c e s

Qi (i = 1 , 2 , . . . , n), Eqs. ( 6 . 2 5 ) a n d (6 .26 ) b e c o m e

<nv = l5Q}{q} (6.28)

l5l:i'Li:C TION ANALYSIS OF STRUCTURAL SYSTtiMS 1 4 5

a n d

SI' - L < 5 2 | { £ } dV ( 6 . 2 9 )

T h e p r i n c i p l e o f v i r t u a l f o r c e s 2 4 ( p r i n c i p l e o f v i r t u a l w o r k ) d i c t a t e s t h a t a n

e l a s t i c s t r u c t u r a l s y s t e m i s i n a c o m p a t i b l e s t a t e o f d e f o r m a t i o n if f o r e v e r y

a r b i t r a r y v i r t u a l f o r c e &Q, t h e c o m p l e m e n t a r y v i r t u a l w o r k is e q u a l t o t h e c o m -

p l e m e n t a r y s t r a i n e n e r g y c o n s i s t e n t w i t h s t a t i c c o n d i t i o n s o f e q u i l i b r i u m . M a t h -

e m a t i c a l l y , t h e p r i n c i p l e m a y b e e x p r e s s e d a s

SW = 5U ( 6 . 3 0 )

o r

l'^J!</! = £ L<5£|{£} dV (6.31)

w h e r e t h e d i s p l a c e m e n t s {</] a n d t h e c o r r e s p o n d i n g s t r a i n s { £ } a r e c o m p a t i b l e

a n d t h e v i r t u a l f o r c c s {/>(?} a n d c o r r e s p o n d i n g s t r e s s e s {<5£} s a t i s f y t h e e q u i l i b -

r i u m c o n d i t i o n

!<•>£} = []{<52} ( 6 - 3 2 )

S u b s t i t u t i n g E q . ( 6 . 3 ? ) i n t o E q . ( 6 . 3 1 ) y i e l d s

l ' s ( ? J [ < / i - SQiWiE} dV ( 6 . 3 3 )

I n E q . ( 6 . 3 3 ) . if e v e r y v i r t u a l f o r c c SQr ' ( r = 1, 2 , . . . , n), r i ] i s s e t e q u a l t o

z e r o w i t h t h e ; ' th f o r c e 3 Q t g i v e n a u n i t v a l u e , t h e n t h e e q u a t i o n b e c o m e s

'/; = L d > ; J { E } dV ( 6 . 3 4 )

w h e r e t h e r o w m a t r i x |"<1>] is d e f i n e d b y E q . ( 6 . 3 2 ) , w h i c h u n d e r t h e c o n d i t i o n

( 0 f o r r # i = '

b e c o m e s

0 forr*/ . ^ . y , ,

11 f o r r — i

L ^ J = 1 < 5 S J ( 6 . 3 5 )

E q u a t i o n ( 6 . 3 4 ) is r e f e r r e d t o b y A r g y r i s a n d K e l s e y a s t h e unit-load-method. T h e p r i n c i p l e o f s t a t i o n a r y t o t a l c o m p l e m e n t a r y p o t e n t i a l m a y b e d c d u c e d

f r o m l i q . ( 6 . 1 8 ) a n d is g i v e n b y

i ) { p f U ) = d V = 0 ( 6 . 3 6 )

w h e r e

n fiW=-8P= I q,5Q, (6.37)

E q u a t i o n ( 6 . 3 6 ) m a y b e s h o w n t o b e e q u i v a l e n t t o

dV — = 0 i = 2 , n ( 6 . 3 8 ) <>Qi

w h i c h r e p r e s e n t s t h e c o m p a t i b i l i t y c o n d i t i o n s a t a l l c o o r d i n a t e s /.

If w e c o n s i d e r E q . ( 6 . 3 0 ) , f r o m E q . ( 6 . 2 8 ) t h e c o m p l e m e n t a r y s t r a i n e n e r g y c a n

b e e x p r e s s e d a s

a { 7 = | > 5 e . J t e } ( 6 - 3 9 )

E x p a n d i n g E q . ( 6 . 3 9 ) i n a T a y l o r s e r i e s y i e l d s

" fiU 1 " " r>2U

o r , in m a t r i x f o r m ,

S V = + \ [ < 5 e ] [ « ] { < 5 Q } + • • • ( 6 . 4 0 )

w h e r e t h e m a t r i x [ a ] is t h e f l e x i b i l i t y m a t r i x w h o s e c o e f f i c i e n t s a r e d e f i n e d b y

d2U dQ; (~)Qj

If o n l y f i r s t - o r d e r v a r i a t i o n i n SQ is r e t a i n e d i n E q . ( 6 . 4 0 ) , t h e n f r o m E q . ( 6 . 3 9 )

= ( W = 1, 2 , . . . , n ) ( 6 . 4 1 )

i m l ' l } = L«5QJ { f ^

L < 5 2 j ( k } - { ^ } ) = 0 ( 6 . 4 2 )

S i n c e [_^<2J a r c i n d e p e n d e n t a r b i t r a r y f o r c e s , f o r E q . ( 6 . 4 2 ) t o b e s a t i s f i e d , t h e

f o l l o w i n g m u s t h o l d t r u e :

o r

{cQ

F o r i = r , E q . ( 6 . 4 3 ) i s e q u i v a l e n t t o

{<?} = ® (6"43>

OQr

w h i c h is t h e s e c o n d t h e o r e m o f C a s t i g l i a n o .

PU q , = ( 6 . 4 4 )

l5l:i'Li:C T I O N A N A L Y S I S O F S T R U C T U R A L SYSTtiMS 1 4 7

6.5 LINEAR ELASTIC STRUCTURAL SYSTEMS10

F o r s t r u c t u r a l s y s t e m s w h o s e b e h a v i o r is l i n e a r l y e l a s t i c , a s s h o w n i n F i g . 6 . 2 , t h e

w o r k d o n e o n t h e s y s t e m is e q u a l t o t h e c o m p l e m e n t a r y w o r k ; l i k e w i s e , t h e s t r a i n

e n e r g y s t o r e d i s e q u a l t o t h e c o m p l e m e n t a r y s t r a i n e n e r g y . A s c a n b e s e e n e a s i l y

f r o m F i g . 6 . 2 , t h e e x p r e s s i o n s f o r t h e w o r k a n d s t r a i n e n e r g y a n d t h e i r c o u n t e r -

p a r t s a r e

W = W =l - l Q . Qz ... QJ

<7I

1z

o r . in c o m p a c t m a t r i x f o r m ,

a n d

W —W — Ml{q)

U = U =-2

o r , i n c o m p a c t m a t r i x f o r m ,

a..

Gxz

U — U = L £ J { Z } dV

dV

( 6 . 4 5 )

( 6 . 4 5 a )

( 6 . 4 6 )

( 6 . 4 6 a )

w h e r e Q,- (i — 1 , 2 , . . . , n) a r e t h e e x t e r n a l a p p l i e d l o a d s a n d qt ( i = 1, 2 , . . . , n ) a r e

1,-j

4

All*

4 V -AIL' =• AIR

A(7

/ / /

u / ^AU=AU

/ u

l-"ii;iirc 6.2 I ine.ir clastic behavior .


( h e c o r r e s p o n d i n g n o d a l c o o r d i n a t e d i s p l a c e m e n t s ; t h e m a t r i c e s [ > ' J a n d { £ } a r e t h e s t r a i n a n d s t r e s s fields, respectively.

B y u t i l i z a t i o n o f H o o k e ' s l a w , E q . ( 6 . 4 6 ) m a y b e e x p r e s s e d in t e r m s o f s t r e s s e s o r s t r a i n s a l o n e a s f o l l o w s :

o r

U = U =4

U = V =4

LZJIXKE} dV r

L E J [ 0 ] { £ } dV

T h e m a t r i c e s [ / V ] a n d [ 0 ] a r e d e f i n e d b y

1

1

— V I s y m m e t r i c

— V — v 1

0 0 0 2(1 + v) 0 0 0 0 2(1 + v) 0 0 0 0 0 2 ( 1 + v ) J

a n d

[ © ] = '7

1 - V

V 1 - V s y m m e t r y V V 1 - V

0 0 0 (1 - 2 v ) / 2 0 0 0 0 (1 - 2 v ) / 2

0 0 0 0 0 (1

( 6 . 4 7 )

( 6 . 4 8 )

( 6 . 4 9 )

( 6 . 5 0 )

w h e r e i] = E( 1 + vX I — 2v )

E = e l a s t i c m o d u l u s o f e l a s t i c i t y

v = P o i s s o n ' s r a t i o

I n s t r u c t u r a l s y s t e m s w h i c h a r e c o n s t r u c t e d f r o m a n a s s e m b l a g e o f b a r e l -

e m e n t s , it is m o r e c o n v e n i e n t t o e x p r e s s t h e s t r a i n e n e r g y i n t e r m s o f t h e e l e m e n t

i n t e r n a l l o a d s a n d t h e n s u m t h e c o n t r i b u t i o n s o f e a c h e l e m e n t t o t h e t o t a l s t f a i n

e n e r g y o f t h e s y s t e m .

T o d e r i v e t h e s t r a i n e n e r g y e x p r e s s i o n , c o n s i d e r a l i n e a r c l a s t i c b a r f o r w h i c h ,

a t a n y p o i n t £ a l o n g i t s l e n g t h , t h e i n t e r n a l l o a d s a r e g i v e n b y M . ( 0 , T ( ( ) , ^ . ( f ) ,

a n d Si0, w h e r e M. — b e n d i n g m o m e n t a b o u t £ a x i s , T = t o r q u e , Vy = s h e a r f o r c e

in r d i r e c t i o n , a n d S = a x i a l f o r c e . T h e n o r m a l s t r e s s e s i n d u c e d in t h e b a r b y t h e

i n t e r n a l f o r c e s y s t e m m a y b e o b t a i n e d f r o m C h a p . 5 a n d a r e g i v e n a s

ff.vo = n o r m a l s t r e s s d u e t o a x i a l f o r c e = ( a ) A(Q

a x x h = n o r m a l s t r e s s d u e t o b e n d i n g = 2 (b)

DEFLECTION ANALYSIS OF STRUCTURAL SYSTEMS 149

If w c a s s u m e t h a t crxx is t h e o n l y n o n z e r o s t r e s s , t h e n E q . ( 6 .47 ) r e d u c e s t o

w h e r e U„ = s t r a i n e n e r g y d u e t o n o r m a l s t r e s s e s o n l y .

S u b s t i t u t i n g E q s . (a ) a n d (/>) i n t o E q . (c) y i e l d s

to

U„ v.. + = 2 M%Q

c A ™2(0

v EA2(Q

J

2E J

2E J K ElUQ y d V

— j — rf/l 4 + - [J C JA Ell(Qy aA ^

b u t

a n d

T h e r e f o r e ,

1 it A — A(Q = c r o s s - s e c t i o n a l a r e a

L> r 2 d A = /_(£) — m o m e n t o f i n e r t i a

,, • 1 f ^ 1 ,r (4) T h e s t r a i n e n e r g y d u e t o t r a n s v e r s e s h e a r m a y b e o b t a i n e d b y c o n s i d e r i n g

F i g . 6 .3 . T h e a v e r a g e s h e a r s t r e s s a c t i n g o n t h e r i g h t f a c e o f t h e b a r c a n b e e x p r e s s e d a s

_ vjtQ _ VM) AM) kA(Q

(e)

w h e r e A $ ) is t h e e f f e c t i v e s h e a r a r e a , /1(C) is t h e a c t u a l b a r c r o s s - s e c t i o n a l a r e a ,

a n d k is t h e s h e a r - f o r m f a c t o r w h i c h a c c o u n t s f o r t h e s h e a r s t r e s s d i s t r i b u t i o n

a c r o s s t h e b a r d e p t h .

F r o m H o o k e ' s l a w ,

U t i l i z i n g (p), E q . ( / ) b e c o m e s

- G

m) GkA(Q

i f )

to)

Figure 6.3

T h e w o r k d o n e o n t h e d i f f e r e n t i a ! e l e m e n t b y t h e f o r c e Ky(Q i s e q u a l t o t h e

s t r a i n e n e r g y s t o r e d , o r ,

dW = dUs = i Vy e.y d(

_ 1 VyiQ 2 GkA{Q

di

T h e r e f o r e ,

"A dC CkA(Q

T h e s t r a i n e n e r g y d u e t o a t o r q u e T c a n b e s h o w n t o b e g i v e n b y

" T2(Q U. = -GJ

dC

(h)

( 0

w h e r e J = t o r s i o n a l c o n s t a n t .

T h u s , f o r a s t r u c t u r a l s y s t e m w h i c h i s m a d e o f m s t r u c t u r a l b a r e l e m e n t s , t h e

s t r a i n e n e r g y e x p r e s s i o n m a y b e w r i t t e n a s

z i = 1 J

s 2 < 0 ^ M2(0 ^ K2(0 , T2(0 + •

kGA(Q GJ{Q J dQ ( 6 . 5 1 ) IEA(0 EI(Q

w h e r e m = t o t a l n u m b e r o f e l e m e n t s i n s t r u c t u r e

S = a x i a l i n t e r n a l l o a d

M = i n t e r n a l b e n d i n g m o m e n t

V — i n t e r n a l t r a n s v e r s e s h e a r

T = i n t e r n a l t o r q u e

AE, EI, AG. GJ = e x t e n s i o n a l , b e n d i n g , s h e a r a n d t o r s i o n a l r i g i d i t y , r e s p e c t i v e l y

A, / , ,/ = c r o s s - s e c t i o n a l a r e a , m o m e n t o f i n e r t i a , a n d t o r s i o n a l

c o n s t a n t , r e s p e c t i v e l y

G = s h e a r m o d u l u s

k — s h e a r - f o r m f a c t o r w h i c h a c c o u n t s f o r d i s t r i b u t i o n o f s h e a r i n g

s t r e s s e s a c r o s s b a r d e p t h ( f o r w i d e f l a n g e s e c t i o n s , k 1,

w h i l e f o r r e c t a n g u l a r s e c t i o n s , k ^ 0 . 8 3 3 )

C = g e n e r a l i z e d c o o r d i n a t e a x i s J

6.6 CASTIGLIANO'S SECOND THEOREM IN DEFLECTION ANALYSIS O F STRUCTURES

F o r l i n e a r e l a s t i c s t r u c t u r a l s y s t e m s , t h e s e c o n d t h e o r e m o f C a s t i g l i a n o m a y b e

w r i t t e n a s

9 , = - S r (V = V) ( f i - 5 2 > vQr

T h i s e q u a t i o n s t a t e s t h a t t h e d i s p l a c e m e n t q a t a n y p o i n t r o n t h e s t r u c t u r e a n d

l5l:i'Li:C TION ANALYSIS OF STRUCTURAL SYSTtiMS 151

i n a n v d i r e c t i o n i s e q u a l t o t h e f i r s t p a r t i a l d e r i v a t i v e o f t h e t o t a l s t r a i n e n e r g y

w i t h r e s p e c t t o a c o r r e s p o n d i n g f o r c e Q a t r a n d i n t h e s a m e d i r e c t i o n a s q. T h e

f o l l o w i n g e x a m p l e s i l l u s t r a t e a p p l i c a t i o n s o f t h i s w i d e l y u s e d t h e o r e m i n t h e

d c f l e c t i o n a n a l y s i s o f s t r u c t u r e s .

E x a m p l e 6 . 1 F i n d t h e v o r t i c a l d e f l e c t i o n o f p o i n t 3 o n t h e s t r u c t u r e s h o w n i n

t h e l i g u r e . A s s u m e a l l m e m b e r s t o b e t h e s a m e m a t e r i a l a n d t o h a v e t h e s a m e

c r o s s - s e c t i o n a l p r o p e r t i e s . P o i n t s 2, 3 , a n d 4 a r c p i n n e d .

S O L U T I O N T h e t o t a l s t r a i n e n e r g y s t o r e d i n t h e s t r u c t u r e i s

U=Ul + U2 + U3

w h e r e L / , , a n d (./., a r e t h e s t r a i n e n e r g i e s i n m e m b e r s 1, 2 , a n d 3 ,

r e s p e c t i v e l y . B y n o t i n g t h a t m e m b e r s 2 a n d 3 a r e a x i a l r o d e l e m e n t s , t h e i r

s t r a i n e n e r g i e s m a y b e e x p r e s s e d a s

* / .

c / a = 5

L ' ,

A AE

dx.

C2L

AE : dx,

w h e r e t h e c o o r d i n a t e .vl;- d e n o t e s t h e a x i s b e i n g t a k e n a l o n g t h e d i r e c t i o n

f r o m i t o j . F o r c l e m e n t I , t h e r e e x i s t i n t e r n a l l o a d s i n t h e f o r m o f b e n d i n g ,

s h e a r , a n d a x i a l l o a d s . T h e r e f o r e t h e s t r a i n e n e r g y e x p r e s s i o n f o r t h i s e l e m e n t

is

S f 1 r Z'-Mj , 1

W dX21 + 2 vl

AG dx2

T h e i n t e r n a l l o a d s i n c a c h e l e m e n t m a y b e f o u n d f r o m s t a t i c s a n d a r e g i v e n


b y

V 5 2Fy

">/3 Fy

V 5 M i = F J > x 2 1

M a k i n g t h e a p p r o p r i a t e s u b s t i t u t i o n s y i e l d s

= •

i r 2 f - I t i . \ ' 2 1 +

2 £ 7 2 / 1 G Fl dx21

o r

r , / 1 1 L 4 L 3 L \ 2 U = ( ttt^: + — + — I Fl VGAE"1" 3EI~r AGj'

B y u s i n g C a s t i g l i a n o ' s t h e o r e m , E q . ( 6 . 5 2 ) y i e l d s

1 ~ PFy ~ V 3 / 4 E + 3 E l + ACJ "

If s h e a r d e f o r m a t i o n is n e g l e c t e d , t h e n t h e d e f l e c t i o n b e c o m e s

( 1 1 L 8 L 3 \ 3 y = {jTE+3E,)F> S

E x a m p l e 6 . 2 F i n d t h e s l o p e a n d v e r t i c a l d e f l c c t i o n a t t h e t i p o f t h e c a n t i l e -

v c r c d b e a m s h o w n i n t h e figure. N e g l e c t s h e a r d e f o r m a t i o n .

\ i \ \

Dlit-'LECTION ANALYSIS OF STRUCTURAL SYSTEMS 1 5 3

SOI.UTION R e c a l l t h a t i n o r d e r t o find a p a r t i c u l a r c o o r d i n a t e d i s p l a c e m e n t

b y u s i n g C a s t i g l i a n o ' s s e c o n d t h e o r e m , t h e r e m u s t e x i s t a c o r r e s p o n d i n g

f o r c e a t t h e p o i n t w h e r e t h e d i s p l a c e m e n t is s o u g h t a n d i n t h e d i r e c t i o n o f t h e

d i s p l a c e m e n t . H e n c e , i n o r d e r t o find t h e s l o p e a t t h e t i p o f t h e c a n t i l e v e r e d

b e a m s h o w n i n t h e figure, a fictitious m o m e n t m u s t b e a p p l i e d a t t h a t p o i n t ,

K(.v)

Mn

0 M„

a s s h o w n . A f t e r t h e s l o p e i s f o u n d , t h i s m o m e n t i s s e t e q u a l t o z e r o , w h i c h

i m p l i e s i t s n o n e x i s t e n c e o n t h e o r i g i n a l s t r u c t u r e .

T h e s t r a i n e n e r g y e x p r e s s i o n f o r t h i s c a s e i s

EI 2

LM\2 ) E I 12

dx + 2 Jo

M 23

EI, dx

F r o m t h e s k e t c h a b o v e .

Mi 2 = M„ = M0 + Fyx

T h e r e f o r e

U = (M0 + F, x)2 dx + — | (M0 + Fyxf dx 2EI

4 i ? l M j L + 2

r M0Fyl3 F^L3

- U s i n g E q . ( 6 . 5 2 ) y i e l d s

(5y = dU 8F„

s l o p e = 0 =

m 0=o

8U

L R 2 EI

dM0 M O = 0

3 FVL2

4 EI

E x a m p l e 6 . 3 F i n d t h e h o r i z o n t a l m o t i o n o f t h e r i g h t s u p p o r t o f t h e s e m i -

c i r c u l a r a r c h s h o w n i n t h e f i g u r e ( p . 154) . A s s u m e b e n d i n g d e f o r m a t i o n s o n l y ,

a n d a s s u m e EI t o b e c o n s t a n t .

SOLUTION I n o r d e r t o find t h e h o r i z o n t a l m o t i o n a t 1, a fictitious h o r i z o n t a l

f o r c e m u s t b e a p p l i e d a t t h a t p o i n t , a s s h o w n i n t h e figure. T h e finding

m o m e n t c a n b e e a s i l y c a l c u l a t e d a n d is g i v e n b y

P it Ml2 = R( 1 - cos 0) j + sin 0)FX 0<0<-

M32 = Ml2

T h e r e f o r e t h e s t r a i n e n e r g y i s

EI \2EI

i t /2

M\ZR dO

EI K ( l - c o s 0) — + FXR s i n 0 R dO

U s i n g C a s t i g l i a n o ' s s e c o n d t h e o r e m y i e l d s

DU <*>.x :

BFr

2 *Jt/2

0 R( 1 - c o s 0) R2 s i n 0 dO

Dlit-'LECTION ANALYSIS OF STRUCTURAL SYSTEMS 155

o r

PR3

2 EI

E x a m p l e 6 . 4 F i n d I h e v e r t i c a l d e f l e c t i o n a t 1 a n d t h e a n g u l a r t w i s t a t 2 o f t h e

s t r u c t u r e s h o w n in ( l i e f i g u r e . N e g l e c t s h e a r d e f o r m a t i o n .

S O L U T I O N T h e s t r a i n e n e r g y s t o r e d i n t h e s t r u c t u r e i s

U = 1 Mil , , 1

~ e T 2

M h . i

o ~ETdx + 2

c-

CJ dx

w h e r e M l 2 = Fr

M 2 3 Ffx

T23 = Tf + 7^1 - - ) + aLF

7> = f i c t i t i o u s t o r q u e a p p l i e d a t 2

> PFy

r « i. M l z 8 M l 2 ^

EI (IF.

+ M2i dM23

EI BF.. dx +

T23 5T23

GJ (IF, dx

a n d

M l 2 d M l 2

EI BTF dz

+ •V/23 *1M2

EI 8TF

dx + T23 3T23

o GJ dTF

T h e p a r t i a l s i n t h e a b o v e e q u a t i o n s m a y b e e a s i l y c a l c u l a t e d a n d a r c g i v e n b y

dM12 dM23 (IT, 3

sry ~z = x Jir = «L

<>M1 2 _ SM23 _ a r 2 3 _ ( T r (-37). ( ?T f

S u b s t i t u t i n g i n t h e i n t e g r a l s a n d c a r r y i n g o u t t h e i n t e g r a t i o n s y i e l d

OLL2T0 .... fa.3 1 a2\ , (

W - ^ + aL'F,

6.7 RAYLEIGH-RITZ METHOD IN DEFLECTION ANALYSIS O F STRUCTURES

T h e R a y l e i g h - R i t z m e t h o d o f o b t a i n i n g d i s p l a c e m e n t s o f c o n s e r v a t i v e e l a s l i c

s t r u c t u r a l s y s t e m s is b a s e d o n t h e p r i n c i p l e o f c o n s e r v a t i o n o f e n e r g y , a s s h o w n

p r e v i o u s l y . T h e m e t h o d a s s u m e s t h a t a d e f l e c t i o n s h a p e o f t h e s t r u c t u r e is k n o w n

a n d m a y b e t a k e n a s

I A ® . ( 6 - 5 3 ) * 1

w h e r e q i s t h e g e n e r a l i z e d c o o r d i n a t e d i s p l a c e m e n t , /?„ a r e u n d e t e r m i n e d c o n s t a n t

p a r a m e t e r s a n d a r e c h o s e n s u c h t h a t t h e t o t a l p o t e n t i a l i n t h e s y s t e m b e c o m e s

m i n i m u m , a n d <D„ a r e a c h o s e n s e t o f f u n c t i o n s w h i c h m u s t s a t i s f y o n l y t h e

d i s p l a c e m e n t b o u n d a r y c o n d i t i o n s o f t h e s t r u c t u r e a t h a n d .

T h e u n d e t e r m i n e d c o n s t a n t s j8„ a r e o b t a i n e d b y s o l u t i o n o f a s e t o f s i m u l -

t a n e o u s l i n e a r a l g e b r a i c e q u a t i o n s w h i c h r e s u l t w h e n t h e p r i n c i p l e o f m i n i m . u m

p o t e n t i a l is a p p l i e d in c o n j u n c t i o n w i t h t h e a s s u m e d d c f l e c t i o n s h a p e g i v e f i b y

E q . ( 6 . 5 3 ) . T h e m e t h o d i s b e s t e x p l a i n e d b y c o n s i d e r i n g t h e f o l l o w i n g e x a m p l e s .

E x a m p l e 6 . 5 U s i n g t h e R a y l e i g h - R i t z m e t h o d , find t h e d e f l e c t i o n of t h e s i m p l e

b e a m s h o w n b e l o w . C o n s i d e r b e n d i n g d e f o r m a t i o n o n l y .

SOLUTION T h e b o u n d a r y c o n d i t i o n s o f t h e b e a m a r e

D e f l e c t i o n = w ( 0 ) = <a(L) = 0

B e n d i n g m o m e n t = M{0) = M(L) = 0 = El , 2

d2(o dx2 (6.5.4)

T h e r e f o r e , if „ in F .q . ( 6 . 5 3 ) i s t a k e n a s s i n ( T I K X / L ) , w h e r e N i s a n i n t e g e r , t h e n i t i s

o b v i o u s t h a t b o t h s e t s o f b o u n d a r y c o n d i t i o n s a r e s a t i s f i e d . H e n c e , t h e a p p r o p r i -

a t e d e f l e c t e d s h a p e o f t h e b e a m m a y b e t a k e n a s

«{-*) = E Pn s i n n= I

T h e p r i n c i p l e o f m i n i m u m t o t a l p o t e n t i a l s t a t e s t h a t

d ( [ / + P ) = 0

w h e r e

u = -2

M 2 ,

>

W

(b)

(c)

a n d

p = -W = P0{x)co(x) dx W

T h e b e n d i n g m u m e n t m a y b e e x p r e s s e d i n t e r m s o f co a s f o l l o w s :

nn \

B y m a k i n g t h e a p p r o p r i a t e s u b s t i t u t i o n s , E q s . (c ) a n d (d ) b e c o m e , r e s p e c t i v e l y ,

U EI

a n d

P =

ZI ITIT I „ . 717LX

.„=, I r J s m t

" „ . nnx , I'u L Pn sin — dx

dx (e)

(/)

P e r f o r m i n g t h e i n t e g r a t i o n s y i e l d s

a n d ( " = 1, 3 , 5 , . . . ) n n

U')

F r o m E q . (b )

T h e d e f l e c t i o n f r o m E q . ( a ) b e c o m e s

, 4P(J L4 £ 1 . nitx

T h e c o n v e r g e n c e o f t h e a b o v e s e r i e s i s v e r y r a p i d , a n d in m o s t c a s e s f e w t e r m s a r e

n e e d e d t o o b t a i n s u f f i c i e n t a c c u r a c y f o r t h e d e f l e c t i o n . F o r e x a m p l e , if o n l y t h e

f i r s t t e r m in t h e s e r i e s i s c o n s i d e r e d , t h e m a x i m u m d e f l c c t i o n wi l l b e 0 . 0 1 3 1

/ ' „ Z,4/(/- ' /). a s c o m p a r e d t o t h e e x a c t v a l u e o f 0 . 0 1 3 0 / \ , L 4 / ( E / ) .

6.8 FINITE DIFFERENCE METHOD IN DEFLECTION ANALYSIS OF STRUCTURES

C l o s e d - f o r m s o l u t i o n s o f d i f f e r e n t i a l e q u a t i o n s r e p r e s e n t i n g t h e d e f o r m a t i o n s o f

s t r u c t u r a l s y s t e m s a r e n o t a l w a y s p o s s i b l e . T h e r e f o r e , t h e a p p l i c a t i o n o f s o m e

a p p r o x i m a t e n u m e r i c a l m e t h o d s is n o t o n l y p e r m i s s i b l e b u t , i n m a n y i n s t a n c e s ,

d e s i r a b l e . T h e m o s t c o m m o n l y u s e d i s t h e m e t h o d o f finite d i f f e r e n c e s . T h e

m e t h o d a p p r o x i m a t e s d i f f e r e n t i a l s b y finite d i f f e r e n c e s ; a s a r e s u l t , i t t r a n s f o r m s a

s e t o f d i f f e r e n t i a l e q u a t i o n s t o a s e t o f s i m u l t a n e o u s a l g e b r a i c e q u a t i o n s , w h i c h

t h e n c a n b e s o l v e d e a s i l y .

T h e f u n d a m e n t a l r e l a t i o n s o f f i n i t e d i f f e r e n c e s m a y b e e s t a b l i s h e d b y c o n -

s i d e r i n g t h e e x i s t e n c e o f a r e a l , c o n t i n u o u s s i m p l e f u n c t i o n

0> = / ( . x ) ( 6 . 5 5 )

F i g u r e 6 . 4 s h o w s a g r a p h i c a l r e p r e s e n t a t i o n o f t h e f u n c t i o n a t e q u i d i s t a n t v a j t i e s

o f Ax. T h e f i r s t d e r i v a t i v e o f ' I > f o r s m a l l v a l u e s o f A.v a t s o m e p o i n t i m a y b e

<M>

dx

w h e r e A<!>, c a n b e f o u n d b y c o n s i d e r i n g e i t h e r p o i n t s i a n d i -I- 1 o r i a n d i — 1 :

A<f>K> a : c l \ - , , - (I),- ( 6 . 5 7 )

o r

A O } w ^ < I ' i - ( D i . ! ( 6 . 5 8 )

T h e d i f f e r e n c e Afl 'S^' r e p r e s e n t s t h e first f o r w a r d d i f f e r e n c e a p p r o x i m a t i o n , w h i l e

A<h<M r e p r e s e n t s t h e first b a c k w a r d d i f f e r e n c e a p p r o x i m a t i o n . T h e m e a n v a l u e o f

A<J>, - 7 - 1 < 6 - 5 6 > Ax

<1>=Ax)

Ax ' Av- ' Av 1 Ax I AT 1 Av

Figure 6.4 One-dimensional finite-difference approximaiion.

t h e s e t w o e x t r e m e s m a y b e e x p r e s s e d a s

A O S 1 a 1 — - —

o r

<l>i+ i - (I>i i

w h e r e A<J>/r> is r e f e r r e d t o a s t h e first central difference approximation. B y u t i l i z i n g E q . ( 6 . 5 9 ) , E q . (6 .56 ) b e c o m e s

<M>

dx 2 Ax

( 6 . 5 9 )

(6.60)

E q u a t i o n ( 6 . 6 0 ) g i v e s t h e a p p r o x i m a t e first d e r i v a t i v e o f <J> a t p o i n t / in t e r m s o f

c e n t r a l d i f f e r e n c e s .

t M >

dx2

d3<t>

dx3

d4V

dx4

[ + , - 2 d ) i + <D, _ 1

(Ax)2

'Dr f 2 - 2 i + 1 + 2 0 , , ! - < t > i - 2

2 (AJC)3

1\ + 2 ~ 40) ,^- ! + 6CH; - 4 ( 1 ) ; - ! + $1-2

(A-x)4

(6.61)

e t c .

F o r a t w o - d i m e n s i o n a l f u n c t i o n s u c h a s s h o w n i n F i g . 6 . 5 , t h e p a r t i a l d e r i v a -

t i v e s m a y b e o b t a i n e d a s i n t h e c a s e o f o n e - d i m e n s i o n a l f u n c t i o n :

M )

dx . ,.i

2 Ax r*l> 0 z

<Kj + i - ft;.;-. 2 Az

d2<t> dx2

j a3® fix3

J

53<J> Pz3

j c1* (p

•.j

340> <3z4

••j d2<$>

dx dz '.j d3<J>

<V dy i.j d3d>

dx dy2 >•}

( A x ) 2 3z2

- 2<!>i+t.j + 2 Q , - _ l w - - <t>i_2. j 2 ( A x ) 3

- 23>,-. m + 2 P , . - <P„ j _ 2

2 ( A z ) 3

m 2

(6.62)

' - r - i . J - - 1 1- i . J • • t - t . j

( A x ) 4

' * I • — 1 . /

' h j + 2 ~ 4r- y + 1 + <J> l W + I - 4 ® ,

( A z ) 4

G . - M . / - M i . j + i 4 A x A z

ft.M.j-M - ^.^l.j- 1 ~ 2<t>i.J+l + 2<t>j j_ 1 - <!>,•-I j-t +

2 ( A x ) 2 A z

2 A x ( A z ) 2

E x a m p l e 6 . 6 U s i n g t h e f i n i t e d i l T e r e n e e m e t h o d , find t h e v e r t i c a l d e f l e c t i o n a t

•i>=/U> r )

Az typical

r iVt typical

Figure 6.5 Two-dimensional finite-dilTercnce approximat ion.

DI-H .P.RTION ANALYSIS OF STRUCTURAL SYSTEMS 161

m i d s p a n a n d t h e s l o p e a t t h e l e f t s u p p o r t ( s e e t h e figure). A s s u m e El = c o n -

s t a n t .

-1.12- - A B -

SOLUTION I n o r d e r t o u s e t h e f i n i t e d i f f e r e n c e m e t h o d i n s t r u c t u r a l a n a l y s i s ,

t h e f o l l o w i n g s t e p s m u s t b e t a k e n :

1. F o r m u l a t e a l l t h e g o v e r n i n g d i f f e r e n t i a l e q u a t i o n s w h i c h d e s c r i b e t h e b e -h a v i o r o f t h e s t r u c t u r e .

2 . E x p r e s s t h e g o v e r n i n g d i f f e r e n t i a l e q u a t i o n s a n d t h e e q u a t i o n s w h i c h d e -s c r i b e t h e b o u n d a r y c o n d i t i o n s in t e r m s o f finite d i f f e r e n c e s .

3 . O b t a i n a s e t o f s i m u l t a n e o u s a l g e b r a i c e q u a t i o n s b y s a t i s f y i n g t h e r e -s u l t i n g f i n i t e d i f f e r e n c e e q u a t i o n s i n s t e p 2 a t finite d i s c r e t e p o i n t s o n t h e s t r u c t u r e . T h i s is e q u i v a l e n t t o i d e a l i z i n g t h e s t r u c t u r e t o a finite n u m b e r of e l e m e n t s c o n n e c t e d a t d i s c r e t e p o i n t s .

4 . S o l v e f o r t h e f i n i t e d i f f e r e n c e m e t h o d s . I n t h i s e x a m p l e , t h e g o v e r n i n g d i f f e r e n t i a l e q u a t i o n o f d e f o r m a t i o n f o r a

s i m p l e b e a m s t r u c t u r e m a y b e o b t a i n e d f r o m t h e s t r e n g t h o f t h e m a t e r i a l a n d is g i v e n b y

EI <fw dx4 "

(a)

w h e r e w = b e a m v e r t i c a l d c f i c c t i o n EI = b e a m b e n d i n g r i g i d i t y ;

p = d i s t r i b u t e d n o r m a l a p p l i e d l o a d

T h e s l o p e a t a n y p o i n t a l o n g t h e b e a m is

o J w dx

E x p r e s s i n g E q s . (n) a n d (/J) in t e r m s of finite d i f f e r e n c e s y i e l d s

PI (A.X)4

« ' H , - , + 6ii'f — 4wt., + iv,• _2 = -(El),

(b)

V )

a n d

0, = • 2 Ax

ib')

w h e r e i = 1, 2 , 3 N a r e p o i n t s a l o n g t h e b e a m s p a n .

B y e x a m i n i n g E q . ( a ' ) , i t c a n b e c o n c l u d e d t h a t o n l y a f i n i t e n u m b e r o f

p o i n t s m a y b e s a t i s f i e d if a f e a s i b l e s o l u t i o n h a s t o b e r e a c h e d . T h u s , f o r

s i m p l i c i t y , d i v i d e ( i d e a l i z e ) t h e b e a m i n t o f o u r e q u a l e l e m e n t s o n l y , a s s h o w n

in t h e f i g u r e b e l o w .

PHU4)

•UA-

- Z . / 4 - - £ / 4 - - Z / 4 - - £ / 4 - -1.14- L/4 H 5 6

N o t e t h a t t h e a c c u r a c y i n c r e a s e s w i t h a n i n c r e a s e in t h e n u m b e r o f

e l e m e n t s t a k e n . T h e fictitious e x t e n s i o n o f t h e b e a m b e y o n d i t s o u t e r s u p -

p o r t s is n e c e s s a r y a s c a n b e s e e n f r o m E q . (« ' ) . T h e d i s p l a c e m e n t s o f p o i n t s 0

a n d 6 a r e a s s u m e d t o b e e q u a l t o t h e n e g a t i v e d i s p l a c e m e n t s o f p o i n t s 2 a n d

4 , r e s p e c t i v e l y . F o r fixed s u p p o r t s , t h e s e d i s p l a c e m e n t s a r e e q u a l i n s i g n a n d

m a g n i t u d e . S e e s k e t c h e s («) a n d (/>) b e l o w . A l s o n o t e t h a i t h e c o n c e n t r a t e d

l o a d P i s a s s u m e d t o b e d i s t r i b u t e d u n i f o r m l y o v e r h a l f t h e l e n g t h o f e a c h

e l e m e n t o n b o t h s i d e s o f t h e p o i n t w h e r e t h e l o a d i s c o n c e n t r a t e d . T h i s is

n e c e s s a r y b e c a u s e t h e l o a d f u n c t i o n in E q . ( a ' ) i s a d i s t r i b u t e d l o a d .

(a)

K + M-,

(ft)

W r i t i n g E q . (<7') f o r p o i n t s 2 , 3 , a n d 4 y i e l d s

w4- — 4U'3 + 6W2 — 4 w , + vi'o = 0

PL3 iv j — 4 u ' 4 + 6 i v , — 4 w 2 + iv , = •

/

6 4 EI

vv„ — 4IV5 + 6 U ' 4 — 4U>3 + u ' 2 = 0

F r o m b o u n d a r y c o n d i t i o n s a n d s y m m e t r y , t h e f o l l o w i n g d e f l e c t i o n s a r e

k n o w n :

H ' | = U ' ; = 0 V I ' , = U ' 4 .

A l s o , for t h e d e f o r m a t i o n t o b e c o n t i n u o u s o v e r t h e s u p p o r t s ,

iv„ = - • ii '2 a n d iv,, = — u ' 4

N o t e t h a t t h e first a n d t h e l a s t e q u a t i o n s in ( r ) a r c i d e n t i c a l b e c a u s e o f

s y m m e t r y ; t h e r e f o r e , b y d e l e t i n g o n e a n d u t i l i z i n g t h e a b o v e c o n d i t i o n , E q .

(c) b e c o m e s

6vv2 4 w , = 0

— 4 t v , + 3vv, = PL1

128£i

o r , in m a t r i x f o r m ,

6 — 4~ w 2

— 4 3_ -W 3 _

^ e x a c t :

S o l v i n g E q . (</) f o r w 2 a n d iv 3 y i e l d s

0 . 0 1 5 6 P L 3

H', = 2 FJ

PL3

'Vj ~ 42.1 HI

T h e s l o p e f r o m E q . { h ' ) is

u ' 2 - w 0 w 2 0 . 0 6 2 4 P L 2

e x a c t :

P L

128£/

0 . 0 1 4 3 P L :

EI

PL3\

(d)

<h = 2 A.x A.v EI

4 8 £ / /

e x a c t : 0.0625PI/

EI

E x a m p l e 6 .7 U s e t h e f i n i t e d i f f e r e n c e a p p r o a c h t o find t h e m a x i m u m b e n d i n g m o m e n t s a n d v e r t i c a l d e f l e c t i o n o n t h e p l a t e s t r u c t u r e s h o w n . A s s u m e t h a t t h e p l a t e is s i m p l y s u p p o r t e d a n d a c t e d o n b y a u n i f o r m p r e s s u r e p. A l s o a s s u m e t h a t t h e p l a t e is o f u n i f o r m t h i c k n e s s f.

T h e g o v e r n i n g d i f f e r e n t i a l e q u a t i o n o f a f l a t p l a t e i s g i v e n b y 25

.4 + 2 a ! , , , 2 + a „ 4 - n ( a ) 3 4 w 3 4 w 3 4 W p

+ a x 2 + V = ^

w h e r e D = Er3/[12(1 _ v2 ) ] a n d v = P o i s s o n ' s r a t i o . T h e b e n d i n g m o m e n t s

a r c g i v e n b y 2 5

Jd2™ 82A ' D \ d f + V d S j W

F r o m F.q. (6 .62 ) , F .q . ( a ) m a y b e e a s i l y e x p r e s s e d in t e r m s o f finite d i f f e r -e n c e s :

+ CWi,j-2

+ I . j - l + +

Awi_2j + BWi-u + wLj + Bw,nj+ Awn2.j = P

+ Ew-t _ i , j +1 + F w { J + 1 + E w j + 1 . j + 1 (c)

+ CwLj+2

w h e r e A = H

4 ( 1 + a 2 )

H

U

2a2

4 « 2 ( 1 + a 2 ) F = •

P =

H

P(Ax Ay a)2

DH

Ax

H = 6 + 8 a 2 + 6 * 4

A s i n t h e b e a m e x a m p l e , f o r s i m p l i c i t y , d i v i d e t h e p l a t e i n t o f o u r e q u a l

p a r t s a l o n g t h e <c a n d y a x e s , a s s h o w n i n t h e f i g u r e . F r o m s y m m e t r y , o n l y

p o i n t s I , 2 , 4 , a n d 5 n e e d t o b e c o n s i d e r e d . T h u s , w r i t i n g R q . (c) f o r t h e s e

p o i n t s y i e l d s

l)i:i l.liC I I()N ANAI.YSIS ()!•' STRUCTllltAI. SYSTEMS 163

s y m m e t r y ; t h e r e f o r e , b y d e l e t i n g o n e a n d u t i l i z i n g t h e a b o v e c o n d i t i o n , E q .

(c) b e c o m e s

6 w 2 — 4 w 3 = 0

PL3

—4U>2 + 3 w 3 = 1 2 8 EI

o r , i n m a t r i x f o r m .

r e —4 L - 4 3 .

S o l v i n g E q . (r/) f o r w 2 a n d iv 3 y i e l d s

0 . 0 1 5 6 P L 3

PL3

w 2

L W 3 j

^ e x a c t :

( e x a c t : 4 2 . 7 EI

T h e s l o p e f r o m E q . (/>') is

IV, - u 'n vv, 0 . 0 6 2 4 P L 2

PI} 128 EI

0 . 0 1 4 3 PL

EI

PL3\ 4 8 £ / y

(<0

0, = 2 Ax Ax EI

e x a c t : 0 . 0 6 2 5 P L 2

EI

E x a m p l e 6 . 7 U s e t h e finite d i f f e r e n c e a p p r o a c h t o find t h e m a x i m u m b e n d i n g

m o m e n t s a n d v e r t i c a l d e f l e c t i o n o n t h e p l a t e s t r u c t u r e s h o w n . A s s u m e t h a t

t h e p l a t e is s i m p l y s u p p o r t e d a n d a c t e d o n b y a u n i f o r m p r e s s u r e p. A l s o

a s s u m e t h a t t h e p l a t e is o f u n i f o r m t h i c k n e s s t.

T h e g o v e r n i n g d i f f e r e n t i a l e q u a t i o n o f a fiat p l a t e i s g i v e n b y 25


d*w (14w d*W p dx'1 + 2 T s U f + [ a )

w h e r e D fit3/[ 1 2 ( 1 - v 2 ) ] a n d v = P o i s s o n ' s r a t i o . T h e b e n d i n g m o m e n t s a r e g i v e n b y 2 5

F r o m E q . ( 6 . 6 2 ) , E q . ( a ) m a y b e e a s i l y e x p r e s s e d i n t e r m s o f finite d i f f e r -e n c e s :

+ Cwi.j-2

+ Ew,. { j , + Fwu_i + Ewu , j ,

A w ; . 2J + Bw; _ j _ j + j + Bwt,, A iv,., j j = P

+ i.j+i + Fw;,j+i + + 1J+l (c)

+ Cwi.j+2

w h e r e A = — H

g = _ 4 ( 1 + a 1 )

H

_ 4 a 2 ( l + a 2 )

H

- P(Ax Ay a)2

DH

Ax

" T y H = 6 + 8 a 2 + 6 a 4

A s i n t h e b e a m e x a m p l e , Tor s i m p l i c i t y , d i v i d e t h e p l a t e i n t o l o u r e q u a l

p a r t s a l o n g t h e x a n d y a x e s , a s s h o w n i n t h e figure. F r o m s y m m e t r y , o n l y

p o i n t s 1, 2 , 4 , a n d 5 n e e d t o b e c o n s i d e r e d . T h u s , w r i t i n g E q . (c ) f o r t h e s e

p o i n t s y i e l d s

1.0 - 0 . 1 4 9 - 0 . 5 9 7 0.0597 " w t 4.664" - 0 . 1 4 9 0.5 0.0597 - 0 . 2 9 9

10 ~*PL* D

2.332

- 0 . 5 9 7 0.0597 0.5 - 0 . 0 7 5 2.332 (d) 0.0597 — 0.299 - 0 . 0 7 5 0.25 .1 .166 _

(d)

S o l v i n g F q . (</) y i e l d s

iv, = 0.00559

»v2 = 0.00723

w4 = 0.00778

PL* D

PL* D

PI* D

0 . 0 1 0 0 8 PL*

D e x a c t : 0 . 0 1 0 1 3

P L ;

D

S i n c e w e . h a v e t h e d i s p l a c e m e n t s , t h e m o m e n t s a r e c a l c u l a t e d a s f o l l o w s :

M , = M.x. s = — " i - M + v w ^ . - ^ + w , , . , ^

V Axz Ay /

16 D L}

— (h-6 - 2vv5 -I- w4) + 0.3(ws — 2w5 + w2)

= 0 . 0 4 5 4 P L 2 ( e x a c t : 0.0419 PL;)

L i k e w i s e ,

M. /V/,,5 = .0.966PLy (exact: 0.094%PL\)

6.9 REDUNDANT STRUCTURES AND THE UNIT-LOAD METHOD

A r e d u n d a n t ( . s t a t i c a l l y i n d e t e r m i n a t e ) s t r u c t u r a l s y s t e m i s o n e f o r w h i c h t h e

e x t e r n a l r e a c t i o n s o r i n t e r n a l l o a d s c a n n o t b e c o m p l e t e l y d e t e r m i n e d f r o m t h e

c o n d i t i o n s f o r s t a t i c e q u i l i b r i u m . A s t a b l e a n d s t a t i c a l l y d e t e r m i n a t e s t r u c t u r a l

s y s t e m c o n t a i n s o n l y e n o u g h e x t e r n a l s u p p o r t r e a c t i o n s o r s t r u c t u r a l m e m b e r s

f o r s t a b i l i t y , a n d t h e e q u a t i o n s o f s t a t i c e q u i l i b r i u m a r e s u f f i c i e n t t o d e t e r m i n e

c o m p l e t e l y t h e r e a c t i o n s o r t h e m e m b e r i n t e r n a l l o a d s . I f o n e m e m b e r o r r e a c t i o n

i s r e m o v e d , t h e s t r u c t u r e b e c o m e s u n a b l e a n d h e n c e i n c a p a b l e o f r e s i s t i n g a p p l i e d

l o a d s . I f o n e m e m b e r o r r e a c t i o n i s a d d e d , t h e s t r u c t u r e b e c o m e s s i n g l y r e -

d u n d a n t , a n d t h e reactions a n d m e m b e r i n t e r n a l l o a d s m u s t b e o b t a i n e d b y


c o n s i d e r i n g t h e d e f o r m a t i o n o f t h e s t r u c t u r e i n a d d i t i o n t o t h e c o n d i t i o n s o f

s t a t i c e q u i l i b r i u m .

N o r m a l l y , a r i g i d c o p l a n a r s t r u c t u r e r e q u i r e s t h r e e e x t e r n a l r e a c t i o n s f o r

s t a b i l i t y , a n d t h e y m a y b e c a l c u l a t e d f r o m t h e t h r e e e q u a t i o n s o f s t a t i c s . T h e

n u m b e r o f r e a c t i o n s , h o w e v e r , i s n o t t h e o n l y c r i t e r i o n f o r s t a b i l i t y , a n d i t i s

n e c e s s a r y t o e x a m i n e e a c h p a r t i c u l a r s t r u c t u r e i n o r d e r t o d e t e r m i n e w h e t h e r i t i s

s t a b l e , u n s t a b l e , o r s t a t i c a l l y i n d e t e r m i n a t e . F o r e x a m p l e , a h o r i z o n t a l s i m p l e

b e a m n o r m a l l y r e q u i r e s t h r e e r e a c t i o n c o m p o n e n t s . H o w e v e r , if t h e b e a m i s

s u p p o r t e d o n r o l l e r s a t t h r e e p o i n t s a l o n g t h e s p a n , t h e b e a m w i l l b e u n s t a b l e f o r

r e s i s t i n g h o r i z o n t a l f o r c e f a n d s t a t i c a l l y i n d e t e r m i n a t e f o r v e r t i c a l f o r c e s . S i m -

i l a r l y , if t h e t h r e e r e a c t i o n s o f a s i m p l e b e a m a c t t h r o u g h a n y c o m m o n p o i n t i n

t h e p l a n e , t h e m o m e n t s a b o u t t h a t p o i n t w i l l b e z e r o r e g a r d l e s s o f t h e m a g n i t u d e s

o f t h e f o r c e s , a n d t h e m o m e n t e q u a t i o n c a n n o t b e u s e d t o o b t a i n t h e r e a c t i o n s .

S u c h a s t r u c t u r e , s h o w n i n F i g . 6 . 6 , i s a m e c h a n i s m t h a t i s f r e e t o r o t a t e t h r o u g h

a s m a l l a n g l e a b o u t p o i n t O a s a n i n s t a n t a n e o u s c e n t e r , a n d it i s u n s t a b l e i n

r e s i s t i n g a n y l o a d w h i c h d o e s n o t a c t t h r o u g h p o i n t 0 . I f a l o a d a c t s t h r o u g h

p o i n t 0 , t h e s t r u c t u r e i s s t a t i c a l l y i n d e t e r m i n a t e .

I n m o s t s t r u c t u r e s , t h e n u m b e r o f e q u a t i o n s o f s t a t i c s m a y b e c o m p a r e d w i t h

t h e n u m b e r o f r e d u n d a n t s t o d e t e r m i n e t h e c o n d i t i o n s o f s t a b i l i t y . F o r s u c h

s p e c i a l s t r u c t u r e s a s t h a t s h o w n i n F i g . 6 . 6 , a n a t t e m p t t o find t h e t h r e e u n k n o w n

r e a c t i o n s f r o m t h e t h r e e e q u a t i o n s o f s t a t i c s w i l l y i e l d e q u a t i o n s w h i c h a r e n o t

i n d e p e n d e n t ( o n e o f t h e e q u a t i o n s c a n b e d e r i v e d f r o m t h e o t h e r s ) . W h e n a n

a t t e m p t t o a n a l y z e a s t r u c t u r e b y t h e e q u a t i o n s o f s t a t i c s r e s u l t s i n s u c h a

c o n d i t i o n , t h e s t r u c t u r e m u s t b e e x a m i n e d f o r i n s t a b i l i t y o r r e d u n d a n c y . F o r

s i m p l e r e d u n d a n t s t r u c t u r e s , t h e u n i t - l o a d m e t h o d i s o f t e n u s e d , a n d i t i s d e m o n -

s t r a t e d h e r e .

I f a s t r u c t u r e h a s o n e m o r e m e m b e r o r r e a c t i o n t h a n is r e q u i r e d f o r s t a b i l i t y ,

t h e s t r u c t u r e h a s s i n g l e r e d u n d a n c y . I n m a n y c a s e s , a n y o n e o f s e v e r a l m e m b e r s

o r r e a c t i o n s m a y b e r e m o v e d w i t h o u t c a u s i n g i n s t a b i l i t y . T h e n o n e d e l l e c t i o n

e q u a t i o n m u s t b e u s e d i n a d d i t i o n t o t h e e q u a t i o n s o f s t a t i c s i n o r d e r t o a n a l y z e

t h e s t r u c t u r e . I f a s t r u c t u r e h a s s e v e r a l m o r e m e m b e r s o r r e a c t i o n s t h a n a r e

r e q u i r e d f o r s t a b i l i t y , i t h a s m u l t i p l e r e d u n d a n c y . T h e d e g r e e o f r e d u n d a n c y i s

e q u a l t o t h e n u m b e r o f r e d u n d a n t m e m b e r s , a n d it i s e q u a l t o t h e n u m b q r o f

d e f l e c t i o n c o n d i t i o n s w h i c h m u s t b e u s e d i n t h e a n a l y s i s . ^

7 /

/ s /

/

Figure P5.7

DEFLECTION ANALYSIS OF STRUCTURAL SYSTEMS 1 6 7

6.10 STRUCTURES WITH SINGLE REDUNDANCY

A t r u s s w h i c h i s c o m p o s e d o f c l a s t i c m e m b e r s a n d w h i c h h a s s i n g l e r e d u n d a n c y i s

c o n s i d e r e d first. A t y p i c a l t r u s s o f t h i s t y p e i s s h o w n i n F i g . 6 . 7 a . T h e s u p p o r t s

a r e a s s u m e d t o b e r i g i d , a n d t h e m e m b e r s a r e a s s u m e d t o b e u n s t r e s s e d b e f o r e

l o a d P i s a p p l i e d . T h e r e a r e f o u r e x t e r n a l r e a c t i o n s , a n d o n l y t h r e e w o u l d b e

r e q u i r e d f o r s t a b i l i t y . T h e h o r i z o n t a l r e a c t i o n c o m p o n e n t A! -! i s c o n s i d e r e d a s t h e

r e d u n d a n t , a n d t h e d c f l e c t i o n e q u a t i o n w i l l b e o b t a i n e d f r o m t h e c o n d i t i o n t h a t

t h e h o r i z o n t a l s u p p o r t d e f l e c t i o n 5 i s z e r o . F r o m E q . ( 6 . 3 4 ) i t c a n b e s h o w n t h a t

w h e r e S r e p r e s e n t s t h e f o r c e i n a n y m e m b e r o f t h e s t r u c t u r e o f F i g . 6 . 7 a a n d s

r e p r e s e n t s t h e f o r c e in a n y m e m b e r o f t h e s t r u c t u r e d u e t o a u n i t l o a d a p p l i e d i n

t h e d i r e c t i o n o f t h e d e s i r e d d e f l e c t i o n , a s s h o w n i n F i g . 6 . 7 c .

T h e f o r c e S i n a n y m e m b e r i s f o u n d b y s u p e r i m p o s i n g t h e l o a d i n g c o n d i t i o n s

s h o w n i n F i g . 6 . 7 b a n d d. I f t h e r e d u n d a n t f o r c e i s r e m o v e d , t h e r e s u l t i n g s t a t i -

c a l l y d e t e r m i n a t e s t r u c t u r e i s s h o w n i n F i g . 6 . 7 b , a n d t h e a p p l i e d l o a d s p r o d u c e a

f o r c e S0 i n a n y m e m b e r . I f t h e r e d u n d a n t f o r c e X t i s a c t i n g a l o n e , i t p r o d u c e s a

f o r c e .Y(.v i n a n y m e m b e r , a s s h o w n i n F i g . 6 . 7 d , s i n c e t h e s f o r c e s r e s u l t f r o m

a u n i t v a l u e o f . Y j . T h e t o t a l f o r c e S i s o b t a i n e d a s t h e s u m o f t h e f o r c c s f o r t h e

t w o c o n d i t i o n s :

( 6 . 6 3 )

. S = S0 + XiS

S u b s t i t u t i n g f r o m F .q . ( 6 . 6 4 ) i n t o E q . ( 6 . 6 3 ) y i e l d s

( 6 . 6 4 )

r

Li^iiri- (t.7

o r , f o r /> = 0 ,

_ IS0sL/(AE) X l ~ ~ Zs>L/(AE) {6M)

w h e r e a l l t h e t e r r a s o n t h e r i g h t s i d e o f t h e e q u a t i o n m a y b e o b t a i n e d f r o m t h e

l o a d i n g a n d g e o m e t r y o f t h e s t r u c t u r e .

E q u a t i o n ( 6 . 6 6 ) i s a p p l i c a b l e t o a n y e l a s t i c t r u s s w i t h s i n g l e r e d u n d a n c y i n

w h i c h t h e d e f l e c t i o n i n t h e d i r e c t i o n o f t h e r e d u n d a n t i s z e r o . I f t h e d e f l e c t i o n i n

t h e d i r e c t i o n o f t h e r e d u n d a n t <5 i s a k n o w n v a l u e o t h e r t h a n z e r o , t h i s d e f l e c t i o n

m a y b e s u b s t i t u t e d i n t o E q . ( 6 . 6 5 ) a n d t h e v a l u e o f X0 d e t e r m i n e d f o r t h i s

c o n d i t i o n .

T h e p h y s i c a l s i g n i f i c a n c e o f t h e t e r m s i n E q s . ( 6 . 6 5 ) a n d ( 6 . 6 6 ) is d i s c u s s e d i n

o r d e r t o v i s u a l i z e t h e a c t i o n o f a r e d u n d a n t s t r u c t u r e . I f t h e r e d u n d a n t f o r c e i s

r e m o v e d , t h e s t r u c t u r e i s s t a t i c a l l y d e t e r m i n a t e , a n d t h e d e f l e c t i o n i n t h e d i r e c t i o n

o f t h e r e d u n d a n t d u e t o t h e a p p l i e d l o a d s h a s t h e f o l l o w i n g v a l u e :

- 5 1 0 = L ^ f ( 6 . 6 7 )

T h i s d e f l c c t i o n i s a s s u m e d p o s i t i v e i n ( h e d i r e c t i o n o f t h e r e d u n d a n t . A u n i t v a l u e

o f t h e r e d u n d a n t d e f l e c t s t h e s t r u c t u r e a d i s t a n c e

(6.68)

w h i c h i s a l s o p o s i t i v e i n t h e d i r e c t i o n o f t h e r e d u n d a n t . T h e v a l u e o f A", r e q u i r e d

t o g i v e a z e r o d e f l e c t i o n i s o b t a i n e d b y d i v i d i n g t h e d e f l e c t i o n r e s u l t i n g f r o m t h e

a p p l i e d l o a d s b y t h e d e f l e c t i o n r e s u l t i n g f r o m t h e u n i t l o a d , a n d i t w i l l b e n e g a -

t i v e w i t h t h e a s s u m e d s i g n c o n v e n t i o n s :

( 6 - 6 9 )

1 5 1 1 /

I t i s e a s i e r t o v i s u a l i z e t h e d e f l e c t i o n t e r m s o f E q . ( 6 . 6 9 ) t h a n t h e s u m m a t i o n

t e r m s o f E q . ( 6 . 6 6 ) . T h e f o r m o f E q . ( 6 . 6 9 ) w i l l a p p l y i d e n t i c a l l y f o r r i g i d f r a m e

s t r u c t u r e s w i t h s i n g l e r e d u n d a n t ? e x c e p t <5,, a n d <5 ( 0 w i l l h a v e d i f f e r e n t d e f i n i - .

t i o n s .

E x a m p l e 6 . 8 F i n d t h e r e a c t i o n s a n d t h e s t r e s s e s i n t h e m e m b e r s o f t h e s t r u c -

t u r e o f F i g . 6 . 8 . T h e a r e a s o f m e m b e r s AB, BC, a n d BD a r e 4 i n 2 , a n d t h e

a r e a s o f m e m b e r s AD a n d DC a r e 3 - 6 i n 2 . ; A s s u m e r i g i d s u p p o r t s a n d

E = 1 0 , 0 0 0 k i p s / i n 2 .

SOLUTION T h e n u m e r i c a l s o l u t i o n o f E q . ( 6 . 6 6 ) is d i s p l a y e d i n T a b l e 6 . 1 , w i t h

t h e n o t a t i o n a s s h o w n i n F i g . 6 . 7 . T h e h o r i z o n t a l r e a c t i o n c o m p o n e n t -YX i s


c o n s i d e r e d t h e r e d u n d a n t . T h e f o r c c s S0 i n t h e t r u s s w i t h t h e r e d u n d a n t

r e m o v e d a r e t a b u l a t e d i n c o l u m n 1. T h e l e n g t h s L o f t h e v a r i o u s m e m b e r s

a r e c a l c u l a t e d f r o m t h e d i m e n s i o n s s h o w n i n F i g . 6 . 8 a n d a r e d i v i d e d b y t h e

a r e a s a n d m o d u l u s o f e l a s t i c i t y . T h e v a l u e s o f L/(AE) a r e t a b u l a t e d i n c o l u m n

2 . T h e f o r c e s r e s u l t i n g f r o m a u n i t v a l u e o f A ' , a p p l i e d a s s h o w n i n F i g .

6 . 7 c , a r c t a b u l a t e d i n c o l u m n 3 . P o s i t i v e s i g n s i n d i c a t e t e n s i o n , a n d n e g a t i v e

s i g n s i n d i c a t e c o m p r e s s i o n . T h e s u m m a t i o n t e r m s o f E q . ( 6 . 6 6 ) a r e e v a l u a t e d

a s t h e s u m o f t h e v a l u e s f o r t h e i n d i v i d u a l m e m b e r s t a b u l a t e d in c o l u m n s 4

a n d 5 .

£ S0sL/{AE) 0 .5150

- ' V l = - Zs>L/(AE) = 0 ^ = 25"6klpS

T h i s v a l u e i s m u l t i p l i e d b y t h e t e r m s i n c o l u m n 3 t o o b t a i n v a l u e s o f X i S f o r

a l l m e m b e r s . T h e f i n a l f o r c e s S i n t h e m e m b e r s a r e o b t a i n e d a s t h e a l g e b r a i c

s u m o f t e r m s i n c o l u m n s 1 a n d 6 a n d a r e l i s t e d i n c o l u m n 7 .

E x a m p l e 6 . 9 F i n d t h e f o r c e s in t h e m e m b e r s o f t h e s t r u c t u r e o f E x a m p l e 6 . 8

if t h e s u p p o r t a t p o i n t C i s d e f l e c t e d 0 . 2 5 i n t o t h e r i g h t a n d t h e t e m p e r a t u r e

is d e c r e a s e d 4 0 " F . A s s u m e a t e m p e r a t u r e c o e f f i c i e n t o f e x p a n s i o n a = 1 0 " 5

i n / ( i n • ° F ) .

S O L U T I O N A t e m p e r a t u r e d e c r e a s e w o u l d c a u s e t h e r i g h t e n d o f t h e s t a t i c a l l y

d e t e r m i n a t e t r u s s o f F i g . 6 . 7 b t o m o v e t o t h e l e f t a d i s t a n c e ( a L ) A T , w h e r e L

T a b l e 6 . 1

I. S2L

Men; hers • V kips A A' s AE * AE' S in /k ip in in /k ip

(I) (21 (3) (4) 15) (6) O)

AH - 56.6 0.001414 1.414 - 0 . 1 1 3 0 0.00283 36.3 - 2 0 . 3 nc - 5 6 . 6 0.00)414 1.414 - 0 . 1 1 3 0 0.00283 36.3 - 2 0 . 3

AD 44.8 0.001245 - 2 . 2 3 3 - 0 . 1 2 4 5 0.00621 - 5 7 . 3 - 1 2 . 5

DC 44.X 0.001245 - 2 . 2 3 3 - 0 . 1 2 4 5 0.00621 - 5 7 . 3 - 1 2 . 5

pn 40.0 0.000500 - 2 . 0 - 0 . 0 4 0 0 0.00200 - 5 1 . 3 - 1 1 . 3

To ta l - 0 . 5 1 5 0 0.02008


T a b l e 6 . 2

Member S„ Xts S

AB BC AD DC BD

- 5 6 . 6 16.4 - 4 0 . 2 - 5 6 . 6 16.4 —40.2

44.8 - 25.9 18.9 44.8 - 2 5 . 9 18.9 40.0 —23.2 16.8

is t h e d i s t a n c e b e t w e e n s u p p o r t s ( 8 0 i n ) :

« L ( A T ) = 1 0 - 5 x 8 0 x 4 0 = 0 . 0 3 2 i n

T h e s u p p o r t o f p o i n t C i s d i s p l a c e d a n a d d i t i o n a l 0 . 2 5 in t o t h e r i g h t ;

t h e r e f o r e , t h e t o t a l d i s p l a c e m e n t 5, w h i c h m u s t b e g i v e n t o t h e r i g h t s u p p o r t

o f t h e t r u s s b y t h e s t r a i n s i n t h e m e m b e r s , i s

w h e r e t h e n e g a t i v e s i g n i n d i c a t e s t h a t t h e d e f l e c t i o n d u e t o t h e s t r e s s e s i s

o p p o s i t e t o A" , . F r o m E q . (6 .65 ) ,

w h e r e t h e s u m m a t i o n t e r m s a r e o b t a i n e d i n T a b l e 6 . 1 a n d

- 0 . 5 1 5 + 0 . 0 2 0 0 8 Z , = - 0 . 2 8 2

X , = 1 1 . 6 k i p s

T h e v a l u e s o f t h e r e s u l t i n g f o r c e s i n t h e m e m b e r s a r e f o u n d i n T a b l e 6 . 2 a s

S = S0 + Xis, w h e r e v a l u e s o f S0 a n d s a r e t h e s a m e a s f o r E x a m p l e 6 .8 .

E x a m p l e 6 . 1 0 F i n d t h e b e n d i n g m o m e n t a t a n y p o i n t o f t h e s e m i c i r c u l a r

a r c h o f F i g . 6 . 9 if t h e s u p p o r t s d o n o t m o v e . T h e v a l u e o f EI i s c o n s t a n t f o r

a l l c r o s s s e c t i o n s . N e g l e c t a x i a l d e f o r m a t i o n . J

5 = 0 0 . 0 3 2 - 0 . 2 5 = - 0 . 2 8 2 in

S O L U T I O N T h e s t r u c t u r e i s s y m m e t r i c a l a b o u t a v e r t i c a l c e n t e r l i n e , a n d a l l

i n t e g r a l s a r e e v a l u a t e d f o r t h e l e f t h a l f o f t h e s t r u c t u r e a n d m u l t i p l i e d b y 2 .

T h e h o r i z o n t a l r e a c t i o n X t i s c o n s i d e r e d t h e r e d u n d a n t , a n d t h e v a l u e o f M 0

i s c a l c u l a t c d f o r t h e s t a t i c a l l y d e t e r m i n a t e s t r u c t u r e f o r m e d b y s u p p o r t i n g

o n e e n d o f t h e a r c h o n f r i c t i o n l e s s r o l l e r s :

PR M o = — (1 - c o s P )

T h e v a l u e o f m i s c a l c u l a t e d f o r a u n i t l o a d a c t i n g i n t h e d i r e c t i o n o f

m = — R s i n /?

U t i l i z i n g E q . ( 6 . 3 4 ) a n d n o t i n g t h a t M = M 0 + Xym a n d = R dp, y i e l d

Y - V'R/(.2ED)Jl - cos P)(-R sin P)R dp P

' ' ~ ~ 2 f t ' 2 [ ( - « sin P)2R/iEl)] dp

T h e final b e n d i n g m o m e n t i s o b t a i n e d b y s u p e r i m p o s i n g t h e v a l u e s o f M 0 f o r

t h e a p p l i e d l o a d s a n d Xtm f o r t h e r e d u n d a n t :

. M = M0 + X^m

PR PR M = — - ( 1 - c o s / ? ) - — s i n p

2 n

T h i s e q u a t i o n a p p l i e s f o r 0 < (i < n/2, a n d t h e b e n d i n g - m o m e n t d i a g r a m i s

s y m m e t r i c a l a b o u t a v e r t i c a l c e n t e r l i n e .

E x a m p l e 6 . 1 1 T h e s t r u c t u r e s h o w n i n F i g . 6 . 1 0 a c o n s i s t s o f a r o u n d t u b e i n a

h o r i z o n t a l p l a n e , b e n t a t a n a n g l e o f 9 0 ° . T h e f r e e e n d s u p p o r t s a l o a d o f 2

k i p s , a n d i t i s a l s o s u p p o r t e d b y a v e r t i c a l w i r e . F i n d t h e t e n s i o n i n t h e w i r e

a n d m a k e t h e b e n d i n g - m o m e n t a n d t o r s i o n d i a g r a m s f o r t h e t u b e .

S O L U T I O N T h e w i r e i s c o n s i d e r e d t h e r e d u n d a n t m e m b e r . T h e v e r t i c a l d e f l e c -

t i o n o f t h e f r e e e n d o f t h e t u b e i s c a l c u l a t c d b y a s s u m i n g t h e w i r e t o b e

r e m o v e d a n d t h e s t a t i c a l l y d e t e r m i n a t e t u b e t o s u p p o r t t h e l o a d o f 2 k i p s .

T h e b e n d i n g m o m e n t a n d t o r s i o n i n t h e t u b e u n d e r t h e l o a d i n g a r e d e -

s i g n a t e d a s A / 0 a n d T 0 , r e s p e c t i v e l y , a n d a r e p l o t t e d i n F i g . 6 . 1 0 6 . T h e

d e f l e c t i o n r e s u l t i n g f r o m b e n d i n g a n d t o r s i o n a l d e f o r m a t i o n o f t h e t u b e i s ,

f r o m E q . ( 6 . 3 4 ) . m

Tn b\C\ =

M0m i r d s +

T h e v a l u e s o f m a n d i n , , t h e b e n d i n g m o m e n t a n d t o r s i o n a l m o m e n t , r e s p e c t -

i v e l y , i n t h e s t a t i c a l l y d e t e r m i n e d s t r u c t u r e f o r X 1 = 1, a r e p l o t t e d i n F i g .

6 . 1 0 c . T h e i n t e g r a l s o f t h e a b o v e e q u a t i o n b e c o m e

3 6 x 4 + 8 1 x 6 1 2 x 6 x 9 t A / . , ' ) . „ = — = — 1 . 4 4 i n 10 1000 800

T h e n e g a t i v e s i g n i n d i c a t e s a d e f l e c t i o n i n t h e o p p o s i t e d i r e c t i o n t o t h e u n i t

l o a d , o r a d o w n w a r d d e f l e c t i o n .

T h e d e f l e c t i o n <5 , , , r e s u l t i n g f r o m a u n i t f o r c e A" , i n t h e s e p a r a t e w i r e ,

c o n s i s t s o f p a r t s d u e t o t o r s i o n a n d b e n d i n g o f t h e t u b e a n d t e n s i o n i n t h e

w i r e :

f in2 , f mf S2L

T h e t e n s i o n s i n t h e w i r e i s 1, a n d t h e v a l u e s o f m a n d m, a r e s h o w n i n H g .

7"(, = 1 2 i n • k i p

Figure 5.22

6 . 1 0 c :

18 x 4 + 4 0 . 5 x 6 6 x 6 x 9 l 2 x 4 0

' = i o ^ + - M O - + ~ m ~ " ° " 9 2 l n / k , p

T h i s is h o w f a r a 1 - k i p t e n s i o n f o r c e i n t h e w i r e w o u l d m o v e t h e c u t e n d s t o g e t h e r . T h e f o r c e X , r e q u i r e d t o m o v e t h e c u t e n d s o f t h e w i r e t h e d i s t a n c e — (5 j 0 is f o u n d f r o m E q . ( 6 . 6 9 ) :

v 1 - 4 4

" V l = - ^ = 0 ^ 2 = L 5 6 5 k , p S

N o w t h e b e n d i n g - m o m c n t a n d t o r s i o n d i a g r a m s f o r t h e t u b e c a l c u l a t e d f r o m

t h e e q u a t i o n s o f s t a t i c s a n d a r e s h o w n in F i g . 6 . 1 0 ^ .

6 . 1 1 S t r u c t u r e s w i t h M u l t i p l e R e d u n d a n c y

T h e p r o c e d u r e f o r a n a l y s i s o f s t r u c t u r e s w i t h t w o o r m o r e r e d u n d a n t s is s i m i l a r

t o t h a t u s e d f o r a s t r u c t u r e w i t h o n e r e d u n d a n t m e m b e r o r r e a c t i o n . T h e f i r s t

s t e p is t o r e m o v e t h e r e d u n d a n t m e m b e r s o r r e a c t i o n s i n o r d e r t o o b t a i n a

s t a t i c a l l y d e t e r m i n a t e b a s e s t r u c t u r e . T h e n t h e d e f l e c t i o n s o f t h e s t a t i c a l l y d e t e r -

m i n a t e b a s e s t r u c t u r e i n t h e d i r e c t i o n s o f t h e r e d u n d a n t s a r e c a l c u l a t e d i n t e r n i s

o f t h e r e d u n d a n t f o r c e s a n d a r e e q u a t e d t o t h e k n o w n d e f l e c t i o n s , w h i c h a r e

u s u a l l y z e r o . T h e n u m b e r o f k n o w n d e f l e c t i o n c o n d i t i o n s m u s t b e e q u a l t o t h e

n u m b e r o f r e d u n d a n t s . F o r a s t r u c t u r e w i t h n r e d u n d a n t s , t h e d e f l e c t i o n c o n d i -

t i o n s y i e l d n e q u a t i o n s w h i c h m u s t b e s o l v e d s i m u l t a n e o u s l y f o r t h e v a l u e s o f t h e

r e d u n d a n t s .

T h e t r u s s s h o w n in F i g . 6 . 1 1 a h a s o n l y t h r e e r e a c t i o n s a n d is s t a t i c a l l y

. d e t e r m i n a t e e x t e r n a l l y , b u t t h e r e a r e t w o m o r e m e m b e r s t h a n a r e r e q u i r e d f o r

s t a b i l i t y ; t h e r e f o r e , i t i s s t a t i c a l l y i n d e t e r m i n a t e i n t e r n a l l y . T h e d e f l e c t i o n c o n d i -

t i o n s w h i c h w i l l b e s p e c i f i e d a r e t h a t t h e r e a r e n o s t r e s s e s i n t h e s t r u c t u r e w h e n i t

i s n o t l o a d e d o r if t w o m e m b e r s o f t h e u n l o a d e d s t r u c t u r e a r e c u t , t h e r e l a t i v e

d e f l e c t i o n s <5, a n d 52 o f t h e c u t e n d s w i l l b e z e r o . T h e d e f l e c t i o n s a r e n o w

e x p r e s s e d in t e r m s o f t h e f o r c e s X i a n d X 2 i n t h e r e d u n d a n t m e m b e r s . A l l

d e f l e c t i o n s a r e a s s u m e d t o b e e l a s t i c .

T h e s t a t i c a l l y d e t e r m i n a t e b a s e s t r u c t u r e , s h o w n i n F i g . 6 . 1 1 f r , i s f o r m e d b y

c u t t i n g o r r e m o v i n g t h e r e d u n d a n t m e m b e r s . T h e a p p l i e d l o a d s p r o d u c e f o r c e s S0

i n t h e m e m b e r s o f t h e b a s e s t r u c t u r e . T h e f o r c e s st i n t h e m e m b e r s a r e p r o d u c e d

b y a u n i t v a l u e o f X t a p p l i e d t o b a s e s t r u c t u r e , a s s h o w n i n F i g . 6 . 1 1 c . S i m i l a r l y ,

a f o r c e X2 = 1 p r o d u c e s f o r c e s s2 i n t h e m e m b e r s w h e n a p p l i e d t o t h e b a s e

s t r u c t u r e , a s s h o w n in F i g . 6.11c/ . T h e f i n a l f o r c e S i n a n y m e m b e r m a y b e

o b t a i n e d b y s u p e r i m p o s i n g t h e f o r c e s d u e t o t h e a p p l i e d l o a d s a n d t h e r e d u n d a n t

f o r c e s :

S - S o + ^ i S i + X 1 S 2 ( 6 . 7 0 )

\ \ s 0 ,

Figure 6.11

T h e d e f l e c t i o n s (5, a n d <52 o f t h e c u t e n d s o f t h e m e m b e r s a r e n o w e q u a t e d t o

z e r o : /

T h e s e e q u a t i o n s m a y b e s o l v e d s i m u l t a n e o u s l y f o r X , a n d X 2 • T h e final v a l u e s

o f t h e f o r c e s S m a y b e o b t a i n e d f r o m E q . ( 6 . 7 0 ) .

T h e p r e c e d i n g e q u a t i o n s h a v e b e e n d e r i v e d f r o m t h e s u p e r p o s i t i o n o f t h e

s t r e s s c o n d i t i o n s , a s s t a t e d in F.q. (6 .70) . T h e y m a y a l s o b e o b t a i n e d f r o m a

s u p e r p o s i t i o n o f t h e d c l l e c l i o n c o n d i t i o n s . T h e a p p l i e d l o a d s a r e a s s u m e d t o

p r o d u c e d e f l e c t i o n s < ) 1 0 a n d <>20 o f t h e r c d u n d a n t s . A u n i t v a l u e o f X l p r o d u c e s

d e f l e c t i o n s (5,, a t A ' , a n d <>2I a t X2 - A u n i t v a l u e o f A f 2 p r o d u c e s d e f l e c t i o n s 5 2 2

a t X 2 a n d r ) I 2 a t A , . T h e t o t a l d e f l e c t i o n s i n t h e d i r e c t i o n s o f t h e r e d u n d a n t s c a n

b e o b t a i n e d b y s u p e r i m p o s i n g t h e e f f e c t s o f t h e v a r i o u s l o a d s :

<5, = 5 I 0 + AR151 , + X25l2

<h = <>iO + + X 2 5 2 2

In m o s t p r o b l e m s , t h e d e f l e c t i o n s <5, a n d <52 a r e z e r o , b u t i n s o m e c a s e s , k n o w n

v a l u e s o f s u p p o r t d e f l e c t i o n s o f s i m i l a r d e f o r m a t i o n s m a y b e s u b s t i t u t e d .

F o r s t r u c t u r e w i t h n r e d u n d a n t s , a n u m b e r o f d e f l e c t i o n s c o n d i t i o n s n m u s t

b e u s e d . T h e s e m a y b e w r i t t e n a s

T>, = <5I0 + A", ( • ) , , + X 2 <5,2 + • • • + X „ 6 U

52 = <S2(, -f A , ^ , , +X2S22+ • •• + X„b2n (6 .71)

t>„ = (>„[} + X 1 1 + A 2 <5,I2 + ' ' • + A ndrnt

o r , in m a t r i x f o r m .

3 . " ^10 " S i . " •

<>> = + Sn <522 ' •d2„ Xi

Jn 1_ J * '

T h e t e r m s in E q . (6 .71) m a y b e d e f i n e d in t h e f o l l o w i n g m a n n e r f o r t r u s s

s t r u c t u r e s :

_ ~ Sa s„ I- x _ Sms„L "no — A L- —

AE AE (6 .72)

F o r r i g i d - f r a m e s t r u c t u r e s w h e r e o n l y b e n d i n g d e f o r m a t i o n is c o n s i d e r e d , t h e - t e r m s in E q . ( 6 . 7 1 ) a r e d e f i n e d a s

0„„ = I El € d„ = I El dc (6 .73)

F r o m M a x w e l l ' s r e c i p r o c a l t h e o r e o T , <>,7 = d y .

T h e i n t e r n a l l o a d s in e a c h m e m b e r j m a y b e o b t a i n e d f r o m t h e e q u a t i o n s

Sj -= S() + -Y,.s, + X2s2 + • • • + X„s„

o r Mj --- A f „ -1 A ' | m , + X2m2 + ••• + Xnm„ (6 .74)

T h e a p p l i c a t i o n s o f t h e p r e c e d i n g e q u a t i o n s a r e i l l u s t r a t e d b y n u m e r i c a l

e x a m p l e s .

1 7 6 AIRC'RAIT STRUC TURES

© (?) Pi

•<F) Pi

© \ r y

3 ] i

hl— 0 - — -

W

10 kips

- 1 0

-0.707 / >v—0.707 \ / /

>v—0.707 \ 1 /

u \ o / 0 \ / L 0 \

0.707

-0 .707

(c)

(d) Figure 6.12

Example 6.12 Find (he forces in the members of the truss shown in Fig. 6.12a if P, = P2 — 10 kips, /i = /i t , and L/(AE) is the same for each member of the structure. The members are unstressed when Pi = P2 = 0. a n ( i stresses do not exceed the elastic limit.

SOLUTION: The numerical values of Lj(AE) are not required for the members, since only relative values are important. For 5 t = d2 = •' • = = 0, in Eqs. (6.71) the summation terms may be multiplied by any constant value. If any

DI:M .I;CTION ANALYSIS OF S T R U C T U R A L SYSTEMS 1 7 7

5n is not zero, it is necessary to know the numerical values. Thus it is assumed that L/{AE) is unity for all members. The calculations of the sum-mation terms are made in Table 6.3. The forces S0 in the members of the statically determinate base structure are shown in Fig. 6.12fc and tabulated in column 1. The values of st and s2 due to unit values of the redundant are shown in Fig. 6.12c and d and are tabulated in columns 2 and 3. The terms for the summations of Eqs. (6.72) are obtained as the totals of columns 4, 5, 6, 7, and 8. Substituting these totals into Eq. (6.71) yields

KM-42-421 70.70J

4.0 0.5" _0.5 4.0_

By carrying out the matrix inversion, the unknowns are obtained:

Xi = -8 .53 and X2 = -16 .60

Hence the internal load in each member is given by

Sj = (S0 - 8.53sL - 16.6s2)j

where j is the member number. The values of S} are shown in column 9 of Table 6.3.

Example 6.13 Find the forces in the truss of Fig. 6.13a. Assume that L/(AE) is 0.01 ill/kip for all members. The right-hand support deflects 0.5 in from the unstressed position of the truss. Assume h =

SOLUTION The redundants A'j and X2 are chosen as shown in Fig. 6.13a,

Table 6.3

S 0 s , S„ s2 L sfL s | L S,S2 L S,

S0 s, s , AE AE AE AE AE kips

(1) (2) (3) (4) (5) (6) (7) (8) (9)

1 0 1.000 0 0 0 1.0 0 0 - 8 . 5 3 2 0 0 1.000 0 0 0 1.0 0 - 1 6 . 6 0

3 - 1 0 - 0 . 7 0 7 0 7.07 0 0.5 0 0 - 3 . 9 7 4 0 - 0 . 7 0 7 0 0 0 0.5 0 0 + 6.03

5 14.14 1.00 0 14.14 0 1.0 0 0 + 5.61 6 - 10 - 0 . 7 0 7 0 7.07 0 0.5 0 0 - 3 . 9 7 7 - 2 0 - 0 . 7 0 7 - 0 . 7 0 7 14.14 14.14 0.5 0.5 0.5 - 2 . 2 4 8 10 0 - 0 . 7 0 7 0 - 7 . 0 7 0 0.5 0 + 21.73

9 28.2S t) 1.000 0 28.28 0 1.0 0 + 11.68 10 - 3 0 0 - 0 . 7 0 7 0 21.21 0 0.5 0 - 1 8 . 2 7

11 - 2 0 0 - 0 . 7 0 7 0 14.14 0 0.5 0 - 8 . 2 7 12 30 0 0 0 0 0 0 0 + 30.00

Tota l 42.42 70.70 4X) 1 5 OS


leaving the^rame statically determinate base structure as for Example 6.12. The values for SQ and .s2 are the same as for Example 6.12, but values o f ^ must be calculated as shown in Fig. 6.l3/>. The terms involving S0 and.^ are therefore the same as those calculated in Table 6.3, except that the value of L/(AE) is now 0.01 for each member, whereas a value of 1.0 was used in Table 6.3. For constant values of L/(AE) for all members, Eq. (6.7i) may be written as

" 16.0 — 3.535~j ["A1 n _ T 190.0" 3.535 4.0 J b r J ""L-70.70_

Table 6.4

V i • V j S, kips

(1) (2) (3) (4) 15)

2 0 0 0 0 - 8 . 9 0 3 0 0 0 0 - 10.00

4 0 0 0 0 0

5 - 1 . 4 1 4 - 2 0 0 2.0 U09 6 1.0 - 1 0 0 t.0 - 0 . 0 7 7 1.0 - 2 0 - 0 . 7 0 7 1.0 - 3 . 7 7 8 - 1 . 0 - 1 0 0.707 1.0 6.23 9 - 1 . 4 1 4 - 4 0 - 1 . 4 1 4 2.0 5.33

10 2.0 - 6 0 - 1 . 4 1 4 4.0 - 3 . 8 4 11 1.0 - 2 0 - 0 . 7 0 7 1.0 - 3 . 7 7 12 - 2 . 0 - 6 0 0 4.0 10.14

Total - 2 4 0 - 3 . 5 3 5 T & o

DEFLECTION ANALYSIS OF STRUCTURAL SYSTEMS 1 7 9

V -8.90

Solving for ,Y| and X , yields

X , = 9 . 9 3 and

where Sj = (S0 + 9.93.S, - 8.90s2);

Table 6.4 shows the calculations for the quantities used in Eq. (6.71). It also gives the final results of the internal loads in each member.

Example 6,14 Find the bending-moment diagram for the frame shown in Fig. 6.14a. The value of EI is constant, and the members are fixed against ro-tation at the supports.

SOLUTION The structure has three redundant reactions. If the frame is cut at

10 kips 10 kips

•V.,

- ' ) f l -

6 ft 6 ft

7-rrr

A = 180

•v, i 7777

I 9

m 3 m 3 m 3 m 3

* 2 = »

(c)

i r

-V, = I

12 f t-kip

V . f t -k ip{

r77T \ » • i s rt-kip

(e)

r 10 kips

~ 7 7 7 Z

18 f t-kip

V)

F igure 6.14

AIRCRAFT STRUCTURES

the left support, the remaining structure is stable and statically determinate. The two force components and the couple, Xu X2, and X3, are assumed to be redundant reactions. The bending-moment diagram for the base structure under the action of the applied load is shown in Fig. 6.14/?. All bending moments are plotted on the compression side of the members and are not designated as positive or negative. The product of two bending moments is positive if they are both plotted on the same side of the member. The bending moments mi, m 2 , and m3 due to unit values of the redundants are plotted in Figs. 6.14c, d, and e. The various deflection terms are evaluated semi-graphically, by reference to the moment diagrams.

EI&10 =

Eld 20 = JA/„I

M0mi dx = 180 x 2 = 360

,m2 dx = - 180 x 9 = - 1620

E!5i0 = jAfo/nj dx = - 180 x 1 = - 180

EISt, = wj dx = 2 x 18 x 4 + 6 x 9 x 6 = 468

EI312 =

33

EI5

m22 dx = 40.5 x 6 + 54 x 9 = 729

= J / h j dx = 1 x 1 x 2 1 = 2 1

l2 = Jm,m2 dx = - 6 x 40.5 - 1 8 x 9 = -

l-;iSs 3 = in, hi , dx = —1 x 1 8 x 2 — 1 x 5 4 = — 90

Eld23 = »>2»h dx = 1 x 40.5 + 1 X 54 = 94.5

Upon substituting in Eq. (6.71), the following is obtained:

468.0 -405.0 -90 .0 -405.0 729.0 94.5 - 9 0 . 0 94.5 21.0

Inverting the matrix and solving for the unknowns yield:

Xy = 5.0 X2 = 2.667 X3 = 18.0

-360.0 = 1620.0

180.0

DEFL ECTION ANALYSIS OF S T R U C T U R A L SYSTEMS 1 8 1

Now, the final bending moments in each member can be obtained from Eq. (0.74) and are shown in Fig. 6.14f.

6.12 Shear Lag

It was pointed out in Chap. 5 that many of the assumptions made in deriving the simple beam flexure theory arc somewhat in error. The assumptions that plane sections remain plane after bending and that bending stresses are proportional to the distance from the neutral axis are less accurate for semimonocoque structures than they are for heavy structures, because the shearing deformations in thin webs are not always negligible.

The effect of shearing deformations in redistributing the bending stresses in a box beam is commonly known as shear lag. The effect may be illustrated by considering the cantilever box beam shown in Fig. 6.15. For simplicity, it is assumed that the beam cross section is symmetrical about a vertical centerline and that the load is applied along this centerline, so that there is no torsional deformation. An analysis, using the simple beam theory shows that all the stringers on the upper surface are the same for all cross sections. As a result of these shearing stresses, an originally plane cross section will deform to the posi-tion indicated by line a' h' c'.

Al the support, however, the cross scction is restrained from warping out of its original plane, and line ahc of Fig. 6.15 remains straight. Since the distance cc' is greater than the distance aa', the stringer at c resists a smaller compressive stress than the stringer at a. Thus, the bending stress at a must be greater than that calculated .by the simple flexure theory, and the bending stress at c must be less than indicated by the simple theory. In this case, all the cross sections at some distance from the support warp the same amount, and thus all the stringers have approximately the same bending stress and strain. The shear-lag effect is greatest at the support and is something of a local effect.


Many wing structures are spliced only at the spars, so that the stringers resist no bending stress at the splice. The box beam shown in Fig. 6.16 is spliced in this manner, so that only the corner danges resist axial loads at the left-hand support. In this case, the cross section at the support deforms as indicated by line a" b" c", in an opposite direction to the deformation a' b' c' of cross sections some distance from the support. The middle stringer has a final length c' c" which is consider-ably greater than the final length «' a" of the corncr stringer. The shear-lag effect is greater in this beam than in the beam with the entire cross section restrained. The elTect is also localized near the support, as the stringers at some distance from the support resist bending stresses, which are approximately as calculated by the simple flexure theory.

The effect of shear iag may be desirable, since it permits a structure to resist higher ultimate bonding moments than are calculated from the simple flexure theory. The allowable bending stresses for the stringers between the spars are smaller than the allowable stresses for the corner flanges, or spar caps. The stringers tend to fail as columns with lengths equal to the rib spacing. The spar caps are supported vertically by the spar web and horizontally by the skin, and they usually resist high compressive stresses.

When a rectangular box beam is subjected to torsion, a cross section tends to warp from its original plane. When one end is restrained against warping, axial loads are induced in the flanges, and the shear flows are redistributed near the fixed end. This also is an elTect of the shear deformation and is sometimes referred to as a shear-lag effect.

The extent of the shear-lag effect can be studied by considering the stresses resulting from a few simple conditions of loading and then superimposing them with other stress conditions. The structure with two webs and three stringers, shown in Fig. 6.17a, is assumed to extend for an indefinite length in the x direction and to be loaded as shown. The distribution of the loads and defor-mations of the x direction are investigated.

The force P in the center stringer is a function of the distance x. In a length f/.v, (he force changes an amount dP, and the web shearing deformation changes an amount dy, as shown in Fig. 6.176. From the spanwise equilibrium of a stringer, the load increase dP results from the shearing stress fs in the web: ,-

The change in the angle y results from the axial elongation of the stringers:

dP= - 2 f s t dx

The deformation y results from the web shearing stress:

L = S7

(6.75)

(6.76)

The variables / j and y may be eliminated from these three equalions in order to obtain a differential equation for P as a function of x. Differentiating Eq. (6.75) and substituting from Eq. (6.76), we have

DEFL E C T I O N ANALYSIS OF STRUCTURAL SYSTEMS 1 8 3

a.x dx

The value of dy/dx may be substituted from Eq. (6.77):

d2P dx2 = k2P

where

k 2 =

2 iG • + •

1

bE \A 2A i

Equation (6.79) may be integrated as follows:

+ C2e~kx

(6.78)

(6.79)

(6.80)

(6.81) where C, and C2 are constants of integration. The load P approaches zero at a large value of ,v; thus for = oo, P = 0 and C\ = 0. At the loaded end, JC = 0 and P — P0 or C2 = Pu. Therefore, Eq. (6.81) has the following value:

P = P0e~kx (6.82)

Equation (6.82) may be differentiated and equated to Eq. (6.75) in order to obtain an expression forf s :

The displacement 6 of the force P„ is equal to yb, or fsb/G for x = 0:

Pokh 6 =

2tG

(6.83)

(6.84)

t

•H

y + dy

L

dx

Kipire 6.17

1 8 4 AIRCRAFT STRUCTURES DEFLECTION ANALYSIS OF S T R U C T U R A L SYSTEMS 1 8 5

The structure shown in Fig. 6.18 may be analyzed in a similar manner to the structure of Fig. 6.17. The flange forces P at any cross section are defined by

P = P o e - * *

where k2 = 4G(l/6) + 1 /c AE{l/t) + 1 ftx

(6.85)

(6.86)

It is assumed that the ribs are closely spaced for Eq. (6.85) to be valid. The shear flows q may be similarly defined by

9 - f t , * " * (6.87)

where qQ is the shear flow when x = 0. In the shear-lag portion only, the shear flow must be the same in all four webs to satisfy the equilibrium condition for torsional moments. The warping displacement 5 of each of the forces PQ from the plane of the original cross section is measured as shown in Fig. 6.18b:

8 = <?o((l/0 + Vh PokMt+i/h) 4G(l/6 + 1/c)

(6.88) 2G(l /6+ 1/c)

To illustrate shear-lag calculations, consider the box beam of Fig. 6.19a. All the webs have thickness t = 0.02 in, and the material has the properties E = 107

lb/in2 and G — OAE. The simple beam theory yields shear flows which are con-stant for the length of the span, with the values shown, and values of the axial stringer loads as shown. For this theory to apply, however, the cross section at-the support must warp so that the middle stringer is displaced the distance 50

from the original plane:

s - f ' h - 2 0 0 * 1 0 0 G 0.020 x 4,000,000 = 0.025 in

If the cross section at the support is restrained from warping, the center stringer resists a compression force smaller than 40,000 Ib and the corner string-ers resist compression forccs larger than 20,000 lb. The force P0 acting as shown

s'

Figure 6.18

Figure 6.19

in Fig. 6.196, which is required to displace the structure a distance 5 0 , is calcu-lated from Eq. (6.84). Then the system of forces shown in Fig. 6.196 is superim-posed on those obtained by the simple flexure theory in Fig. 6.19a.

The structure of Fig. 6.196 is equivalent to that of Fig. 6.17. From Eq. (6.80)

k2 = 2 x 0.020 x 0.4 / 1 1

10 2 + 2 0.0016

or k = 0.04

Substituting <5 = 0.025 in into Eq. (6.84) and solving for Pu, we have

P « = 2?G<5 2 x 0.020 x 4 x 106 x 0.025

kb

From Eq. (6.82)

0.04 x 10

P = 10,000e~

= 10,000 lb

and from Eq. (6.83)

f s = 1 0 , 0 0 0 c "


Table 6.5

x e-o.ot* P = I0,000e-°oi' q = 200e~ooix

0 1 10,000 200 5 0.817 8,170 163

10 0.670 6,720 134 20 0.450 4,500 90 40 0.202 2,020 40

100 0.019 190 4

or q — fst = 200e _ 0 04jc. Thus, at the support, the corner stringers each resist compression forces of 25,000 lb, and the center stringer resists a compression force of 30,000 lb. The shear flow is zero at this cross section, which is obviously necessary for the assumed condition of no shearing deformation.

The values of P and q at various distances x from the fixed support are calculated in Table 6.5. These tabulated values must be superimposed on the values shown in Fig. 6.19a. It is observed that the correction forces at a cross section 20 in from the support are less than one-half the values at the support.

The loading condition shown in Fig. 6.19 requires larger corrections for the effects of shear lag than are required in the normal airplane wing. The shear loads in an airplane wing usually are resisted at the side of the fuselage, but the cross section at the center of the fuselage is prevented from warping. Thus the cross section at the side of the fuselage, which has the maximum shear flows, is permit-ted to warp and distribute the shear flows in almost the same manner as pre-dicted by the simple flexure theory.

PROBLEMS

6.1 Find the displacements of point 3 or the truss structure shown in Fig. P6.1. Assume I in2 for the area of each member and £ = 107 lb/in2. y

100 in

Figure P6.1

DEFL ECTION ANALYSIS OF S T R U C T U R A L SYSTEMS 1 8 7

6.2 Find the rotat ion of member 4-5 of the truss s tructure shown in Fig. P6.2. Area = 1 in2 and £ = 107 lb/in2 for all members.

20 kips

Figure P6.2

6.3 Find the displacements (translation and rotation) of points 2 and 3 of the beam structure shown in Fig. P6.3. Assume that the beam has a uniform area.

Figure P 6 J

6.4 You are given a simply supported beam under the action of a uniformly distributed load, as shown in Fig. P6.4. Find the midspan deflection, using Castigliano's theorem.

E! = constant

P lb/in

Figure P6.4

6.5 Solve Prob. 6.4 by using the Rayleigh-Ritz method.

6.6 Solve Prob. 6.4 by using the fmile difference method.

6.7 Solve Prob. 6.4 by solving the classical beam differential equation EI d^w/dx* = p(x). 6.8 Using the finite difference method, solve for the vertical displacements a t points a, h, and c and the slopes at the left and right supports of the multispan beam shown in Fig. P6.8. EI = 10s.


10 Ih/in

6.9 Use the finite difference method to find the moments and shears at the locations indicated on the beam structure shown in Fig. P6.9.

6.10 A square plate of uniform thickness t is fixed at two opposite edges and simply supported at the other two opposite edges. Using the finite difference method, find the maximum displacements and bending moments. Assume that the plate is under the action of a uniform normal pressure P. Use Poisson's ratio v = 0.25.

6.11 Analyze the truss of Fig. 6.8. assuming that L/(AE) is constant for all members. Analyze the truss of Fig- 6-8 for a horizontal load of 20 kips at B, in addition to the vertical load shown. Assume AE is constant for all members. 6.12 Analyze the truss of Fig. 6.8 assuming there is an additional member AC with a cross-secticin area of 3 in2. Take B'D as the redundant.

6.13 Repeat Prob. 6.12, assuming BC is the redundant member. 6.14 Repeat Prob. 6.12, assuming that member BD is 0.1 in too long because of manufacturing tolerances. Assume no external loads on the structure.

6.15 Analyze the structure of Fig. 6.9 if load P is acting horizontally at the same point. Assume a constant value of EI.

6.16 Analyze the structure of Fig. 6.9 for the loading shown, assuming the supports to be spread horizontally a distance or0.5 in. Use R = 50 in, P = 2 kips, I = 1.0 in*, and E = 10,000 kips/in2. . 6.17 Analyze the structure of Fig. 6.10, assuming the 2-kip load to be applied at the point whtfre the tube is bent. 6.18 Repeal Example 6.12, assuming It — 40 in, /il = 3 0 in, and AE = 10,000 kips for each member.

6.19 Repeat Example 6.13, if h = 40 in, /i, = 3 0 in, and AE = 10,000 kips for each member. Member 12 has a length of 10 in.

6.20 Repeat Prob. 6.19 assuming member 12 is a redundant in place of the reaction X,. 6.21 Analyze the frame of Fig. 6.14a, assuming the left support to be pin-connected and the right support to be fixed. 6.22 Analyze the frame of Fig. 6.14u. assuming an additional load of 20 kips to act down at the center of the structure. *

6.23 Obtain the bending-momcnt diagram for the frame shown ill Fig. P6.23 to P6.25 if P - 0 and = 4.0 kips/ft.

6.24 Obtain the bending-momcnt diagram for the frame shown in Fig. P6.23 to P6.25 if W — 0, P = 10 kips, and a = h 4.5 ft.

D E F L E C T I O N A N A L Y S I S O F S T R U C T U R A L S Y S T E M S 1 8 9

6 ft

T 1 i t l l t l • 1 = 0.15

/ = 1.0 / = 1.0

W kips/ft

6 ft

- 9 f t - Figure P6.23 lo P6.25

6.25 Obtain the bending-moment diagram for the f rame shown in Fig, P623 to P6.25 if W = 0, P = 10 kips, a = 3 ft, and b = 6 ft. 6.26 The frame for a fuselage of rectangular cross section is shown in Fig. P6.26 to Pfi.28. Calculate the bending moments if W = 3 kips and a = 1.5 ft. Assume EI is cons tant 6.27 Repeat Prob. 6.26 if EI for the bottom frame member is 4 times as large as the value of EI for the other members. 6.28 Repeat Prob. 6.26 if 2W = 6 kips and a = 3.0 ft.

-6 ft-

f 6 ft

Figure P6.26 to P 6 J 8

6.29 Use the principle of virtual displacements to establish the relationship between the applied forces and corresponding displacements on the beam structure shown in Fig. P6.29. This relationship is of the form {Q = [S]{<?}.

/•.'/. AE

Qs'ls

Figure P6.29

6.30 Repeat Prob. 6.29 except the beam is as shown in Fig. P6.30.

Qi.1i

<2l, 12

EI, A E, L

Qs.?s Q4.94

Figure P 6 J 0

C H A P T E R

SEVEN FINITE ELEMENT STIFFNESS METHOD IN STRUCTURAL ANALYSIS

7.1 INTRODUCTION

In real design cases, generally structural systems are composed of a large assem-blage of various structural elements such as beams, plates, and shells or a combi-nation of the three. Their overall geometry becomes extremely complex and cannot be represented by a single mathematical expression. In addition, these built-up structures are intrinsically characterized as having material and structur-al discontinuities such as cutout, thickness variations of members, etc., as well as discontinuities in loading and support conditions. Given these factors, relative-to the structure geometry and discontinuities, it becomes apparent that the classical methods can no longer be used, particularly those whose prerequisites are the formulation and the solution of governing differentia! equations. Thus, for com-plex structures, the analyst has to resort to more general methods of analysis where the above factors place no difficulty in their application. These methods are the finite element stifincss method and the finite element flexibility method.

With the advent of high-speed, large-storage-capacity digital computers, the finite clement matrix methods have become one of the most widely used tools in the analysis of complex structural systems. These methods are based on the formulations of a simultaneous set of linear algebraic equations relating forces to corresponding displacements (stiffness method) or displacements to correspond-ing forces (flexibility method) at discrete, preselected points on the structure.

190

F I N I T E E L E M E N T S T I F F N E S S M E T H O D I N S T R U C T U R A L A N A L Y S I S 1 9 1

These methods offer many advantages:

1. Capability for complete automation. 2. The structure geometry can be described easily. 3. The real structure can be represented easily by a mathematical model com-

posed of various structural elements. 4. Ability to treat anisotropic material. 5. Ability to treat discontinuities. 6. Ability to implement residual stresses, prestress conditions, and thermal load-

ings. 7. Ability to treat nonlinear structural problems. 8. Ease of handling multiple load conditions.

Figure 7.1


The finite element matrix methods have gained great prominence throughout the industries owing to their unlimited applications in the solution of practical design problems of high complexity. For further information on the finite element matrix methods, the reader is encouraged to consult Refs. 10, 24, and 27 to 35.

7.2 MATHEMATICAL MODEL OF THE STRUCTURE

The basic concept of the finite clement matrix method in structural analysis is that the real structure can be represented by an equivalent mathematical model which consists of a discrete number of finite structural elements, as shown in Figs. 7.1 to 7.3. The structural behavior of each of these elements may be described by different sets of functions which normally are chosen such that continuity of stresses and/or strains throughout the structure is ensured.

The types of elements which are commonly used in structural idealization are the truss, beam, two-dimensional membrane, shell and plate bending, and three-dimensional solid elements. Figures 7.4 and 7.5 illustrate schematically each el-

Figure 7 J


ement and the typical nodal forces and corresponding displacements associated with each type.

7.3 ELEMENT DISCRETIZATION

The mathematical relationships which govern the structural behavior of an el-ement are derived on the basis of an idealized element model. For example, once the element shape is selected, it is discretized by placing a finite number of nodes at various locations on ihe element surface, as shown in Fig. 7.6. Generally

Figure 7.4 Structural elements, (n) Truss element; (/>) beam element; (o) membrane plate element, (d) plate bending elements.

F I N I T E F . I . E M K N T S T I F F N E S S M E T H O D I N S T R U C T U R A L A N A L Y S I S 1 9 5

speaking, the accuracy increases with an increase in nodal points considered on the element. Likewise, the smaller the element size, the more accurate the analyti-cal results become for a given structural system. Note that the core storage requirement increases rapidly with an increase in the number of element nodes and the number of elements considered in a structure.

7.4 APPLICATIONS OF FINITE ELEMENT MATRIX METHODS

The finite element matrix method has a wide range of applications in structural analysis. Its uses extend to every engineering field, from space structures to land-based and marine structures.

Figure 7.5 Structural elements, (a) She!) elements; (b) three-dimensional elements.


7.5 COORDINATE SYSTEM

The coordinate system used here in conjunction with the finite element matrix method is a set of orthogonal axes, x, y, and z, shown in Fig. 7.6a. These axes are referred to as either the global axes or the local axes, respectively, depending on whether the structure or the element is being treated.

In the formulation of element relationships, the element node geometry, forces, and displacements arc referenced with respect to the local axes of the element. However, when relationships are being established for a structure, th.e node geometry, forces, and displacements are referenced with respect to the struc-ture global axes. See Fig. 7.6b.

Truss structure t x2

(b)

Figure 7.6 Global and local axes elements, (n) Coordina te system: (/i) x, v -- global axes for s t ructure; x , , y i = local axes for element 1: v , . y 2 - local axes for element 2.

7.6 FORCES, DISPLACEMENTS, AND TIIEIR SIGN CONVENTION

The term force is used here to denote both forces and moments; likewise, displace-ment is used to mean both translational and rotational, displacements. The forces and their corresponding displacements are assumed positive if they act in the positive direction of their respective axes, as shown in Fig. 7.7. The right-hand rule is used here to represent the vector notation of moments.

In formulating element relationships, it is convenient to adopt the concept of generalized forces and displacements as follows:

[ 7 . 1 7 / M V f i f y S2 <5, h S-. h mx 54 f s mr 5s o,

_ f b \ J e - Jz_

For instance, consider the element shown in Fig. 7.8. The force and displace-

4 3 F t , 5 y

y

/ r Af s

F : , S2

Mz, 6,

Figure 7.7 Positive forces and displacements.

ment vectors are represented as

/ t fu <h fl h k fj* <53

k ft, <5* dj,

fs fkx <55 $kx k _fky_

7.7 STIFFNESS METHOD CONCEPT

The stillness method, or what is commonly referred to as the displacement method, is based on the principle of superposition of displacements. Consider the plane stress problem shown in Fig. 7.9. At every node of the solid such as and k, there exist two possible translational displacements in the x and y directions. Corresponding to each displacement there exists a set of induced forces at each node. The method of superposition for linear structures states that the total force induced at a given node in a given direction by all the possible nodal displace-ments that the solid element may experience can be obtained by simple algebraic summation:

/• =/! + / i + / ? + / ! + / ? + / ?

h =/i +/i + f i + f i + f i + f l

k = f l + f l + f l + / « + f l + f t

(7.2)

L 6 6

y = y r / i .1 m jLjJmn

n — I n = 1

( m = 1 , 2 , . . . , 6 ) (7.3)

where /„,„ = f m = force induced in the direction of coordinate force m as a result of the nodal coordinate displacement n and fm is the total m coordinate force resulting from all possible coordinate displacements <c/„ (n = 1, 2, . . . , ) , as shown in Fig. 7.9.

/ y

Ffcr>6fer

• f,'kySky

k

Figure 7.8 (''lenient force and d i sp lacement no ta t ion .

F I N I T E F . I .EMKNT S T I F F N E S S M E T H O D I N S T R U C T U R A L A N A L Y S I S 1 9 9

I. 2. . . . . 6 = possible coordinate nodal displacement

Figure 7.9 Plane solid with three nodes showing ail p o s -sible nodal coordinate dis-placements t>„ (/i = 1,2, 6).

For linearly elastic structures, forces are directly proportional to displace-ments:

/=k<5 (7.4) where / = force *

5 = displacement k = proportionality constant

By utilizing the basic relationship in Eq. (7.4), Eq. (7.3) can be written q,s

L kmnS„ (in = 1 , 2 , . . . , 6 ) (7.5)

where k„m is referred to as a constant stiffness coefficient with dimensions of force per unit displacement and is defined as the force induced in the direction of coordinate m due to a unit displacement applied in the direction of nodal coordi-nate n.

In matrix form, Eq. (7.5) may be rewritten as

A k i t

h * 2 1 k22 • A = * 3 1 *32 k33 A k<it ki2

A k5l kS3

A _ kei hi k63

symmetric 4 4

54 ^ 5 5

64 ^ 6 5

<5t <52

'<53

<54

<5s

. V

(7.6)

or, in compact matrix form,

{/} = m{5} ( 7 . 7 )

where { / } and {<5} are the columns of force and displacement vectors, respec-tively, and {k} is a square matrix of the stiffness coefficients.

Equation (7.7) forms the basis of the stiffness method and establishes the conditions of static equilibrium.

For element relationships, Eq. (7.7) takes the form

{/„} = W W (7.8)

where {/„,} = internal nodal forces for member m {<5m} = structural nodal displacements associated with nodes of member in {km} = stiffness matrix for mth member

See Fig. 7.10a. For a structural system such as shown in Fig. 7.10ft, Eq. (7.7) becomes

{ F } = [ K ] {A} (7.9)

where {F } = structure external applied nodal forces and/or reactions {A} = structure nodal displacements [ K ] = structure aggregate stiffness matrix

Equation (7.9) set the sum of the internal element forces equal to generalized forces {F} acting at the nodal points of the structure..The aggregate stiffness matrix [ /v] can be formed by the direct addition of element stiffness matrices ;

[ K ] = £ [ f c j (7.10) HI — 1

where N = total number of elements in the structure under consideration. Since Eq. (7.9) represents the relationship between all nodal point forces and

corresponding displacements, it may be conveniently rewritten in a matrix parti-tioned form as

ss (7.11) AFr}_ IKrrl. _{Ar}_

where {Fa} = specified external nodal forces {Fr} = unknown nodal reaction forces {A„} = unknown nodal displacements {Ar} = specified nodal displacements

Expanding Eq. (7.11) yields - /

{Fa} = [ * „ ] { * . } + [K„J{A,} (7.12) and

{Fr} = [K r J[A. } + [ K J { A r } (7.13)

Equation (7.12) can be solved for the unknown displacement {A„}, and then the reactions may be determined by utilizing Eq. (7.13). Element internal forces are determined from element relationships.

To illustrate, consider the simple truss structure shown in Fig. 7.10ft. Assume that elements 1 and 2 have cross-sectional areas of 4 J l and 4 in2, respectively, and a modulus of elasticity equal to 107 lb/in2. As is shown later, the element relationships are given by

Element 1:

' f i x f y fix

h , fix hy

= 4 x 105

0.5 0.5 0 0 - 0 . 5 - 0 . 5 ~ 0.5 0.5 0 0 - 0 . 5 - 0 . 5 <>ty 0 0 0 0 0 0 $2x 0 0 0 0 0 0 Sly

- 0 .5 - 0 . 5 0 0 0.5 0.5 hx - 0 .5 - 0 . 5 0 0 0.5 0.5

(a)

Element 2 :

7 . , '

f y fix

hy hx

f>y.

= 4 x 105

0 0 0 0 0 0 " M

0 0 0 0 0 0

0 0 0 0 0 0 hx 0 0 0 1 0 - 1 hy 0 0 0 0 0 0 hx 0 0 0 - 1 0 1

(b)

Figure 7.10 Clemen! and s t ruc tu re relationships.


If we carry out the matrix addition of the element stiffness matrices in Eqs. (a) and {b), the stiffness matrix relationship for the entire structure becomes

(c)

0.5 0.5 0 0 - 0 . 5 - 0 . 5 . . 0 R i y 0.5 0.5 0 0 - 0 . 5 - 0 . 5 0 Rlx 0 0 0 0 0 0 0 R2y = 4 x 103 0 0 0 1 0 - 1 0

10 - 0 . 5 - 0 . 5 0 0 0.5 0.5 A3* 0 i- - 0 . 5 - 0 . 5 0.1 0.5 1.5_

Equation (c) may be expanded as

Rlx Riy

"10" "0.5 0.5" "A 3 / = 4 x 105 "A 3 / _ 0 _

= 4 x 105 .0.5 1.5_ LAJ

Ri L ^ j

= 4 x 103

- 0 . 5 - 0 . 5

0

0

- 0 . 5 - 0 . 5

0

- 1 (e)

..From Eq. (d), after the reduced aggregate stiffness matrix is inverted, the "displacements are

A3x = 0.075 in

A3y = 0.025 in Having determined the unknown displacements, we can calculate the reac-

tions from Eq. (e): K l x = - 1 0 k i p s Rix = 0

Rly = - 1 0 kips R2y = 10 kips

From Eqs. (a) and (b), the internal loads acting on each element are Element 1

A , = - 1 0 kips

fly= - 1 0 kips

/ 3 j = 1 0 k i p s

f3y = 10 kips

Element 2

/ ; * = 0

f2y = 10 kips

f3x = 0

f3y = - 1 0 kips

7.8 FORMULATION PROCEDURES FOR ELEMENT STRUCTURAL RELATIONSHIPS

The element stiffness matrix relationship can be derived by several different ap-proaches, all of which will result in a set of algebraic simultaneous equations of

the form {/} = m m (7.14)

where { / } represents the element forces, [£] represents the element stiffness matrix, and {<5} represents the element displacements.

Depending on the type of element being considered, one of three methods may be conveniently selected to establish the basic clement relationship for finite-element analysis: direct method, method of weighted residuals, and energy me-thods in conjunction with variational principles. Although to cover theory devel-opment is not the intent of this text, a brief description of the first and third methods is presented.

Direct Method

The direct method of finite-element formulations consists of the following steps:

1. A set of functions q is chosen to define the element displacements in terms of its nodal displacements.

2. Element strains are expressed in terms of the chosen displacement functions in accordance with the basic strain-displacement relationships of the element.

3. Element stresses are expressed in terms of the displacement functions in ac-cordance with the basic strain-stress relationships of the element.

4. Element node forces arc expressed in terms of element stresses in accordance with the statical equivalence of element boundary stresses.

To illustrate, consider the simple element shown in Fig. 7.11. The displacement of an axial rod clement is described by two nodal dis-

placements 5ix and <5JC, as shown in Fig. 7.11. The displacement function qx(x) may be chosen as a polynomial which must satisfy the nodal boundary condi-tions

qx{ 0) = S!x and qy(L) = Sjx

Hence, the polynomial must be a linear function of x ; that is,

qx{x) = ct + c2x ib)

or

< n d 0,

Figure 7.11 Axial rod element.

The displacement function in Eq. (c) can be expressed in terms of the 'element nodal displacements by using Eq. (a):

or

- h a a (d)

(e)

where N-,= 1 — x/L and Nj ~ x/L are referred to as shape functions, which play a major role in finite-element analysis.

A generalization of Eq. (e) may be written as follows:

or

where

{<?} = IN, N j - ' - N ^ S, Sj

Six

(7.15)

(7.15a)

(7.16)

The linear strain-displacement relationship for an axial rod element is given by

Sqx

dx

By utilizing Eq. (d), Eq. ( / ) becomes

[ J J

(/)

to)

A generalization of Eq. (g) may be written as

{£} = [G]{<5} (7.17)

The matrix [G] can be calculated easily once the shape function matrix [ N ] is determined. y

For isotropic material, the strain-stress relationship Tor uniaxial stress is given by

* E

= Ee, or

For a two-dimensional state of stress, Eq. (h) becomes

{£} = [D]{£}

where {£ } = ay T t jJr

CO

(7.18)

(7.19)

and [£>] is the elasticity matrix. The superscript T in Eq. (7.19) denotes the transpose.

In order to express the stress in terms of the nodal coordinate displacements, Eq. (<•/) is substituted into Eq. (/i):

a a The general form of Eq. (/) may be written as

{£ } = [D][G]{«5} = IS]{5} (7.20) where [SJ is the stress matrix.

As a final step, the stresses are transformed to equivalent nodal forces. This is done by simply multiplying the stress by the cross-sectional area A of the rod:

G H - 3 IOJ (J) LJjxJ L

or, in general, { / } = [ " ] { £ } (7.21)

where [ / / ] is a matrix which relates boundary stresses to equivalent element nodal forces.

Utilizing Eq. (/) in Eq. (J) yields the final result:

[ £ K U "IIS In general, Eq. (k) may be written as

if) = [H][ / ) ] [G]{5}

or { / } = M{«5} (7.22)

and {k} = [H] [D] [G] (7.23)

where {k} = element stiffness matrix f / / l = matrix relating nodal forces to element boundary stresses [D] = matrix relating stresses to strains [G] = matrix relating strains to nodal displacements

Energy Methods The energy methods in finite-element formulations are based on (1) work and strain energy and on (2) complementary work and complementary strain energy in conjunction with calculus of variation. In the first case, the methods yield the element stiffness matrix, while in the second the methods yield the element flexi-bility matrix. In this section, only element stiffness matrix formulation is pre-sented.

From the principle of virtual work and virtual strain energy, the finite-

206 A I R C ' R A I T S T R U C T U R E S

element Formulation proceeds as follows. The element basic relationships are

{q} = m{d}

( D = { £ } = [G]{<5}

The strain energy and work expressions are

t / = [£J{E} dV

and ' W = IA|{ / }

From the principle of virtual displacements,

5U = |_<5£J{2;} dV

8W = | & 5 J { / }

{<?£} = [G]{35}

where 8 denotes the first variation. By utilizing Eqs. (7.17), (7.18), and (7.28), Eq. (7.26) becomes

dU =

Equating the virtual work to the virtual strain energy yields

1A1

or

[ G ] r [ Z > ] [ C ] { 5 M K - { / } ) = 0

W W = { / }

where [&] is the stiffness matrix and is given by

W = J [ G ] t [ D ] [ G ] dV

Asan illustration, consider the axial rod element in Fig. (7.11), where

[GJ =

[G] T =

-. L L j

L

1

[D] = [E]

(7.15a)

(7.18) -

(7.17)

(7.24)

(7.25)

(7.26)

(7.27)

(7.28)

(7.29)

(7.30)

(7.31)

>7.32)

(a)

(b)

(c)

Therefore, the stillness matrix from Eq. (7.32) is

' - 1 ~ *

L L [*] =

V 1 1

L L

dV

~ E E •i. L2 1}

Jo . A -E E

_ V L1

AE 1 - f L -1 1

dA dx

which, again, is the same result as in previous treatments.

7.9 E L E M E N T SHAPE F U N C T I O N S

We showed that the displacement shape functions play an important role in the formulation of finite-element relationships. Although stress or strain shape func-tions may be selected, in this section wc discuss only those functions that arc based on the displacement field.

The criteria for sclccting displacement shape functions are:

1. The continuity of the chosen set of functions must prevail throughout the element design.

2. Free-strain conditions throughout the element must exist under rigid-body motion.

3. Constant strain or stress components must be present under the conditions of the chosen set of functions.

There are two means of arriving at shape functions: polynomial scries and direct formulation through the interpolation technique.

Polynomial Methods:

Any function <J> can be represented by a polynomial series as follows:

® = I P . = L P J W - (7.33)

where the polynomial P„ of any order n may be written in two dimensions as

P„= Z Q . v - y 1 = 0

(7.34)

The coefficients Cnl- are generalized parameters and are chosen such that their total number is at least equal to the total number of assumed nodal coordinate displacements of the element. For example, in case of the axial rod element, only two nodal coordinate displacements are possible. Therefore, for this case, the displacement function must be taken as

qx(x) =P0 + Pi = Coo + Cl0x + <??!>-

= C 0 0 + C 1 0 x = [ l (a)

where C , l is set equal to zero because qx cannot be a function of y. Equation (7.33) may be rewritten in terms of general displacement functions:

{q} = F J M (7.35) The column matrix {c} must be determined so that Eq. (7.35) is satisfied for all nodal displacements {<5}; that is

~ - {5} = [A]{c} (7.36)

and {c } = V r l { 5 } (7.37)

where {5} = nodal coordinate displacements [A] = matrix whose coefficients are constant functions of node coordinates { c } = sought coefficients of assumed polynomial

Hence Eq. (7.35) becomes

{q} = L P J m - l { t } = m { 5 } (7.38)

where [ N ] is the matrix of the shape function. To further illustrate the selection of displacement function polynomials, con-

sider a plane stress triangular element as shown in Fig. 7.12. Consider that the element has only three nodal points, in which case the total number of nodal coordinate displacements is 6. The displacements at any point on the element are

Figure 7.12 Plane stress triangular element.

F I N I T E F . I .EMKNT S T I F F N E S S M E T H O D I N S T R U C T U R A L A N A L Y S I S 2 0 9

described by qx{x, y) and qy(x, y), which are the displacements in the x and y directions, respectively. Since there are six nodal displacements, the polynomials for qx and r/,. must he chosen so' that the total number of coefficient parameters Cy does not exceed three constants in each case; i.e.,

<D = C0 0 + C 1 0 x + Cn}>

Then qx and qy will be of the form

qx=Ci + C2x+C3y

qy= C 4 + C 5 J C + C6y

The shape functions are determined in accordance with Eqs. (7.37) and (7.38).

7.10 ELEMENT STIFFNESS MATRICES

The stiffness matrix for the general beam element shown in Fig. 7.13 can be derived easily by any of the related energy methods discussed previously. One is Castiglia no's second theorem:

dU* = (7-39)

where U* = complementary strain energy = strain energy for linear elastic structures = nodal coordinate displacements

fi — corresponding nodal coordinate forces

fs, 85 / ] 1. 511

/2 .S2 4 fs. S8

/io.«n

/ • j . i ,

J.J.L / i

U-«« fa-

Figure 7.13 General beam element.

For the beam shown, the strain energy is given by

rr 1 fL (Ml Ml Ml S2 V 2 \ ,

where M, S, and V are the internal moments, axial force, and transverse shear force, respectively. These internal loads may be expressed in terms of nodal forces at either end of the beam element:

M=gAfd S = g2(fd V =

Thus

U = G(f<)

The stiffness coefficients are obtained by systematically giving each nodal coordinate displacement, one at a time, a unit value, fixing all others, and then calculating the induced reactions. Thus, if is set equal to 1 and all other <5's are set equal to 0, the resulting reactions are the stiffness coefficients kn(i = 1, 2 , . . . ,12). The coefficient ktl is defined as the force induced in the coordinate direction i due to a unit displacement applied in coordinate 1. By performing this operation for every possible displacement, the beam stiffness matrix can be ob-tained:

f„ k 0 K s>, f>. 0 0 k„ s„ M„ 0 0 0 e,x

0 0 0 ** «* M„ 0 0 0 0 k>, o.. fl, 0 0 0 0 0 ki h, 0 - k 0 0 0 -kz, 0 K, 4 0 0 0 ** 0 0 0 K

0 0 0 -k, 0 0 0 0 0 k, K 0 0 0 0 0 0 kz, 0 Oj, 0 0 0 0 k„, 0 -k2, 0 0 0 k3z_

y'{iA\)

Note-that from Maxwell's reciprocal theorem10 for linear elastic structures, ktJ = kji(i £ j). This can be easily proved by considering

D2U d2U

dqt dqj ~~ dqj Hi

where the order of differentiation with respect to qt and qj is immaterial. In compact form Eq. 7.41 becomes

{ / } = M{<5} (7.41a)

where

f. _ £ ± k a 4 . — A5

/> = (f> <u

a, = •

a2 j a3

b b

1 ^62

a2L -

a5 ^62 b

L - •••• « 2 —

AE « 2 2 EL

L « 5

L

El, « 5 ~ GJ 3/ i / . / ( 0 G

Tlie A',v expressions can he obtained by replacing the subscripts z and y in the above definitions. The bars which appear in Eq. (7.41a) denote relationship in local axes of the element.

Equation (7.41) is valid only when nonlinearities, material and/or geometric, do not exist. When large displacements are encountered, the structure behavior is referred to as geometrically nonlinear behavior. In this case, the force equilibrium equations must be formulated while the structure is in its deformed position. The stiffness matrix for the-geometrically nonlinear beam element behavior can be derived by any of the previously presented methods in conjunction with the nonlinear si rain-displacement relationships.

7.11 F R O M E L E M E N T T O S Y S T E M F O R M U L A T I O N S

Element formulations generally are carried out in reference to an element local set of coordinate axes x, f. and z .The stillness matrix relation for such elements is

{/} = [*]{*} (7 .42)

where the overbar denotes reference to the local axes. Structural systems arc defined as an assemblage of structural elements which

may be randomly oriented in space. In order to formulate the structural system force-dispiacemcnt equations, all element relationships such as Eq. (7.42) must be transformed to a common set of structure global axes x, y, and z.

From vector mathematics it can be shown that any set of three orthogonal axes .y, and z can be expressed in terms of another set of orthogonal axes A", y, and z by

X y - b , 2 _ h :

(7.43)

where /.vv = cosine of angle between _v and x axes, etc. See Fig. 7.14. In general, Eq. (7.43) states that any set of three vectors {T} in local coordi-

Figure 7.14 Local and global axes.

nates can be transformed to a vector {T} in global coordinates by

{r} = M{r} (7.44) where [A] is the 3 x 3 transformation matrix given in Eq. (7.43).

In accordance with Eq. (7.44), the following transformations may be accom-plished on any force and displacement vectors { / } and {<5}:

{ / } = m } / } and {<?} = [r]{<5} (7.45) /

where [ T ] is a matrix whose diagonal elements are the submatrices [A]. The number "of elements in the diagonal submatrices depends on the order of the matrix vectors { / } and {<5}. i

By utilizing Eq. (7.45), Eq. (7.42) becomes

C TVJ} = M[T]{<5}

or { / } = [ 7 T ' M m { < 5 } = M{<5} (7-46)

where [/c] is referred to as the element global stiffness matrix and is defined by

[fc] = [ T ] " ' [ fe ] [T] = [ T ] r [ / c ] [ T ] (7.47)


where Ihe inverse of [ T ] is its transpose:

£ T ] _ 1 = [ T ] r

This property holds true for any orthogonal square matrix. Once the element global stiffness matrix has been found, the global structure

stiffness matrix can be constructed easily by simple algebraic summation: m

[ K ] = I M (7.48) i= 1

where m is the total number of elements which make up the structure. Note that in order to carry out the summation in Eq. (7.48), the order of [fcj must be made compatible with the order of the anticipated structure matrix [JsTJ. This can be done by filling zero arrays corresponding to all nodal forces and corresponding displacements not connected with the nodes which appear on the element being considered.

To illustrate, consider the structure shown in Fig. 7.15. For each element, the stiffness matrix equation is formulated in accordance with Eq. (7.46). Each el-ement is identified by its connectivity. For example, element 1 has connectivities 1. 3, and 4; element 2 has the connectivities 1 and 2; etc. Each element stiffness matrix [A] is placcd in the appropriate location of the global matrix in accord-

© © © 1

I") j 5

Figure 7.15 Construct ion of structure relationship from element relations.

ance with its connectivity. Each dot may represent a single coefficient or a sub-matrix of coefficients if more than one degree of freedom is being considered at the node. Once this is done for each element, a simple algebraic summation is performed which yields the structure stiffness matrix equation. Applying the boundary conditions will result in a reduced aggregate structure stiffness matrix equation which can be solved for the unknown displacements. Back substitution of these displacements into each element equation will yield element internal forces.

7.12 GLOBAL STIFFNESS MATRICES FOR SPECIAL BEAM ELEMENTS

The space truss is shown in Fig. 7.16a Its global stiffness matrix is

<5;* &iy <5fz <5j* <5j,

M „ AE

L

R l l * 1 3

-R\I

—RiiRn

^12 E-12 Rl3

- R ? 2 — ^ 1 2 ^ 1 3

Rl 3 r u r 1 3

R\2 R\i Ri 3

symmetric

" u ^ 1 1 ^ 1 2

RtiRii Rh Ri2RI3

(7.49)

R ] 3

where R11 = " 1 ' 1 LJ

yj - y<-L

zi ~ z<

Rl 2 =

r 1 3 =

L = l(xj - Xi)2 + 0'j - yd2 + ~ Zi)2]1'2

A — cross-sectional area of member

- E = Young's modulus

The plane truss is shown in Fig. 7.16/j. Its global stiffness matrix is

(7.49a)

<5;,

OTp, = : AE

Rh symmetric

R11R12 Rh

-Ri, —RuRn Rl. - Rl 1^12 -Rh RiiRn Rl 2

(7.50)


The plane frame is shown in Fig. 7.16c. Its global stiffness matrix is

8iy Ou Sjy K a symmetric

b d

c e f •

—a -b —c a

-b -d — e b d

c e 9 —c —e f

where a = jR^otj + R21aAz b = R^R^a-i +

c = R2la. 2l d = Rj2 a! + R\2 a4l

e = R22CC2z f=«3; 9=a 6z

R » s s - L R l l ~ L

L = [(Xj - X,-)2 + {yj - y,)1]1'2

AE GEL 4 Elx a ' = T " & 2 z = a 3

12 EIZ 2EI. a 4 z = — — —

/ . = cross-sectional moment of inertia about z axis

Example 7.1 Use the finite element stiffness method to analyze the structures shown in Fig. 7.17.

Case 1 Plane truss problem The joints and their geometries are: x coordinate, y coordinate.

Joint no. in in

1 0 0 2 100 0 3 100 100 4 0 100

100 in

50 kips

50 rt

10 kips

2 © 3

© ©

. \\\

1 - 20 f t -

0>)

Figure 7.17 Plane truss and frame structures.

The following information is given about each member:

Member no. (anil its nodes) .-1, in2 E, lb/ in1 !,, in

1 (1-21 2 !07 100 2 (2 3) 2 to7 100 3 (2-4) 2 7 2 107 100 ^ 2

The rotation matrix coefficients for member 1

and

x2 - x , 100 R l l = ~ l ~ = w d = 1

R i 2 = yi^iL = o

For member 2, the vertical member with — y2 = 100 — 0 > 0,

/?,, = 0 and R l 2 = 1 For member 3,

.v, - x2 l< M

« . 2

= V 4 L

X z = 0 - 1 0 0 / 1 0 0 ^ / 2 = - y i / 2

V4 - J'2 = ( i o o - o ) / i o o N / 2 = s j f y i

The known boundary conditions are:

Displacements'. 5ix = e>4v = r>lr = <5, = 0

External forces: Flx = !0l F2y = - 10k

The global stiffness matrices [Eq. (7.50)] are

Member I: 2 x 105

1 0 - 1 0

1 0 5 0 0 0 0 1 0 5

- 1 0 1 0 _ 0 0 0 0 _

Member 2:

Member 3: 2 x 105

0 0 0 0

105 0 0

1 0

0 0

- 1 0

_0 - 1 0

0.5 - 0 . 5 0.5 0.5 - 0 . 5 0.5 0.5 -0 .5 - 0 . 5 0.5 0.5 -0 .5

0.5 - 0 . 5 - 0 . 5 0.5_

(b)

M

Since tiiere arc four joints on the structure with two degrees of freedom at each joint, the order of the structure stiffness matrix is 8 x 8. Expanding each member matrix to 8 x 8 by adding rows and corresponding columns of zeros associated with joints not appearing in the member itself, thus making all matrices compatible in size with the structure matrix, and then using matrix addition yield

[K R 8 = I M 8 ' 8

o 1 symmetric 0 (}

- 1 0 1

0 0 0 0 + 0 0 0 0 0

0 0 0 0 0 0 _ 0 0 0 0 0 0 0 _

© 0

0 0 s y m m e t r i c

0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 - 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0

Q .

+

0 0 0 symmetric 0 0 0.5 0 0 - 0 . 5 0.5 0 0 0 0 0 0 0 0 0 0 0 0 0 - 0 . 5 0.5 0 0 0.5 0 0 0.5 - 0 . 5 0 — 0.5 0.5

M —

i

o

- l

o

0

0

0

0

0 0 0 0 0 0 0

L.5 - 1 . 5

0 0

- 0 . 5 0 . 5

symmetric

1.5 0

- 1

0 . 5 - 0 . 5

0 0 0

0 - 0 . 5 0.5 0.5

(<0

H e n c e t h e m a t r i x e q u a t i o n f o r t h e s t r u c t u r e b e c o m e s

'Rix ' Ri,

1

0

1

0 1

s y m m e t r i c

10* - I 1.5 - 10* 0

o l

- 0 . 5 1.5

0 o l 0 0 0

0 0 1 0 - 1 0 1

R'ix 0 0 1 - 0 . 5 0 . 5 0 0 0 . 5

0 0 | 0 . 5 - 0 . 5 0 0 - 0 . 5 0 . 5 _

The unknowns arc £ 2 a , <52 and all the reactions Rlx, Rly, etc. Thus, cxtract-

ing the third and fourth equations from the above yields

10<~| = [~ 1.5 -0.5~lp2 . t

10* J L - 0 - 5 1 - 5 1 ^ , .

1

2 x 10 s

Inverting the matrix gives

'2.x

J ' J L - o . :

0 . 0 2 5

0 . 2 5

1 [ "0 .75 0 . 2 5 T 10*

2 x 1 0 s L o . 2 5 0 . 7 5 j L — 10*

The reactions from Equ(e) are as follows (see the sketch below):

Rlx = - { 2 x 1 0 5 ) ( 5 2 x ) = - 5 * Rix = 0

= 0

- 5 ' R, 5k

10 kips

The internal loads arc found from the element relationships:

" 1 0 - 1 0 " " 0 " ~—5k

= 2 x 1 0 s 0 0 0 0 0 0

fix = 2 x 1 0 s

- 1 0 1 0 0 . 0 2 5 5*

fl, 0 0 0 0 - 0 . 0 2 5 0

5 k i p s > kips

©

h . " 0 0

f2> = 2 x 1 0 5 0 1

fix = 2 x 1 0 5

0 0

hy . 0 - 1

0 0 "

0 - 1 0 0

0 1

- 0 . 0 2 5 " " 0 ""

- 0 . 0 2 5 — 5 k

0 0

0 5k


fly h* A ,

= 2 x !05

0.5 -0.5 -0.5 0.5 -0.5 0.5 0.5 -0.5 -0.5 0.5 0.5 -0.5

0.5 -0.5 -0 .5 0.5

~ 0.025" ~ 5*' -0.025 — 5k

0 —5k

0 5*

t 5 kips

Case 2 Plane frame structure The joints and their geometries are as follows:

Joint no. .v coordinate, y coordinate.

0 0

240 240

0 600

600

0

The following information is given for cach member:

Member no. (and its nodes) A, in2 i . , in4 E, lb/ in 2 L, in

i (1-2) 20 360 30 x 106 600 2 (2 3) 20 360 30 x 10" 240 3 (3 4) 20 360 30 x 10® 600

The rotation matrix coefficients are as follows: Member /:

R i . = . \-2 -

R, , = -

L

J'2 - V'l

= 0

2 2 2 A I K C R A I - T S T R U C T U R E S

Member 2:

Member 3:

Rl2 =

* 2 I

A-4 - A-J = 0 R , , >'4 - >'2

* 2 2 = 0

Example 7.2 Determine the displacements and internal loads lor the truss structure shown in the figure. Assume £ = 1 0 7 lb/in2 for both rods; -4(rod 1) = 4^/2 in2 and ,4(rod 2) = 4 in2.

10.000 III

SOLUTION By choosing the system axes as shown, from Eq. (7.49a) the trans-formation coefficients Ry for each element may be obtained:

Element 1: K M = 1

Rn =

U-Va - -x'i)2 + 0 ' 2 - J i ) 2 ] " 2 Y / l

y 2 - y ,

[ ( .v 2 - . v , ) 2 + ( y 2 - > . , ) 2 ] 211/2 1

V 2 Element 2: y 3 - y 2 < 0

R , , = 0 R 1 2 = — 1

From Eq. (7.50) the element relationships are the following

— = 4.12 x \ t f m j l = 4 x 105 L N x

y

Element I:

0.5 8\ F] 0.5 0.5 1*2 - 0 . 5 - 0 . 5 0.5 Si Fl = 4 x 1()5 - 0 . 5 - 0 . 5 0.5 0.5 ,55 F} 0 0 0 0 0 r-Z 0 0 0 0 0 0 _ JU

(a)

F I N I TP. E L E M E N T S T I F F N E S S M E T H O D I N S T R U C T U R A L A N A L Y S I S 2 2 3

Element 2: A 2 E 4 x 10 7

4 x 10s - L2 - 100

4 x 10s

p ' l 1 0

Fl 0 0 symmetric 0 0 0

p ; = 4 x 10s 0 0 0 1.0 0 U 0 0 0 0 0 0 - 1 . 0 0 1.0

<5?

<5! <K (h)

Carrying out the matrix addition of the element stiffness matrices in Eqs. (ci) and (/>), the stiffness matrix relationship for the entire structure becomes

K i , " R i y

\ o k

0 = 4 X 105

L R l y .

0.5 0 0.5 0.5 symmetric 0

- 0 . 5 - 0 . 5 0.5 <51 - 0 . 5 - 0 . 5 0.5 1.5 <5$

0 0 0 0 0 0 0 0 0 - 1 . 0 0 1.0 0

(c)

Extracting the two middle equations from (c) yields

" 1

_ 0 _

'2 x 105

2 x 10s

2 x 105' 6 x 10 s

Solving for the unknown displacements gives

2 x 105

2 x 105

2 x 105" 6 x 105

<5? Si.

10fc"

0 W

where the inverse of the matrix in (e) is

0.75 -0.25" 10" -0.25 0.25

Therefore the displacements arc

SI = 0.075 in

SI = -0 .025 in

Displacements known, the external reactions on the structure and the ele-ment internal loads may be found from Eqs. (c), (a), and (b), respectively:

' - 2 x 105 - 2 x 105" 0.075" ~ -10 4 "

. - 2 x I05 - 2 x 105_ 0.025_ _ —104_ lb

"0 0 0.075" " 0 " _0 - 4 X 10s_ 0.025. - 1 0 \


Fx 0.5 0 - 1 0 4 " F\ 0.5 0.5 symmetric o - 1 0 4

n = 4 x 105 - 0 . 5 - 0 . 5 0.5 0.075 = 104 lb _—0.5 - 0 . 5 0.5 0.5 — 0.025_ 104_

> 3 ~ 0 0.075 ~ 0 " n 0 1 symmetric —0.25 - 1 0 " F's = 4 x 105 0 0 0 0 0 lb

. 0 - 1 0 1 0 104

A free-body sketch of the internal loads is shown below.

104 Hi

IUJ ib

Example 7.2 Assume that the structure of Example 7.1 has rigid joints as shown in the ligure below. Determine the displacements and internal loads. Neglect shear deformation, /(beam 1) = 10^/2 in4, /(beam 2) = 10 in4.

10.000 lb

SOLUTION The transformation coefficients Ru arc obtained from Eq. (7.51): Beam I:

1 1

Beam 2:

* n = 0

Rz i = l

- 1

R21 = 0

From Eq. (7.51) the clcmcnl stiffness matrix relationships are as follows: Beam I:

n 0.2003 symmetric F]i 0.1997 0.2003 symmetric

MT -0 .03 0.03 4 FX

2 = io6 -0.2003 -0.1997 0.03 0.2003 Fi -0.1997 -0.2003 - 0 . 0 3 0.1997 0.2003

_M\_ - 0 .03 0.03 2 0.03 - 0 . 0 3

o-f i

<55

Beam 2:

F -2 ' F>2

M\ n = 1 0 6

Fl -M

symmetric 0.0012 0 0.4 0.06 0 4

-0.0012 0 -0.06 0.0012 0 - 0 . 4 0 0 0.4 0.06 0 2 - 0 . 0 6 0 4

a* "2 <51

J l .

(b)

As in Example 7.1, the overall stilTncss matrix relationship for the whole structure may be easily constructed from Eqs. (a) and (b):

0.200.1 symmetric R\ (I.I 997 0.2003 F A/j 0.53 0.03 4.0

103 0.2003 -0.2003 -0.03 0.2015 0 = 10" 0.1997 -0.2003 -0.036 0.1997 0.6003 0 0.03 0.03 2.0 0.091 -0.03 8.0

0 (1 (1 —0,0012 -0.0012 -0.06 0.0012 0 0 0 -0 .4 0 0

FM% 0 0 0 0.06 0 2.0 -0.06 0.4 0

(C)

where FM denotes fixed end moment reactions. Extracting from (c) the equations corresponding to the nonzero displace-

ments yields

IO3" "0.2015 symmetric Si 0 = 10A 0.1997 0.6003 51 id) 0 0.09 - 0 . 0 3 8.0 _ 01 _

Carrying out matrix inversion produces the unknown displacements:

p l l 7.415 -2.468 -0.0371 " 1 0 4 " 0.07415 in = 10"6 -2 .468 2.488 0.01859 0 = -0.02468 in

-0.0371 0.01859 0.1249 _ 0 -0.000371 rad z

Note that for large structural systems, the matrix order becomes quite large and the conventional matrix inversion approach is avoided. Other means of solving large sets of simultaneous algebraic equations are available.

The internal loads are calculatcd from Eqs. (a) and (h) and arc shown in the sketch below.

9962 Ih

I I 22 Hi-i iS"*-9962 lb

99r,2 i[i

34 lb

34 lb

X 2fi16 It)-ill

99(,2 lb

PROBLEMS

7.1 In ihe following set of algcbraic equations, Rx . <52, and arc the unknowns. Put them in a matrix form, and solve for the unknowns. Use only matrix notation and tminipulalion.

Rl=Sl-S2

P = + 1.5^2 — 0.5^3

2 P = S1- 0.5^2 + 0.5rt3

7.2 A structure is acted on by the forces shown in Fig. P7.2. ir the materia! behavior is given hv a - Kv}. where K is a constant, iind the strain energy stored in the system. Neglect the dead wejglit, and assume AE to be constant. ^

Uniformly distributed load, p lb/in

\ \ W \ Figure P7.2

7 3 Find the matr ix relationship { Q j = [/!]{<?! for the truss s t ructure shown in Fig. P7.3. Note: Assume AE to be constant and the same for both members.

i 90 45°

Figure P 7 J

7.4 Given the truss structures shown in Fig. P7.4, determine the following information for each: (a) The external nodal applied loads to be used (/>) The known displacement boundary conditions (r) The size of the element stiffness matrix (r/) The size of the structure reduced stiffness matrix

Figure P7.4 (</) Planar truss; (/>) planar truss; (r) space truss.

7.5 Using the finite element method, determine the reactions, displacements and internal loads for the planar truss structure shown below. Shmv all the details.

10 kips

7.6 Write mathematical expressions which will determine for any structure (a) the order of the aggregate stiffness matrix and (/>) the order of the reduced aggregate stiffness matrix. Illustrate the use of the derived expressions on structures shown in Fig. P7.6.

r r s

(0)

A

s < \ V V

Figure P7.6 (i/) Plane frame; (h) plane truss; (r) space frame; [J) space truss.

7.7 Find the joint loads which must be used in the finite-element analysis of the frame structure shown in Fig. P7.7.

AT = 100°F

8000 lb

F. = 10* lb/in2

U= 10-5 in/(in)(°F) 8000 lb

Figure P7.7


7.8 S h o w h o w you wou ld go a b o u t analyzing the b e a m o n c o n t i n u o u s elastic founda t ion shown in Fig. P7.8 by using Ihe linite-elemcnl technique.

I team

Soil ' • " • ' ' • • ' ' ••

l-'igim- l'7.H

7 .9 What d isp lacement funct ions would you use to deve lop the stiiTness mat r ix relat ionship for cach of the e lements s h o w n in Fig. P7.y?

Figure 1*7.9 (t/) Axial rod element ideal i /a l into four nodes ; (/;) p la te e lement (in p lane forces only); six nodes.

7.10 F ind the s t r a in -energy expression per unit vo lume of a s t ruc tura l m e m b e r whose stress-strain behavior is related a s

a = ce"

where c and n are cons tan t s .

7.11 F ind the to ta l d isp lacement of point 2 on the truss s t ruc tu re in Fig. P 7 . l l . Assume E = 107

lb / in 2 and cross-secl ional a rea of cach m e m b e r = 2 in 2 .

ioooo Ji ih

w w w Figure P7.11

7.12 T h e d i sp lacements for a two-dimensional solid are given by

Is = "i + azx + a3y q, = b i + b2x + b3y

F ind the s t ra in energy per uni t volume of the solid.

7.13 Show the structure global axes and each element's axes for the structure in Fig. P7.13.

o A \ V s Figure P7.13

7.14 An axial rod element is idealized as sliown in Fig. P7-I4. The displacement function </(v) is assumed lo be

<?M = c, + c2.v + r3x2

How would you derive the element stiffness matrix equat ion?

Figure 1*7.14

7.15 Derive the stiffness coefficients for the element shown in Fig. P7.I5.

Figure P7.15

7.16 Illustrate by means of sketches the physical significance of the stiffness cocfTicicnts correspond-ing lo the actions shown in Fig. P7.I6.

Qi Qi

r v v W U Figure P7.16

7.17 Find the coefficients of the transformation matrices for cach of the truss structure elements shown in.Fig. P7.17.

H N I T K i;i . l -MMNT STI I -TNLSS M E T H O D I N S T R U C T U R A L A N A L Y S I S 2 3 1

7.18 Find the equivalent nodal loads lo be used in the finite-element analysis for each of the struc-tures in Fig. P7.I8. You are given that

A T = 100"F, a = 1 0 " 6 in/(in • °F)

/•; 107lb/in2 A = 10 in2 / = 100 in"

Figure P7.18 (a) Frame.

7.19 The radius of a tapered, solid, tubular shaft element varies as follows:

= R 0 , / ? = const

Assuming that only torsional loads can be transmitted, find the stiffness matr ix of the element. 7.211 The nodal displacements of the truss s tructure shown in Fig. P7.20 were found by the finite element lo bo

t>2x = - 0 . 0 2 in S l s = - 0 . 0 1 in

Find the internal loads on clement 2. AK — 107 lb.

7.21 Determine the displacements and the internal loads for the truss s t ructure shown in Fig. P7.21. Assume a cross-sectional area of 1 in ' for cach rod and a modulus of elasticity of 107 lb/in2. Use the stillness method.

10.000 lb-in

Figure P7.21

7.22 In each of the s t ruc tu res in Fig. P7.22, de te rmine the in te rna l loads a n d the deflections. Use the st iffness mat r ix m e t h o d on ly .

10 kips 10 kips

10 kips 10 kips

t 10 ft

5 kips

t o n

1

10 kips 10 kips

(?)

20 lb/in i » H h 6 i

(<o

Section c-c

(c)

Figure P7.22 (a) Rigid f r a m e : all m e m b e r s a re m a d e o r 2-in tub ing with 3/8-in wall thickness, (b) T r u s s : 2-in t ub ing ; 3/ft-in wall thickness.

CHAPTUR

EIGHT ANALYSIS OF TYPICAL MEMBERS

OF SEMIMONOCOQUE STRUCTURES

8.1 INTRODUCTION

This chapter presents approximate methods for the analysis of typical members of semimonocoque structures. Inherently, these structures are highly redundant, and an accurate analysis would require the use of a computer in conjunction with the matrix methods discussed in Chap. 6. Cutouts, shear lag, warping restraint, discontinuity of loads, etc. are some of the factors which affect the accuracy of the analysis, and their inclusion is what makes computer solutions inevitable.

8.2 DISTRIBUTION OF CONCENTRATED LOADS TO THIN WEBS

Modern aircraft structures arc constructed primarily from sheet metal. The metal is necessary for a covering and this is utilized for structure as well. The thin sheets or webs are very efficient in resisting shear or tension loads on the planes of the webs, but usually they must be stiffened by members more capable of resisting compression loads and loads normal to the web. When no stiffening members are used and the skin or shell is designed to resist all loads, the construction is called monocoque, or full monocoque, from the French word meaning "shell only." Usu-ally it is not feasible to have the skin thick enough to resist compression loads.

2 3 3


and stiffeners are provided to form scmimonocoque structures. In such structures, the thin webs resist tension and shearing forces in the planes of the webs. The stiffeners resist either compression forces in the plane of the web or small, dis-tributed loads normal to the plane of the web.

When semimonocoque structures must resist large, concentrated loads, it is necessary to transmit the loads to the planes of the webs. Since the concentrated loads may have components along three mutually perpendicular axes, it is neces-sary to provide webs in different planes, so that the loads may be applied at the intersection of two planes. A fuselage structure, for example, has closely spaced rings or bulkheads which resist loads in transverse planes, while the fuselage shell resists loads in the fore-and-aft direction. Concentrated loads must be applied at the intersection of the plane of the bulkhead and the shell, or else additional structure members must be provided to span between bulkheads and transfer the loads to two such intersecting planes.

When a concentrated load is applied to the plane of a web, a stiffening member is required to distribute this load to the web, as shown in Fig. 8.1A. This member should be in the direction of the load, or the load should be applied at the intersection of two stiffeners, so that each stiffener resists the load component in its direction. The load P shown in Fig. 8.1 is distributed to the web by the stiffener AB. The shear flows ih and qz in the adjacent webs are approximately constant for the length of the stiffener. The axial load in the stiffener, therefore, varies linearly from P at point B to 0 at point A, as shown in Fig. 8.1c. From the

/ O

O f

O f

O "

•ti-

ll

Figure 8.1

A N A L Y S I S O F TYPICAL M E M B E R S O F S E M I M 0 N 0 C 0 Q U E STRUCTURES 2 3 5

equilibrium of the forces shown in Fig. 8.1/?, P = [q{ + q2)cl. Thus the required length 11 of the stiffener depends on the ability of the webs to resist shear, sincc a longer stilfcncr rcduees shear flows <j, and q2. The end of the stiffencr, point xt, should always be at a transverse stilTener. If a stiffener ends in the ccntcr of a wch, it produces abrupt change in the shear Hows at the end of the stiffener and undesirable concentration conditions.

In this chaptcr, thin wehs are assumed to resist pure shear along their bound-aries. In actual structures, the thin webs may wrinkle in shear, thus introducing tension field stresses in addition to those calculated. The effects of tension field stresses are calculated in later chapters. It is found at that time that the tension field stresses can be superimposed readily on those calculated by the methods used here, and the methods used in this chapter remain valid for obtaining the shear distribution in tension field webs. In some cases, the tension field stresses produced by wrinkling of the webs induce additional axial compression loads in stifieners. These loads should be computed separately and added algebraically to (lie l o a d s o b t a i n e d in t h i s c h a p t c r .

A study of a simple numerical example demonstrates the method by which loads are distributed to shear webs. The beam shown in Fig. 8.2a is similar to a wing rib which is supported by spars at the ends and which resists the load of 3000 lb, as shown. The stiffener AB transmits this load to the two webs in inverse proportion to the horizontal lengths of the webs, since the vertical shear at any cross section of the beam must be in equilibrium with the external reaction on the beam. The axial load in AB is shown in Fig. 8.2c; it varies from 3000 at B to 0 at A. The axial load in the upper flange of the beam can be obtained from cither the

Figure 8.2

bending-moment diagram of the beam or a summation of the shear flows, as shown in Fig. 8.2ft. The compression at point A of 2000 lb can be obtained from the shear flow of 100 lb/in for 20 in or from the shear flow of 200 lb/in for 10 in.

The cantilever beam shown in Fig. 8.3 resists a load R which has a horizontal component of 1500 lb and a vertical component of 3000 lb. The horizontal stiffener AB must be provided to resist the horizontal component of the load, and the vertical stilTcncr CBD must resist the vertical component. The interseclion of these stiffeners, point B, should be on the line of action of R. The shear flows qx

and q2 can be obtained from the equilibrium of these stiffeners. For stiffener AB to be in equilibrium underthe forces shown in Fig. 8.3c,

10qy - 10q2 = 1500

Similarly, for member CBD to be in equilibrium under the forces shown in Fig. 8.3ft,

5qt + I0q2 = 3000

Solving the above two equations simultaneously yields </, =300 and q2 = 150 lb/in. These values also can be derived by analyzing the beam separately for each of the two load components and then superimposing the results. The vertical load alone would produce shear flows of 200 lb/in in each web while the horizontal

+ 5 0 0 0 lb

2 0 0 lb/ in 3 0 0 lb/ in

(c)

Figure P7.7

ANALYSIS OF TYPICAL MEMBERS OF SEMIMONOCOQUE STRUCTURES 2 3 7

load would produce a shear flow of 100 lb/in in the upper web and —50 lb/in in the lower web. The axial loads in the upper flange member, shown in Fig. 8.3d, could not be obtained readily from a bending moment diagram of the member.

The loads considered above were assumed to act in the plane of the web. When loads have components along all three references axes, the structure should be arranged so that the loads act as the intersection of two webs, as shown in Fig. 8.4n. Here each of the three components of the force R is distributed to the webs by a stiffener in the direction of the force component. In some cases, this is not practical and a load normal to a web, as shown in Fig. 8.4b, cannot be avoided. If the load is small, the stiffener may be designed to have enough bending strength to resist the load. In many cases, the loads are such that it is necessary to provide an additional member, such as web ABCD in Fig. 8.4c, to resist the load. This member spans between ribs or bulkheads and can resist any load in its plane by means of the three reactions F,, F2, and F3 shown in Fig. 8.4d. Even small loads such as those from brackets supporting control pulleys should not be applied as normal loads to an unsupported web. Such brackets may be attached to stiffeners or may be located at the intersections of webs.

i,/i

Figure 8.4

8.3 LOADS ON FUSELAGE BULKHEADS

The structural unit which transfers concentrated loads to the shell of an airplane fuselage or wing is commonly called a bulkhead. Bulkheads are attached to the wing or fuselage skin continuously around their perimeters. They may be solid webs with stiffeners or beads, webs with access holes, or truss structures. Fuselage bulkheads usually are open rings or frames, so that the fuselage interior is not obstructed. Normally the chordwise bulkheads in wings are called ribs, while fuselage bulkheads are called rings or frames, fn addition to transferring loads to the skin, wing and fuselage bulkheads supply column support to stringers and redistribute shear flows in the skin. The first step in the design of a bulkhead is to obtain the loads which act on the bulkhead and thus hold it in static equilibrium. In the case of fuselage rings, this step is simpler than the next problem—to obtain the unit stresses from the loads. Unit stresses in fuselage rings and similar struc-tures are analyzed in a later chapter on statically indeterminate structures.

Fuselage shells normally are symmetrical about a vertical centerline and often are loaded symmetrically with respect to the centerline. The fuselage bend-ing stresses can be obtained by the simple flexure formula / = My/l, and the fuselage shear flows can be found from the related expression derived in Chap. 5,

In applying Eq. (5.31) to a symmetrical box structure, often it is convenient to consider only half of the structure, since the shear flow must be zero at the top and bottom centerlines. Thus, each term of Eq. (5.31) applies to only half the fuselage shell. If stringers or longerons are located on the top or bottom center-lines, half of their area is considered to act with each side of the structure.

The fuselage ring shown in Fig. 8.5 is loaded by a vertical load P on the centerline of the airplane. This vertical load P must be in equilibrium with the running loads q which are applied to the perimeter of the ring, as shown in Fig. 8.5c. The present problem is to obtain the distribution of the forces q. The fuselage cross section just forward of the ring has an external shear V„, and the cross section aft of the ring has a shear Vb, as shown in Fig. 8.5« and b. The. load P on the ring must be equal to the difference of these shears: ^

If, for ihe moment, the shear resisted by the in-plane components of the stringer loads is neglected, the shear flows on the two cross sections adjacent to the ring are

1 = ~r y d A (5.31)

V.-V„=P

</„ = y j v dA (8.2)

and V- f ,A qb = -j- .v dA (8.3)

ANAI YSIS C)l TYPICA1. MPMRERS Or SF.MIMONOOOQUR STRUCTURES 239

The load q transmitted to the perimeter of the ring must equal the difference between r/„ and qh, or

From i:qs. (8.1) to (8.4),

V„- V„

(8.4)

y (I A

<1 = V dA (8.5)

When the areas resisting bending of the shell are concentrated as flange areas Af, the integral is replaced by a summation, as in the following equation:

<Z = J Z )'Af (8.6)

liquations (H.5) ami (K.6) are correct even when the relieving effects of the in-plane components of the stringer forces are considered, since this shear resisted by the stringers must be lhe same on both fuselage cross sections adjacent to the ring, if the stringers have no abrupt change in direction at the ring. Thus the difference

' r

Figure 8.5

2 4 0 AIRC'RAiT STRUC TURES

F i g u r e 8 . 6

in total shear forces, VA — VB, must equal the difference in the shears resisted by the webs.

In many cases, a fuselage structure may be symmetrical but the loads may not be symmetrical. Any unsymmetrical vertical load may be resolved into a vertical load at the centerline and a couple. The couple applied to the ring will be resisted by a constant shear flow

where T is the magnitude of the couple and A is the area enclosed by the fuselage skin in the plane of the bulkhead.

A fuselage ring also may resist loads which have horizontal components. In this case it is not possible to find a web with zero shear flow by inspection, as in the case for symmetrical vertical loads. It is necessary first to obtain all the shear flows in terms of one unknown and then to find this unknown from the equilib-rium of moments, as was done in the analysis of box beams. The method^vill be obvious after you study Example 8.2.

Example 8.1 The fuselage bulkhead is shown in Fig. 8.6. Only one-half of the shell is considered, as shown in Fig. 8.6b. The value of P resisted on this half of the structure is 500 lb, and the moment of inertia is found for only one-half of the structure. Of course, the value of P/I in Eq. (8.8) will be the same if both values are obtained for the entire structure, since both will be doubled.

S O L U T I O N The solution is shown in Table 8 . 1 . The areas AF listed in column 2 are the total areas of stringers 2, 3, and 4 but only half the areas of stringers 1 and 5, since the structure shown in Fig. 8.6b is being considered. The


Table 8.1

Slringcr no. A, Afy' y yAf y1AJ X yAf q, lb/in

(I) (2) (3) (4) (5) (6) (7) (8) (9)

1 0.05 34.0 1.7 18.5 0.925 17.12

2 (1.10 2-1.0 2.4 X.5 0.X5 7.23

3 0.10 15.0 1.5 - 0 . 5 - 0.05 0.02

4 0.10 6.0 0.6 - 9 . 5 - 0 . 9 5 9.02

5 0.05 0.0 0.0 - 1 5 . 5 - 0 . 7 7 5 12.01

X 0.4 6.2 45.40

0.925 10.20

1.775 19.55

1.725 19.00

0.775 8.54

eenlroid y is determined from the summations of columns 4 and 2:

_ £ Afy' 6.2 y = „ 1 . = — = 15.5 in

S A, 0.4

It is now necessary to obtain coordinates of the stringers with respect to I he ccntroidal'axis. These values, r = y' = y, obtained by subtracting 15.5 f rom te rms in c o l u m n J , are s h o w n in c o l u m n 5. T h e t e rms y A j and y A j a r e calculated in columns 6 and 7. The summation of column 7 yields the moment of inertia I. Equation (8.6) becomes

P v „ 500 9 = 7 Z - v A f ^ z y A f

The values of q arc calculated in column 9 and are shown in Fig. 8.6a.

Example 8.2 The fuselage bulkhead shown in Fig. 8.7 resists a horizontal load as shown. The stringer coordinates are given in column 3 of Table 8.2.

LLO.S Ih'int llfi.X lb/in)

(1.7 lb/in)

(17.3 lb/ill I (17.3 lb/in I

4 ft

i?(>.2 lb/in i

F igure 8.14

242 AIRC'RAiT STRUC TURES

Table 8.2

Stringer no. .X x>A, 2 xAf <f 2/1 2 A q' 1

(I) (2) (3) (4) (5) (6) (V) (8) (9) (10)

1 0.05 0 0 0 0 0 140 0 + 16.8

2 0.10 8 0.8 6.4 0.8 15.1 100 1,510 + 1.7

3 0.10 10 i.O 10.0 1.8 - 3 4 . 1 100 - 3 , 4 1 0 - 1 7 . 3

4 0.10 10 "l.O 10.0 2.8 - 5 3 . 0 160 - 8 , 4 8 0 - 3 6 . 2

5 0.05 0 0 0

Z 0.4 26.4 500 - 1 3 , 4 0 0

The areas enclosed by the skin segments and the lines to reference point 0 are indicated by the double areas listed in column 8 of Table 8.2. Find the reactions of the skin on the bulkhead.

SOLUTION First the shear flows are obtained with the assumption that web 1-2 resists a zero shear flow. The resulting shear flows q' are

«i = f Z xAf (8.8)

The calculations are performed in Table 8.2. The value of /,. for half the structure is obtained in column 5 as 26.4 in*. If the shear flows are positive clockwise around the ring, the force P is considered as negative in Eq. (8.8), or

, - 1 0 0 0 „ q = 2 Z 2 M Z x A '

The values of q arc calculatcd in column 7. These values are obviously the same for the left half of the structure because of symmetry. Theyshear flows q' produce moments of 2Aq' about point 0. These moments are'calcu-lated in column 9. It is now necessary to superimpose a constant shear flow c/o around the ring so that the external moments on the ring are in equilib-rium. Taking moments about 0 and considering clockwise moments as posi-tive, we have

qQ X 2A + E 2Aq' + 1000 x 10 = 0

or

1000</o - 2 x 13,400 + 1000 x 10 = 0

q0 = 16.8 lb/in

\NAI.YSIS o r TYPICAI. MFMItF.RS OF SF.MIMON'OCOQUF. STRUCTURHS 2 4 3

The resulting shear (lows q are found by adding 16.8 to each value in column 7. These values are shown in column 10 and in Fig. 8.7.

In the simplest type of wing structure, that in which the bending stresses are resisted by only three concentrated flange members, the skin reactions on the ribs can be obtained from the equations of statics. There will be only three unknown shear flows, and these may be obtained readily from the equations for the equilib-rium of forces in the vertical and drag directions and for the equilibrium of moments about a spanwise axis.

Then the internal stresses in the rib are obtained from the shears and bending moments at the various cross sections. Normally, there v/ill be axial loads in the rib in addition to the shears and bending moments; thus it is necessary to calculate bending moments about a point with a vertical position corresponding to the neutral axis of the rib. For the rib analyzed in Example 8.3, we assume that ail bending moments are resisted in the rib flange members and that all shears are resisted by the webs. With these assumptions, it is more convenient to calculate bending moments about the neutral axis.

In the more general case of a wing in which the bending moments are resisted by more than three flange members, it is necessary to determine the section properties or the wing cross section before the shear reactions on the ribs can be found. The problem is similar to that of calculating reactions on fuselage bulkheads, but it differs in the condition that the fuselage cross section usually is symmetrical, whereas the wing cross section seldom is symmetrical. Thus the skin shear flows, and "consequently the skin reactions on the ribs, must be obtained by the more general methods which involve the product of inertia of the cross section.

Example 8.3 Find the shear flows acting on the rib of Fig. 8.8. The wing bending moments arc resisted by the three flange areas shown in Fig. 8.8a to

8.4 ANALYSIS OF WING RIBS

10 in

3000 Ih

Figure 8.8


c. Calculate the loads in the rib flanges and the shear flows in the rib webs at a vertical cross section through flange a and at vertical cross sections a short distance to either side of the applied loads.

SOLUTION The reactions of the wing skin on the rib must be in equilibrium with the applied loads of 9000 and 2000 lb. F rom a summation of moments about point c and a summation of forces in the vertical and drag directions, the following equations are obtained:

Z A f r = 9000 x 10 - 168qoc - 222qba = 0 XF3 = 20qac - 20qb a - 2000 = 0 E F 3 = 9000 - 10qbc - 4qb„ - 6qac = 0

The solution of these equations yields qha = 274, qac = 374, and qhc = 566 lb/in. At a vertical cross section through flange a, the stresses are obtained by

considering the free body shown in Fig. 8.9a. The total shear at the cross section is V — 374 x 6 = 2244 Ib, and the bending moment is M — 2 x 24 x 374 = 17,940 in • lb. The horizontal components of the axial loads in the flanges are found from the bending moment as = Pz = M/6 = 2990 lb. The lower rib flange is horizontal at this point, but the upper member has a slope of 0.4. The shear carried by the flange is therefore

0.2/>, = 1448 Ib

Figure 8.9

ANALYSIS OF T Y P I f A l . M E M B E R S OF S E M 1 M 0 N 0 C 0 0 U E STRUCTURES 2 4 5

Vf = QA x 2990 = 1 1 9 5 lb. The remaining shear, which is resisted by the web, is Vw = V - Vf = 2244 - 1195 = 1049 lb. The shear flow at this section is q = 1049//) = 178 lb/in. These values arc shown in Fig. 8.9a.

The stresses at a vertical section to the left of the applied loads are obtained by considering the free body shown in Fig. 8.9/J. The shear at the cross section is V = 274 x 3 + 374 x 6 = 3066 lb. The bending moment about the lower flange is M = 2 x 45 x 275 + 2 x 54 x 374 = 65,100 in • lb. The horizontal component of the upper flange load is P3 = 65,100/9 = 7240 lb. The lower flange load is obtained from a summation of horizontal forces as P4 = 7240 - 10 x 274 + 10 x 374 = 8240 lb. It is obvious that the bend-ing moment in the rib depends on the vertical location of the center of moments, since there is a resultant horizontal load at the section. If the entire depth of a beam resists bending moment, the centroid of the section is used as the center of moments, and the stresses resulting from the axial load then are uniformly distributed over the area of the cross section. The vertical component of Ihe upper flange load is 0.2 x 7240 = 1448 lb. The shear flow in the web is therefore q = (3066 — 1448)/9 = 180 lb/in. These values are shown in Fig. 8.9h.

The stresses at a vertical cross section just to the right of the applied loads are obtained in similar manner and are shown in Fig. 8.9c. Since the bending moment was computed about the intersection of the two applied loads, the value of R, is the same as for the previous case. The axial load in the lower flange and the web shear flow differ from those shown in Fig. 8.9b.

Example 8.4 The rib shown in Fig. 8.10 transfers the vertical load of 10,000 lb to the wing spars and to the wing skin. Find the reacting shear flows around the perimeter of the rib, the shear flows in the rib web, and the axial loads in the top and bottom rib flanges.

S O L U T I O N The distribution of shear flows depends on the spar-cap areas. The section properties are: /_ = 200 in4,. ly — 800 in4, Iyz = —200 in4. The reac-ting shear flows on the rib are equal to the shear flows in the skin of a box which resists an external shear of 10,000 lb, but have opposite directions. The change in shear flow at each flange area is found from Eq. (5.25):

( = h 2 5 i ,

</„H 5(in I

- 3 7 5 ) ' « / »

(</1

1 i n 2 ? i» I - 1 5 i i i - F i g u r e 8 . 1 0

+ 3 7 5 0 I h ( t o p i

' - - 3 7 5 0 I h ( b o t t o m )

<>25 I h / i n

1 2 5 l b / i n

ft 2 5 ! l i / i n j

o

1

3 7 5 H i / i n

o :

3 7 5 l h , ' i n

I 2 5 I b . ' i n

1 0 . 0 0 0 l b F i g u r e S. l I

If we assume q positive clockwise, the following values are obtained:

Aq = —500 at flanges a and b

Aq — +500 at flanges c and d

If a shear flow q0 is assumed in the top skin, the other shear flows are found in terms of q0, as shown in Fig. 8.10. The final shear flows can be obtained now from the equilibrium of moments about some convenient point, say flange a:

200<]o + 200(q0 + 500) = 10,000 x 5 0 F qQ — — 325 lb/in

The shear flows in the remaining webs can be found now from these values of q0 and are shown in the proper directions in Fig. 8.11. The shear Hows in the rib webs are obtained from the equilibrium of forces on the vertical cross sections and are shown in Fig. 8.11 as 625 lb/in. on the left-hand web and 375 lb/in on the right-hand web. The axial loads in the rib flanges are shown in Fig. 8.11 and are derived by a summation of the shear-flow forces acting on the rib flanges, including shear flows from both the wing and the rib web. A comparison of the rib web shears and flarfge loads shown in Fig. 8.11 with those for a simple beam of the same dimensions shows that the flange loads will be the same in both cases, but the web shears will be different.

8.5 SHEAR FLOW IN TAPERED WEBS

The shear flow in a tapered beam with two concentrated flanges is considered in Sec. 5.9, in the discussion of the unit method of shear-flow analysis for tapered box beams. The distribution of the shear flow in a tapered web is considered now in greater detail, since a large proportion of the shear webs in an airplane struc-

ANALYSIS OF TYPICAL MEMBERS OF SF.M1MONOCOQUE STRUCTURES 2 4 7

lure are tapered rather than rectangular. The shear flows in the web of the beam shown in Fig. 8.12« are obtained in Sec. 5.9. F rom Eq. (5.30) the shear V„ resisted by the web will be

where the notation corresponds to that shown in Fig. 8.12. The shear flow q may be expressed in terms of the shear flow q0 = Vjh0 at the free end by the following equations:

The distribution of the shear flow q along the span of the beam is shown in Fig.

In many problems it is necessary to obtain the average shear flow in a tapered web. The average shear flow between the free end and the point .x of the beam shown in Fig. S. 12a can be found from the spanwise equilibrium of the flange shown in Fig. 8.12c. The horizontal component of the flange load is found by dividing the bending moment qnh0b by the beam depth h. The average shear flow in this length, </nv, is therefore obtained by dividing this force by the hor-izontal length b:

!/. = Vr ^ = K — • h x (8.9)

(8.10)

8 . 1 2 / ) .

h or q() — (8.11)

. V -

\

R \ „ • /<„ • i . / = ' / „ < — • = </n i j i

''i -T 111 I T T r r h f r m

I — I

1'iRiirc 8 .12

If the shear flow on one side of a tapered web is known, the shear flows on the other three sides may be obtained from Eqs. (8.10) and (8.11).

It is assumed in the derivation of Eqs. (8.10) and (8.11) that the stresses existing on ail four boundaries of the tapered plate are pure shearing stresses. It has been shown that pure shearing stresses can exist on only two planes, which must be at right angles to each other. Since the corners of the tapered webs do not form right angles, it is necessary for some normal stresses to act at the boundary of (he web. In order lo estimate the magnitude of these normal stresses, a tapered web in which pure shearing stresses may exist at all the boundaries is considered. -

It can be shown by the theory of elasticity that a sector such as shown in Fig. 8.13 may have pure shearing stresses on all the boundaries. Under these bound-ary conditions, any element such as that shown will have no normal stress in the radial direction a „ and no normal stress in the tangential direction <rM. The shearing stresses on these radial and tangential faces must satisfy the equation

where K is an undetermined constant. This equation is similar to Eq. (8.10) if the taper is small.

By comparing the sector of Fig. 8.13 with the tapered web of Fig. 8.12, it is " seen that the assumption of pure shear on the top and bottom boundaries of the tapered web was correct. The left and right boundaries also must resist some normal stresses, however. The magnitude of these normal stresses may be deter-mined for the Mohr circle of Fig. 8.146. The element under pure shearing stresses has faces A and B which are inclined at an angle 0 with the vertical and hor-izontal. The Mohr circle for the pure shear condition will have a center at the origin and a radius r , . Point A will be at the top of the circle, and point C, representing stresses on the vertical plane, will be clockwise at an angle 20 from point A. The coordinates of point C represent a tensile stress of Ts sin 20 and a shearing stress of r, cos 20 on the vertical plane. The normal stresses obviously are negligible for small values of the angle 8.

The equations for shear flow in tapered webs first were derived for the web of

K ffr« = ~ 2 (8.12)

Figure 8.13


a beam wilh two concentrated flanges and then were shown to be approximately correct for any web that resists no normal loads at its boundaries. It can be shown by examples of other structures containing tapered webs that the shear flows may be applied lo the webs by members other than beam flanges. Tapered webs often are used in torque boxes, such as shown in Fig. 8.15. For this box, all four sides are tapered in such a way that the corners of the box would intersect if extended. The enclosed area at any cross section varies with x according to

A_ A0

- T l j A'o) (8.13)

The shear flow at any cross section, for the pure torsion loading condition shown, is obtained from

4 = 2A (8.14)

From Eqs. (8.13) and (8.14) and from the value of the shear flow at the left end, q0 = T/(2.40), the following expression for q is obtained:

11 = T

2A0 lo

x 0 (8.15)

sin 211

s20

T( sill 10

lc/>

Figure 8.14


This corresponds to the value obtained in Eq. (8.10) for the two-flange beam. The shear flow q given by Eq. (8.15) applies for all four webs of the box. Conse-quently, there is no axial load in the flange members at the corners of the box, since the shear flow at the sides of the box will be transmitted directly to the top and bottom webs.

In the structure shown in Fig. 8.15, all four webs are tapered in the same ratio, so that the shear flow obtained from Eq. (8.15) will be the same for all four webs. Who n the taper ratio for the horizontal webs is not the same as the taper ratio for the vertical webs, the shear flows will not have the same distribution for all webs. If, for example, the top and bottom webs arc rectangular and the side webs are tapered, as shown in Fig. 8.16, the shear in the rectangular web must remain constant for the entire length, while the shear in the tapered web must vary according to Eq. (8.10). The shear flow for this structure which has ribs only at the ends cannot be obtained from Eq. (8.14), although Eq. (8.14) is quite accurate for the common airplane wing structure with closely spaced ribs. The ribs divide the tapered web into several smaller webs and distribute shear flows so that (hey arc approximately equal in the horizontal and vertical webs.

The shear flows in the tapered webs of Fig. 8.16 vary according to Eq. (8.10):

Since the flange members at the corners of the box must be in equilibrium for spanwisc forces, the shear flows q„ in the lop and bottom webs must equal the average shear flows for the tapered webs, as obtained from Eq. (8.11),

and from Eq. (8.16),

The difference in shear flows between two adjacent webs produces axial loads in the flange member between these webs. At any intermediate cross section of the box, the in-plane components of the flange loads must be considered irS addition to the web shears, in order to check the equilibrium with the external torque on the box. At the end cross sections, the shear flows are in equilibrium with the

(8.16)

(8.18)

Figure 8.13

ANALYSIS Of TYPICAL MEMBERS OF SF.M1MONOCOQUE STRUCTURES 2 5 1

Figure 8.16

external torque. For the ieft end of the box,

(<7o + <7oWi0 = T

a n d s u b s t i t u t i n g v a l u e s f r o m E q . (8 .17) g i v e s

T </„ =

«C>i + h o) (8.19)

This equation also can be-derived from the equilibrium of forces at the right end of the box

T = «/!,((/, + qa)

and substituting values from Eq. (8.18) yields

<J„ = r

a{hl +h0)

which checks the previous value. The denominator of Eq. (8.19) represents the average value of 2A for the box, as might be expected from Eq. (8.14). For most conventional wing or fuselage structures, the ribs and bulkheads are closely spaced, and it is seldom necessary to consider the taper of the structure when torsional shear flows are obtained. Equation (8.14) may be used, and the shear flows in all webs will be approximately equal at a cross section. For uncon-ventional structures, however, where the ribs cannot distribute shear flows, it may be necessary to use methods similar to those employed for the structure of Fig. 8.16. If the top and bottom webs also are tapered, but have a different taper ratio than the side webs, then the flows may be obtained by applying Eq. (8.10) to each web, equating the average shear flows for all four webs, and then equating the torsional moments of the shear flows at one end to the external torque on the structure.

8.6 CUTOUTS IN SEMIMONOCOQUE STRUCTURES

Typical aircraft structures which consist of closed boxes with longitudinal stiffe-ners and transverse bulkheads are analyzed in preceding sections. In actual air-


craft structures, however, it is neccssary to provide many openings in the ideal continuous structure. Wing structures usually must be interrupted to provide wheel wells for retraction of the main landing gear. Other openings may be necessary for armament installations, fuel tanks, or engine nacelles. Fuselage structures often must be discontinuous for doors, windows, cockpit openings, bomb bays, gun turrets, or landing-gear doors. It is also necessary to provide holes and doors for access during manufacture and for inspection and mainten-ance in service. These "cutouts" are undesirable from a structural standpoint, but are always necessary. Often they occur in regions where high loads must be resisted, and frequently'considerable structural weight is required for re-inforcements around the cutouts.

A simplified example of a structure with a large cutout is shown in Fig. 8.17. This corresponds to a wing structure with four flange members in which the lower skin is completely removed. In previous sections we stated that a closed torque box was necessary to provide stability for resisting torsional loads. In order for the structure of Fig. 8.17 to be stable, one end must be built in, so that the torsion may be resisted by the two side webs acting independently as cantile-ver beams, as shown in Fig. 8.17/?. The flange members resist axial loads, which have the values P = TL/(bh) at the suppor t The shear (lows q in the vertical webs are double the values obtained in a closed torque box with the same dimensions. The horizontal web resists no shear flow in the case of the pure torsion loading, but it is necessary for stability in resisting horizontal loads. The torque box, with webs on all six faces, is capable of resisting torsion with no axial loads in the flange members. Hence the torque box is much more rigid in torsion, since the shear deformations of the web are negligible in comparison with bend-ing deformations of a cantilever beam.

In a full-cantilever airplane wing, it is not feasible to have an open structure for the entire span, for the wing tip would twist to an excessive angle of attack under some flight conditions. A closed torque box is necessary for most of t f e span, but may be omitted for a short length, such as the length of a wheel-well opening. When the lower skin is omitted for such a region, the torsion is resisted

by "differential bending" of Ihe spars, as indicated in Fig. 8.! lb. The axial loads in the spar Manges usually are developed at both sides of the opening, since the closed torque boxes inboard and outboard of the opening both resist the warping deformation of the wing cross section. For the torsion loading shown, often it would be assumed that flange loads were zero at the midpoint of the opening and that loads of P/2 were developed at both sides of the opening.

The open box with three webs and four flange areas is stable for any loading if one or both ends arc restrained. The shear flows in the three webs may be obtained from three equations of statics. The method of obtaining the shear flows is obvious from a numerical example such as that indicated by Fig. 8.18. From the equilibrium of moments about point C of Fig. 8.18b,

. 10in x 20 = 10,000 x 10 + 2000 x 5 + 40,000

or qi = 750 lb/in

From the equilibrium of vertical forces,

10^3 + 10 x 7 5 0 = 10,000

or qi — 250 lb/in

Similarly, from the equilibrium of horizontal forces,

20 q2 = 2000

or q2 = 100 lb/in

The axial loads in the flange members can be found from a summation of span-wise forces:

pa = 40c/! = 30,000 lb

ph = 40</, + 40t/2 = 34,000 lb

pc = 40</, - 40^2 = 6000 lb

pd = 40q3 = 10,000 lb

While these forces satisfy the conditions for static equilibrium, they cannot be obtained from the flexure formula. For an open beam containing only four flange members, the flange loads are independent of the flange areas. If the beam has more than four flanges, one must consider the flange areas in estimating the distribution of axial loads. This problem is statically indeterminate, and usually it is solved by approximate methods.

A cutout in a short length of the wing structure affects the shear flows in the ' adjacent sections of the wing which have closed torque boxes. First consider a case in which the wing resists pure torsion. The shear flow in a continuous closed box is

I , = ^ ( 8 . 2 0 )

This equation is derived from the assumption that the flange members resist no

axial loads. At the edges of the cutout, however, the flange loads resulting from deferential bending have their maximum values. These flange loads are dis-tributed to the webs, and at some distance from the cutout, the flange loads becomc zero for the box in pure torsion. The distance along the span required for the distribution of the flange loads depends on the relative rigidities of the mem-bers, but it will be approximately equal to the width of the cutout. The shear flows in the torque box are affected considerably by this distribution of load.

The rectangular torque box shown in Fig. 8.19A resists pure torsion. The lower skin is cut out for the entire width of the box, for a length L. The effect of the cutout is assumed to extend a distance L along the span on cither side of the cutout; therefore, it is necessary to consider only the length 3L, which is shown. The shear flows at the section through the cutout are similar to those obtained in Fig. 8.17b, or they will be zero in the upper skin and 2q, in the spar webs, where q, is the shear flow in a continuous box, as obtained from Eq. (8.20). The axial loads P in the spar flanges are assumed to be equal on the inboard and outboard sides of the cutout and thus are half the value shown in Fig. 8.176, or

P = q,L (8.21)

This axial load must be transferred to the webs adjacent to the flange in the assumed length L. From the equilibrium of the flange member showir'in Fig. 8.19b,

q t f , — ii2 L — P (8.22)

or. from Eqs. (8.21) and (8.22),

ql-q2 = q, (8.23)

The shear flows qt and qz must satisfy the conditions of equilibrium of the structure shown in Fig. 8.19c. For the vertical forces to be in equilibrium at a cross section, the shear flows in the spars must have equal and opposite values </2. For horizontal forces to be in equilibrium, the shear flows in top and bottom skins must have equal and opposite values q F o r the shear flows in all four webs to react the torque T, the following condition must be satisfied:

ANALYSIS OI- TY1MCAL VliMllliRS OK SKMIMONOCOQUE STRUCTURE 255

q,A + q2 A = T

or, from F.q. (8.20).

qx + q2 = 2 q, (8.24)

Solving Eqs. (8.23) and (8.24) yields

</, = l.5r/, and q2 = 0 . 5 q ,

Tlie values of these shear flows arc shown in parentheses in Fig. 8.19a. From these resulting values of shear flows qt and q2, the cutout is seen to

have a serious effect on the shear flows in the closed torque boxes adjacent to the cutout. The top and bot tom skins have shear flows of times the magnitude of those for a cont inuous box, while the shear flows in the spars are only one-half as much. The ribs adjacent to the cutout also resist high shear flows. The rib just outboard of the cutout is shown in Fig. SA9d. The rib receives the shear flow of \.5q, from the top and bot tom skins of the to rque box. The spars transfer shear flows of 2q, from ihe cutout section and 0.5r/, from the torque box seciion. Since

i./I

Figure 8.19

the sheaf flows are in opposite directions, the resultant shear flow applied to the rib is h5q,.

The problem of a box beam's resisting a more general condition of loading usually is analyzed by another method. The method employed for the structure resisting pure torsion becomes more difficult when the spar flanges resist axial loads resulting from wing bending in addition to those resulting from the differ-ential bending. The common procedure for the general case is first to analyze the continuous wing structure as if there were no cutout. Then a system of correcting shear (lows must be obtained and superimposed on the original shear flows found for the continuous structure. In finding the correcting shear flows, only a short length on either side of the cutout need be considered, since the loads applied to the wing in arriving at these shear flows are in equilibrium with themselves. One of the established principles of mechanics, formulated by Saint Venant, states that the stresses resulting from such a system of forces will be negligible at a distance from the forces. The distance is approximately equal to the width of the opening.

The method of obtaining correcting shear flows is illustrated for the wing structure shown in Fig. 8.20. The wing is assumed to have a constant shear of 30,000 lb in the vertical direction and —9000 lb in the chordwise direction for the entire length from station 30 (30 in from the airplane centerlinej to station 120. The lower skin is removed for the entire width between the spars from station 60 to station 90. The wing bending moments affect the flange loads but not the shear flows; therefore, the bending moments are not considered. The dimensions of the cross section arc shown in Fig. 8.20b. The shear flows in the continuous closed box with no cutout are shown in Fig. 8.20c. These are computed by the methods discussed in Chap. 7, and the computations are not discussed here. Since the external shear is constant, the shear flows in all webs between stations 30 and 120 would have the values shown in Fig. 8.20c if there were no cutout. '

The correcting shear flows are now found by applying the loads of 660 lb/in in the cutout region, as shown in Fig. 8.21«, and finding the shear flows in the remaining webs. It is obvious that the loads of 660 lb/in around all four sides of the cutout are in equilibrium with one another. The shear flows at the cross section through the cutout are assumed to be </,, q2, and q 3 , as shown in Fig. 8.21ft, and must have a resultant equal to the applied load of 660 lb/in,at the lower skin. From a summation of horizontal forces,

30<72 = 30 x 660

or

q2 = 660 lb/in

From a summation of moments about point O of Fig. 8.20b,

2 x 90 x 660 = 2 x 75 x </, - 2 x 200 x 660 + 2 x 90 x q3

Solving these equations simultaneously yields qt — 1340 lb/in and q3 = 1010 lb/in. These correcting shear flows are shown in Fig. 8.2!fl for the structure between stations 60 and 90. The final shear flows in the cutout region are now

#

ANALYSIS Ol- TYPICAL MI-MHERS OF SEMIMONOCOQUIi STRUCTURES 2 5 7

obtained by superimposing [he values shown in Fig. 8.20c and 8.21a. This super-position yields a shear flow of 300 lb/in in the upper skin, 1930 lb/in in the front spar web, and 940 lb/in in the rear spar web, as shown in parentheses on Fig. 8.20a.

The correcting shear flows between stations 90 and 120 arc found from the equilibrium of the forces on a cross section. The shear flows are shown in Fig. 8.21c, and the following equations are derived from the equilibrium of the shear flows on the cross section:

Z F . = 30f/s - 30<7? = 0

£ F , - 10</4 + 2,h - 12(j(, = 0

£Af 0 = 2 x 75f/4 + 2 x 200qs + 2 x 90q5 + 2 x 90<b = 0

One additional equation may be derived from the spanwise equilibrium of


forces on one of the flange areas. For the flange member shown in Fig. 8.2Id, the axial load at station 90 is obtained by assuming no axial loads at the centcr of the cutout, station 75. From the shear flows shown in Fig. 8.2 In, P = 15(1340 + 660) = 30,0001b. From Fig. 8.2 Id,

30q7 - 30q4 = 30,000

Solving these four equations simultaneously yields q4 = —670, q5 = 330, q„ = - 5 0 5 , and q7 — 330 lb/in. These values of the correcting shears are shown in Fig. 8.2la. The final shear flows arc obtained by superimposing the correcting shear

Kignre 8 . 2 1

ANALYSIS Ol- TYI'ICAI. MI.'MBERS OH SEMiMONOCOQUIi STRUCTURE 2 5 9

flows and those for the continuous structure, shown in Fig. 8.20c. The corrected values are shown in parentheses on Fig. 8.20a.

The loads acting on the rib at station 90 are found from the differences in shear flow on the two sides of the rib and shown in Fig. 8.21 e. The shear flows Iransfcrrctl to the rib by lhc wing skin arc seen to be greater than the shear flows in the skin, since the skin shears act in the same direction on the rib and must be added. The rib at station 60 will resist the same loads as the rib at station 90, but the directions of all loads will be reversed.

Cutouts in fuselage structures are treated in essentially the same manner as cutouts in wing structures. Fuselage structures usually have lighter stringers and skin and resist smaller loads, particularly torsional loads. The torsional rigidity of fuselages is not as important as the torsional rigidity of wings, although flutter problems may develop in high-speed aircraft if the fuselage is too flexible tor-sional!}'. Fuselage structures often are open for a large proportion of their length in order lo pennil long cockpit openings or long bomb bays. These structures are able to resist lhc torsional loads by differential bending of the sides of the fuse-lage.

Fuselages of large passenger airplanes often contain rows ot windows, as shown in Fig. 8.22. If these windows arc equally spaced and have equal sizes, the shear flows in webs adjacent to the windows can be obtained in terms of the average shear flow q0 which would exist in a continuous structure with no windows. If the windows have a spacing u> and the webs between them have a width it',, the shear in these webs </, can be derived from a summation of forces on a horizontal section through the windows:

ii7

q i = — io (8-25)

Figure 8.22

Similarly, if the effect of the cutouts is assumed to extend over a vertical distance h, as shown, the shear flows in webs above and below the windows may be obtained by considering a vertical cross section through a window:

The shear flows q3, shown in Fig. 8.22, can be found by considering either a horizontal section through the webs or a vertical cross section through the webs. The two equations are

<}2 "'2 + qi = (Jo w

and 1ih2 + q3fh = q0h

Either of these two equations, when values from Eqs. (8.25) and (8.26) are substi-. tutcd, reduces to

The notation is shown in Fig. 8.22. Openings for large fuselage doors may be analyzed in the same manner as

wing cutouts. Sometimes it is difficult to provide rigid fuselage bulkheads on either side of an opening, bccausc of interior space limitations. In such cases, a rigid doorframe can be provided so that "the doorframe itself resists the shear loads in place of the cutout structure. If such a structure is provided, it is no longer necessary to have the heavy bulkheads adjacent to the opening. A fuselage doorframe usually must follow the curvature of the fuselage and hence does not lie in a plane. Thus, the structure of the doorframe must be capable of resisting torsion as well as bending, and the frame must be a closed-box structure.

In the analysis of semimonocoque vehicle structures, the shear stress distribution is of great importance. Much of the classical theory of statically i n d e t e r m i n a t e

structures has been developed for the analysis of heavy structures in which shear-ing deformations are of minor importance. Consequently, much of the published work on structural deflections and indeterminate structures docs not t r e a t shear-ing deformations. The deflections caused by shearing deformations can be deter-mined by the method of virtual work, in the same manner as other types of dcflcctions are analyzed.

The shearing deformation of an elastic rectangular plate with thickness f, width Lx, and length Ly is indicated in Fig. 8.23ci. The shearing strain exy is obtained from the relation

h q2 = h[q° (8.26)

(8.27)

8.7 SHEARING DEFORMATIONS

(8.28)

/ 1 ~7

r v r

u 7/ // ''«

It

I/')

Figure 8.2.1 Plate deformat ion under shear action.

where G = shear modulus crx), = shearing stress

q = <7vr / = shear flow

In finding the dcflecfions of a structure resulting from shear deformations of the webs, it is convenient to use a unit virtual load applied at the point of the desired deflection A. Thus from Eq. (6.31), replacing the generalized displacement q by A yields

i m w = L<>IJ{E'} dV

or 1 • A

Utilizing Eq. (8.28) in Eq. (8.29) yields

1 • A =

&<JX>exy dV (8.29)

^ d V t tG

or M Gt2 dV (8.30)

where qu — <5 = shear flow due to the unit applied load and q = real shear flows which produce the deformation. Integrating Eq. (8.30) gives

A = Gt

(8.31)

For a structure which has N webs affecting its deflection, Eq. (8.31) becomes

A = £ q»!CIiL*iL>' i = 1,2,..., N (832) C,t,

Equation (8.32) applies only lo clastic deformations which satisfy Eq. (8.28).

8.8 TORSION OF BOX BEAMS

One of the most common applications of Hq. (8.32) is finding the angle of twist of box beams, such as that shown in Fig. 8.24. The shear flows q may result f rom


Figure 8.24 Bo \ beam.

any condition of loading, and they are obtained by the methods used in Chap. 5. Since an angular deflection is required, a unit virtual couple (torque) will be applied as shown in Fig. 8.25 and the resulting virtual shear flows are 17,, - I/(2A), where A is the enclosed area of the box. The webs are assumed to have dimen-sions Lx = As and Ly = L. The angle of twist 0 is obtained by substituting these values into Eq. (8.32):

The summation includes all webs ol the structure.

Example 8.5 The box beam shown in Fig. 8.26 has front spar-flange areas which arc 3 times the rear spar-flange areas. Find the angle of twist at the free end. Assume C = 4 x 10* lb/in2.

SOLUTION The shear flows q are shown in Fig. 8.27. All these shear (lows are positive except the shear flow in the right-hand web, which tends to produce

Figure S.25 Box beam under Iht* action of a unit loique.

ANALYSIS OK TYl'iC'Ai. MEMBERS OF SEMIMONOCOQUE STRUCTURES 2 6 3

I (..000 III 16.000 111 Rear spar

0 . 0 4 0 - A ,!,. o

— 0.040 ]

40 in -1 1 20 in

un (/>)

Figure 8.26

1.400 Ih/iii 200 Ili/in

:00 il>/in .

200 Ili/in Figure 8.27

a counterclockwise rotation. Hence, from Eq. (8.33) the twist is

O = Y - = y L 2AiG 2AG t

120 /1400 x 10 200 x 40 x 2 200 x 10 ~ 2.x~400 x 4 x 106 V 0.081 + 0.040 ~ 0.032

= 0.020 rad

8.9 ELASTIC AXIS OR SHEAR CENTER

The elastic axis of a wing is defined as the axis about which rotation will occur when the wing is loaded in pure torsion. For the wing shown in Fig. 8.28«, in which the cross section is uniform along the span, the elastic axis is a straight line. Points on the elastic axis do not deflect in the torsion loading, but points forward of the elastic axis arc deflected upward. It is necessary to calculate the position of the clastic axis in order to make a flutter analysis of the wing.

The .shear center of a wing cross section is defined as the point at which the resultant shear load must act to produce a wing deflection with no rotation. The shear force shown in Fig. 8.28/) deflects the wing in translation, but causes no rotation of the cross section about a spanwise axis. If the wing is an elastic structure, then the shear center of a cross section must lie on the elastic axis, since a force at the shear center produces no rotation at the point of application of the couple and the couple must therefore produce no vertical deflection at the point

2 6 4 AIRCRAFT STRUCTUR1S

F i g u r e 8 . 2 8

of application of the force. Practical wings deviate slightly from conditions of elasticity because the skin wrinkles and becomes ineffective in resisting com-pression loads but, for practical purposes, the elastic axis may be assumed to coincide with the line joining the shear centers of the various cross sections.

The shear center of a cross section may be calculated from Eq. (8.33) by finding the position of the resultant shear force which yields a zero angle of twist. The shear-center location depends on the distribution of the flange areas and the thickness of the shear webs. The procedure can be studied best by means of an illustrative example.

Example 8.6 Find the shear center for the wing cross section shown in Fig. 8.29. Web 3 has a thickness of 0.064 in, and the other webs have thicknesses of 0.040 in. Assume C is constant for all cross sections. The cross section is symmetrical about a horizontal axis.

S O L U T I O N The position of the shear center does not depend on the mag-nitude of the shear force. Thus a shear force V — 400 lb is assumed arbi-trarily. The shear-flow increments are obtained by the methods used in Chap. 5 and are as shown in Fig. 8.30. The shear flow in web 1 is assumed to have a value of </0, and the remaining shear flows are expressed in terms of as shown. In previous problems, the shear flow q0 was obtained from the equi-librium of torsional moments, but the external torsional moment is not known now. So we assume that the 400-lb shear force acts at a distance x from the right side, as shown, and that this point is the shear center. The shear flow q„ is found from the condition that the angle of twist 0 be zero. From Eq. (8.33), where L. = 1 and 2 AG is constant for all webs,

^ 2AtG

be taken outside the summation sign and then canceled:

0 = 1 ^ = 0 (8.34)

ANALYSIS OF TYPICAL MEMbeRS OF SF.M1MONOCOQUE STRUCTURES 2 6 5

The total shear flow q is the sum of the component q0 and the component q\ where q' is the shear flow if web 1 is cut:

q = q0 + q'

Substituting this value into l-q. (8.34) and taking q0 outside the summation sign (because it is constant for all webs) produce:

'/o Z Y + Z y = 0 (8-35)

The numerical solution is tabulated below. Column 1 lists values of A.s\ the circumferential lengths for the various webs, as shown in Fig. 8.29. These values are divided by the web thickness in column 2. The values of q', the shear flows when web 1 is cut, are tabulated in column 3. The shear flows are considered positive when they are clockwise around the outboard face of the element of I ig. X.30. The values ofcy' A.sy'f arc calculatcd in column 4.

W e b A s

( t )

As

1

(2) 13)

A s

" ' 7

(4)

2A

(5)

2Aq'

(6) 1 (7)

I 8 2 0 0 0 0 8 0 0 - 0 . 5 3

2 8 2 0 0 - 1 0 — 2 0 0 0 8 0 - 8 0 0 - 1 0 . 5 3

3 10 156 - 2 0 - 3 1 2 0 0 0 - 2 0 . 5 3

4 8 2 0 0 - 1 0 — 2 0 0 0 0 0 - 1 0 . 5 3

5 8 2 0 0 0 0 0 0 - 0 . 5 3

6 15.7 3 9 2 + 2 0 7 8 4 0 2 3 9 4 7 8 0 + 19 .47

T o t a l 1348 7 2 0 3 9 9 3 9 8 0

Substituting the totals of columns 2 and 4 into Eq. (8.35) yields

1348<z0 + 720 = 0

c/o = - 0.53 lb/in

The final shear flows in the web q are calculated in column 7 by adding the value of q0 to the value of q' in column 3.

The position of the shear center is now calculated from the equilibrium of torsional moments. The moment about any point can be obtained from the relation i.

T = 22Aq

or, since q = q0 + q',

T = qa"L2A + L2Aq' (8.36)

where A is the area enclosed by a web and the lines joining the endpoints of the web and the center of moments. The center of moments is taken as the lower right-hand corner of the box, and values of 2A are tabulated in column 5 and values of 2Aq' in column 6. The totals of columns 5 and 6, when substituted into Eq. (8.36), yield

•400.x = - 0 . 5 3 x 399 + 3980

x = 9.42 in

This value of x determines the horizontal location of the shear center. From symmetry, the vertical location is on the line of symmetry. For cross sections which are not symmetrical about a horizontal axis, the vertical location of the shear center can be found by considering a horizontal shear force to act on the section and then proceeding in the same manner as above to find the shear flows for a zero twist. The location of the resultant of these shear flows, obtained by equating torsional moments, gives the vertical posi-tion of the shear center.

8.10 WARPING OF BEAM CROSS SECTIONS

When a rectangular box beam is subjected to torsional moments, it deforms as shown in Fig. 8.31. If the cross section is square and the web thickness is the same on all sides, the cross sections will remain plane after the box is twisted. Similarly, if lhc box beam is subjected to bending with no torsion, the plane cross sections will remain plane after bending. In the usual case, however, the box is rectangular and resists some torsion; therefore the cross sections do not remain plane, but warp. In the analysis of box beams in Chap. 5, we assume that torsional moments do not aflcct the distribution of bending stresses, or that cross sections arc not restrained against warping. The shear flows computed from these assumptions are accurate for all cross sections excepl those which are very close to a fixed cross section.

ANALYSIS Ol' TYl 'K'AL MEMBERS Ol-' SKMIMONOCOQUE STRUCTUKUS 2 6 7

The amount of warping of a cross section can be measured by the angle 0 between spar cross sections, as shown in Fig. 8.32a. This angle is calculated by applying unit virtual couples, as shown, and calculating the relative rotation from E<.j. (8.32). The warping of cross sections of the beam of Fig. 8.26 is calculated by assuming that the shear flows a are as shown in Fig. 8.27. A unit spanwise length of the beam is considered, as shown in Fig. 8.32. The unit couples acting on the spars are represented by forces of 0.1 at distances 10 in apart, and the values ofg„ must be 0.05 for all webs, as shown, in order to satisfy all conditions of static equilibrium. Substituting values from Figs. 8.27 and 8.32 into Eq. (8.32) yields

0 V q " l l " h

IG 0.05 x 200 x 40 x 1 x 2

0.040 x 4 x 106

+ • 0.05 x 200 x 10 x 1 0.05 x 1400 x 10 x 1

0.032 x 4 x 106 0.081 x 4 x 106 = 0.00362 rad

This warping of the cross section is the same for all cross sections on which the shear flows are as shown in Fig. 8.27. The I-in length along the span was selected arbitrarily, but any other length b might be used. The values ofg„ shown in Fig. 8.326 would be divided by an assumed length b; then the terms in the above summation would be multiplied by b instead of the unit length, in order to yield the same final result.

At the fixed support shown in Fig. 8.26, obviously the warping of the cross

>ju = 0 . 0 5

U)

< f „ = - 0 . 0 5

Figure 8.32

16.000

I.Mill </, ( I S I ' M

'/1 (81 ) F i g u r e 8 .33

section is prevented; therefore the values of q cannot be as shown in Fig. 8.27. The values o f g which are required to prevent warping are calculated now for this beam. At any cross section, the shear Rows must be as shown in Fig. 8.33 in order to satisfy conditions of equilibrium. From the equilibrium of moments about a spanwise axis through one corner, and from the equilibrium of horizontal and vertical shearing forces, the shear flows q, must be equal and in the directions shown for three webs. For the front spar web, the shear flow is 1600 — q^ Substituting the values of q from Fig. 8.33 and the values of g„ from Fig. 8.32/) into Eq. (8.32) gives

At the fixed support, 0 = 0, or qt = 81 lb/in. The final shear flows are shown in parentheses in Fig. 8.33. These are seen to be considerably different from those shown in Fig. 8.27.

The shear flows of Fig. 8.27 apply a running load of 400 lb/in to the rear spar flanges and a running load of 1200 lb/in to the front spar flanges. Ne^r the support, the shear flows apply a running load of 162 lb/in to the rear spar flanges and a funning load of 1438 lb/in to the front spar flanges. Near the support, the bending stresses arc therefore higher in the front spar than in the rear spar, the axial strains in the front spar flanges are greater, and the cross sections change from plane sections to warped cross sections. The spanwise distance required for the transition depends on the flange areas and web gages. In this problem, the shear flows at a section 30 in from (he support have approximately the values shown in Fig. 8.27. The bending stresses outboard of this cross section arc ap-proximately equal for the two spars, since all cross sections warp the same amount. The effect of a fixed cross section is to increase the bending stresses and shear flows in the loaded spar for a spanwise distance which is approximately equal to the average of the cross-sectional dimensions.

0 = 1 quqab 0 .05,1x 40 x 1 x 2 0.05, t x 10 x 1

tG ~ 0.040 x 4 x 10" + 0.032 x 4 x 106 +

ANALYSIS OF TYPICAL MFMHERS OF SEM1MONOCOOUE STRUCTURES 2 6 9

Figure 8 3 4

8.11 REDUNDANCY OF BOX BEAMS

The only type of box beam which is stable and statically determinate consists of three flange areas and three webs, as shown in Fig. 8.34. In this beam, the three unknown flange forces Pt, P2, and P , and the three unknown shear flowsq^ q2, and q3 may be obtained from the six equations of static equilibrium, XFx = 0, ZF y = 0, E F . = 0, XA/V = 0, SA/j. = 0, and ZM: = 0. As in any other statically determinate structure, the internal forces are independent of the areas of stiffness properties of the members. In any statically indeterminate structure, the areas and elastic properties affect the distribution of the internal forces in the members.

The internal bending stress distribution in all common beams is statically indeterminate, and the deformations are considered in deriving the flexure for-mula a = My/I. This equation is so common that it is not customary to think of such beams as statically indeterminate. The bending stress distribution in a box beam containing more than three flanges, such as that shown in Fig. 8.35, de-pends on the area and elastic properties of the flanges. The shear-flow dis-tribution, which is derived from the bending stress distribution, also depends on the areas of the flanges and is therefore statically indeterminate.

In this and the following section, we assume that the bending stress is ob-tained by the simple formula and that a pure torsion load shown in Fig. 8.35 produces no axial stresses in the flanges. These assumptions have been used in previous shear-flow analyses and have been shown to be accurate in most cases. The shear-flow distribution in a single-cell box then can be obtained from the conditions of statics, and such a box is considered as statically determinate for shear-flow calculations. With the assumption that a torsional moment produces no axial stresses in the flanges, the equation q = T/{2A) for the shear flows is obtained from the conditions of statics.

A two-cell box such as that shown in Fig. 8.36 cannot be analyzed by the equation of statics. It is assumed that the wing ribs have sufficient rigidity that the two cells deflect through the same angle 0. Then this deflection condition and the equations of statics are sufficient for the shear-flow analysis, and the structure has a single redundancy. It is, of course, assumed that the torsion produces no axial load in the flanges; hence the flanges are not shown in the sketch. A box structure with several cells has one less redundant than the number of cells, since webs in all but one cell may be cut to leave a single cell as a statically determi-nate base structure.

The angle of twist of a box beam was found by Eq. (8.33):

^ 2AttGi '

where the terms arc indicated in Fig. 8.35. This equation can be used for the angle of twist of a multicel! structure, if the summation is evaluated around any closed path and the area A is enclosed by this closed path. Thus, for a three-cell structure, the summation can be evaluated around the entire perimeter enclosing the three cells, the entire perimeter enclosing any one cell, or the area enclosing two cells. This procedure is sometimes defined as a line integral, as follows:

J 2At iG ;

Where the integral represents an evaluation along a closed path, returning t,a the starting point. The values of the summation or integral are considered positive in going clockwise around the enclosed areas.

8.12 TORSION OF MULTICELL BOX BEAMS

For the two-cell box of Fig. 8.36, the angle of twist 0, for cell 1 must equal the angle 02 for cell 2. A unit length L may be considered, since L is always the same for the two cells:

y </,• As, ^ y q; A j 2 ^ i f , G ; j 2 / 4 2 f ; G i

(8.37)

ANALYSIS OF TYPICAI MFMRHRS OF SFMIMONOCOQUE STRUCTURES 2 7 1

The first summation must be evaluated around the total perimeter of cell 1, including the interior web, and llie second summation must include all webs or ccll 2 and the interior web.

The value of q for any exterior web of ccll 1 is <7,,, and for the interior vvcb it is '/i, - q2<. Similarly, the value of q for an exterior web of cell 2 is q2n and for the interior web it is q2, — qu. liquation (8.37) now can be rewritten by making these substitutions, moving constant terms outside the summation signs, and assuming C to be constant in all webs:

The following abbreviations are used for the terms in Eq. (8.38):

S u - I . ^ = = (8-39) i ' i 2 h \ I / 1 - 2

The term <5,, represents a summation around the entire perimeter of cell 1, $22 a summation around the entire perimeter of cell 2, and S i 2 the value of the interior web. The terms 5tl and <522 both include the term 5l2 for the interior web. The <5 terms do -not have quite the same significance as the similar terms used in previous structures, because the constants are eliminated for simplicity and the redundants 'arc taken as shear flows.

By utilizing Eq. (8.39), Eq. (8.38) becomes

"7" ('/IAI ~ '12,'hi) = 4 " (h^ii ~ qi,S12) (8.40) A1 A2

The equation for equilibrium of moments about a torsional axis can be found by reference to Fig. 8.36:

T = 2Alqu + 2A2q2, (8.41)

Equations (8.40) and (8.41) can be solved simultaneously for the two unknowns, qlt and q2l.

The shear flows resulting from pure torsion may be obtained in a similar manner for a box beam with n cells. From the conditions of continuity of defor-mations between adjacent cells,

0 , = 0 , = 0 j = • • • = 0„

or

T " t ' 7 > i ' , i 1 ~ < / 2 , ' 5 i 2 ) = 4 " ^2,^22 - hi <5i2 - 4 3 , < 5 2 3 ) /li A 2

= 7 " U / 3 : ^ 3 3 - <72, ^ 2 3 ~ ? 4 , ( 8 - 4 2 )

= "7" Ulnl — - 1 )i i )„) fi

2 7 2 AIRC 'RAiT STRUC TURES

The terms in Eq. (8.42) are defined as

5n« = Yj~r < 5 ™ = < L „ = a < i \ 1 / m-n and A„ = nth cell enclosed area (8.43)

The summation of Eq. (8.43) includes all webs around the circumference of the cell, and the term 5mn applies to the interior web between cells m and n. The equation for equilibrium of torsional moments is

T = 2A{qu + 2.4j q2, + 2A3 q3t + • • - + 2A„q„ (8.44)

Equations (8.42) and (8.44) form a set of n simultaneous linear algebraic equa-tions which can be solved for the n unknowns, <7,,, q2l,..., qnl.

8.13 BEAM SHEAR IN MULTICELL STRUCTURES

Box beams usually resist transverse shearing forces in addition lo the torsional moments already considered. Often it is convenient to consider the two effects separately, as a shearing force applied at the shear center and as a torsional moment about the shear center.

In a multicell box such as that shown in Fig. 8.37, the increments of flange loads AP can be calculated from the bending stresses at two cross sections or from the shear equations as used in Chap. 5 for a single-cell box. If one web is cut in each cell, the shear flows q' can be obtained from the equilibrium of spanwise forces on the stringers. The structure shown in Fig. 8.37 is unstable for torsional moments, but the system of shear flows q' will be in equilibrium with external shear forces acting at the shear center of the open section.

The shear flows qu, q2s, and q3s in the cut webs can be obtained and superimposed on the shear flows q' to give a system of shear flows which have a resultant equal to the external shearing force acting at the shear center of the closed multicell box. A superposition of the conditions shown in Figs. 8.37 and 8.38 yields the shear flows in a closed multicell box with no twist. The values of qls, q2s, and qis are found from the condition that the angles of twist Gu Qj.-, and 03 for each ceil must be zero. After q!s, q2s, and q3s are obtained, the equation of torsional moments yields the position of the shear center of the closed box. Then the external torque about the shear center can be computed and the shear flows resulting from this torque calculated by the methods of Sec. 8.11.

a n a l y s i s o f t y p i c a l m e m b e r s o f s F. m 1 m o n o c o q u e s t r u c t u r e s 2 7 3

The conditions that the angles of twist for each cell be zero yield the follow-ing equation for each cell:

^ 2AGt

Since G usually is constant and 2/1 is always constant, these terms may be canceled from the equation. Then

— q As 1 ^ = 0 ( 8 . 4 5 )

For cell one Eq. (8.45)'yields

• th As, — As, (As\ I J ~ + <?2.i — = 0 ( 8 . 4 6 )

I ' I 1 ' i \ £ / I - 2

in which summations are evaluated around the entire perimeter of the cell, in-cluding the interior web, and the last term applies to the interior web only. The terms in Eq. (8.46) may be abbreviated, and similar equations may be written for the other cells:

<>10 + < J l A l - <?2S<5l2 = 0

(520 f (Iis$12 - tfiAi - q i J i i = o (8.47)

• I j o + </-U<5JJ - qis<>ii = o

The following abbreviations are used:

„ a] As, , , a'. As, v-, q', As, ^ o = L - , T - i ^ 3 0 = I i 7 - 1 (8.48)

I [ i 1 - 3 ' i

in addition to the abbreviations given in Eq. (8.43). Equations (8.47) may now be solved simultaneously for qls, and qZi.

These equations are applicable to a two-coll structure if all terms containing the subscript 3 are dropped. Similar equations may also be written for any number of cells.

Example 8.7 Find the shear flows in the two-cell box of Fig. 8.39. The horizontal webs have gages of / = 0.040 in. Assume G is constant for all webs. The cross section is symmetrical about n horizontal centerline.'

4000 lb

( - 1 0 - 4 1 20 ^ Figure 8.39

SOLUTION The shear flows may be obtained by superposition of the values of q for the structure shown in Fig. 8.40a, and the values of <?j and q2 are shown in Fig. 8.406. The shear flows and q2 are computed as the sum of values for a load at the shear center and for a pure torsion loading. First, if we consider the shear flows qls and q2s required to produce no twist of the structure, the following equation is obtained for cell 1:

V fr 20

0.040,

+ ( < J i . - h .

+ ( ? . , - 100)1 10

0.040 + q .J 20

0.040

100) 10

0.050 0

And so

1450q ls - 200q2s - 5000 = 0

A similar equation is written for cell 2:

(a)

<7, As,-ti

10 10 \ 10 : H o ^ J + + 2 0 0 ) \ m j + 92'<ao40

+ te,-<?,-100)(^) = 0

and

(b) 825q2s - 200c,. + 5000 = 0

Fquations (o) and (/;) can now be solved simultaneously, yielding qu = and q2s= —5.4 lb/in. The minus sign indicates that the shear flow q2 is opposite to the assumed direction, or counterclockwise around the box. These values represent the shear flows for a load applied at the shear center.

q -t 10U

>li ~ <li> ' <h, <12 •" <hi ' ''21

q I U U

2.1 ' 200

l/>l Figure 8,40

a n a l y s i s oK t y l ' i c ' a i . m e m b e r s o f s e m i m o n o c o q u e s t r u c t u r e s 2 7 5

1

i l l i / i t i

(.1..! Ih,'i.i ( • ' « l!i/in Figure 8.41

The torsional moment about the reference point O of the shear flows is

!L2Aq' + 2A1qls + 2A2q2s

where Al and A2 represent the enclosed areas of the cells, and

100 x 200 - 100 x 200 + 2 x 200 x 2.7 - 2 x 100 x 5.4 = 0

The external shearing force of 4000 lb acting at the shear center produces no torsional moment about point O, or the shear center of the cross section is at point O.

The actual load of 4000 lb acting at the left-hand web has a moment arm of 10.0 in -about the shear center. The shear flows qt, and q2t must now be obtained for a pure torque of T = 4000 x 10 = 40,000 in • lb. From Eq. (8.44)

40,000 = 400(j I r + 200q2,

and from Eq. (8.42)

g,,/2 x 20 10 10 \ q2, 10 200\ 0.040 + 0.040 + 0.050/ ~ 200 0.050

q 2 J 2 x 10 10 10 \ q u 10 ~ 100V 0.040 + 0.050 + 0 .080/ 100 0.050

These two equations are solved simultaneously and yield qlt — 66.7 and <lit = 66.7 lb/in. The final shear flows in the cut webs are obtained from Eqs. (a) .and {b) as q , , = 69.4 and q2, = 61.3 lb/in. These values are now superim-posed on the values of q\ and the final shear flows are shown in Fig. 8.41.

PROBLEMS

8.1 Find Ihe shear flow in each web of Ihc beam shown in Fig. P8.1 and P8.2, and plot the

* s

j \ j \ ; \ \ S \ ! \ ! \ 1 \

* s

j \ j \ ; \ \ S \ ! \ ! \ 1 \

j .

* s

j \ j \ ; \ \ S \ ! \ ! \ 1 \

1 I I I i n

1 1 \l I N — - l O i n — | ~ - 1 0 i n — H - - 1 0 i n — -

! \ 1

J N

2 7 6 a i r c ' r a i t s t r u c t u r e s

dis t r ibut ion of axia l load a l o n g each st i ffening m e m b e r . Solve for each of Ibe following load ing condi t ions ;

(«) P , = 3000 lb P2 = = 0 (b) P, = 6000 lb ps = p3 = 0 (c) P3 = 6000 lb P2 = 6000 lb P3 = 6000 lb

8.2 Repeat P rob . 8.1 for the fol lowing load ing cond i t ions : P , =- 2400, P2 = 1200, and P3 = 18001b.

8.3 T h e pulley b racke t shown in Fig. P8.3 and P8.4 is a t t ached to \vcbs a l o n g the three sides. F ind the react ions R „ J J j . a n d R3 of the webs if P = 1000 lb and 0 — 45°.

8.4 Repeat P rob . 8.3 fo r P = 2000 !b and 0 = 60".

8.5 Find the shea r flows appl ied by the skin to the fuselage r ing s h o w n in Fig. P8.5 to P8.8 if / ' , = 2000 lb and P2 = M = 0.

8.6 Find the skin react ions on the fuselage ring in Fig. P8.S to PS.8; P2 = 1000 lb, and P , = Af = 0.

8.7 Find the skin react ions on the fuselage ring if P , = 2000 lb, P, = 1000 ib, and M = 10,000 in • lb.

8.8 Find the skin react ions on the fuselage ring if P , = 15001b. P a = 5001b, a n d M = 8000 in • lb.

8.9 Find the skin react ions on the r ib shown in Fig. P8.9 and P 8 . I 0 if t he rib is loaded by the dis t r ibuted load of 20 lb/in. Calcula te the shear f lows in the r ib web and Ihe axial loads in the rib llangcs a! vertical sect ions 10 and 20 in forward of Ihe spar .

8.1(1 Urpeal P rob . 8.9 ir the rih is loaded by a coiiccntrnU'd upward force of MK1 |h, applied nl a point 20 in forward of the spar , instead of the dis t r ibuted load.

8.11 Find (he skin react ions on the rih shown in Fig. P 8 . l l to P8.16. Analyze vertical cross sect ions a I 10-in intervals, ob ta in ing the web shear flows and the axial loads in t h e rib flanges. Assume the loads / ' , = 40.000 lb and P2 = 0. T h e spar flange areas are u = b = c = d = 1 in2 .

Figure P 8 3 and P8.4

All si ringers A. - 0.2 in-'

F igure PS.5 t o P8.8

a n a l y s i s oK t y l ' i c ' a i . m e m b e r s o f s e m i m o n o c o q u e s t r u c t u r e s 287

20 Hv'in

t t m t U H t H t

t » - h ill

=3-—-T t

20 ill — Figure P8.9 lo P8.10

Figure P8.I I to P8.16

8.12 Repeal P r o b . 8.1! Tor / ' , = 0 and 1 \ « 80001b.

8.13 Repeat P r o b . 8.11 for P , « 20,000 tb a n d P 2 =• 8000 lb.

„ . , „ s „ t „ . , f ( ) r _ 4 ( 1 0 0 0 lb and P 2 = 0 if the spa r flange areas are a = 3 a n d

8.15 Repeat P r o b . X.I I for / ' , ••= 0 and P j ^ 8000 lb ir Ihe spar l lange areas are a = 3 and /» = c = </ - I in 2 .

8.16 Repeat P rob . 8.11 for / ' , ^ 2(MKX) and P2 ^ 8000 lb if the s pa r l lange areas are a = 3 a n d b = c = d = 1 in 2 .

8.17 The s t ruc tu re of Fig. X.16 has the d imensions in inches h0 = 5, hj = 15, a = 20, length L = 100. F o r a t o rque T of 40,000 in • lb, find q 0 . q u and<;„. F i n d the axial loads in the c o m e r flanges a n d the shear flows a t a c ross sect ion 50 in f rom o n e end. Check the values by the equi l ibr ium of tors ional moments , inc luding the in-p lane c o m p o n e n t s of the f lange loads .

8.18 Repeat P rob . S . ^ i f / i , = 10 in.

8.19 Find the shear flows and the flange loads for the s t ruc tu re of Fig. 8.1S if only the horizontal load of 2000 lb is a c t i n g : - — ' ~ '

8.20 Find the shear flows and flange loads for the nacel le s t ruc tu re shown in Fig. PS.20 and P8.21.

10.00011' 10.000 Ih

\ \

\

\

\

\

\

\

\

\

•30 in-

8.21 Find the shear f lows a n d f lange loads Tor the nacelle s t ruc ture s h o w n in Fig. P8.20 and P8.2I i f a clockwise couple load of 200,000 in • lb is ac t ing in addi t ion to the loads shown .

8.22 Find the shear flows in all webs of the s t ruc tu re shown in Fig. 8.20 if a clockwise couple load of HK).(KX) in • lb is act ing in add i t ion to the loads shown.

8.23 A cantilever wing spa r is 10 in deep between centroids of the flange areas . T h e bending stresses in the flanges are 30,000 lb / in 2 , and the shear stresses in the web a re 15,000 lb/ in 2 at ail points . Find the detlection result ing f r o m shear and bend ing de fo rma t ions as well as the percentage of the deflec-tion con t r ibu ted by the shea r at (<i) 20 in f rom the fixed suppor t , (ft) 40 in f r o m the fixed suppor l , and (<•) 100 in f rom the lixed s u p p o r t . Use E = 101 and G = 3,000.000 lb / in 2 .

8.24 F ind the angle of twist of Ihe wing shown in Fig. 8.26 iTall four f langes have equal areas. Use G = 4 x 106 lb/ in2 .

8.25 Find the angle of twist of the wing shown in Fig. 8.26 if all webs have a thickness of 0.040 in. Use G = 4 x 10* lb / in 2 .

8.26 Calcula te the loca t ion or Ihe shear center , o r elastic axis, of the wing in Fig. 8.26.

8.27 Find the shear flows in the webs of the s t ructure shown in Fig. 8.39. T h e hor izonta l webs have gages of 0.064 in and G is cons t an t .

8.28 Find the shear flows in the webs of the s t ructure shown in Fig. 8.39 if all f lange a reas are 1 in2 . The hor izonta l webs have gages of 0.064 in, and G is cons tan t .

8.29 Find the shear Hows in the webs of the s t ructure shown in Fig. P8.29 lo P8.32 if all flanges have areas or I in2 and all webs have gages of 0.040 in. Assume V — 3000 lb, e -- 8 in, and G is cons tan t for all webs.

A r e a

1 0 0 i n - '

y % / /

AS - 3 2 i n

S y m n u ' t r i e : i l a b o u t

c e n t e r l i n e

•->> / -'".I

- l O i n - - 1 0 i n -

• 1 0 i n

_ L

Figure P8.29 to P8.32

8.30 Repeal P rob . 8.29 if c = 0.

8.31 Find the shear f lows in the webs of the s t ruc tu re shown in Fig. P8.29 to P8.32 if /4, = /4., = 1 in2 , A, = 2 in 2 , t , = t2 = 0.064 in, V - 4000 lb, and e = 10 in. T h e o the r webs have gages of 0.040 in, and G is constant for all webs.

8.32 Repeal Prob . 8.31, a s suming an addi t ional vertical web of0.064-in gage at f lange / I j .

8.33 Assume the box b e a m s h o w n in Fig. 6.19ti to be loaded by a tors ional couple of 160,000 in • lb at the free end instead of the vertical load shown . Assume the cross section t o be symmetrical about a hor izonta l centerl ine and all web gages to be ( = 0.020 in. Calcula te Ihe w a r p i n g displacements or a cross section which is free 10 warp , and calcula te the axial llange loads a n d web shear flows at the wall and at cross sect ions at 10-in intervals a long the span . No te that the s t r inger of 2-in2 area resists no load and does no t a l lec t the analysis. Assume E = 10" lb / in 2 and G - 0 . 4 £ .

C H A P T E R

NINE THERMAL STRESSES

9.1 INTRODUCTION

Temperature changes in structural system components are accompanied by a change in length which results in what is called thermal stresses. The subject of thermal stress analysis encompasses a wide range of structural systems and com-ponents, with applications in airframe vehicle structures, nuclear reactors, jet and rocket engines, oil refining lines, and some civil engineering structures. In this chapter, we introduce thermal stress analysis.

9.2 THERMAL STRESS PROBLEM: PHILOSOPHY

A temperature change in a given solid material causes fibers to expand and contract in different amounts. For the solid to remain continuous, a system of thermal strains and corresponding thermal stresses may be induced, depending on the characteristics of the solid and its temperature distribution. A homoge-neous solid with no physical external restraints is free of thermal stresses if the temperature distribution is uniform throughout the solid. This condition is re-ferred to as free expansion or contraction of the solid. Upon imposing external restraints, the free expansion or contraction is prevented, and thus thermal stresses arc introduced to the body. To illustrate, consider the beam of Fig. 9.1 to be heated uniformly from a datum temperature T0 to a final temperature T degrees Fahrenheit. The elongation of the beam due to the temperature change

279

aj:,cc

Figu re 9.1

T — T0 is given by

Sb = a(T - T0)L (9.1)

where a = material thermal coefficient of expansion X, in • °F L = beam length, in

T, T0 = final and da tum temperatures, respectively, °F

The elongation of a beam due to a uniform tensile stress a is

>.-•-£ M From compatibility conditions, the total deformations given by Eqs. (9.1) and (9.2) must equal the connecting spring deformation <>b:

a(T~T0)L + ~ = Ss (9.3)

From equilibrium conditions, it may be seen easily that

Fh + Fs = a A + K5b = 0 (9.4)

where Fb = beam internal force = a A and Fs = spring force = K5S. Thus

W )

Upon substituting Eq. (9.5) into Eq. (9.3) and solving for the stress a, the follow-ing is obtained:

a (T - T0)L

711 (9.6)

Ljli + A/K

aKEUJ - Ta) KL + AE

By examining Eq. (9.6), it can be shown that if K = 0, which is equivalent to the free expansion of the beam, the thermal stress is zero. In the case where K = oo, which is equivalent to having the beam supported between two rigid walls, the thermal stress is

a = —aE(T — T0) (9.7)

In general, Eq. (9.6) may be written as

a = -R,zE(T - T0) (9.8)

t h e r m a l s t r e s s e s 2 8 1

where RS is a nondimensional restraint coefficient whose value ranges between 0 and 1. For R„ — 0, the beam is completely free to expand, whereas for RS = 1, the beam is completely prevented from expansion.

From the preceding illustration, it may be concluded that the formulation of the thermal stress problem is identical lo (hat of the isothermal stress problem in that it requires consideration of the following conditions:

1. Equilibrium of forces 2. Compatibility of deformations 3. Stress-strain and strain-displacement relationships 4. Boundary conditions

9.3 F O R M U L A T I O N O F E Q U A T I O N S F O R T H E R M A L STRESS A N A L Y S I S

Most structural problems encountered in engineering are commonly in a three-dimensional stale of stress. Quite often, three-dimensional problems are difficult to solve and hence through valid simplifying assumptions are normally reduced to two- or one-dimensional problems. For instance, in the case of a plate struc-ture, if we assume that the thickness is small compared to its other dimensions, then the stresses through the thickness direction usually are neglected, thus re-ducing a three-dimensional problem to a two-dimensional one. In the case of a beam, all stresses are neglected but one normal stress and one shearing stress, thus reducing the problem to a one-dimensional one. Therefore, in the rest of this book, all formulations of equations are in one or two dimensions.

Equilibrium Equations

The equilibrium equations of a two-dimensional solid shown in Fig. 9.2 are derived in Chap. 3:

"xx, X + "xy, y + X = 0 (9.9)

(9.10)

where a x x , ayy = normal stresses ax y — shearing stress

A", Y = body forces

In polar coordinates, Eq. (9.10) becomes

"rr — "w r

+ R = 0 (9.11)

"oo. a r

where the stresses and body forces are shown in Fig. 9.3.

2 8 2 a f r c r a f t s t r u c t u r e s

CT + (7rv x<i.V ".xv + "vi". x'ix

Oxv + f / v l . , flv

Figure 9.2

Strain-Displacement Relationships

If we consider small deformations, the strain-displacement relationships are

• = qx% , f„, = qy_ y exy = qx_ v + qy% x (9.12)

or, in polar coordinates, _ „ <?r . 5o. o

rr ~ Qr.r ^110 = I r r

(Jr. 0 , <?0 ert> = + 9(l . r

r r

where q = generalized displacement function and e denotes strain components.

Thermoclastic Strain-Stress Relationships

The thermoelastic strain-stress relationships for plane stress problems are ex-pressed as follows:

= g frxx - VCyy) + aT

= ^ - VFFJCJ + A T (9.14)

Figure 9.3

where a = coefficient of thermal expansion and T = temperature above the datum temperature. For plane strain problems Eq. (9.14) becomes

« « = ( V „ - y y - ^ + a n 1 + v)

1 - v 2 / v — K - — v = — r ~ I a r r - : J + a T ( l + V) ( 9 . 1 5 )

In polar coordinates, Eqs. (9.14) and (9.15) become, respectively,

1 err = T. ( < V ~ ™0a) + ^ T h

1 emi = y ( "ou — v<!„) + a ! p l a n e s t r e s s ( 9 . 1 6 )

Crf ~ G

1 - v 2 ( v = — j - ( arr - j—- am j + <xT( 1 + v)

I - v 2 / V \

e<w = — y — I - (TrrJ + aT( 1 + v) plane strain (9.17)

G

Compatibility Equations

The compatibility equations can be expressed in terms of strains alone:

or, in polar coordinates,

(9.18)

, £ r r . 80 , 2 e o f l t , Srr r g,g. rg C f 9 g e «0 . rr + ~ 2 r ~ = 1 2 l*" ' r r r r r

In terms of stresses alone for plane stress problems, Eq. (9.18) becomes

or, for plane strain problems,

8 d \ f a E T \ 1 (ex 8 Y \

Boundary Conditions

For a specific thermal stress problem, the applied surface loads Sx and Sy must be in equilibrium with the induced stresses at the boundaries of the solid:

where axx, axy and oxy arc stresses at the boundary surfaces and I and m are direction cosines. ~


9.4 S O L U T I O N M E T H O D S FOR T H E R M O E L A S T I C P R O B L E M S

In general, two-dimensional thermoelastic problems involve six unknowns of stresses and strains. The method of solution quite often is dependent on the type of structure under consideration. The following are some typical techniques used in the solution of thermoelastic problems.

Direct Solution Using Equilibrium and Compatibility Conditions

A number of simple thermal stress problems can be solved by merely seeking a solution for the differential equations which describe them. Consider, for exam-ple, a circular solid plate having uniform thickness t and subjected to a temper-ature distribution T = T{r). Equations (9.11) and (9.19) become (note that the stresses and strains are independent of 0)

Sx = axx I + ahxym

(9.22) Sy = (Tyy in + a%l

Sr = <jhrr I + <Jh

r0 m

(9.23) S0e = Coo in + a%'

dlrr "rr ~

dr r (924a)

e,w -€rr + r — = 0 (9.24/))

Equation (9.24h) may be expressed in terms of stresses as

<1(7,10 _ i d r r r l T ( 1 -| v)(fT„ - (T„g)

dr dr dr r

Solving Eqs. (9.24a) and (9.24//) yields

= 0 (9.24//)

(9.25)

THERMAL STRESSES 2 8 5

A direct integration of Eq. (9.24) yields

aE Q Tr dr + Ct +-£•

o r (9.26)

For the solid plate shown in Fig. 9.4, r = 0 at the ccntcr; hence Eq. (9.26) becomes undefined because of the term C2/r2. To render a feasible solution, C2

must be set to zero; thus Eq. (9.26) becomes

aE o , r = - - Tr dr + Ct (9.27)

F r o m boundary conditions a„ — 0 at r = R0. Hence the constant of inte-gration C j is

aE

R2o

Ro Tr dr

and f7„ becomes

and

aE _2

f7„„ = — 7. ET f

aE

R2oJ

«0 Trdr Tr dr +

•o

a £ ( r „ , a E r " f r + 5 f J

Ro Tr dr (9.28)

For a hollow circular plate Eq. (9.28) becomes

aE ( r - Rf Or, '

r2 \R2O - Rf J Ro

Tr dr - Tr dr Ri

000 aE ( r2 + Rf

Ro-Rf J

Ho Tr dr + Tr dr - Tr1 (9.29)

<rr0 = 0

Equations (9.26) and (9.27) are cases of generalized plane stress or plane strain • problems (body forces are zero) and thus apply equally to unrestrained open-end

cylindrical shells.

Stress Function Solution

If a stress function O = (.r, v) is chosen such that the stresses arc defined by

d2<t> d 2 0 > 0 2 d> =

c V >T a . x 2 Oxdy ( 9 . 3 0 )

then (he equilibrium equations, Eqs. (9.9) and (9.10), are satisfied with the body forces set equal to zero. If <1> has to describe the true stress field, it must satisfy not only equilibrium but also compatibility and boundary conditions. Thus, substituting Eq. (9.30) into JEq. (9.20) for the plane stress problem yields

d40 „ d4<!) fa2T d2T. ~ + 2 2 - , 2 + T 7 + c f E T T + T T = 0 ( 9 - 3 1 ) Ox4 dx dy dy \ dx By2 J


L l i . l i l V ^ 2 ^ i a 2 \ 1 + r , V + r J 0 0 2 J W + ~ r J r + 7 2 W 2 r

(d2 I d 1 d2\ + + + r = 0 ( 9 - 3 2 )

where the stresses are defined by

1 a2

~ r dr + r2 001

£?2C[) = T J C -33) or

d \ <1<1> ar0 ~ ~ dr\r d0

For a given specific thermal stress problem, Eq. (9.31) or (9.32) can be solved by using techniques of solving differential equations.

Equivalent Load Solution

Thermal stress problems may be handled in the same manner as isothermal problems if the actual thermal loads are converted to what is called equivalent static loads. In two-dimensional problems, the conversion is accomplished by making the following substitution:

Ecr.T = ff« + 7 3 7

where a x x , a v y , ff(v = equivalent stresses (displacement stresses)

a x x , <ryy, frvr = actual thermal stresses

E'xT = local uniform pressure counterbalancing frcc-cxpansion

1 — v , . s t r a i n

Equation (9.34) may be written as

EOLT °xx — axx ] _ J)

Ea T

Substituting Eq. (9.35) into Eqs. (9.9) and (9.10) and assuming that body forces A' and Y do not exist yield

axx. x + ffxy. y + X = 0

rr,.,. ,, + <Txy , -l- y = 0

where .V and Y arc equivalent body forces defined as

Eac dT

(9.36)

X = -

y .

1 - V dx

Ea dT 1 — v dy

(9.37)

If we assume the applied surface tractions are zero, then the boundary condi-tions, Eq. (9.22), become

\ = < 4 ' + S%m

S y = cryy m + a x y I

where Sx and S arc equivalent surface tractions defined by

Ey.T ,

Ey.T Sy = m f 1 A*

(9.38)

(9.39)

In polar coordinates, Eqs. (9.35) to (9.39) become

EOLT "rr = "rr ~

1 — V

EuT "oo = "00 - y — ^ ( 9 . 4 0 )

"rO = "rtl

- , "rO.Q , rr "OO . n n "rr r H 1 h K = 0 r r

"eo.o , - , 2"rg = n + + © = 0 r r

(9.41)

where

R =

and

EaT dr 1 - v dr

(9.42) E * T d T

I - v 80

Sr = ah„l + d*„m

(9.43) Se = "m "i + "rol

1 ~ v (9.44) ^ E<xT S „ = m

1 — v

By examining Eqs. (9.36), (9.38), (9.41), and (9.43), it is apparent that for any given temperature distribution, with the use of Eqs. (9.37), (9.39), (9.42), and,-(9.44) the thermal stress problem can be converted to an equivalent isothermal stress problem with conventional loads. Thus, solutions of thermal stress problems become identical to those of the isothermal ones. As an illustration, consider the unrestrained beam of Fig. 9.5 to be heated to a uniform temperature T0. The surface tractions at the ends of the beam, if we consider axial direction only, arc <xET0. as shown.

These surface tractions will induce a constant equivalent axial stress — aET0 throughout the beam. Hence from Eq. (9.35) for the one-dimensional prob-lem, the thermal stress is

(Trv = "xx — Ey.T = aET0 — aET0 = 0

which is lhc same result as was obtained in See. 9.2.

- cv/-.T„ a : / : T „ = ? '

Figure 9.5

Now let us assume that the beam of Fig. 9.5 is heated according to the following temperature distribution:

(<0 r = 7 H 5 ;

The surface tractions at the ends will now take the form = «ET0 (y/0)3, as shown in Fig. 9.6,

These surface tractions produce equivalent end bending moments:

M =

bEaTn , , 3EaT01 - r y i y — z - Cb)

where b = beam width and I = moment of inertia of beam area. From beam theory, the equivalent stress due to the equivalent moment M is

Afr 3EOLT0 ff- = T " I T -

Thus, the actual thermal stress in the beam is

(c)

(Txx = (Txx — CiET = CCETq 5 c Vc

A 3 "

9.5 T H E R M A L STRESSES IN U N R E S T R A I N E D B E A M S W I T H T E M P E R A T U R E V A R I A T I O N T H R O U G H T H E D E P T H O N L Y

Consider the beam of Fig. 9.7 to be subjected to a temperature gradient T = T(v). From elementary beam theory it may be concluded that

(Tyy = ff„ = 0X. = ay: = 0

and ".XX = AX .>•(-»')

i

Figure 9.6


Figure 9.7

By assuming body forces are zero, it can be seen from Eqs. (9.9) and (9.10) that equilibrium is identically satisfied. The compatibility equation [Eq. (9.20)] becomes

A2

dy J (CRXX + CTET) = 0 ( 9 . 4 5 )

Therefore, the thermal stress can be obtained by simply integrating Eq. (9.45):

AXX = — TUET + ATY + A2 ( 9 . 4 6 )

The constants of integration At and Az may be chosen such that for any temper-ature gradient T(r), the resultant force and moment induccd by a x x are zero over the ends'of the beam:

axx dA = axx y dA = 0 (9.47)

Utilizing Eq. (9.46) in Eq. (9.47) and solving for the constants yield the following expression for the thermal stress in the beam:

a x x — — a ET + aE 2c

3a Ey 17" Ty dy (9.48)

Equation (9.48) may be written in a more conventional form as

r-r*. ST M r

where

ST = a E T dA M T = otE Ty dA

(9.49)

(9.50)

with A and / being the cross-scctional area and moment of inertia of the beam, respectively.

For beams of arbitrary cross section, Eq. (9.49) becomes

SR /)• A'ff- — Ir. M \ I M?R — /,.. My-+ 7 + + (9.51)


where

Sr = aE T dA M?r = aE Tz dA MZT = aE | Ty dA (9.52) 1 JA

Here Iy and I . arc moments of inertia about the y and z axes, respectively, and Iy . is the product moment of inertia.

As an illustration, consider the beam shown in Fig. 9.6, where T = T0(Y/C)3. From Eq. (9.50),

Substituting the above expressions for S r and MT into Eq. (9.49) yields the beam thermal stress:

which is the same as obtained previously by using the equivalent load method. Thus, F.qs. (9.49) and (9.51) represent the equivalent load solution for beam problems.

The thermal stress of built-up structures can be analyzed easily by using the equivalent load method. Techniques such as the stiffness matrix method, energy methods, etc. that arc used in the analysis of isothermal problems can be used identically for thermal problems once the equivalent thermal loads are calculated. To illustrate, consider the determinate truss structure shown in Fig. 9.8. Assume that member I is heated to a uniform temperature T, above a certain datum

9,6 T H E R M A L STRESSES IN B U I L T - U P S T R U C T U R E S

Figure 9.6

temperature. Also assume that all members are of the same material and each has a cross-sectional area A. The equivalent thermal load caused by the temperature change in member 1 is AEOLT and is shown in Fig. 9.8. Using the equations of static equilibrium in conjunction with the joint method yields the forces in mem-bers 1 and 2:

F2 = 0 F1 = AECCT

Hence from F.q. (9.35) for the one-dimensional problem, the (hernial stresses

Ft „ AEctT ff! = - aET| = — aET = 0

A A

Fi 0 <72 = - aET2 = - - «£(0) = 0 A A

It is appropriate to state at this point that determinate truss structures in which any member undergoes a uniform change in temperature are stress-free if we assume no conventional loads exist.

To illustrate the use of the stiffness matrix method in the solution of thermal problems, consider the indeterminate truss shown in Figure 9.9. Assume that member 1 has an area A and is heated to a constant temperature T. The area of members 2 and 3 is y/2A each. From Chap. 6 the element stiffness matrix relationships are:

Element 1:

0 symmetric ' f t = 0 F\ AE 0 1 5 - 1 = 0

(fl) FL L 0 0 0 K

(fl)

0 - 1 0 1 SI _

Element 2:

FX2 0.5

F>2 AE 0.5 L - 0 . 5

- 0 . 5

0.5 -0.5 -0.5

0.5 0.5

symmetric

0.5 S3 — 0 5 5 = 0

Figure 9.9

t h h u m a i . s t r l i s s i - s 2 9 3

Element 3:

F\ 0.5 F\ AE - 0 . 5 0.5 FJ L - 0 . 5 0.5 0.5

. n . 0.5 - 0 . 5 - 0 . 5

symmetric

0.5

[<5-2 1

<5i <55 = 0

<5£ = 0

(c)

Tlie overall reduced matrix can be obtained easily, as was done in Chap. 6, and is given by

(d) 0 AE "l.O 0 V

AEaT L 0 2.0 A Inverting and solving for the unknown thermal displacements yield

~sr I.

~ AE

1 .0 0 0 ~ 0

Si I.

~ AE 0 0.5 AEaT — aLT

_ 2 _

The equivalent thermal loads on each element may be obtained from Eqs. (a) to (c):

Element 1:

F\ = F? = 0

AEaT

or

Element

F"i= — Fy2 — — •

_ _ _ AEaT f - \ = C(f ' I)2 + ( P ; ) 2 ] " 2 = — T — ( t e n s i o n )

or

Element 3:

F * = - Fi =

P 2

Fz AEaT

AEaT 4

AEaT

2j2

F% •F i = -

4

(tension)

AEaT

AEzT

4

AEaT

or F 3 = • 2J2

4

(tension)

B E


Hcnce the actual thermal stress in each rod is

r T A E a T r-r E a T

Pi P_ EaAT E*T a2 = ——— — a t ' 2 = — T = — 7 = — — U =

V2/1 IsJl-JlA 4 " 3 = f r 2

The preceding problem can be solved by using the energy method. This may be accomplished by making the structure redundant, as shown in Fig. 9.10. For compatibility of deformation, the relative displacement in the direction of must be zero. Thus from Castigliano's theorem,

5U

where U is the total strain energy stored in the structure and Rj is the unknown equivalent internal load in member 1.

From the equations of static equilibrium in conjunction with the joint method, the equivalent internal loads in members 2 and 3 can be calculatcd:

- - EaAT-R,

Therefore, the strain energy U is

n _ i v ! ( ^ A T - R . V r-2 , = i A;Ei ~ 2 AE 2y/2AE \ ^ 2

1 (E*AT-Rty F

+ 2v/2 AE\ 72 /

Performing the differentiation with respect to Rj and setting the result equal to zero yield

EaAT _ Ri = - j - = Ft

J* - _ EaAT F 2 = F 3 = 7 -

2^2

. I /:'o r

v ©

© Figure 9.11)


or

EaT 2

Ex AT ffZ 2^/2 J2 A

which arc the same results obtained by using the stiffness matrix technique. As may be seen from the two preceding problems, the advantage of the stiffness matrix method is that it yields not only the internal equivalent thermal loads but also the actual thermal deflections.

Rigid-frame structures under the action of thermal loads are analyzed in the same manner as truss structures. First the equivalent thermal loads arc calcu-lated, and then any convenient technique employed in conventionally loaded structures can be used.

9.1 Find Ihc radiii] d isp lacement of a thin, c i rcular , solid plate which is subjected to a t empe ra tu r e dis t r ibut ion V =• 7'(r).

9.2 Find the radial d isp lacement of a circular cyl inder whose ends are held between two rigid walls. Assume a t empe ra tu r e d is t r ibu t ion 7' — 7 (r).

9.3 Find the the rmal stresses in s tr ips I and 2 of the compos i t e beam shown in Fig. P9.3. Assume uniform t empera tu re s T, a n d 7 \ for strips 1 and 2, respectively. Also a s s u m e tha t no slippage takes place a t the bond line.

9.4 Find the thermal stresses in the idcali/cd s t ruc tu re s h o w n in Fig. P9.4. Assume the skin and the str inger to be hea ted to un i fo rm tempera tures Ts a n d Tb, respectively.

P R O B L E M S

Figure P9 .3

] Skin


9.5 T h e inner a n d o u t e r surface t empera tu res of a circular cylindrical shell (inner rad ius = R, and ou te r rad ius = Rc) a r e k e p t cons t an t a t T, and T„, respectively. T h e end surfaces a re perfectly insula-ted. Find the s t eady-s ta te t e m p e r a t u r e d is t r ibut ion in the cylinder, a n d p lo t the results. Assume the d a t u m t empe ra tu r e to be 7^ .

Hint: See G a t e w o o d , Jr . , Aero. Sri., vol. 21, no . 9, 1954, pp. 645- 646 .

d*T 1 dT „

9.6 Refer to P r o b . 9.5. F i n d the stresses in the cylinder. Assume unres t ra ined condi t ions . P lo t the results.

9.7 F o r the riveted s t r u c t u r e s h o w n in Fig. P9.7, find the loads on each r ivet . T h e critical rivet load

© - C 7 "

- 2 0 i n - - 4 0 i n - - 2 0 i n -Figure P9 .7

and deflection a re 1200 lb and 0.006 in, respectively. Assume s teady-s ta te t empera tu re d is t r ibut ion. The fol lowing d a t a a r e g iven :

£ , = E 2 = I0 1 lb / in 2

2A, = A2 — 2.0 i n 2

a , = ot2 = 10~ 7 in/(in • °F)

2T, T j = 400°F above d a t u m t e m p e r a t u r e

T0 = 8 0 ° F

Him: See G a t e w o o d , Jr . . Arm. Sci., vol. 21, no. 9, 1954, pp . 645-646.

9.8 T h e general equa t ion which describes the general s ta te of stress in a one-dimensional thermally

loaded beam is

oxx = « £ [ - T(y) + Ay + B]

where A and B a r e a rb i t r a ry cons t an t s which depend on the end cond i t i ons of the beam. Fjn'd the stress in an unres t ra ined b e a m whose cross section is shown in Fig. P9.8 a n d subjected t o a temper-a tu re d is t r ibu t ion given by T = T0ye~'*.

Figure P9.8

THERMAL STRESSES 2 9 7

9.9 In P r o b . 9.8, cons ide r a b e a m whose cross section is shown in Fig . P9.9 . T h e skin is m a d e ou t of a l u m i n u m and is at a c o n s t a n t t empera tu re 7 j , while the web and f lange a r e m a d e ou t of different mater ia l and at a c ons t a n t t empera tu re 7^., F ind the stresses in the skin, web , and flange.

(«) Assume that the beam is completely unres t ra ined. (b) A s s u m e tha t the beam is restrained in axial compress ion .

Skin j .'1 > = web ;irra~]

- t * u . - web area

A J = flange area Figure P9 .9

9.10 T h e t e m p e r a t u r e of a b e a m of rec tangular cross sect ion, as s h o w n in Fig. P9.10, is T = Tae'-X+M. Us ing t h e Airy stress function iji = i'(x)fi(y), find the stress d is t r ibut ion t h r o u g h o u t the beam. Assume the b e a m is unres t ra ined.

21., A", / Figure P9.10

9.11 Design the t russ s t ruc tu re shown in Fig. P 9 . l l . Assume 2024-T42 a luminum-al loy tub ing con-struct ion is used.

20.000 Ih Figure 1*9.11

9.12 Find Ihe thermal stresses a n d deflections for each of the s t ruc tu res s h o w n in Fig. P9.12.

100 j„ 45

i l l 45°

©

A, = , t j = / I , = 2 in 2

= t '2 = E} = 107 l b / in 2

Zi = a 2 = <xi = 10~ 7 in/in "F

r , = 100°F, Tz = 200"F

T } = 400"F

Figure P9.12

A | = A2 — 4 in2

/ , = / j = 10 in 4

Et = E2 = 10 1 lb / in 2

a , = « 2 = 1 0 " 7 in/in " F

T, = 400 " F

T2 = 200 T

9.13 Find Ihe d isplacements for the s t ruc ture shown in Fig. P9.13. A s s u m e the beam is subjected lo a t empera tu re d is t r ibu t ion given by T = 400j-//i, as shown. Assume E = 107 lb/in2 and a = 10~ 7

in/(in • "I r) for both rod and b e a m .

Figure P9.13

C H A P T E R

TEN DESIGN OF MEMBERS IN TENSION,

BENDING, OR TORSION

10.1 T E N S I O N M E M B E R S

Tension members are analyzed and designed more readily than other types of members. The stress conditions existing in tension members at the ultimate load condition are accurately known and are not subject to the uncertainties which exist in joints, fittings, and other types of structural members. The allowable tensile stress for a structural material is easy to determine, and a single value of the allowable stress applies to members of any shape. It is shown that the allow-able stresses for structural members in bending, torsion, or compression depend on the shapes of the members and on other factors which are not considered for tension members.

For a conccntric tension load P on a member with a net area A, the tensile stress is found from the equation a, — P/A. The allowable tension stress rr,„ is the minimum guaranteed value for the material. The margin of safety may be calcu-latcd in the usual manner as a j u , — 1.

Tension members frequently must resist bending and compression stresses under other loading conditions, and these other conditions often determine the shape of a member, even when the tension load is the largest load.

The primary tension structure in a semimonocoque wing consists of the skin, stringers, and spar caps on the under surface. Although the positive bending moments in a wing arc about twice as large as the negative bending moments, the compression loads from negative bending determine the design of most of the

299

7 1

V

" 0 <'l <'2 ''.1 <'•>

Figure 10.1

structure on the underside of (he wing. The wing skin resists tension stress, but buckles and becomes ineffective for compression stress. Thus the compressive area is less than the tensile area, and the allowable compressive stress is consider-ably less than the allowable tensile stress. Even though the compressive loads are smaller, they must always be considered when the shape of the stiffening members is determined and frequently they determine the required areas.

In the previous calculations of bending stresses, we assume that the stresses are below the elastic limit. In most types of machine design and structural design, the strength at the yield stress is the important criterion for design, and the conven-tional elastic stress distribution is satisfactory for use in design, fn airframe structures, however, the ultimate strength of a member is the design criterion. Before failure, the stress exceeds the elastic limit and is said to be in the plastic range. The assumptions used in deriving the flexure formula <rb = My/I no longer apply.

The initially straight beam shown in Fig. 10-la has bending stresses exceed-ing the elastic limit. Plane sections remain plane after bending, and thus the strain distribution is proportional to the distance from the neutral axis, as in elastic beams. If the beam is deflected so that the extreme fiber has a strain e 4 , there will be a stress cr4 at this point, as shown by the stress-strain curve of Fig. 10.1b. For other strains eu and the corresponding stresses o^, tr2 ,«^and tr3

do not vary linearly with the strains above the elastic limit, and the stress dis-tribution on the beam cross section varies as shown in Fig. 10.2.

The ultimate resisting moment of a beam depends on both the shape of the

10.2 PLASTIC B E N D I N G

••a Figure 10.2

PJISIGN o r MHMHHRS IN TENSION, RENDING, OR TORSION 301

b e a m c r o s s s e c t i o n a n d t h e s h a p e o f t h e s t r e s s - s t r a i n c u r v e . S i n c e t h e r e i s n o s i m p l e t h e o r e t i c a l r e l a t i o n s h i p w h i c h a p p l i e s t o a g e n e r a l c a s e , f r e q u e n t l y a n e m p i r i c a l m e t h o d i s u s e d t o d e t e r m i n e t h e u l t i m a t e b e n d i n g s t r e n g t h . A fictitious s t r e s s ffb, t e r m e d t h e bending modulus of rupture, i s d e f i n e d b y t h e e q u a t i o n a h = M c / I , w h e r e M is t h e u l t i m a t e b e n d i n g m o m e n t , a s d e t e r m i n e d f r o m t e s t s o f s i m i l a r b e a m s , 1 i s t h e m o m e n t o f i n e r t i a o f t h e c r o s s s e c t i o n , a n d c is t h e d i s t a n c e f r o m t h e b e a m n e u t r a l a x i s t o t h e e x t r e m e fiber. T h e t r u e s t r e s s d i s t r i b u t i o n i s s h o w n i n F i g . 10 .2 , a n d t h e fictitious s t r a i g h t - l i n e s t r e s s d i s t r i b u t i o n w h i c h y i e l d s a n e q u a l b e n d i n g m o m e n t a n d h a s a m a x i m u m v a l u e o f <rB i s s h o w n b y t h e d o t t e d l i n e . F o r g e o m e t r i c a l l y s i m i l a r s e c t i o n s s u c h a s r o u n d t u b e s w i t h t h e s a m e r a t i o o f o u t s i d e d i a m e t e r t o w a l l t h i c k n e s s D / t , t h e b e n d i n g m o d u l u s o f r u p t u r e m a y b e f o u n d f o r a n y m a t e r i a l b y m e a n s o f t e s t s .

T h e b e n d i n g m o d u l u s o f r u p t u r e f o r r o u n d t u b e s o f c h r o m e - m o l y b d e n u m s t e e l is s h o w n i n F i g . 10 .3 f o r v a r i o u s v a l u e s o f D / t a n d o f e r , „ , t h e u l t i m a t e t e n s i l e

i>:t

Figure 103

3 0 2 A I R C ' R A i T STRUC TURES

s t r e s s t o w h i c h t h e m a t e r i a ! is h e a t - t r e a t e d . F o r t h e l a r g e r v a l u e s o f D / t , t h e t u b e

w a l l s a r e t h i n a n d t e n d t o c r i p p l e l o c a l l y . T h e l o c a l c r i p p l i n g s t r e s s o f t h e t u b e

w a l l , w h i c h i n i t s e l f is d i f f i c u l t t o c o m p u t e t h e o r e t i c a l l y , c o r r e s p o n d s t o t h e s t r e s s

<jA s h o w n i n F i g . 1 0 . 1 . T h u s t h e t e s t s t a k e i n t o c o n s i d e r a t i o n t h e e f f e c t s o f l o c a l

c r i p p l i n g , a s w e l l a s t h e e f f e c t s o f t h e s h a p e o f c r o s s s e c t i o n a n d o f t h e s t r e s s -

s t r a i n c u r v e . T h e b e n d i n g m o d u l u s o f r u p t u r e i s p r o p o r t i o n a l t o t h e b e n d i n g

m o m e n t , a n d t h e m a r g i n o f s a f e t y m a y b e c o m p u t e d f r o m t h e u s u a l r e l a t i o n ,

( I o / ( T b ~ T h e t r u e m a x i m u m s t r e s s cr 4 o f F i g . 1 0 . 1 i s n o t p r o p o r t i o n a l l o t h e

b e n d i n g m o m e n t a n d c a n n o t b e u s e d i n o b t a i n i n g t h e m a r g i n o f s a f e t y .

E x a m p l e 1 0 . 1 A l j b y 0 . 0 S 3 - i n s t e e l t u b e r e s i s t s a b e n d i n g m o m e n t o f 2 5 , 0 0 0

i n • l b . W h a t i s t h e m a r g i n o f s a f e t y if t h e m a t e r i a l i s h e a t - t r e a t e d t o a n

u l t i m a t e t e n s i l e s t r e s s <?,„ o f 1 8 0 , 0 0 0 l b / i n 2 ?

S O L U T I O N T h e p r o p e r t i e s o f a 1- | b y 0 . 0 8 3 - i n s t e e l t u b e a r e D / t = 1 8 . 0 8 a n d

I/c = 0 . 1 2 4 1 i n 3 . F r o m F i g . 1 0 . 3 , aB = 2 2 0 , 0 0 0 l b / i n 2 . T h e f i c t i t i o u s b e n d i n g

s t r e s s o h is o b t a i n e d f r o m t h e s i m p l e flexure f o r m u l a :

M c 2 5 , 0 0 0 , <7" = ~ = ^ = 2 ° 1 ' 0 0 0 ! b / i n

T h e m a r g i n o f s a f e l y ( M S ) i s n o w o b t a i n e d i n t h e u s u a l m a n n e r :

, „ „ 2 2 0 , 0 0 0 , _ M S = - 1 = 0 . 0 9

201,000

I t i s a l s o n e c e s s a r y t o d e t e r m i n e a y i e l d m a r g i n o f s a f e t y . F o r t h i s m a t e r i a l ,

t h e y i e l d s t r e s s i s 1 6 5 , 0 0 0 l b / i n 2 . T h e a p p l i e d o r l i m i t b e n d i n g m o m e n t is

f x 2 5 , 0 0 0 = 1 6 , 6 7 0 i n • l b a n d s o

5 1 5 ? ' , b > " ' '

o r t h e m a r g i n o f s a f e l y f o r y i e l d i n g i s

= 1 = 0 . 2 3 1 3 4 , 0 0 0 ^

10.3 CONSTANT BENDING STRESS

F o r s o m e m a t e r i a l s , t h e s t r e s s - s t r a i n c u r v e r e m a i n s a l m o s t h o r i z o n t a l a f t e r t h e

e l o n g a t i o n e x c c c d s a v a l u e c o r r e s p o n d i n g t o t h e y i e l d p o i n t . I f a b e a m o f s u c h a

m a t e r i a l i s s u b j e c t e d t o b e n d i n g b e y o n d t h e y i e l d s t r e s s , t h e b e n d i n g s t r e s s e s w i l l

a p p r o x i m a t e t h o s e s h o w n i n F i g . 1 0 . 4 . B o t h t h e t e n s i o n a n d t h e c o m p r e s s i o n

s t r e s s e s m a y b e a s s u m e d t o h a v e c o n s t a n t v a l u e s o f <Tn o v e r t h e e n t i r e a r e a . T h e

b e n d i n g m o m e n t is o b t a i n e d b y t a k i n g t h e s u m o f t h e m o m e n t s o f i n f i n i t e s i m a l

f o r c e s ov, d A a b o u t t h e n e u t r a l a x i s :

Df.SIGN OF Mi'MRf.rS IN TENSION, BENDING, OR TORSION 3 0 3

It

Kijjurf 10.4

c M = a0 y dA (10.1)

I n t h e c a s e o r a c r o s s - s e c t i o n a l a r e a w h i c h is s y m m e t r i c a l w i t h r c s p c c t t o a

h o r i z o n t a l a x i s , t h e b e n d i n g m o m e n t b e c o m e s

F o r t h e s y m m e t r i c a l a r e a , t h e n e u t r a l a x i s c o r r e s p o n d s w i t h t h e a x i s o f s y m m e t r y ,

a s i n t h e c a s e o f c l a s t i c b e n d i n g . F o r a n u n s y m m e t r i c a l a r e a , t h e n e u t r a l a x i s i s

n o t a t t h e c c n t r o i d , b u t i s l o c a t e d s o t h a t t h e c r o s s - s e c t i o n a l a r e a a b o v e t h e

n e u t r a l a x i s i s e q u a l t o t h e a r e a b e l o w i t , s i n c e t h e t o t a l t e n s i o n f o r c e m u s t e q u a l

t h e l o t a ! c o m p r e s s i o n f o r c e .

F o r a r e c t a n g u l a r b e a m o'" w i d t h b a n d d e p t h h, A = bh2/&. S u b s t i t u t i n g i n

F .q . ( 1 0 . 2 ) y i e l d s

T h e b e n d i n g m o d u l u s o f r u p t u r e <rb c a n b e f o u n d b y e q u a t i n g t h e b e n d i n g

m o m e n t o f E q . ( 1 0 . 4 ) t o t h e e x p r e s s i o n w h i c h d e f i n e s ab, o r M = ubI/c. F o r t h e

r e c t a n g u l a r s e c t i o n , / / c = bh2/6, a n d <rh = I .5<7 0 . T h e p a r a b o l i c s h e a r s t r e s s d i s t r i b u t i o n f o r a r e c t a n g u l a r b e a m i n w h i c h t h e

s t r e s s e s a r e b e l o w t h e c l a s t i c l i m i t w a s o b t a i n e d f r o m t h e b e n d i n g s t r e s s d i s -

t r i b u t i o n , a n d d o e s n o t a p p l y f o r o t h e r d i s t r i b u t i o n s o f b e n d i n g s t r e s s . F o r a

r e c t a n g u l a r c r o s s s c c t i o n , o r f o r o t h e r s i m i l a r c r o s s s e c t i o n s i n w h i c h t h e b e n d i n g

m o d u l u s o f r u p t u r e is c o n s i d e r a b l y l a r g e r t h a n t h e a c t u a l s t r e s s , t h e s h e a r i n g

s t r e s s e s a r e s e l d o m v e r y h i g h a n d m a y b e a p p r o x i m a t e d w i t h s u f f i c i e n t a c c u r a c y .

T h e p l a s t i c b e n d i n g o f a b e a m in w h i c h t h e c r o s s s e c t i o n i s n o t s y m m e t r i c a l

a b o u t t h e n e u t r a l a x i s i s c o n s i d e r e d b y a n a l y z i n g a n u m e r i c a l e x a m p l e . T h e a r e a

s h o w n i n F i g . 1 0 . 5 h a s i t s c c n t r o i d a l a x i s 0 . 3 i n a b o v e t h e b a s e . F o r p l a s t i c

b e n d i n g w i t h a c o n s t a n t s t r e s s c r 0 , t h e n e u t r a l a x i s w i l l b e 0 . 2 i n a b o v e t h e b a s e ,

i n o r d e r f o r t h e t e n s i o n a r e a t o b e e q u a l t o t h e c o m p r e s s i o n a r e a . T h e t e n s i o n

a n d c o m p r e s s i o n f o r c e s w i l l b e e q u a l t o 0 . 1 2 < r 0 a n d w i l l r e s i s t a b e n d i n g m o m e n t

M = 2Qa0 (10.2)

w h e r e

( 1 0 . 3 )

0.2 — -

t 0.6

- r h > r = 0.3

0.12 o 0

T 0.4

I .

[« 0.6

to)

Figure 10.5

0.12u„ M = 0.04So0

CO

.1/, - 0 . 0 2 7 2o„ M 4 = 0 .04St r 0

o f 0 . 0 4 8 c r o , a s s h o w n i n F i g . 10.5b. T h e e l a s t i c b e n d i n g s t r e s s f o r t h i s a r e a i s

o b t a i n e d f r o m t h e e q u a t i o n M = a b 1/c = 0 . 0 2 1 a b . T h e b e n d i n g m o d u l u s o f r u p -

t u r e i s t h e r e f o r e e q u a l t o a b = ( 0 . 0 4 8 / 0 . 0 2 7 2 ) c r 0 = l . 7 6 5 c r 0 . T h e s t r e s s d i s -

t r i b u t i o n o f v a r i o u s b e n d i n g m o m e n t s i s s h o w n i n F i g . 1 0 . 5 c . F o r b e n d i n g m o -

m e n t s l e s s t h a n M = 0 . 0 2 7 2 c r o , t h e s t r e s s e s a r e b e l o w t h e e l a s t i c l i m i t a n d h a v e a

s t r a i g h t - l i n e d i s t r i b u t i o n w i t h t h e n e u t r a l a x i s a t t h e c e n t r o i d o f t h e a r e a , a s

s h o w n b y c u r v e 1, F o r l a r g e r v a l u e s o f t h e b e n d i n g m o m e n t , t h e s t r e s s e s w i l l

e x c e e d t h e e l a s t i c l i m i t a t t h e u p p e r s i d e o f t h e b e a m , b u t r e m a i n b e l o w t h e e l a s t i c '

l i m i t o n t h e l o w e r s i d e , w i t h t h e n e u t r a l a x i s s h i f t i n g d o w n w a r d , a s s h o w n b y

c u r v e 2 . F o r f u r t h e r i n c r e a s e s i n b e n d i n g m o m e n t , t h e s t r e s s e s a p p r o a c h t h e

c o n s t a n t v a l u e s s h o w n b y c u r v e 4 , w i t h t h e n e u t r a l a x i s b e t w e e n t h e t w o

r e c t a n g l e s .

10.4 TRAPEZOIDAL DISTRIBUTION OF BENDING STRESS

T h e s t r e s s - s t r a i n c u r v e s f o r m o s t a i r c r a f t m a t e r i a l s c a n b e a p p r o x i m a t e d a c c u -

r a t e l y b y a t r a p e z o i d a l c u r v e , a s s h o w n i n F i g . 1 0 . 6 a . T h e i d e a l i z e d b e n d i n g s t r e s s

d i s t r i b u t i o n i s s h o w n i n F i g . 1 0 . 6 b . T h i s a p p r o x i m a t i o n f o r o b t a i n i n g t h e b e n d i n g

s t r e n g t h i n t h e p l a s t i c r a n g e w a s p r o p o s e d b y C o z z o n e . 7 T h e b e n d i n g m o m e n t

f o r t h e t r a p e z o i d a l s t r e s s d i s t r i b u t i o n i s r e a d i l y o b t a i n e d a s t h e s u m o f t h e W n d -

<j "M

"u A \ h - H

ifc)

Figure 10.6

DESIGN OF MEMBERS IN TENSION, BENDING, OR TORSION 3 0 5

i n g m o m e n t f o r a c o n s t a n t s t r e s s t r , , , a s g i v e n b y E q . ( 1 0 . 2 ) , a n d t h e b e n d i n g

m o m e n t f o r a l i n e a r s t r e s s d i s t r i b u t i o n v a r y i n g f r o m 0 t o a k l :

M = 2 Q<R0 + ^ ( 1 0 . 5 ) c T h e t e r m <T,„ m a y b e i n t r o d u c e d i n s t e a d o f <rM b y s u b s t i t u t i n g c r b l = <rm — cr0 i n t o

E q . ( 1 0 . 5 ) . M a k i n g t h i s s u b s t i t u t i o n a n d d i v i d i n g b y I j c g i v e

T h e t e r m i n p a r e n t h e s e s d e p e n d s o n t h e s h a p e o f t h e c r o s s s e c t i o n a n d m a y v a r y

f r o m 0 f o r c o n c e n t r a t e d f l a n g e a r e a s t o 1 . 0 f o r a d i a m o n d s h a p e . I f t h i s t e r m i s

d e s i g n a t e d b y K , E q . ( 1 0 . 6 ) b e c o m e s

AT = AM + K<70 ( 1 0 . 7 )

2 O w h e r e K — - y - — 1 ( 1 0 . 8 ) I/c S o m e v a l u e s o f K f o r v a r i o u s c r o s s s e c t i o n s a r e s h o w n i n F i g . 1 0 . 7 .

T h e v a l u e o f a a s h o u l d b e d e t e r m i n e d i n s u c h a w a y t h a t t h e b e n d i n g

m o m e n t r e s i s t e d b y t h e a s s u m e d t r a p e z o i d a l s t r e s s d i s t r i b u t i o n i s e q u a l t o t h e

b e n d i n g m o m e n t r e s i s t e d b y t h e a c t u a l s t r e s s e s . T h e r e f o r e , t h e c o r r e c t v a l u e o f cr0

w o u l d d e p e n d s o m e w h a t o n t h e c r o s s - s e c t i o n a l a r e a . I f t h e v a l u e o f cr0 w e r e

c a l c u l a t e d f o r e a c h a r e a , t h e r e w o u l d b e n o a d v a n t a g e i n a s s u m i n g a t r a p e z o i d a l

s t r e s s d i s t r i b u t i o n , s i n c e i t w o u l d b e n e c e s s a r y t o c a l c u l a t e t h e t r u e r e s i s t i n g

m o m e n t o f t h e b e a m i n o r d e r t o c a l c u l a t e a 0 . C o z z o n e h a s s h o w n t h a t i t is

s u f f i c i e n t l y a c c u r a t e t o c a l c u l a t e a 0 f o r a r e c t a n g u l a r c r o s s s e c t i o n a n d t o u s e t h i s

v a l u e f o r a l l c r o s s s e c t i o n s .

E x a m p l e 1 0 . 2 A b e a m w i t h t h e c r o s s s e c t i o n s h o w n i n F i g . 1 0 . 8 i s m a d e o f

a l u m i n u m - a l l o y f o r g i n g . T h e t r u e s h a p e o f t h e f o r g i n g i s s h o w n b y t h e d o t t e d

l i n e s , b u t t h e t r a p e z o i d s s h o w n b y t h e s o l i d l i n e s a r e a s s u m e d . C a l c u l a t e t h e

S o c l i o n K S i ^ ' l i t m K

( >

0 © 0 J 5 t o 0 . 7

s N

a s s 0 . 5 1 . 0

E 1 0 i . i 0 . 5 I i

Figure 10.7

3 0 6 A IRC 'RAiT STRUC TURES

o . : o o . i :

Figure 10.8

u l t i m a t e b e n d i n g s t r e n g t h a b o u t a h o r i z o n t a l a x i s if a „ = 6 5 , 0 0 0 , a 0 =

6 0 , 0 0 0 , a n d t h e y i e i d s t r e s s a , y = 5 0 , 0 0 0 l b / i n 2 .

S O L U T I O N T h e v a l u e s o f / a n d Q a r e c a l c u l a t e d f o r t h e a s s u m e d a r e a , w h i c h

is c o m p o s e d o f e i g h t o f t h e t r i a n g l e s ( 1 ) a n d f o u r o f t h e r c c t a n g i e s ( 2 ) :

r 8 x 0 . 1 2 x l 3 4 x 0 . 2 0 x l 3 I = + = Q J 4 7 j n 4

2Q

12 ' 3

x 0 . 0 6 x 0 . 3 3 3 + 4 x 0 . 2 0 x 0 . 5 = 0 . 5 6 i n 1

T h e b e n d i n g m o d u l u s o f r u p t u r e i s n o w c a l c u l a t e d f r o m E q s . ( 1 0 . 7 ) a n d (10.8):

„ 2 0 , 0 . 5 6 ,

7 / c = 0 3 4 7 -

c„ = nM + Ka0 = 6 5 , 0 0 0 + 0 . 6 1 x 6 0 , 0 0 0 = 1 0 1 , 6 0 0 l b / i n 2

T h e u l t i m a t e b e n d i n g s t r e n g t h i s

M = ^ j i = 1 0 1 , 6 0 0 — - = 3 5 , 3 0 0 i n • l b c 1 .0

I n t h i s c a s e , i t w o u l d n o t b e p o s s i b l e t o u t i l i z e t h e f u l l u l t i m a t e b e n d i n g

s t r e n g t h b e c a u s e t h e s t r e s s a t t h e a p p l i e d l o a d c o n d i t i o n w o u l d e x c e e d t h e

y i e l d s t r e s s . T h e e x a c t a m o u n t o f p e r m a n e n t s e t p e r m i t t e d a t t h e a p p l i e d o r

l i m i t l o a d is n o t s p e c i f i e d c l e a r l y f o r a m e m b e r i n b e n d i n g , b u t d e p e n d s

s o m e w h a t o n t h e j u d g m e n t o f t h e d e s i g n e r . I n s o m e c a s e s , t h e b e n d i n g m o -

d u l u s o f r u p t u r e is n o t p e r m i t t e d t o e x c e e d t h e y i e l d s t r e s s ; o r , f o r t h i s

p r o b l e m , t h e v a l u e o f M c / I w o u l d n o t e x c e e d 5 0 , 0 0 0 l b / i n 2 a t t h e a p p l i e d

l o a d a n d c o n s e q u e n t l y c o u l d n o t e x c e e d 7 5 , 0 0 0 l b / i n z a t t h e u l t i m a t e o r

d e s i g n l o a d . E v e n a t t h e y i e l d s t r e s s , h o w e v e r , s o m e p l a s t i c b e n d i n g e f f e c t s

m a y b e c o n s i d e r e d . T h e s t r e s s - s t r a i n d i a g r a m f o r t h e m a t e r i a l o f F i g . 1 0 . 8 , i n

w h i c h t h e s t r e s s d o e s n o t e x c e e d t h e y i e l d s t r e s s s h o w n i n F i g . 1 0 . 9 , m a y b e

r e p r e s e n t e d b y t h e t r a p e z o i d w i t h cr0 = 2 1 , 2 0 0 a n d <rm = 5 0 , 0 0 0 l b / i n 2 . T h e

Df.SIGN OF Mi'MRf.RS IN TENSION, BENDING, OR TORSION 307

5 0 . 0 0 0

21.200

Figure 10.9

b e n d i n g m o d u l u s o f r u p t u r e a t t h e l i m i t l o a d c a n b e c a l c u l a t e d f r o m E q . ( 1 0 . 7 ) :

GN =• o m -1 A'rr,, = 5 0 , 0 0 0 + 0 . 6 1 X 2 1 , 2 0 0 = 6 3 , 0 0 0 l b / i n 2

T h e a l l o w a b l e v a l u e o f M r / / w o u l d t h e n b e 6 3 , 0 0 0 f o r t h e l i m i t l o a d a n d

1.5 x 6 3 , 0 0 0 = 0 4 , 5 0 0 l b / i n 2 a t t h e u l t i m a t e o r d e s i g n l o a d .

10.5 CURVED BEAMS

M o s t b e a m s t r u c t u r e s a r e a n a l y z e d b y t h e m e t h o d s p r e v i o u s l y c o n s i d e r e d , i n

w h i c h a n y i n i t i a l c u r v a t u r e o f t h e a x i s o f t h e b e a m i s n e g l e c t e d . H o w e v e r , w h e n

t h e r a d i u s o f c u r v a t u r e i s o f t h e s a m e o r d e r o f m a g n i t u d e a s t h e d e p t h o f t h e

b e a m , t h e s t r e s s d i s t r i b u t i o n d i f f e r s c o n s i d e r a b l y f r o m t h a t f o r s t r a i g h t b e a m s .

T h e s t r e s s e s o n t h e c o n c a v e s i d e o f t h e b e a m a r e h i g h e r t h a n t h o s e f o r a s i m i l a r

s t r a i g h t b e a m , a n d t h e s t r e s s e s o n t h e c o n v e x s i d e a r e l o w e r . W h e n t h e m a x i m u m

s t r e s s e s e x c e c d t h e e l a s t i c l i m i t , l o c a l y i e l d i n g o c c u r s , w h i c h p e r m i t s a r e -

d i s t r i b u t i o n o f s t r e s s . A t t h e u l t i m a t e b e n d i n g m o m e n t , t h e s t r e s s e s a p p r o a c h t h e

s a m e d i s t r i b u t i o n a s f o r t h e p l a s t i c b e n d i n g o f a s t r a i g h t b e a m . T h u s t h e b e a m

c u r v a t u r e h a s t h e e f f e c t o f r e d u c i n g t h e y i e l d s t r e n g t h b u t o f n o t a p p r e c i a b l y

c h a n g i n g t h e u l t i m a t e b e n d i n g s t r e n g t h .

T h e b e a m s h o w n i n F i g . 1 0 . 1 0 h a s a n i n i t i a l r a d i u s o f c u r v a t u r e R m e a s u r e d

t o t h e c e n t r o i d o f t h e c r o s s s e c t i o n . A p l a n e c r o s s s e c t i o n pp r e m a i n s p l a n e a f t e r

b e n d i n g , a n d i t s r e l a t i v e p o s i t i o n a f t e r b e n d i n g i s s h o w n b y t i n . A l o n g i t u d i n a l

l i b e r o f t h e b e a m o f i n i t i a l l e n g t h L i s e x t e n d e d a d i s t a n c e <5. S i n c e <5 i s m e a s u r e d

b e t w e e n t h e s t r a i g h t l i n e s pp a n d n i l , it v a r i e s l i n e a r l y w i t h t h e d i s t a n c e y f r o m t h e

c e n t r o i d t o t h e fiber:

<5 = ki + kzy T h e t e r m s k { , k 2 . a n d k } a r c d e t e r m i n e d c o n s t a n t s . T h e l e n g t h o f t h e fiber L is

p r o p o r t i o n a l t o i t s d i s t a n c e f r o m t h e c e n t c r o f c u r v a t u r e :

L = k,(R + >•)

308 MRCRAl'T STRUCTURES

fa) U>)

Figure 10.10

T h e u n i t s t r e s s i s n o w o b t a i n e d a s t h e p r o d u c t o f t h e u n i t s t r a i n b j b a n d t h e

e l a s t i c m o d u l u s £ :

= R S _ r k l + k i y

* L J c 2 ( R + y )

T h i s e x p r e s s i o n m a y b e s i m p l i f i e d b y d i v i d i n g t h e n u m e r a t o r b y t h e d e n o m i n a t o r

a n d g r o u p i n g t h e c o n s t a n t s i n t o t w o n e w , u n d e t e r m i n e d c o n s t a n t s « a n d b:

iJ = A + R + y (10.9)

T h i s s t r e s s d i s t r i b u t i o n is p i c t u r e d i n F i g . 1 0 . 1 0 c .

If a r e s u l t a n t t e n s i o n f o r c e P a c t s a t t h e c e n t r o i d o f t h e a r e a , i t m u s t e q u a l t h e

s u m o f t h e i n t e r n a l f o r c e s , t h a t i s , j" a d A . F r o m E q . ( 1 0 . 9 ) w e h a v e

P = dA = aA + b dA R + y (10.10)

w h e r e A i s t h e t o t a l c r o s s - s e c t i o n a l a r e a a n d a a n d b a r e u n d e t e r m i n e d c o n s t a n t s .

F o r t h e c a s e o f p u r e b e n d i n g , ( h e f o r c e P v a n i s h e s . y

T h e e x t e r n a l b e n d i n g m o m e n t M a b o u t t h e c e n t r o i d a ! a x i s m u s t e q u a l t h e

m o m e n t o f t h e i n t e r n a l f o r c e s J /Y d A :

M = r DA y dA i- h j1 DA R + y

T h e f i r s t i n t e g r a l o n t h e r i g h t s i d e o f t h e e q u a t i o n i s z e r o b e e a u s c y i s m e a s u r e d

f r o m t h e c e n t r o i d a l a x i s . T h e s e c o n d i n t e g r a l m a y b e s e p a r a t e d i n t o t w o t e r m s b y

division:

f — = ( 7

R J V R

R + y dA = A - R dA R + y

DESIGN OF MEMBERS tN TENSION, BENDING, OR TORSION 3 0 9

T h e b e n d i n g - m o m e n t e q u a t i o n m a y n o w b e w r i t t e n a s

dA M = bA — bR R+y (10.11)

T h e u n k n o w n c o n s t a n t h c a n b e f o u n d f r o m E q . ( 1 0 . 1 1 ) s i n c e a l l o t h e r t e r m s a r e

k n o w n f r o m t h e g e o m e t r y a n d l o a d i n g o f t h e b e a m . T h e o t h e r u n k n o w n c o n -

s t a n t , a , i s o b t a i n e d f r o m E q . ( 1 0 . 1 0 ) . T h e s t r e s s d i s t r i b u t i o n i s t h e n f o u n d f r o m

E q . ( 1 0 . 9 ) . T h e e f f e c t o f t h e a x i a l l o a d P i s t o c h a n g e t h e c o n s t a n t a a n d t h e s t r e s s

a b y a n a m o u n t P j A . H e n c e t h e s a m e s t r e s s d i s t r i b u t i o n c a n b e o b t a i n e d b y

s u p e r i m p o s i n g t h e b e n d i n g s t r e s s e s f o r P = 0 a n d t h e s t r e s s e s P j A r e s u l t i n g f r o m

t h e a x i a l l o a d P a t t h e c e n t r o i d o f t h e a r e a .

T h e e x t r e m e f i b e r s t r e s s e s f o r v a r i o u s c u r v e d b e a m s c a n b e d e t e r m i n e d a s

r a t i o s o f t h e s t r e s s e s c o m p u t e d b y t h e flexure f o r m u l a . T h e s e r a t i o s h a v e b e e n

c o m p u t e d f o r v a r i o u s c r o s s s e c t i o n s . T h e t e r m s K f o r t h e e q u a t i o n cr = KMcjl a r e p l o t t e d i n F i g . 1 0 . 1 1 f o r a f e w c o m m o n c r o s s s e c t i o n s . I t i s o b s e r v e d t h a t t h e

s t r e s s e s a l w a y s b e c o m e i n f i n i t e o n t h e c o n c a v e s i d e w h e n R / c = 1 , w h i c h c o r r e -

s p o n d s t o a s h a r p r e e n t r a n t a n g l e o n t h e c o n c a v e s u r f a c e o f t h e b e a m . S u c h

r e e n t r a n t a n g l e s s h o u l d b e a v o i d e d i n a n y s t r u c t u r e o r m a c h i n e p a r t .

A n o t h e r e f f e c t o f b e a m c u r v a t u r e w h i c h c a n n o t b e a n a l y z e d b y s i m p l e t h e o r y

o c c u r s i n b e a m s w i t h t h i n f l a n g e s , a s i n d i c a t e d i n F i g . 1 0 . 1 2 . I f t h e c o n c a v e s i d e o f

a b e a m is i n c o m p r e s s i o n , t h e f l a n g e s t e n d t o d e f l e c t t o w a r d t h e n e u t r a l a x i s , a s

s h o w n i n F i g . 1 0 . 1 2 / j . T h e b e n d i n g s t r e s s i s n o t d i s t r i b u t e d u n i f o r m l y a l o n g t h e

K 2.0

8 10

Figure 10.11


A

Figure 10.! 2

h o r i z o n t a l b e a m flange, b u t i s m u c h h i g h e r n e a r t h e w e b t h a n i t is i n t h e o u t -

s t a n d i n g l e g s , a s s h o w n i n F i g . 1 0 . 1 2 6 . W h e n t h e b e n d i n g p r o d u c e s c o m p r e s s i o n

o n t h e c o n v e x s i d e o f t h e b e a m , t h e flanges d e f l e c t a w a y f r o m t h e n e u t r a l a x i s , b u t

t h e s t r e s s d i s t r i b u t i o n a l o n g t h e h o r i z o n t a l w i d t h o f t h e flange i s e s s e n t i a l l y t h e

s a m e a s s h o w n .

10.6 TORSION OF CIRCULAR SHAFTS

T h e s t r e s s e s r e s u l t i n g f r o m t o r s i o n a l m o m e n t s a c t i n g o n e l a s t i c c y l i n d r i c a l m e m -

b e r s o f c i r c u l a r c r o s s s e c t i o n c a n b e o b t a i n e d r e a d i l y . I t h a s b e e n f o u n d e x p e r -

i m e n t a l l y t h a t t h e r e i s n o d i s t o r t i o n o f a n y c r o s s s e c t i o n o f t h e s h a f t e i t h e r i n t h e

d i r e c t i o n n o r m a ! t o t h e p l a n e o f t h e c r o s s s e c t i o n o r i n t h e p l a n e o f t h e c r o s s

s e c t i o n . A n y t w o c r o s s s e c t i o n s o f t h e s h a f t , s u c h a s t h o s e s h o w n i n F i g . 1 0 . 1 3 6 ,

h a v e a r e l a t i v e r o t a t i o n a b o u t t h e a x i s o f t h e s h a f t . S i n c e t h e t w o c r o s s s e c t i o n s

h a v e n o r e l a t i v e d i s p l a c e m e n t r a d i a l l y , o r a l o n g t h e a x i s o f t h e s h a f t , t h e o n l y

s t r e s s e s a r e t h e s h e a r i n g s t r e s s e s i n t h e c i r c u m f e r e n t i a l a n d a x i a l d i r e c t i o n s , a s

s h o w n i n F i g . 1 0 . 1 3 6 .

r

T Figure 10-13

d f . s i g n o f m i ' m r f . r s i n t e n s i o n , b e n d i n g , o r t o r s i o n 311

T h e s h e a r i n g s t r a i n y a n d c o n s e q u e n t l y t h e s h e a r i n g s t r e s s T s m u s t v a r y

l i n e a r l y in p r o p o r t i o n t o t h e d i s t a n c e r f r o m t h e c e n t e r o f t h e s h a f t :

x , = Kr (10.12)

T h e t e r m K i s a c o n s t a n t o f p r o p o r t i o n a l i t y w h i c h i s d e t e r m i n e d l a t e r . T h e

e x t e r n a l t o r s i o n a l m o m e n t T m u s t e q u a l t h e s u m c f t h e m o m e n t s o f t h e i n t e r n a l

s h e a r i n g f o r c e s o n t h e c r o s s s e c t i o n :

T = : , r d A = K J r 2 d A ( 1 0 . 1 3 )

T h e i n t e g r a l o f E q . ( 1 0 . 1 3 ) r e p r e s e n t s t h e p o l a r m o m e n t o f i n e r t i a o f t h e c r o s s -

s e c t i o n a l a r e a , a n d u s u a l l y it i s d e s i g n a t e d a s J o r I P . T h e v a l u e o f K m a y b e

F o u n d f r o m E q . ( 1 0 . 1 3 ) a s K = T j J . S u b s t i t u t i n g t h i s v a l u e i n t o E q . ( 1 0 . 1 2 ) y i e l d s

t h e f o r m u l a f o r t o r s i o n a l s h e a r s t r e s s e s i n c i r c u l a r s h a f t s :

-s = - J ( 1 0 . 1 4 )

T h e a n g l e o f t w i s t o f a c i r c u l a r s h a f t m a y b e d e t e r m i n e d f r o m t h e a n g l e o f

s h e a r i n g s t r a i n y . T h e s h e a r i n g m o d u l u s o f e l a s t i c i t y G i s d e f i n e d a s t h e r a t i o c f

s h e a r s t r e s s t o s h e a r s t r a i n :

G = — ( 1 0 . 1 5 ) 7

A n d G is r e l a t e d t o E f o r a n i s o t r o p i c m a t e r i a ! o n l y b y t h e f o l l o w i n g e q u a t i o n :

c = 2 [ r b (10-16)

T h e s h a f t o f r a d i u s ;-0 s h o w n i n F i g . 1 0 . 1 3 t w i s t s t h r o u g h a n a n g l e <[) i n l e n g t h

L. A p o i n t o n t h e c i r c u m f e r e n c e o f t h e u p p e r c r o s s s e c t i o n m o v e s a d i s t a n c e <pr0

, d u r i n g t h e d e f o r m a t i o n . T h i s p o i n t is a l s o d i s p l a c e d a d i s t a n c e y L , a s s h o w n :

f/)ru = yL T h e a n g l e o f t w i s t cf> m a y b e e x p r e s s e d in o t h e r f o r m s b y s u b s t i t u t i n g v a l u e s f r o m

E q s . ( 1 0 . 1 5 ) a n d ( 1 0 . 1 4 ) :

<t> = — — — ( 1 0 . 1 7 ) r 0 Gr0 JG

T h e s e e q u a t i o n s a p p l y o n l y t o t o r s i o n m e m b e r s w i t h s o l i d c i r c u l a r c r o s s s e c t i o n s ,

o r l o t u b u l a r m e m b e r s w i t h h o l l o w c i r c u l a r c r o s s s e c t i o n s .

10.7 TORSION OF A NONC1RCULAR SHAFT

T h e s t r e s s e s i n a t o r s i o n m e m b e r o f a r b i t r a r y c r o s s s e c t i o n c a n n o t b e o b t a i n e d b y

m e a n s o f a s i m p l e , g e n e r a l e q u a t i o n . A f e w s p e c i a l c a s e s m a y b e a n a l y z e d , h o w -


e v e r , a n d s o m e g e n e r a l p r o p e r t i e s o f t h e s h e a r s t r e s s d i s t r i b u t i o n a r e e x a m i n e d . A

c y l i n d r i c a l t o r s i o n m e m b e r o f a r b i t r a r y c r o s s s e c t i o n i s s h o w n i n F i g . 1 0 . 1 4 a . T h e

s m a l l c u b i c a l e l e m e n t a t t h e s u r f a c e o f t h e m e m b e r i s e n l a r g e d i n F i g . 1 0 . 1 4 6 . I n

g e n e r a l , t h r e e p a i r s o f s h e a r i n g s t r e s s e s T J , T 2 , T 3 e x i s t o n a n y s u c h e l e m e n t ; b u t

i n t h e c a s e w h e r e o n e f a c e o f t h i s e l e m e n t i s a f r e e s u r f a c e o f t h e m e m b e r , t h e

s h e a r s t r e s s e s T 2 a n d T 3 o n t h e f r e e s u r f a c e m u s t b e z e r o . T h u s t h e e l e m e n t a t t h e

b o u n d a r y h a s o n l y t h e s h e a r s t r e s s e s t l 5 w h i c h a r e p a r a l l e l t o t h e b o u n d a r y .

H e n c e t h e s h e a r s t r e s s e s o n a c r o s s s e c t i o n n e a r a n y b o u n d a r y a r e p a r a l l e l t o t h e

b o u n d a r y , a s s h o w n i n F i g . 1 0 . 1 4 a . T h e r e s u l t a n t o f a l l s h e a r i n g f o r c e s o n t h e

c r o s s s e c t i o n m u s t b e e q u a l t o t h e e x t e r n a l t o r q u e .

T h e e x a g g e r a t e d d e f o r m a t i o n s o f a s q u a r e s h a f t in t o r s i o n a r e s h o w n i n F i g .

1 0 . 1 5 a . A n e l e m e n t a t a c o r n e r o f t h e c r o s s s e c t i o n m a y b e c o m p a r e d t o t h e

e l e m e n t s h o w n in F i g . 1 0 . 1 4 6 , a n d s i n c e t w o p e r p e n d i c u l a r f a c e s o f t h e e l e m e n t

c a n h a v e n o s h e a r s t r e s s e s , a l l t h e s h e a r s t r e s s e s t j , t 2 , a n d T 3 m u s t b e z e r o a t a

c o r n e r o f t h e m e m b e r . T h e s h e a r s t r e s s e s o n t h e c r o s s s e c t i o n h a v e m a x i m u m

v a l u e s a t t h e c e n t e r o f e a c h s i d e a n d h a v e d i r e c t i o n s a p p r o x i m a t e l y a s s h o w n i n

Dl SICIN OF Mi;.MUl:RS IN TENSION, BENDING, OR TORSION 313

ft

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ib)

m

Figure 10.16

F i g . 1 0 . 1 5 b . T h e c u b i c a l e l e m e n t s a t t h e c o r n e r s r e m a i n c u b i c a l , a s i n d i c a t e d i n

F i g . 1 0 . 1 5 a , b u t e l e m e n t s n e a r t h e c e n t e r s o f t h e s i d e s h a v e r a t h e r l a r g e s h e a r i n g

d e f o r m a t i o n s . T h e c r o s s s e c t i o n s t h e r e f o r e d o n o t r e m a i n p l a n e , b u t w a r p a s

s h o w n .

O n e t y p e o f m e m b e r w h i c h i s u s e d f r e q u e n t l y o n flight v e h i c l e s t r u c t u r e s h a s

a n a r r o w , r e c t a n g u l a r c r o s s s e c t i o n . W h i l e s u c h c r o s s s e c t i o n s a r e i n e f f i c i e n t f o r

t o r s i o n m e m b e r s , o f t e n t h e y m u s t r e s i s t s o m e t o r s i o n a l s t r e s s e s . F o r t h e c r o s s

s e c t i o n s h o w n i n F i g . 1 0 . 1 6 o f l e n g t h b a n d w i d t h f , t h e s h e a r s t r e s s e s m u s t b e

p a r a l l e l t o t h e b o u n d a r y . I f t h e l e n g t h h i s l a r g e c o m p a r e d t o t h e t h i c k n e s s r, t h e

e n d e f f e c t s a r e s m a l l , a n d t h e s h e a r s t r e s s e s m a y b e a s s u m e d t o b e d i s t r i b u t e d a s

s h o w n i n F i g . 1 0 . 1 6 b f o r t h e e n t i r e l e n g t h b . I t c a n b e s h o w n t h a t t h e s h e a r s t r e s s

h a s t h e f o l l o w i n g v a l u e : 3 J

bt (10.18)

E q u a t i o n ( 1 0 . 1 8 ) i s a c c u r a t e w h e n t h e w i d t h b i s l a r g e c o m p a r e d t o t h e

t h i c k n e s s t . F o r r e c t a n g u l a r c r o s s s e c t i o n s i n w h i c h t h e d i m e n s i o n s a r e o f t h e

s a m e o r d e r , t h e m a x i m u m s t r e s s , w h i c h o c c u r s a t t h e m i d d l e o f t h e l o n g e s t s i d e , i s

f o u n d b y J r

a bt ( 1 0 . 1 9 )

V a l u e s o f a a r e g i v e n i n T a b l e 1 0 . 1 . T h e s e v a l u e s h a v e b e e n c a l c u l a t e d b y t h e o r -

e t i c a l m e t h o d s w h i c h a r e n o t w i t h i n t h e s c o p e o f t h e p r e s e n t d i s c u s s i o n . F o r l a r g e

r a t i o s o f b / t , t h e v a l u e o f a i s 0 . 3 3 3 , w h i c h c o r r e s p o n d s t o E q . ( 1 0 . 1 8 ) . F o r s m a l l e r

v a l u e s o f b / t , t h e e f f e c t s o f t h e e n d s a r e m o r e n o t i c e a b l e , a n d t h e v a l u e s o f a a r e

s m a l l e r t h a n 0 . 3 3 3 .

T h e a n g l e o f t w i s t o f a rectangular s h a f t o f l e n g t h L c a n b e o b t a i n e d f r o m

TL pbt3G

w h e r e ( j j is i n r a d i a n s a n d [I i s a c o n s t a n t w h i c h is g i v e n i n T a b l e 1 0 . 1 .

T a b l e 10 .1 C o n s t a n t s f o r E q s . ( 1 0 . 1 9 ) a n d ( 1 0 . 2 0 )

b/t 1.00 1.50 1.75 2.00 2.50 3.00 4 6 8 10 00

a 0.208 0.231 0.23y 0.246 0.258 0.267 0.282 0.299 0.307 0.313 0.333

P 0.141 0.196 0.214 0.229 0.249 0.263 0.281 0.299 0.307 0.313 0.333

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(c) Figure 10.17

T h e t o r s i o n a l p r o p e r t i e s o f a r e c t a n g u l a r p l a t e a r e n o t a p p r e c i a b l y a f f e c t e d if

t i i c p l a t e i s b e n t t o s o m e c r o s s s e c t i o n , s u c h a s t h o s e s h o w n i n F i g . 1 0 . 1 7 ,

p r o v i d e d t h a t t h e e n d c r o s s s e c t i o n s o f t h e m e m b e r a r e f r e e t o w a r p . T h u s t h e s e

s e c t i o n s m a y b e a n a l y z e d b y E q s . ( 1 0 . 1 9 ) a n d ( 1 0 . 2 0 ) . T h e a n g l e o f t w i s t f o r a

s h a f t o f a n y c r o s s s e c t i o n m a y b e e x p r e s s e d i n t h e f o r m

(10.21)

w h e r e K i s a c o n s t a n t w h i c h d e p e n d s o n o n l y t h e c r o s s - s e c t i o n a l a r e a . B y c o m -

p a r i n g E q s . ( 1 0 . 2 0 ) a n d ( 1 0 . 2 1 ) , t h e v a l u e o f K f o r a r e c t a n g u l a r c r o s s s c c t i o n i s

f o u n d t o b e j)bt3. F o r a c r o s s s c c t i o n m a d e u p o f s e v e r a l r e c t a n g u l a r e l e m e n t s ,

s u c h a s t h a t s h o w n i n F i g . 1 0 . 1 8 , t h e v a l u e o f K i s g i v e n a p p r o x i m a t e l y b y

K = ( i . b A + b 2 i l + / ? 3 b 3 t j ( 1 0 . 2 2 )

E x a m p l e 1 0 . 3 T h e r o u n d t u b e s h o w n i n F i g . 1 0 . 1 9 « h a s a n a v e r a g e r a d i u s R

a n d a w a l l t h i c k n e s s t . C o m p a r e t h e t o r s i o n a l s t r e n g t h a n d r i g i d i t y o f t h i s

t u b e w i t h t h a t o f a s i m i l a r t u b e w h i c h i s s l i t f o r i t s e n t i r e l e n g t h , a s s l n 5 w n i n

F i g . iO. 19/? a n d c . A s s u m e R / t = 2 0 .

S O L U T I O N A p p r o x i m a t e v a l u e s o f t h e a r e a a n d m o m e n t o f i n e r t i a a r e s a t i s -

!" h i — I f , Figure 10.18

' 1 DESIGN OF MEMBERS IN TENSION, BENDING, OR TORSION 315

f a c t o r y f o r t h i n - w a l l e d t u b e s . T h e a r e a is e q u a l t o t h e p r o d u c t o f t h e c i r c u m -f e r e n c e 2KR a n d t h e w a l l t h i c k n e s s t. T h e p o l a r r a d i u s o f g y r a t i o n is a p p r o x i -m a t e l y e q u a l t o R , a n d t h e p o l a r m o m e n t o f i n e r t i a is t h e n o b t a i n e d a s f o l l o w s :

J = 2nRh ( 10 .23 )

I t is s u f f i c i e n t l y a c c u r a t e t o u s e t h e a v e r a g e r a d i u s R i n p l a c e o f t h e o u t s i d e r a d i u s in c o m p u t i n g t h e m a x i m u m s h e a r i n g s t r e s s . T h e v a l u e s o f t h e c l o s e d t u b e a r e o b t a i n e d f r o m E q s . (10 .14 ) a n d ( 1 0 . 1 7 ) :

_ 7 > _ T

2kR2I ( 1 0 . 2 4 )

TL JG :

TL 2nR tG

( 1 0 . 2 5 )

T h e s l i t t u b e is a n a l y z e d b y E q s . (10 .18) a n d (10 .20) , a s s u m i n g b = 2nR a n d

a = f} = 0 . 3 3 3 : 3 T

t . = - — r r (10 .26 ) 2 izRt2

3TL 2nRtiG ( 1 0 . 2 7 )

T h e r a t i o o f s h e a r i n g s t r e s s e s f o r t h e t w o m e m b e r s is o b t a i n e d b y d i v i d i n g E q . (10 .26 ) b y E q . (10 .24) . T h e v a l u e s o f t h e s t r e s s a n d a n g l e f o r t h e c l o s e d t u b e a r e d e s i g n a t e d t s l > a n d t p n , r e s p e c t i v e l y :

*S0

3 R

t 60

T h e s h e a r i n g s t r e s s e s a r c t h e r e f o r e 6 0 t i m e s a s h i g h in t h e s l i t t u b e a s in t h e

c l o s e d t u b e , o r t h e c l o s e d t u b e w o u l d b e 6 0 t i m e s a s s t r o n g if t h e a l l o w a b l e

s h e a r i n g s t r e s s e s w e r e e q u a l .

T h e t o r s i o n a l s t i f f n e s s o f t h e t w o m e m b e r s m a y b e c o m p a r e d b y d i v i d i n g

E q . ( 1 0 . 2 7 ) b y E q . ( 1 0 . 2 5 ) :

 o 1200 (10.28)

to) (/'i

(O Figure 10.19


F o r a g i v e n a n g l e o f t w i s t , t h e c l o s e d t u b e r e s i s t s 1 2 0 0 t i m e s t h e t o r s i o n o f t h e

o p e n s e c t i o n ; o r , f o r a g i v e n t o r q u e , t h e o p e n s e c t i o n t w i s t s t h r o u g h a n a n g l e

1 2 0 0 t i m e s a s g r e a t a s t h e c l o s e d t u b e . W e a s s u m e d t h a t t h e o p e n t u b e w a s

f r e e t o d i s t o r t a s s h o w n i n F i g . 1 0 . 1 9 c , o r t h a t t h e e n d c r o s s s e c t i o n s w e r e n o t

r e s t r a i n e d a g a i n s t w a r p i n g .

10.8 END RESTRAINT OF TORSION MEMBERS

I n S e c . 1 0 . 7 w e a s s u m e t h a t t h e e n d c r o s s s e c t i o n s o f t h e t o r s i o n m e m b e r s a r e f r e e

t o w a r p f r o m t h e i r o r i g i n a l p l a n e a n d t h a t t h e r e a r e n o s t r e s s e s n o r m a l t o t h e

c r o s s s e c t i o n s . I t h a s b e e n p o i n t e d o u t t h a t m a n y a i r c r a f t s t r u c t u r a l m e m b e r s

m u s t b e c o n s t r u c t e d w i t h t h i n w e b s a n d t h a t s u c h m e m b e r s a r e v e r y i n e f f i c i e n t i n

r e s i s t i n g t o r s i o n a l l o a d s u n l e s s t h e y f o r m a c l o s e d b o x . I n s o m e c a s e s , i t i s

n e c e s s a r y t o u s e o p e n s e c t i o n s w i t h t h i n w e b s ; a n d i n m o s t o f t h e s e c a s e s , t h e

e n d s s h o u l d b e r e s t r a i n e d t o p r o v i d e a d d i t i o n a l t o r s i o n a l rigidity a n d s t r e n g t h .

T h e 1 b e a m s h o w n i n F i g . 1 0 . 2 0 r e s i s t s p a r t o f t h e t o r s i o n b y m e a n s o f t h e

s h e a r s t r e s s e s d i s t r i b u t e d f o r t h e i n d i v i d u a l r e c t a n g l e s , a s s h o w n i n F i g . 1 0 . 1 6 .

T h e r e m a i n d e r o f t h e t o r s i o n i s r e s i s t e d b y h o r i z o n t a l b e n d i n g o f t h e b e a m

f l a n g e s , a s s h o w n i n F i g . 1 0 . 2 0 b . T h e p r o p o r t i o n o f t h e t o r s i o n w h i c h i s r e s i s t e d

b y e a c h o f t h e t w o w a y s d e p e n d s o n t h e d i m e n s i o n s o f t h e c r o s s s e c t i o n a n d t h e

l e n g t h o f t h e m e m b e r . T h i s p r o p o r t i o n a l s o v a r i e s a l o n g t h e m e m b e r , a s m o r e o f

t h e t o r s i o n i s r e s i s t e d b y f l a n g e b e n d i n g n e a r t h e fixed e n d t h a n n e a r t h e f r e e e n d .

F o r m e m b e r s i n w h i c h t h e w e b s a r e t h i n a n d t h e l e n g t h is n o t g r e a t , a i l t h e

t o r s i o n m a y b e a s s u m e d t o b e r e s i s t e d b y f l a n g e b e n d i n g . I n t h e c a s e o f l o n g

m e m b e r s w i t h t h i c k w e b s , a l l t h e t o r s i o n m a y b e a s s u m e d t o b e r e s i s t e d b y t h e

t o r s i o n a l r e s i s t a n c e o f t h e r e c t a n g u l a r e l e m e n t s . I n s o m e c a s e s , h o w e v e r , i t m a y

b e n e c e s s a r y t o c a l c u l a t e t h e p r o p o r t i o n o f t h e t o r s i o n r e s i s t e d b y e a c h m e t h o d .

A m e m b e r w h i c h i s n o t r e s t r a i n e d a t t h e e n d s t w i s t s a s s h o w n i n F i g . 1 0 . 2 1 .

T h e a n g l e o f t w i s t v a r i e s u n i f o r m l y a l o n g t h e l e n g t h a n d m a y b e c o m p u t e d f r o m

E q . ( 1 0 . 2 1 ) . F o r a m e m b e r r e s i s t i n g t o r s i o n b y m e a n s o f f l a n g e b e n d i n g , a s s h o w n

i n F i g . 1 0 . 2 0 b , t h e f l a n g e b e n d i n g s t r e s s e s v a r y f r o m z e r o a t t h e f r e e e n d t o

m a x i m u m v a l u e s a t t h e f i x e d e n d . T h e a n g l e o f t w i s t a n d t h e a m o u n t o f c r o s s

s e c t i o n w a r p i n g v a r y a l o n g t h e s p a n . F o r t h e / - b e a m c r o s s s e c t i o n , T i m o s ' h e n k o

a n d G e r e 9 g i v e t h e f o l l o w i n g e q u a t i o n s f o r t h e m a x i m u m f l a n g e b e n d i n g m o m e n t

niSKiN OF MEMBERS IN TENSION, BENDINCi, OR TORSION 3 1 7

Figure 10.21

M m „ , f o r t h e m a x i m u m t o r s i o n r e s i s t e d b y w e b s h e a r s a n d f o r t h e a n g l e o f

t w i s t 0 i n l e n g t h L ( t h e c o n s t a n t a i s a r a t i o o f t h e r e l a t i v e flange b e n d i n g r i g i d i t y

t o t h e t o r s i o n a l r i g i d i t y ) :

= T a t a n h -h a

( 1 0 . 2 9 )

A . = T ( 1 - s e c h - J ( 1 0 . 3 0 )

0 = - ^ \ L - a tanh-) ( 1 0 . 3 1 ) KG V «

2 IJ R

" " W T c T ( 1 0 . 3 2 )

T h e t e r m I { i s t h e m o m e n t o f i n e r t i a o f o n e flange o f t h e b e a m a b o u t a v e r t i c a l

a x i s , K i s d e f i n e d b y E q . ( 1 0 . 2 2 ) , a n d h i s t h e b e a m d e p t h b e t w e e n c e n t e r s o f

f l a n g e s . T h e a n a l y s i s f o r t h e / b e a m s h o w n i n F i g . 1 0 . 2 0 a l s o a p p l i e s f o r a b e a m

o f l e n g t h 2 L w h i c h h a s a t o r q u e o f 2 T a p p l i e d a t t h e c e n t e r a n d w h i c h r e s i s t s h a l f

IN)

Figure 10.22


t h e t o r q u e a t c a c h e n d , if t h e e n d s a r e n o t r e s t r a i n e d a g a i n s t w a r p i n g . I n t h i s c a s e ,

a c r o s s s c c t i o n a t t h e c e n t e r o f t h e b e a m is p r e v e n t e d f r o m w a r p i n g b e c a u s e o f t h e

s y m m e t r y o f l o a d i n g .

A c h a n n c l c r o s s s c c t i o n , s u c h a s t h a t s h o w n i n F i g . 1 0 . 2 2 K , f r e q u e n t l y is u s e d

a s a n a i r c r a f t s t r u c t u r a l m e m b e r . I f a c h a n n e l m e m b e r i s r e s t r a i n e d a t t h e e n d , i t

w i l l r e s i s t t o r s i o n b y b e n d i n g o f t h e f l a n g e s i n m u c h t h e s a m e m a n n e r a s t h e

/ - b e a m f l a n g e s s h o w n i n F i g . 1 0 . 2 0 b . T h e a n a l y s i s f o r f l a n g e b e n d i n g i s s l i g h t l y

m o r e c o m p l i c a t e d t h a n f o r t h e / b e a m , b e c a u s e t h e v e r t i c a l w e b o f t h e c h a n n e l

a c t s w i t h t h e f l a n g e s i n r e s i s t i n g f l a n g e b e n d i n g , a s s h o w n i n F i g . 1 0 . 2 2 « . T h e

c r o s s s c c t i o n s h o w n i n F i g . 1 0 . 2 2 / ; a c t s i n t h e s a m e w a y a s t h e c h a n n e l s e c t i o n ,

b u t t h e s t r e s s d i s t r i b u t i o n i s s t i l l m o r e d i f f i c u l t t o a n a l y z e .

10.9 TORSIONAL STRESSES ABOVE TIIE ELASTIC LIMIT

I t h a s b e e n a s s u m e d t h a t t h e s t r e s s e s a r c b e l o w t h e e l a s t i c l i m i t i n t h e p r e v i o u s

a n a l y s i s o f t o r s i o n a l s t r e s s e s . I n m a n y d e s i g n a p p l i c a t i o n s , t h e u l t i m a t e t o r s i o n a l

s t r e n g t h i s d e s i r e d . W h i l e t h e r e i s n o t m u c h p u b l i s h e d i n f o r m a t i o n c o n c e r n i n g

s t r e s s - s t r a i n c u r v e s f o r s p e c i m e n s i n p u r e s h e a r , t h e s e c u r v e s w i l l h a v e t h e s a m e

g e n e r a l s h a p e a s t h e t e n s i o n s t r e s s - s t r a i n c u r v e s a n d w i l l h a v e o r d i n a t e s a p p r o x i -

m a t e l y 0 . 6 o f t h o s e f o r t h e t e n s i o n c u r v e s . T h u s t h e t o r s i o n a l s t r e s s e s i n a r o u n d

b a r a r e d i s t r i b u t e d a s s h o w n i n F i g . 1 0 . 2 3 w h e n t h e s t r e s s e s e x c e e d t h e e l a s t i c

l i m i t .

A s i n t h e p l a s t i c b e n d i n g o f b e a m s , i t is c o n v e n i e n t t o w o r k w i t h a fictitious

s t r e s s i n s t e a d o f t h e e x a c t s t r e s s d i s t r i b u t i o n s . T h i s s t r e s s i s d e s i g n a t e d a s t h e

t o r s i o n a l m o d u l u s o f r u p t u r e x r , w h i c h i s d e f i n e d b y

r r = y ( 1 0 . 3 3 )

w h e r e T i s t h e u l t i m a t e t o r s i o n a l s t r e n g t h o f t h e m e m b e r . F o r s t e e l t u b e s , t h e

v a l u e o f x T d e p e n d s o n t h e p r o p o r t i o n s o f t h e c r o s s s c c t i o n . T h e v a l u e s o f t h e

r a t i o TR/RR,„ a r e s h o w n i n F i g . 1 0 . 2 4 f o r v a r i o u s v a l u e s o f t h e r a t i o o f o u t s i d e

d i a m e t e r t o w a l l t h i c k n e s s Dji. T h e s e c u r v e s a r e t a k e n f r o m M I L - H D B K - 5 .

E q u a t i o n ( 1 0 . 3 3 ) a p p l i e s t o o n l y c i r c u l a r o r h o l l o w c i r c u l a r c r o s s s e ' e t i o n s ,

d e s i g n (>! m e m b e r s in t e n s i o n , b e n d i n g , o r t o r s i o n 3 1 9

a n d F i g . £ 0 . 2 4 s u p p l i e s i n f o r m a t i o n f o r a l l o w a b l e s t r e s s e s o n a l l s u c h c r o s s s e c -

t i o n s f o r v a r i o u s a i r c r a f t s t e e l s . T h e p l a s t i c t o r s i o n a l s t r e s s d i s t r i b u t i o n i n n o n -

c i r c u l a r s e c t i o n s u s u a l l y c a n n o t h e o b t a i n e d b y a n y s i m p l e a n a l y s i s . O f t e n i t m a y

b e n e c e s s a r y t o m a k e s t a t i c t e s t s o n t o r s i o n m e m b e r s o f n o n c i r c u l a r c r o s s s e c -

t i o n s t o d e t e r m i n e t h e a l l o w a b l e t o r s i o n a l m o m e n t s t o u s e f o r d e s i g n .

E x a m p l e 1 0 . 4 A r o u n d s t e e l t u b e w i t h 1 - i n o u t e r d i a m e t e r ( O D ) 0 . 0 6 5 i n

t h i c k r e s i s t s a d e s i g n t o r s i o n a l m o m e n t o f 5 0 0 0 i n • l b . F i n d t h e m a r g i n o f

s a f e t y if t h e u l t i m a t e t e n s i l e s t r e s s <T,„ = 1 0 0 , 0 0 0 l b / i n 2 .

S O L U T I O N T h e v a l u e s D/t = 1 5 . 3 8 a n d l / y = 0 . 0 4 1 9 3 i n 3 a r e o b t a i n e d f o r t h i s

t u b e . F r o m F i g . 1 0 . 2 4 , t h e r a t i o T T j a l u = 0 . 6 i s o b t a i n e d . T h e m a r g i n o f s a f e t y

is t h e r e f o r e c a l c u l a t c d a s f o l l o w s :

2 T = 0 . 6 X 1 0 0 , 0 0 0 = 6 0 , 0 0 0 l b / i n

= 5 0 0 0 - 5 9 , 6 0 0 l b / i n 2

' J 21/y 2 x 0 . 0 4 1 9 3 '

rT 60,000 M S = — - 1 = r - 1 — - 1 = 0 . 0 0 7

t , 5 9 , 6 0 0 •

E x a m p l e 1 0 . 5 D e s i g n a r o u n d t u b e t o r e s i s t a t o r s i o n a l m o m e n t o f 8 0 0 0

i n • l b . T h e m i n i m u m p e r m i s s i b l e w a l l t h i c k n e s s i s 0 . 0 4 9 i n , a n d t h e m a t e r i a l

h a s a n u l t i m a t e t e n s i l e s t r e s s a l u o f 1 0 0 , 0 0 0 l b / i n 2 .

S O L U T I O N A t o r s i o n t u b e m u s t b e d e s i g n e d b y a t r i a l - a n d - e r r o r p r o c e s s ,

b e c a u s e a l l o w a b l e s t r e s s d e p e n d s o n t h e D/t r a t i o a n d c a n n o t b e d e t e r m i n e d

e x a c t l y u n t i l t h e t u b e i s s e l e c t e d . F o r t u b e s o f a p p r o x i m a t e l y t h e s a m e w e i g h t ,

t h e l a r g e r l / y v a l u e s a r e o b t a i n e d f o r l a r g e r d i a m e t e r s , b u t t h e h i g h e r a l l o w -

a b l e m o d u l i o f r u p t u r e a r c o b t a i n e d f o r t h i c k e r t u b e w a l l s . T h u s s e v e r a l t u b e s

m a y h a v e a b o u t t h e s a m e w e i g h t a n d s t r e n g t h , a l t h o u g h t u b e s w i t h h i g h e r

D / t r a t i o s u s u a l l y h a v e a s t r e n g t h - w e i g h t a d v a n t a g e . A s s u m e a n a v e r a g e r a t i o

1)1, Figure 10.24


Table 10.2

cot D /

Tube , in iOO in / y iT T , M S

l j x 0.049 6.32 30.60 0.07847 53,000 51,000 0.04 If x 0.049 5.78 28.05 0.06534 54,000 61,500 - 0.12 l | x 0.058 6.15 21.55 0.06187 56,000 64,700 - 0.12

t T M u o f 0 . 6 0 a s a f i r s t a p p r o x i m a t i o n , o r T t = 6 0 , 0 0 0 l b / i n 2 . T h e r e q u i r e d

t u b e p r o p e r t i e s a r e n o w o b t a i n e d a s f o l l o w s :

n . J 2 1 T 8 0 0 0 R e q u i r e d — = — = — —

r y t T 6 0 , 0 0 0 o r

R e q u i r e d - = 0 . 0 6 6 7 i n 3

y T h e l i g h t e s t t u b e w h i c h m e e t s t h e r e q u i r e m e n t o f 0 . 0 6 6 7 i n 3 i s a l j - b y

0 . 0 4 9 - i n t u b e w i t h D / t = 3 0 . 6 0 , 1 / y = 0 . 0 7 8 4 7 , a n d a w e i g h t o f 6 . 3 2 l b / 1 0 0 i n .

F r o m F i g . 1 0 . 2 4 , r T / a , u = 0 . 5 3 , o r r r = 5 3 , 0 0 0 l b / i n 2 . T h e m a r g i n o f s a f e t y i s

T 8 0 0 0

M S = I l _ l = ^ _ l , 0 . 0 4 t , 5 1 , 0 0 0

O t h e r l u b e s a r e c o m p a r e d i n T a b l e 1 0 . 2 . T h e b y 0 . 0 4 9 - i n t u b e i s t h e o n l y

o n e w h i c h h a s a p o s i t i v e m a r g i n o f s a f e t y . A n y o t h e r t u b e s w h i c h a r e l i g h t e r

t h a n b y 0 . 0 4 9 i n a n d w h i c h h a v e a m i n i m u m w a l l t h i c k n e s s o f 0 . 0 4 9 i n a r e

o b v i o u s l y u n d e r t h e r e q u i r e d s t r e n g t h .

10.10 COMBINED STRESSES AND STRESS RATIOS

T h e d e s i g n o f m e m b e r s r e s i s t i n g t e n s i o n , b e n d i n g , o r t o r s i o n i s d i s c u s s e d i n

e a r l i e r s e c t i o n s . M a n y s t r u c t u r a l m e m b e r s , h o w e v e r , m u s t r e s i s t t h e s i m u l t a n e o u s

a c t i o n o f t w o o r m o r e o f t h e s e l o a d i n g c o n d i t i o n s . I f s t r e s s e s a r e b e l o w t h e e l a s t i c

l i m i t o f t h e m a t e r i a l , t h e n o r m a l a n d s h e a r s t r e s s e s a t a n y p o i n t m a y b e c o m b i n e d

b y u t i l i z i n g M o h r ' s c i r c l e . W h e n w o r k i n g s t r e s s e s b a s e d o n t h e e l a s t i c l i m i t a r e

u s e d , a s i n m o s t t y p e s o f s t r u c t u r a l d e s i g n , i t i s c u s t o m a r y t o d e t e r m i n e t h e

p r i n c i p a l s t r e s s e s a n d t h e m a x i m u m s h e a r i n g s t r e s s e s a t a p o i n t a n d t o c o m p a r e

t h e s e t o t h e a l l o w a b l e w o r k i n g s t r e s s e s .

W h e n m e m b e r s a r e d e s i g n e d o n t h e b a s i s o f u l t i m a t e s t r e n g t h , i t i s n o t

f e a s i b l e t o c a l c u l a t e t h e t r u e p r i n c i p a l s t r e s s e s i n t h e c a s e o f p l a s t i c b e n d i n g o r

t o r s i o n . E v e n if ( h e t r u e s t r e s s e s a r e k n o w n , i t i s d i f f i c u l t t o p r e d i c t t h e l o a d s a t

DESIGN OF MEMBERS IN TENSION, BENDING, OR TORSION 3 2 1

w h i c h f a i l u r e w o u l d o c c u r u n d e r c o m b i n e d l o a d i n g c o n d i t i o n s . A t e n s i o n - m e m b e r

f a i l s w h e n t h e a v e r a g e s t r e s s . r e a c h e s J h e ^ U i m a t e . t e n s i l e s t r e n g t h j r , , , f o r t h e

m a t e r i a l , b u t a m e m b e r r e s i s t i n g c o m b i n e d s t r e s s e s m a y f a i l b e f o r e t h e m a x i m u m

p r i n c i p a l s t r e s s r e a c h e s t h e v a l u e o f a l u . F a i j u r g j n p u r e s h e a r , f o r e x a m p l e , o c c u r s

w h e n t h e p r i n c i p a l t e n s i o n a n d c o m p r e s s i o n s t r e s s e s a r e a b o u t 0 . 6 f f „ , . V a r i o u s

t h e o r i e s o f f a i l u r e o f m a t e r i a l s u n d e r c o m b i n e d l o a d i n g h a v e b e e n c f e v e l o p e d , b u t

n o n e g i v e s a s i m p l e m e t h o d o f p r e d i c t i n g t h e f a i l u r e o f a l l m a t e r i a l s .

S h a n l e y a n d R y d e r h a v e p r o p o s e d a m e t h o d w h i c h p r o v i d e s a p r a c t i c a l

m e a n s o f c o n s i d e r i n g t h e c o m b i n a t i o n o f t h e fictitious s t r e s s e s o f b e n d i n g o r

t o r s i o n a n d o f o b t a i n i n g t h e a l l o w a b l e u l t i m a t e l o a d s f o r c o m b i n e d l o a d i n g s . T h i s

m e t h o d c o n s i s t s o f u s i n g s t r e s s r a t i o s a n d h a s b e e n e x t e n s i v e l y a d o p t e d i n t h e

a n a l y s i s o f flight v e h i c l e s t r u c t u r e s . T h e s t r e s s r a t i o m e t h o d m a y b e a p p l i e d t o

a l m o s t a n y c o m b i n a t i o n o f t w o o r m o r e t y p e s o f l o a d i n g , a l t h o u g h i n s o m e c a s e s

i t m a y b e n e c e s s a r y t o t e s t s o m e s p e c i m e n s i n o r d e r t o a p p l y t h e method. T h e

m e t h o d o f s t r e s s r a t i o s i s a p p l i e d first t o s o m e s p e c i a l c a s e s o f l o a d i n g a n d i s

s t a t e d l a t e r i n a g e n e r a l f o r m .

O n e o f t h e s i m p l e s t t y p e s o f c o m b i n e d l o a d i n g s i s t h a t o f t e n s i o n a n d b e n d -

i n g , a s s K o w n i n F i g . 1 0 . 2 5 . T h e s t r e s s e s m a y b e a d d e d a l g e b r a i c a l l y , a n d f o r

s m a l l l o a d s t h e s t r e s s i s P / A + M y / I a t a n y p o i n t i n t h e c r o s s s e c t i o n , a s s h o w n

i n F i g . 1 0 . 2 5 a . W h e n t h e s t r e s s e s e x c e e d t h e e l a s t i c l i m i t , h o w e v e r , t h e d i s -

t r i b u t i o n b e c o m e s s i m i l a r t o t h a t s h o w n i n F i g . 1 0 . 2 5 6 . T h e t r u e s t r e s s d i s -

t r i b u t i o n i s d i f f i c u l t t o c a l c u l a t e , a n d it i s c o n v e n i e n t t o u s e t h e m e t h o d o f s t r e s s

r a t i o s i n p r e d i c t i n g t h e s t r e n g t h . F o r p u r e b e n d i n g w i t h n o t o r s i o n , f a i l u r e w i l l

o c c u r w h e n t h e r a t i o o f t h e a p p l i e d b e n d i n g s t r e s s crb t o t h e a l l o w a b l e b e n d i n g

s t r e s s aB a p p r o a c h e s u n i t y (Rb = <Jb/aB = 1). S i m i l a r l y , f o r t e n s i o n w i t h n o b e n d -

i n g , f a i l u r e w i l l , o c c u r w h e n t h e t e n s i o n - s t r e s s r a t i o R, = <st/<Jtu a p p r o a c h e s u n i t y .

S i n c e t h e s t r e s s e s b e l o w t h e e l a s t i c l i m i t a d d d i r e c t l y , i t s e e m s l o g i c a l t o a d d t h e

s t r e s s r a t i o s , a n d t e s t s s u b s t a n t i a t e t h i s m e t h o d . T h e f a i l u r e u n d e r c o m b i n e d

t e n s i o n a n d b e n d i n g t h e r e f o r e o c c u r s u n d e r t h e f o l l o w i n g c o n d i t i o n :

Rb + R,= 1 ( 1 0 . 3 4 )

^ir5 to) Figure 10.25

T h e m a r g i n o f s a f e t y i s d e f i n e d b y t h e f o l l o w i n g e q u a t i o n , w h i c h c o r r e s p o n d s

w i l l i p r e v i o u s l y u s e d e x p r e s s i o n s w h e n c i t h e r o f t h e s t r e s s r a t i o s i s z e r o :

F o r r o u n d t u b e s i n c o m b i n e d b e n d i n g a n d t o r s i o n , t h e s t r e s s e s d o n o t a d d

a l g e b r a i c a l l y . F o r s t r e s s e s b e l o w t h e e l a s t i c l i m i t , t h e m a x i m u m s t r e s s a t a n y

p o i n t m a y b e o b t a i n e d r e a d i l y b y u s i n g M o h r ' s c i r c l e . T h e m a x i m u m t e n s i o n a n d

s h e a r i n g s t r e s s e s i n t h e t u b e s h o w n i n F i g . 1 0 . 2 6 a w i l l o c c u r a t t h e s u p p o r t a n d

o n t h e u p p e r s u r f a c e , a s s h o w n . T h e t e n s i o n s t r e s s i n t h e d i r e c t i o n o f t h e a x i s o f

t h e t u b e r r b is f o u n d f r o m t h e b e n d i n g m o m e n t M a s f o l l o w s :

<r4 = - - p ( 1 0 . 3 6 )

T h e s h e a r i n g s t r e s s T s o n t h e p l a n e s s h o w n i s o b t a i n e d f r o m t h e t o r s i o n a l m o -

m e n t T : = ( 1 0 . 3 7 )

T h e s m a l l e l e m e n t a t t h e t o p o f t h e t u b e is e n l a r g e d i n F i g . 1 0 . 2 6 6 . T h e p r i n c i p a l

s t r e s s e s a n d t h e m a x i m u m s h e a r i n g s t r e s s e s a t t h i s p o i n t o n t h e t u b e a r e o b t a i n e d

f r o m t h e s t r e s s e s a b a n d r , b y M o h r ' s c i r c l e , a s s h o w n i n F i g . 1 0 . 2 6 c . T h e m a x i -

m u m s h e a r i n g s t r e s s T m a j i s e q u a l t o t h e r a d i u s o f t h e c i r c l e

S u b s t i t u t i n g v a l u e s <rb a n d T, f r o m E q s . ( 1 0 . 3 6 ) a n d ( 1 0 . 3 7 ) i n t o E q . ( 1 0 . 3 8 ) y i e l d s

± M L ± J L = JFR (10.39) 2l/y 2l/y

Figure 11-37 (a) Two edges free; (b) one edge free; (c) no edge free.

Df.SIGN OF Mi'MRf.rS IN TENSION, BENDING, OR TORSION 3 2 3

w h e r e T c i s a n e q u i v a l e n t t o r q u e d e f i n e d b y E q . ( 1 0 . 3 9 ) . I f t h e t o r s i o n i s l a r g e r

t h a n t h e b e n d i n g m o m e n t , t h e m a x i m u m s h e a r i n g s t r e s s m a y b e u s e d t o p r e d i c t

t h e s t r e n g t h o f t h e t u b e . I f t h e b e n d i n g m o m e n t i s l a r g e i n c o m p a r i s o n t o t h e

t o r s i o n , t h e p r i n c i p a l s t r e s s e s a r e m o r e i m p o r t a n t t h a n t h e s h e a r i n g s t r e s s e s . I n

m a c h i n e d e s i g n p r a c t i c e , s h a f t s i n b e n d i n g a n d t o r s i o n a r e d e s i g n e d s o a s t o k e e p

t h e s h e a r i n g s t r e s s o f E q . ( 1 0 . 3 9 ) a n d t h e p r i n c i p a l s t r e s s e s s m a l l e r t h a n t h e

c o r r e s p o n d i n g a l l o w a b l e w o r k i n g s t r e s s e s , w h i c h a r e a c e r t a i n f r a c t i o n o f t h e

y i e l d s t r e s s .

W h e n a t u b e i s s t r e s s e d b e y o n d t h e e l a s t i c l i m i t , E q s . ( 1 0 . 3 6 ) a n d ( 1 0 . 3 7 ) d o

n o t y i e l d t h e t r u e s t r e s s e s , b u t y i e l d t h e fictitious s t r e s s e s d e f i n e d a s t h e b e n d i n g

m o d u l u s o f r u p t u r e a n d t h e t o r s i o n a l m o d u l u s o f r u p t u r e . E q u a t i o n s ( 1 0 . 3 8 ) a n d

( 1 0 . 3 9 ) , t h e r e f o r e , a r e n o t e x a c t w h e n t h e s t r e s s e s a r e b e y o n d t h e e l a s t i c l i m i t .

W h e n t h e t o r s i o n i s l a r g e i n c o m p a r i s o n t o t h e b e n d i n g m o m e n t , h o w e v e r , E q .

( 1 0 . 3 9 ) m a y b e u s e d t o p r e d i c t t h e u l t i m a t e f o r t h e a l l o w a b l e s t r e s s . A m o r e

a c c u r a t e m e t h o d o f d e s i g n i n g t u b e s i n t o r s i o n o r b e n d i n g i s b y s t r e s s r a t i o s . T h e

b e n d i n g a n d t o r s i o n a l m o d u l i o f r u p t u r e a h a n d T, , r e s p e c t i v e l y , a r e c a l c u l a t e d

f r o m E q s . ( 1 0 . 3 6 ) a n d ( 1 0 . 3 7 ) , a n d t h e a l l o w a b l e v a l u e s crB a n d T T a r e f o u n d b y t h e

s a m e m e t ! - ~ d s a s f o r t u b e s i n b e n d i n g o r t o r s i o n o n l y . T h e s t r e s s r a t i o i n b e n d i n g

Rh = ahjaR i s c o m b i n e d w i t h t h e s t r e s s r a t i o i n t o r s i o n Ra = x j x r i n t h e s a m e

m a n n e r a s t h e l o a d s a r c c o m b i n e d i n E q . ( 1 0 . 3 9 ) . F a i l u r e o c c u r s f o r t h e f o l l o w i n g

c o n d i t i o n :

R l + R l = 1 ( 1 0 . 4 0 )

T h e m a r g i n o f s a f e t y i s

M S = * ( 1 0 . 4 1 ) J R L + RL

W h e r e e i t h e r t h e b e n d i n g m o m e n t o r t h e t o r s i o n a l m o m e n t i s z e r o , E q s . ( 1 0 . 4 0 )

a n d ( 1 0 . 4 1 ) y i e l d v a l u e s w h i c h w e r e p r e v i o u s l y o b t a i n e d f o r b e n d i n g o r t o r s i o n

o n l v .

/? ; + '<] = l l . o - i - — ' 2

K y + iv: = i

\ V V

1 . 0 A' | Figure 10.27

3 2 4 AIRC'RAiT s t r u c t u r e s

E q u a t i o n s ( 1 0 . 3 4 ) a n d ( 1 0 . 4 0 ) r e p r e s e n t t w o w a y s i n w h i c h s t r e s s r a t i o s m a y

b e c o m b i n e d . T h e s e e q u a t i o n s m a y b e p l o t t e d w i t h t h e t w o s t r e s s r a t i o s a s

c o o r d i n a t e s , a s s h o w n i n F i g . 1 0 . 2 7 . T h e g r a p h f o r E q . ( 1 0 . 3 4 ) i s s h o w n b y t h e

s t r a i g h t l i n e , a n d t h a t f o r E q . ( 1 0 . 4 0 ) b y t h e c i r c l e . O t h e r c o m b i n a t i o n s o f l o a d i n g

c o n d i t i o n s m a y b e r e p r e s e n t e d b y a m o r e g e n e r a l e q u a t i o n :

Ri + Ryi= 1 ( 1 0 . 4 2 )

w h e r e t h e e x p o n e n t s x a n d y u s u a l l y m u s t b e f o u n d e x p e r i m e n t a l l y b y p l o t t i n g

t e s t r e s u l t s , a s s h o w n i n F i g . 1 0 . 2 7 , a n d b y w r i t i n g a n e q u a t i o n f o r a c u r v e p a s s i n g

t h r o u g h t h e p o i n t s .

E x a m p l e 1 0 . 6 A 2 - b y 0 . 0 9 5 - i n s t e e l t u b e i s h e a t - t r e a t e d t o a n u l t i m a t e t e n s i l e

s t r e n g t h a , u = 1 8 0 , 0 0 0 l b / i n 2 .

( a ) F i n d t h e m a r g i n o f s a f e t y if t h e l u b e r e s i s t s a d e s i g n t e n s i o n l o a d o f 5 0 , 0 0 0

l b a n d a d e s i g n b e n d i n g m o m e n t o f 3 0 , 0 0 0 i n • l b .

(h) F i n d t h e m a r g i n o f s a f e t y if t h e t u b e r e s i s t s a b e n d i n g m o m e n t o f ' 3 0 , 0 0 0

i n • l b a n d a t o r s i o n a l m o m e n t o f 5 0 , 0 0 0 i n • l b .

S O L U T I O N T h e t u b e p r o p e r t i e s a r e = 0 . 5 6 8 5 i n 2 , D / t = 2 1 . 0 5 , a n d / / V =

0 . 2 4 8 6 i n 3 .

(« ) F r o m F i g . 1 0 . 3 , a „ = 2 1 1 , 0 0 0 l b / i n 2 . T h e s t r e s s e s a n d s t r e s s r a t i o s a r c .

P 5 0 , 0 0 0 ,

0488 1 a,u 180,000

T h e m a r g i n o f s a f e t y is o b t a i n e d f r o m E q . ( 1 0 . 3 5 ) : S •

M S = — — " — r ~ — 1 = — 0 . 0 3 5 RT + R,

T h e n e g a t i v e m a r g i n o f s a f e t y i n d i c a t e s t h a t t h e l u b e i s u n s a t i s f a c t o r y .

(/>) T h e s t r e s s r a t i o f o r b e n d i n g i s t h e s a m e a s c o m p u t e d i n p a r t ( a ) . T h e

t o r s i o n a l m o d u l u s o f r u p t u r e is o b t a i n e d f r o m F i g . 1 0 . 2 4 , f o r D/t = 2 1 . 0 5

a n d T t = 0 . 5 8 x 1 8 0 , 0 0 0 = 1 0 4 , 0 0 0 l b / i n 2 . T h e s t r e s s r a t i o f o r t o r s i o n i s

7 ^ 5 0 , 0 0 0

' 21 2 x 0 . 2 5 8 6

r , 9 6 , 5 0 0 „ „ „ r = - L = — f — = 0 . 9 2 5

T T 1 0 4 , 0 0 0

DESIGN o r MEMBERS IN TENSION, BENDING, OR TORSION 3 2 5

T h e m a r g i n o f s a f e t y i s n o w c a l c u l a t e d f r o m E q . ( 1 0 . 4 1 ) :

M S = . J » - 1 = - 0 . 0 7 yjRb + Ra

T h e t u b e i s u n s a t i s f a c t o r y .

10.11 FAILURE THEORIES IN STRUCTURAL DESIGN

S i n c e t h e s t r e s s - r a t i o t e c h n i q u e w a s d e v e l o p e d b y S h a n l e y a n d R y d e r 3 5 y e a r s a g o ,

m a n y t h e o r i e s o f f a i l u r e i n s t r u c t u r a l d e s i g n h a v e b e e n p r e s e n t e d . T h e t h r e e m o s t

c o m m o n l y u s e d a r e t h e m a x i m u m s t r e s s t h e o r y , m a x i m u m s h e a r t h e o r y , a n d

m a x i m u m d i s t o r t i o n e n e r g y t h e o r y . 3 7

I n t h e m a x i m u m s t r e s s t h e o r y , f a i l u r e i s s a i d t o o c c u r i n a s t r u c t u r a l m e m b e r

u n d e r t h e a c t i o n o f c o m b i n e d s t r e s s e s w h e n o n e o f t h e p r i n c i p a l s t r e s s e s r e a c h e s

t h e f a i l u r e v a l u e ( y i e l d s t r e s s , u l t i m a t e s t r e s s , e t c . ) i n s i m p l e t e n s i o n c r 0 . I n a

t w o - d i m e n s i o n a l s t a t e o f s t r e s s , f a i l u r e i s d e f i n e d a s f o l l o w s :

F a i l u r e o c c u r s w h e n a l = + cr0 ( o ^ > er2)

F a i l u r e o c c u r s w h e n <x2 = +cr0 (a2 > c r j )

o r

• = + 1 ( f f , > <r2) ffo

^ = ± 1 (<r2 > <T0

( 1 0 . 4 3 )

I n t h e m a x i m u m s h e a r t h e o r y , f a i l u r e i s s a i d t o o c c u r i n a s t r u c t u r a l m e m b e r

u n d e r t h e a c t i o n o f c o m b i n e d s t r e s s e s w h e n t h e m a x i m u m s h e a r s t r e s s r e a c h e s t h e

v a l u e o f s h e a r f a i l u r e s t r e s s i n t e n s i o n , T m „ = e r 0 / 2 . U n d e r c o m b i n e d s t r e s s e s i n

t w o d i m e n s i o n s , t h r e e p o s s i b l e s h e a r v a l u e s e x i s t :

c r 0 01—02 R I = J =

t 2 = f = ± f ( 1 0 - 4 4 )

°D , ffl R 3 = J = ± J

T h e r e f o r e , f a i l u r e is d e f i n e d b y o n e o f t h e f o l l o w i n g :

cr i — cr 2 = ± £ r 0

A I -

0\ = ± c r 0

C H A P T E R

ELEVEN BUCKLING DESIGN OF STRUCTURAL MEMBERS

11.1 BEAM-DEFLECTION EQUATIONS

T h e m e t h o d s u s e d i n t h e d e s i g n o f c o m p r e s s i o n m e m b e r s , o r c o l u m n s , a r e b a s e d o n b e a m - d e f l e c t i o n e q u a t i o n s . C o l u m n s d o n o t fa i l a s a r e s u l t o f t h e d i r e c t c o m p r e s s i o n s t r e s s e s , o n l y a s a r e s u l t o f t h e c o m b i n e d c o m p r e s s i o n a n d b e n d i n g s t r e s s e s . S i n c e t h e m a g n i t u d e s o f t h e b e n d i n g s t r e s s e s d e p e n d o n t h e b e n d i n g d e f l e c t i o n s , i t is n e c e s s a r y t o d e r i v e t h e c o l u m n e q u a t i o n s f r o m b e a m - d e f l e c t i o n e q u a t i o n s .

T h e e q u a t i o n s Tor b e a m d e f l e c t i o n s a r c d e r i v e d f r o m t h e c u s t o m a r y a s s u m p -t i o n s t h a t s t r e s s i s p r o p o r t i o n a l t o s t r a i n a n d t h a t t h e d e f l e c t i o n s a r e srarall i n c o m p a r i s o n t o t h e o r i g i n a l d i m e n s i o n s . O n l y d e f o r m a t i o n s r e s u l t i n g f r o m b e n d -i n g s t r e s s e s i s u s u a l l y c o n s i d e r e d , b u t if s h e a r i n g d e f o r m a t i o n s a r e a p p r e c i a b l e , t h e y m a y b e c o m p u t e d s e p a r a t e l y a n d s u p e r i m p o s e d . A n i n i t i a l l y s t r a i g h t b e a m i s

- s h o w n w i t h e x a g g e r a t e d d e f l e c t i o n s i n F i g . 11 .1 . T h e t w o c r o s s s e c t i o n s a d i s t a n c e dx a p a r t a r e p a r a l l e l t o t h e u n s t r e s s e d c o n d i t i o n , b u t h a v e a r e l a t i v e a n g l e dO i n t h e s t r e s s e d c o n d i t i o n . T h e a n g l e dO m a y b e o b t a i n e d b y c o n s i d e r i n g t h e s m a l l t r i a n g l e b e t w e e n t h e n e u t r a l a x i s a n d a p o i n t a d i s t a n c e c b e l o w t h e n e u t r a l a x i s . T h e s t r e s s a a t t h i s p o i n t i s o b t a i n e d f r o m t h e f l e x u r e f o r m u l a :

« = f ( 1 1 . D

T h e l o n g i t u d i n a l f i b e r a d i s t a n c e c f r o m t h e n e u t r a l a x i s h a s a n e l o n g a t i o n a dx/E

328

BUCKLING DESIGN OF STRUCTURAL MEMBERS 3 2 9

F i g u r e 11.1

i n t h e l e n g t h d x . T h e a n g l e dO i s o b t a i n e d b y d i v i d i n g t h i s e l o n g a t i o n b y t h e

d i s t a n c e c , a s s h o w n i n F i g . 1 1 . 1 :

T h e d c f i c c t i o n c u r v e o f l h c b e a m c a n b e r e p r e s e n t e d b y x a n d y c o o r d i n a t e s ,

a s s h o w n i n F i g . 1 1 . 2 . T h e b e a m is a s s u m e d t o b e i n i t i a l l y s t r a i g h t a n d p a r a l l e l t o

t h e x a x i s . T h e d e f l e c t i o n s a r e s m a l l , a n d t h e a n g l e 0 b e t w e e n a t a n g e n t t o t h e

d e f l e c t i o n c u r v e a n d t h e .x a x i s i s s m a l l e n o u g h t h a t i t i s s u f f i c i e n t l y a c c u r a t e t o

a s s u m e t h a t t h e a n g l e i n r a d i a n s , t h e s i n e o f t h e a n g l e , a n d t h e t a n g e n t o f t h e

a n g l e a r e a l l e q u a l :

(11.2)

F r o m F q s . ( 1 1 . 1 ) a n d ( 1 1 . 2 ) ,

( 1 1 . 3 )

0 = s i n 0 = t a n 0 ( 1 1 . 4 )

( 1 1 . 5 )

Figure 11.40


F r o m E q s . ( 1 1 . 3 ) a n d ( 1 1 . 5 ) ,

d2y M (11-6) dx2 EI

A l l c o n v e n t i o n a l m e t h o d s o f o b t a i n i n g b e a m d e f l e c t i o n s a r e b a s e d o n E q . ( 1 1 . 6 ) .

I t s h o u l d b e n o t e d t h a t t h e d e f l e c t i o n s y a r e m e a s u r e d p o s i t i v e u p w a r d , s o t h a t a

p o s i t i v e b e n d i n g m o m e n t M p r o d u c e s a p o s i t i v e c u r v a t u r e d2y/dx2. E q u a t i o n

( 1 1 . 6 ) o f t e n i s d e r i v e d f r o m t h e a s s u m p t i o n t h a t y i s p o s i t i v e d o w n w a r d , a n d a

m i n u s s i g n i s i n t r o d u c e d b e c a u s e a p o s i t i v e b e n d i n g m o m e n t w o u l d t h e n p r o d u c e

a n e g a t i v e c u r v a t u r e .

11.2 LONG COLUMNS

C o m p r e s s i o n m e m b e r s t e n d t o f a i l a s a r e s u l t o f t h e l a t e r a l b e n d i n g i n d u c e d b y

t h e c o m p r e s s i o n l o a d , a n a c t i o n w h i c h i s c o m m o n l y t e r m e d buckling. I n t h e c a s e

o f c o l u m n s w h i c h a r e l o n g i n c o m p a r i s o n t o t h e i r o t h e r d i m e n s i o n s , e l a s t i c b u c k -

l i n g o c c u r s , o r t h e c o l u m n s b u c k l e w h e n t h e c o m p r e s s i v e s t r e s s e s a r e b e l o w t h e

e l a s t i c l i m i t . S u c h c o l u m n s a r e t e r m e d long columns. T h e i n i t i a l l y s t r a i g h t c o l u m n s h o w n i n F i g . 1 1 . 3 i s a s s u m e d t o b e h e l d i n t h e

d e f l e c t e d p o s i t i o n b y m e a n s o f t h e c o m p r e s s i v e f o r c e s P . T h e b e n d i n g m o m e n t a t

a n y c r o s s s e c t i o n i s f o u n d f r o m

M=-Py ( 1 1 . 7 )

I t i s a s s u m e d t h a t t h e m a t e r i a l d o e s n o t e x c e e d t h e e l a s t i c l i m i t a t a n y p o i n t , a n d

t h e r e f o r e E q . ( 1 1 . 6 ) i s a p p l i c a b l e . T h e d i f f e r e n t i a l e q u a t i o n o f t h e d e f l e c t i o n c u r v e

i s o b t a i n e d f r o m E q s . ( 1 1 . 6 ) a n d ( 1 1 . 7 ) :

d2y _ P

dx2 + EI 2 + ~ y = o ( i i . 8 )

T h e g e n e r a l s o l u t i o n o f E q . ( 1 1 . 8 ) i s

y = C t s i n / - £ - x- + C 2 c o s x tf1-9) \ LI \J

T h i s s o l u t i o n m a y b e v e r i f i e d b y s u b s t i t u t i o n a n d m u s t b e a g e n e r a l s o l u t i o n

o f t h e s e c o n d - o r d e r d i f f e r e n t i a l e q u a t i o n b e c a u s e i t c o n t a i n s t h e t w o a r b i t r a r y

c o n s t a n t s C i a n d C 2 . I n o r d e r t o s a t i s f y t h e e n d c o n d i t i o n s s h o w n i n F i g . 1 1 . 3 ,

t h e d c f l e c t i o n c u r v e m u s t p a s s t h r o u g h the points (x = 0 , y = 0 ) a n d {x = L ,


y = 0 ) . S u b s t i t u t i n g t h e first c o n d i t i o n i n t o E q . ( 1 1 . 9 ) y i e l d s C 2 = 0 . T h e s e c o n d

c o n d i t i o n , t h a t v = 0 w h e n .x = L , m a y b e s a t i s f i e d w h e n C l = 0 , w h i c h i s a

t r i v i a l s o l u t i o n , c o r r e s p o n d i n g t o t h e c o n d i t i o n s o f s m a l l l o a d s w h e n t h e c o l u m n

r e m a i n s s t r a i g h t . T h e o n l y s o l u t i o n o f i n t e r e s t i n c o l u m n a n a l y s i s i s t h a t f o r

w h i c h t h e c o l u m n i s d e f l e c t e d a n d C , i s n o t z e r o . T h i s s o l u t i o n , i n w h i c h t h e

c o l u m n i s b u c k l e d , i s o b t a i n e d w h e n t h e v a l u e o f P s a t i s f i e s t h e c o n d i t i o n

/-— L = it, 2n. 3n,..., nn V El n27i2EI

or P = ^ (11.10)

T h e v a l u e o f P i s o b v i o u s l y a m i n i m u m w h e n n = 1, a n d h i g h e r v a l u e s o f n h a v e

n o s i g n i f i c a n c e i n t h i s c a s e , s i n c e t h e c o l u m n w i l l f a i l a t t h e s m a l l e s t v a l u e o f P

t h a t w i l l p r o d u c e b u c k l i n g . T h e c r i t i c a l , o r b u c k l i n g , l o a d P „ i s t h e r e f o r e d e f i n e d

a s f o l l o w s :

( H . i i )

O f t e n i t i s m o r e c o n v e n i e n t t o w o r k w i t h a b u c k l i n g s t r e s s crc r = P J A . T h i s

m a y b e f o u n d b y i n t r o d u c i n g t h e r a d i u s o f g y r a t i o n o f t h e c r o s s - s e c t i o n a l a r e a ,

p =r J T J A , i n t o E q . (1 I . I 1) :

7i2E

T h e a n a l y s i s o f l o n g c o l u m n s w a s first p u b l i s h e d b y t h e S w i s s m a t h e m a t i c i a n

E u l e r . E q u a t i o n ( 1 1 . 1 1 ) [ o r E q . ( 1 1 . 1 2 ) ] i s c o m m o n l y c a l l e d t h e Euler equation, a n d t h e b u c k l i n g l o a d i s o f t e n c a l l e d t h e Euler load.

T h e v a l u e o f C l c a n n o t b e o b t a i n e d a t t h e c r i t i c a l l o a d . T h i s v a l u e i s e a u a l t o

t h e m a x i m u m d e l l c c t i o n 5 a t t h e c e n t e r o f t h e c o l u m n , w h i c h i s i n d e t e r m i n a t e f o r

t h e a s s u m e d c o n d i t i o n s . F o r l o a d s s m a l l e r t h a n P c r , t h e d e f l e c t i o n C , o r 5 m u s t

b e z e r o , o r e l s e t h e c o l u m n r e m a i n s s t r a i g h t . A t t h e c r i t i c a l l o a d , a n y d e f l e c t i o n 5

f o r w h i c h t h e m a x i m u m s t r e s s is b e l o w t h e e l a s t i c l i m i t w i l l s a t i s f y c o n d i t i o n s o f

e q u i l i b r i u m . T h i s m a y b e s h o w n e x p e r i m e n t a l l y b y l o a d i n g a l o n g c o l u m n i n a

s t a n d a r d t e s t i n g m a c h i n e . A s t h e e n d s o f t h e c o l u m n a r e m o v e d t o g e t h e r , t h e

c o l u m n r e m a i n s s t r a i g h t u n t i l t h e E u l e r l o a d i s o b t a i n e d . A s t h e e n d s c o n t i n u e t o

m o v e t o g e t h e r , t h e l o a d r e m a i n s c o n s t a n t a t t h e E u l e r l o a d , b u t t h e l a t e r a l

d e f l e c t i o n 5 i n c r e a s e s . If t h e e l a s t i c l i m i t i s n o t e x c e e d e d , t h e c o l u m n r e t u r n s t o i t s

i n i t i a l s h a p e w h e n t h e l o a d i s r e m o v e d .

II.3 ECCENTRICALLY LOADED COLUMNS

I n a n r s c t u a l s t r u c t u r e , it is n o t p o s s i b l e f o r a c o l u m n t o b e p e r f e c t l y s t r a i g h t o r t o

b e l o a d e d e x a c t l y a t t h e c e n t r o i d o f t h e a r e a . T h e a c t i o n o f a p r a c t i c a l l o n g

c o l u m n m a y b e a p p r o x i m a t e d b y t h e m e m b e r s h o w n i n F i g . 11 .4 , in w h i c h t h e

y

S

R ^ I N V

X

L Figure 11.4

c o l u m n is i n i t i a l ! ) ' s t r a i g h t b u t t h e l o a d s b o t h h a v e e c c e n t r i c i t y a . T h e a x e s o f c o o r d i n a t e s a r e t a k e n a s ^ h o w n . T h e e q u a t i o n o f t h e d c f l e c t i o n c u r v e is s t i l l r e p r e s e n t e d b y E q . ( 1 1 . 9 ) s i n c e E q s . ( 1 1 . 7 ) a n d ( 1 1 . 8 ) a r e a p p l i c a b l e , b u t c o n s t a n t s C j a n d C 2 m u s t b e f o u n d f r o m t h e c o n d i t i o n t h a t t h e d e f l e c t i o n c u r v e s a t i s f y t h e t w o c o n d i t i o n s .x = 0 , y = <5 + a a n d x = 0, dyjdx = 0 . S u b s t i t u t i n g t h e s e t w o c o n d i t i o n s i n t o E q . ( 1 1 . 9 ) p r o d u c e s

T h e v a l u e o f S c a n b e f o u n d n o w f r o m t h e c o n d i t i o n t h a t y = a f o r .x = L / 2 . B y

s u b s t i t u t i n g t h e s e v a l u e s i n t o E q . ( 1 1 . 1 3 ) t h e f o l l o w i n g v a l u e o f <5 is o b l a i n e d :

T h e d e f l e c t i o n S o f a n e c c e n t r i c a l l y l o a d e d c o l u m n t h u s i n c r e a s e s w i t h a n i n c r e a s e i n t h e l o a d P . A s l h c v a l u e o f P reaches t h e E u l e r l o a d P c r , a s d e f i n e d b y E q . ( 1 1 . 1 1 ) , t h e d e f l e c t i o n b e c o m e s i n f i n i t e , s i n c e s e c (JI /2) = o o . F i g u r e 11 .5 s h o w s t h e r e l a t i o n s h i p b e t w e e n P a n d <5 f o r v a r i o u s e c c e n t r i c i t i e s a, a s d e t e r m i n e d f r o m E q . ( 1 1 . 1 4 ) . A l l c u r v e s a r e a s y m p t o t i c t o t h e l i n e P = P c r , f o r t h i s is t h e t h e o r e t i c a l b u c k l i n g l o a d r e g a r d l e s s o f t h e e c c e n t r i c i t y o f l o a d i n g . A l a r g e d e f l e c t i o n m a y s t r e s s t h e m a t e r i a l b e y o n d t h e e l a s t i c l i m i t a n d c a u s e f a i l u r e b e f o r e t h e E u l e r l o a d is o b t a i n e d , s i n c e t h e n t h e l o n g - c o l u m n e q u a t i o n s w o u l d n o l o n g e r a p p l y .

C o l u m n s o f a n y s p e c i f i c m a t e r i a l a r e c l a s s i f i e d a c c o r d i n g t o t h e i r s l e n d e r n e s s

r a t i o L / p . F o r a s l e n d e r n e s s r a t i o g r e a t e r t h a n a c e r t a i n c r i t i c a l v a l u e , t h e c o l u m n

is a l o n g c o l u m n a n d i s a n a l y z e d b y E q . ( 1 1 . 1 2 ) . S h o r t c o l u m n s h a v e a s l e n d e r n e s s

r a t i o l e s s t h a n t h i s c r i t i c a l v a l u e . T h e c r i t i c a l L i p c o r r e s p o n d s t o t h e v a l u e f o r

w h i c h t h e m a x i m u m c o m p r e s s i v e s t r e s s in t h e c o l u m n is e q u a l t o t h e s t r e s s a t

w h i c h t h e c o m p r e s s i v e s t r e s s - s t r a i n c u r v e d e v i a t e s f r o m a s t r a i g h t l i n e , a s s h o w n

b y p o i n t B i n F i g . 11 .6 . U s u a l l y t h i s s t r e s s i s c o n s i d e r a b l y s m a l l e r t h a n t h e y i e l d

s t r e s s , p o i n t C o f F i g . 11 .6 a t w h i c h t h e m a t e r i a l h a s a p e r m a n e n t u n i t e l o n g a t i o n

M o s t f l i g h t v c h i c l e m a t e r i a l s h a v e s t r e s s - s t r a i n c u r v e s s i m i l a r t o t h a t s h o w n

in F i g . 11.6 , i n w h i c h t h e s t r e s s - s t r a i n c u r v e h a s a p o s i t i v e s l o p e a t a l l p o i n t s . T h e

( 1 1 . 1 3 )

( 1 1 . 1 4 )

11.4 SHORT COLUMNS

o f 0 . 0 0 2 .

mjc 'Ki inc ; d i s k j n oi- s t r u c t u r a l mi-:miu;ks 3 3 3

/

6 Figure 1 1 5

c o n s t a n t s l o p e o f t h i s s t r e s s - s t r a i n c u r v e b e l o w t h e e l a s t i c l i m i t is e q u a l t o t h e

m o d u l u s o f e l a s t i c i t y E , a n d t h e v a r i a b l e s l o p e a b o v e t h e e l a s t i c l i m i t i s t e r m e d

t h e tangent modulus of elasticity E,. D u c t i l e m a t e r i a l s , s u c h a s m i l d s t e e l , m a y

h a v e a 7 ,e ro o r n e g a t i v e v a l u e o f E , n e a r t h e y i e l d p o i n t . I f E , i s p o s i t i v e a t a l l

p o i n t s , a s h o r t c o l u m n m a y r e m a i n p e r f e c t l y s t r a i g h t w h e n l o n d e d t o s t r e s s e s

b e y o n d t h e y i e l d p o i n t . I f s u c h a c o l u m n h a s a s l i g h t l a t e r a l d e f l e c t i o n , t h e

i n t e r n a l r e s i s t i n g m o m e n t is f o u n d f r o m a n e q u a t i o n s i m i l a r t o E q . ( 1 1 . 6 ) , e x c e p t

t h a t E i s r e p l a c e d b y £ , , t h e t a n g e n t m o d u l u s f o r t h e c o m p r e s s i v e s t r e s s :

I f t h i s i n t e r n a l r e s i s t i n g m o m e n t is g r e a t e r t h a n t h e b e n d i n g m o m e n t p r o d u c e d b y

t h e l o a d P , t h e c o l u m n w i l l r e m a i n s t r a i g h t w h e n l o a d e d . I f t h e i n t e r n a l r e s i s t i n g

m o m e n t i s n o t a s l a r g e a s t h e e x t e r n a l b e n d i n g m o m e n t , t h e d e f l e c t i o n w i l l

i n c r e a s e a n d t h e c o l u m n w i l l p r o b a b l y f a i l . W h e n t h e b e n d i n g m o m e n t o f t h e

l o a d P i s e q u a l t o t h e r e s i s t i n g m o m e n t d e f i n e d b y E q . ( 3 1 . 1 5 ) , P m a y b e o b t a i n e d

i n t h e s a m e m a n n e r a s i n t h e E u l e r e q u a t i o n , b u t w i t h £ , s u b s t i t u t e d f o r E :

( 1 1 . 1 5 ) .

(11.16)

o Figure 11.6


T h i s e q u a t i o n i s c a l l e d t h e tangent modulus equation, o r t h e Engesscr equation. T h e t a n g e n t m o d u l u s e q u a t i o n d o e s n o t q u i t e r e p r e s e n t t h e t r u e c o n d i t i o n s

f o r s h o r t c o l u m n s . A t p o i n t C o f t h e s t r e s s - s t r a i n d i a g r a m i n F i g . 1 1 . 6 , a s m a l l

i n c r e a s e i n t h e c o m p r e s s i v e s t r a i n p r o d u c e s a n i n c r e a s e i n c o m p r e s s i v e s t r e s s , a s

d e t e r m i n e d b y t h e p o r t i o n C G o f t h e c u r v e , w h i c h h a s s l o p e E , . A s m a l l d e c r e a s e

i n t h e c o m p r e s s i v e s t r a i n , h o w e v e r , p r o d u c e s a d e c r e a s e i n s t r e s s , a s i n d i c a t e d b y

l i n e C D , w h i c h h a s s l o p e E . I f a s h o r t c o l u m n d e f l e c t s l a t e r a l l y i n s u c h a w a y t h a t

t h e c o m p r e s s i v e s t r a i n o n t h e c o n v e x s i d e i s d e c r e a s e d , t h e r e s i s t i n g m o m e n t w i l l

b e g r e a t e r t h a n t h a t g i v e n . b y E q . ( 1 1 . 1 5 ) b e c a u s e t h e m o d u l u s o f e l a s t i c i t y f o r

p a r t o f t h e c r o s s s e c t i o n i s E r a t h e r t h a n E , . T h u s t h e c o r r e c t m o d u l u s o f

e l a s t i c i t y s h o u l d b e a v a l u e b e t w e e n E a n d E , . V a l u e s o f a n e f f e c t i v e m o d u l u s o f

e l a s t i c i t y s h o u l d b e d e r i v e d o n t h e a s s u m p t i o n t h a t t h e c o l u m n i s s u p p o r t e d

l a t e r a l l y a n d r e m a i n s s t r a i g h t u n t i l t h e u l t i m a t e l o a d i s a p p l i e d a n d t h e n b u c k l e s

w i t h n o c h a n g e i n a x i a l l o a d . T h e c o l u m n f o r m u l a o b t a i n e d b y s u b s t i t u t i n g t h i s

m o d u l u s i n t o t h e E u l e r e q u a t i o n i s t e r m e d t h e reduced modulus equation a n d

f r e q u e n t l y i s r e f e r r e d t o i n t h e l i t e r a t u r e .

S h a n l e y 8 h a s s h o w n t h a t t h e c o r r e c t l o a d r e s i s t e d b y a s h o r t c o l u m n i s

b e t w e e n t h e v a l u e s g i v e n b y t h e t a n g e n t m o d u l u s e q u a t i o n a n d b y t h e r e d u c e d

m o d u l u s e q u a t i o n . T h e t a n g e n t m o d u l u s e q u a t i o n y i e l d s v a l u e s w h i c h a r e s l i g h t l y

l o w , s i n c e s o m e s t r a i n r e v e r s a l m u s t t a k e p l a c e b e f o r e t h e u l t i m a t e c o l u m n l o a d i s •

r e a c h e d . T h e r e d u c e d m o d u l u s e q u a t i o n a l w a y s y i e l d s v a l u e s w h i c h a r e l o o h i g h ,

s i n c e t h e c o l u m n i s n o t l a t e r a l l y s u p p o r t e d w h e n t h e l o a d i s a p p l i e d . T h e t a n g e n t

m o d u l u s e q u a t i o n i s u s e d f r e q u e n t l y , b e c a u s e i t c o r r e s p o n d s c l o s e l y t o t e s t r e s u l t s

a n d i s a l w a y s c o n s e r v a t i v e .

I t i s c u s t o m a r y t o r e p r e s e n t c o l u m n e q u a t i o n s b y p l o t t i n g t h e a v e r a g e c o m -

p r e s s i v e s t r e s s crc = P / A a g a i n s t t h e s l e n d e r n e s s r a t i o L / p . F i g u r e 1 1 . 7 s h o w s s u c h

c u r v e s . T h e s l e n d e r n e s s r a t i o b e y o n d w h i c h t h e m a t e r i a l a c t s a s a l o n g c o l u m n i s

a b o u t 1 1 5 . T h e s t r e s s a t t h i s p o i n t c o r r e s p o n d s t o t h e s t r e s s a t p o i n t B o f F i g .

1 1 . 6 . F o r s l e n d e r n e s s r a t i o s l e s s t h a n 1 1 5 , t h e c o m p r e s s i v e s t r e s s is h i g h e r , a n d t h e

t a n g e n t m o d u l u s o f e l a s t i c i t y E , i s s m a l l e r t h a n E . T h u s t h e p o i n t s o n t h e c o l u m n

c u r v e a r e b e l o w t h e E u l e r c u r v e i n t h e s h o r t - c o l u m n r a n g e . T h e t e s t p o i n t s a r e

s e e n t o f o l l o w t h e t a n g e n t m o d u l u s c u r v e v e r y c l o s e l y . S u c h t e s t l o a d s a r e a l w a y s

s l i g h t l y l o w e r t h a n t h e o r e t i c a l l o a d s b e c a u s e o f u n a v o i d a b l e e c c e n t r i c i t i e s o f J o a d -

i n g . T h e c u r v e s s h o w n i n F i g . 1 1 . 7 r e p r e s e n t v a l u e s f o r a n a c t u a l s p e c i m e n ,

w h e r e a s s i m i l a r d e s i g n c u r v e s a r e b a s e d o n m i n i m u m g u a r a n t e e d p r o p e r t i e s o f

t h e m a t e r i a l a n d g i v e s o m e w h a t l o w e r s t r e s s e s .

11.5 COLUMN END FIXITY

I n t h e p r e v i o u s a n a l y s i s , w e a s s u m e t h a t t h e c o l u m n i s h i n g e d a t b o t h e n d s s o

t h a t i t c a n r o t a t e f r e e l y . I n m o s t c a s e s , h o w e v e r , c o m p r e s s i o n m e m b e r s a r e c o n -

n e c t e d i n s u c h a w a y t h a t t h e y a r e r e s t r a i n e d a g a i n s t r o t a t i o n a t t h e e n d s . I n

o r d e r t o h a v e t h e m e a n s o f d e t e r m i n i n g t h e b u c k l i n g l o a d f o r a c o l u m n w i t h

v a r i o u s e n d fixities, F .q . ( 1 1 . 8 ) i s w r i t t e n a s f o l l o w s :

w h e r e k2 = P/(EI). T h e s o l u t i o n o f E q . ( 1 1 . 1 7 ) c a n b e f o u n d e a s i l y :

l' = C , s i n hx + C2 c o s hx -)- C , x + C 4

( 1 1 . 1 7 )

(11.18) C o n s t a n t s t h r o u g h C 4 a r c d e t e r m i n e d f r o m t h e b o u n d a r y c o n d i t i o n s a t t h e

e n d s o f t h e c o l u m n . F o r i n s t a n c e , i n t h e c a s e o f p i n - e n d e d c o l u m n s , t h e b o u n d a r y

c o n d i t i o n s a r e

y = 0

v- = 0

dzy/dx2 = 0 d2y/dx2 = 0

a t x - 0

a t x = L F o r c o l u m n s w h i c h a r e c o m p l e t e l y f i x e d a t b o t h e n d s , t h e b o u n d a r y c o n d i t i o n s

a r e

,V(0) = 0

y{I-) = 0

J Y _

dx(0) dy

dx(L)

= 0

0

C o l u m n s w h i c h a r e fixed a t x = 0 a n d p i n n e d a t x = L h a v e t h e b o u n d a r y

c o n d i t i o n s

3'<0) = 0

dy d.x(0) = 0

y ( L ) = o

d2y dx (L) = 0

Figure 11.40


U p o n s u b s t i t u t i n g t h e b o u n d a r y c o n d i t i o n s i n E q . ( 1 1 . 1 8 ) , a s e t o f f o u r a l g e -

b r a i c h o m o g e n e o u s e q u a t i o n s i s o b t a i n e d . F o r n o n t r i v i a l s o l u t i o n s , t h e d e t e r m i -

n a n t o f t h e u n d e t e r m i n e d c o e f f i c i e n t s C t t o C A m u s t v a n i s h . T h i s w i l l y i e l d a

t r a n s c e n d e n t a l e q u a t i o n f r o m w h i c h t h e b u c k l i n g l o a d i s d e t e r m i n e d . F o r e x a m -

p l e , s u b s t i t u t i n g t h e l a s t s e t o f b o u n d a r y c o n d i t i o n s ( c o l u m n fixed a t o n e e n d a n d

p i n n e d a t t h e o t h e r ) y i e l d s

C 2 + C 4 = 0

J t C i + C 3 = 0

(sin kL)Ct + (cos kL)Cz + LC3 + C4 = 0 (k2 sin kh)Ci + (k2 cos kL)C-2 = 0

F o r n o n t r i v i a l s o l u t i o n s o f t h e a b o v e s e t o f e q u a t i o n s , t h e d e t e r m i n a n t m u s t

v a n i s h , i .e . ,

0

k s i n kL

k2 s i n kL

1 0 1

0 1 0

c o s kL L 1 k2 c o s kL 0 0

= 0

E x p a n d i n g y i e l d s t h e f o l l o w i n g :

k2{kL cos kL- sin kL) = 0

S i n c e k2 = P/{EI), k c a n n o t b e z e r o . T h e r e f o r e

kL cos kL — s i n kL = 0

o r

t a n k L = k L ' ( t r a n s c e n d e n t a l e q u a t i o n )

T h e c r i t i c a l b u c k l i n g l o a d o c c u r s a t t h e s m a l l e s t v a l u e o f k L w h i c h s a t i s f i e s t h e

t r a n s c e n d e n t a l e q u a t i o n a b o v e . T h i s v a l u e c a n b e s h o w n l o b e e q u a l t o s j l n .

H e n c e t h e c r i t i c a l b u c k l i n g l o a d i s

P„ = In2 El

( 1 1 . 1 9 )

I f a c o m p r e s s i o n m e m b e r i s r i g i d l y fixed a g a i n s t r o t a t i o n a t b o t h e n d s , t h e

d e f l e c t i o n c u r v e f o r e l a s t i c b u c k l i n g m a y b e d e t e r m i n e d b y u s i n g t h e a b o v e

p r o c e d u r e a n d w i l l h a v e t h e s h a p e s h o w n i n F i g . 1 1 . 8 6 . A t t h e q u a r t e r p o i n t s o f

t h e fixed c o l u m n , t h e r e w i l l b e p o i n t s o f r e v e r s e c u r v a t u r e , o r p o i n t s o f c o n t r a f l e x -

u r e . A t p o i n t s o f c o n t r a f i e x u r e t h e r e i s n o c u r v a t u r e a n d h e n c e n o b e n d i n g

m o m e n t . T h e p o r t i o n o f t h e c o l u m n b e t w e e n p o i n t s o f c o n t r a f i e x u r e t h u s m a y b e

t r e a t e d a s a p i n - e n d e d c o l u m n . T h e l e n g t h L ' b e t w e e n t h e p o i n t s o f c o n t r a f i e x u r e

i s u s e d i n p l a c e o f L i n t h e c o l u m n e q u a t i o n s p r e v i o u s l y d e r i v e d , a n d t h e s l e n d e r -

n e s s r a t i o i s d e f i n e d a s L j p . A n e n d - f i x i t y t e r m c i s u s e d o f t e n a n d i s d e f i n e d i n t h e

[UJCKLiNG DESIGN OF STRUCTURAL MEMBERS 337

f o l l o w i n g e q u a t i o n :

C, K 2E CK2E

(li/p)2 ~ (L/f>)2 f o r l o n g c o l u m n s o n l y (11.20)

L = L

o r fc

f o r a l l c o l u m n s (11.21) v

F o r t h e f i x e d - e n d c o n d i t i o n o f F i g . 11.8/ ) , L = L / 2 a n d c = 4 . T h u s t h e f i x e d - e n d

c o l u m n w i l l r e s i s t 4 t i m e s t h e l o a d o f a s i m i l a r p i n - e n d e d c o l u m n , if b o t h a r e i n

t h e l o n g - c o l u m n range. T h i s s a m e r e l a t i o n d o e s n o t h o l d i n t h e s h o r t - c o l u m n

r a n g e , b c c a u s c t h e v a l u e o f E, i n E q . ( 1 1 . 1 6 ) is s m a l l e r f o r t h e s m a l l e r v a l u e s o f U .

T h i s f a c t i s e v i d e n t f r o m F i g . 11 .7 , w h e r e i t m a y b e s e e n t h a t a r e d u c t i o n i n L ' / p

h a s a m u c h s m a l l e r e f f e c t o n a c i n t h e s h o r t - c o l u m n r a n g e t h a n i t h a s i n t h e E u l e r

c o l u m n r a n g e .

I n o r d e r t o o b t a i n c o m p l e t e e n d fixity, t h e c o m p r e s s i o n m e m b e r m u s t b e

a t t a c h e d t o a s t r u c t u r e o f i n f i n i t e r i g i d i t y a t b o t h e n d s . T h i s c o n d i t i o n i s a p -

p r o a c h e d l e s s f r e q u e n t l y in p r a c t i c e t h a n t h e c o n d i t i o n o f h i n g e d e n d s . M o s t

p r a c t i c a l c o l u m n s h a v e e n d c o n d i t i o n s s o m e w h e r e b e t w e e n h i n g e d a n d fixed, a s

s h o w n i n F i g . 1 1 . 8 c . T h e e n d s a r e r i g i d l y a t t a c h e d t o a s t r u c t u r e w h i c h d e f l e c t s

a n d p e r m i t s t h e e n d s t o r o t a t e s l i g h t l y . T h e t r u e e n d - f i x i t y c o n d i t i o n s s e l d o m c a n

b e d e t e r m i n e d c x a c t l y , a n d s o c o n s e r v a t i v e a s s u m p t i o n s m u s t b e m a d e . F o r t u -

n a t e l y , s h o r t c o l u m n s u s u a l l y a r e u s e d , a n d t h e e f f e c t o f e n d fixity o n t h e a l l o w -

a b l e c o m p r e s s i v e s t r e s s i s m u c h s m a l l e r t h a n i t w o u l d b e f o r l o n g c o l u m n s .

O t h e r c o m m o n e n d c o n d i t i o n s f o r c o l u m n s a r e s h o w n i n F i g . 11 .9 . F o r t h e

c o l u m n f i x e d a t o n e e n d a n d f r e e t o b o t h r o t a t e a n d m o v e l a t e r a l l y a t t h e o t h e r

e n d , a s s h o w n i n F i g . 1 1 . 9 « , l e n g t h L ' is t w i c e l e n g t h L , s i n c e t h e c o l u m n i s s i m i l a r

t o o n e - h a l f o f t h e c o l u m n w i t h t w o h i n g e d e n d s . T h e c o l u m n w i t h o n e e n d fixed

a n d t h e o t h e r e n d f r e e t o r o t a t e b u t n o t f r e e t o m o v e l a t e r a l l y h a s a n e f f e c t i v e

l e n g t h L ' s 0 . 7 L , a s s h o w n in F i g . 1 1 A

W e l d e d t r u s s e s m a d e o f s t e e l t u b e s f r e q u e n t l y a r e u t i l i z e d i n v e h i c l e s t r u c -

i/>) Figure 11.8


(a) (hi Figure U . 9

l u r e s . T h e e n d s o f a c o m p r e s s i o n m e m b e r i n s u c h a t r u s s c a n n o t r o t a t e w i t h o u t

b e n d i n g a l l t h e o t h e r m e m b e r s a t t h e e n d j o i n t s . S u c h a m e m b e r i s s h o w n i n F i g .

1 1 . 1 0 . T h e p r o b l e m o f o b t a i n i n g t h e t r u e e n d fixity o f s u c h a c o m p r e s s i o n

m e m b e r i s d i f f i c u l t , s i n c e t h e m e m b e r m a y b u c k l e e i t h e r h o r i z o n t a l l y o r v e r t i c a l l y

a n d i s r e s t r a i n e d b y t h e t o r s i o n a l a n d b e n d i n g r i g i d i t i e s o f m a n y o t h e r m e m b e r s .

F o r a s t e e l - t u b e f u s e l a g e t r u s s , u s u a l l y i t i s c o n s e r v a t i v e t o a s s u m e c = 2 . 0 f o r a l l

m e m b e r s . I f a v e r y h e a v y c o m p r e s s i o n m e m b e r i s r e s t r a i n e d b y c o m p a r a t i v e l y

l i g h t m e m b e r s , a s m a l l e r e n d f i x i t y m i g h t b e o b t a i n e d . S i m i l a r l y , a l i g h t c o m -

p r e s s i o n m e m b e r r e s t r a i n e d b y h e a v y m e m b e r s m a y a p p r o a c h t h e fixity c o n d i t i o n

c = 4 . I f a l l t h e m e m b e r s a t a j o i n t a r e c o m p r e s s i o n m e m b e r s , t h e y m a y a l l h a v e a

t e n d e n c y t o r o t a t e i n t h e s a m e d i r e c t i o n , s o t h a t n o n e h e l p s r e s t r a i n t h e o t h e r s ,

a n d a l l s h o u l d b e d e s i g n e d a s p i n - e n d e d . W h e r e t h i s r a r e c a s e e x i s t s w i t h m e m -

b e r s i n a n y p l a n e , t h e m e m b e r s p e r p e n d i c u l a r t o t h i s p l a n e p r o b a b l y w o u l d

s u p p l y t o r s i o n a l r e s t r a i n t t o t h e j o i n t . T e n s i o n m e m b e r s t h a t c o n n e c t t o t h e e n d s

o f c o m p r e s s i o n m e m b e r s s u p p l y g r e a t e r r e s t r a i n t t h a n s i m i l a r c o m p r e s s i o n m e m -

b e r s . O f t e n s t e e l - t u b e e n g i n e m o u n t s a r e d e s i g n e d w i t h t h e c o n s e r v a t i v e a s s u m p -

t i o n o f p i n - e n d e d m e m b e r w i t h c = 1.0.

S t r i n g e r s w h i c h a c t a s c o m p r e s s i o n m e m b e r s i n s e m i m o n o c o q u e w i n g o r

Figure 11. U

f u s e l a g e s t r u c t u r e s o f t e n a r e s u p p o r t e d b y c o m p a r a t i v e l y f l e x i b l e r i b s o r b u l k -

h e a d s . S u c h a s t r i n g e r i s s h o w n i n F i g . 1 1 . 1 1 . S i n c e t h e r i b s o r b u l k h e a d s a r e f r e e

t o t w i s t a s s h o w n , t h e i r r e s t r a i n i n g e f f e c t i s n e g l e c t e d a n d t h e e f f e c t i v e c o l u m n

l e n g t h L ' i s a s s u m e d e q u a l t o t h e l e n g t h L b e t w e e n b u l k h e a d s . W h e r e t h e b u l k -

h e a d s a r e r i g i d e n o u g h t o p r o v i d e r e s t r a i n t a n d c l i p s a r e p r o v i d e d t o a t t a c h t h e

s t r i n g e r s t o t h e b u l k h e a d s , a v a l u e c = 1 .5 , c o r r e s p o n d i n g t o a n e f f e c t i v e l e n g t h L '

o f 0 . S 1 5 L , i s s o m e t i m e s u s e d .

11.6 EMPIRICAL FORMULAS FOR SHORT COLUMNS

O n e d i s a d v a n t a g e o f t h e t a n g e n t m o d u l u s f o r m u l a f o r s h o r t c o l u m n s i s t h a t t h e

r e l a t i o n b e t w e e n t h e a l l o w a b l e c o l u m n s t r e s s a c a n d L ' j p c a n n o t b e e x p r e s s e d b y

a s i m p l e e q u a t i o n . I t i s o f t e n m o r e c o n v e n i e n t t o e x p r e s s t h i s r e l a t i o n s h i p b y a

s i m p l e a p p r o x i m a t e e q u a t i o n w h i c h i s r e a s o n a b l y c l o s e to t h e p o i n t s o b t a i n e d

d i r e c t l y f r o m c o l u m n t e s t s o r f r o m t h e t a n g e n t m o d u l u s e q u a t i o n . T h e s h o r t -

c o l u m n c u r v e s f o r m a n y m a t e r i a l s a p p r o x i m a t e t h e p a r a b o l a

rrc = a c 0 - K ^ J ( 1 1 . 2 2 )

T h e c o n s t a n t s cr,.0 a n d K m u s t b e c h o s e n s o t h a t t h e p a r a b o l a f i t s t h e t e s t d a t a

a n d i s t a n g e n t t o t h e F u l e r c u r v e . B y e q u a t i n g t h e s l o p e o f t h i s p a r a b o l a t o t h a t

o f t h e E u l e r c u r v c a t I h e p o i n t o f t a n g e n c y a n d s u b s t i t u t i n g t h e r e s u l t i n g v a l u e o f

K i n t o E q . ( 1 1 . 2 2 ) , t h e f o l l o w i n g e q u a t i o n i s o b t a i n e d :

= <rco QAVP)1

4 j I 2 E ( 1 1 . 2 3 )

T h e t e r m ac0 i s c a l l e d t h e column yield stress. I t h a s l i t t l e p h y s i c a l s i g n i f i c a n c e ,

s i n c e v e r y s h o r t c o l u m n s { L / p < 12) f a i l b y b l o c k c o m p r e s s i o n r a t h e r t h a n

c o l u m n a c t i o n , a n d E q . ( 1 1 . 2 3 ) is n o t a p p l i c a b l e i n t h i s r a n g e . T h e v a l u e o f t r c 0 i s

d e t e r m i n e d s o t h a t E q . ( 1 1 . 2 3 ) w i l l fit s h o r t - c o l u m n t e s t d a t a f o r v a l u e s o f L / p

a b o v e t h e b l o c k c o m p r e s s i o n r a n g e .

T h e g e n e r a l s e c o n d - d e g r e e p a r a b o l a e q u a t i o n i s s h o w n i n F i g . 1 1 . 1 2 w i t h t h e

c o r r e s p o n d i n g E u l e r e q u a t i o n . T h e v a l u e o f (Tc0 r e p r e s e n t s t h e i n t e r c e p t o f t h i s

0<

N, 4 ]

( T / p ?

0 v / 2 n \ / T Figure 11.12

c u r v e a t t h e p o i n t C / p = 0 . T h e p a r a b o l a i s a l w a y s t a n g e n t t o t h e E u l e r c u r v e

w h e n <rc0 = <x c 0 / 2 , a s i s f o u n d b y s o l v i n g E q . ( 1 1 . 2 3 ) s i m u l t a n e o u s l y w i t h t h e E u l e r

e q u a t i o n , i n t h e s a m e w a y , t h e c r i t i c a l s l e n d e r n e s s r a t i o , w h i c h d i v i d e s t h e l o n g -

c o l u m n a n d s h o r t - c o l u m n r a n g e s , i s f o u n d t o b e Ljp = yj2 n y/E/<jc0 . T h e s h o r t - c o l u m n c u r v e s f o r m o s t a l u m i n u m a l l o y s a n d f o r s e v e r a l o t h e r

m a t e r i a l s a r e r e p r e s e n t e d m o r e a c c u r a t e l y b y s t r a i g h t l i n e s . A s t r a i g h t l i n e

t a n g e n t t o t h e E u l e r c u r v e h a s t h e e q u a t i o n :

T h e c o o r d i n a t e s o f t h e p o i n t o f t a n g e n c y o f t h i s c u r v e a n d t h e E u l e r c u r v e a r e

L/p = yfi n y/E/aco a n d ac = < r c 0 / 3 ( s e e F i g . 1 1 . 1 3 ) . T h e v a l u e o f L/p a t t h i s

p o i n t i s t h e c r i t i c a l v a l u e d i v i d i n g t h e s h o r t - c o l u m n a n d l o n g - c o l u m n r a n g e s .

O t h e r m a t e r i a l s h a v e c o l u m n c u r v e s w h i c h m a y b e r e p r e s e n t e d b y a s e m i -

c u b i c e q u a t i o n . A 1 . 5 - d e g r e e e q u a t i o n w h i c h i s t a n g e n t t o t h e E u l e r c u r v e h a s t h e

f o r m

T h e c o o r d i n a t e s o f t h e p o i n t o f t a n g e n c y a r e a g a i n f o u n d b y s o l v i n g t h e . ^ h o r t -

( 1 1 . 2 4 )

( 1 1 . 2 5 )

0.385/,','/>

0 L'h>

BUCKLING DP-SIGN OF STRUCTURAL MEMBERS 3 4 1

c o l u m n a n d l o n g - c o l u m n e q u a t i o n s s i m u l t a n e o u s l y . T h e c r i t i c a l s l e n d e r n e s s r a t i o

i s ILjp = \.SThtyjK/(Tc0 c o r r e s p o n d i n g t o a s t r e s s o f t r c = 0 . 4 2 9 C T c O .

E q u a t i o n s ( 1 1 . 2 3 ) t o ( 1 1 . 2 5 ) a n d t h e E u l e r e q u a t i o n m a y b e e x p r e s s e d i n

d i m c n s i o n l e s s f o r m b y u s i n g c o o r d i n a t e s B a n d R a a s d e f i n e d b y

A V '

B = — ( 1 1 . 2 6 )

R . = — ( 1 1 . 2 7 )

cO

T h e E u l e r e q u a t i o n , E q . ( 1 1 . 1 2 ) , t h e n b e c o m e s

= ( 1 1 - 2 8 )

E q u a t i o n s ( 1 1 . 2 3 ) t o ( 1 1 . 2 5 ) w i l l h a v e t h e f o l l o w i n g f o r m s , r e s p e c t i v e l y :

R „ = 1.0 - 0 . 2 5 J 5 2 ( 1 1 . 2 9 )

R „ = 1.0 - 0 . 3 0 2 7 B 1 - 5 ( 1 1 . 3 0 )

R a = 1 .0 — 0 . 3 8 5 / 3 ( 1 1 . 3 1 )

T h e s e e q u a t i o n s a r c p l o t t e d in F i g . 1 1 , 1 4 . T h e d i m c n s i o n l e s s f o r m o f e x p r e s s i n g

c o l u m n c u r v e s h a s t h e a d v a n t a g e o f s h o w i n g c o l u m n c u r v e s f o r a l l m a t e r i a l s o n a

0.5 1.0 1.5 3.0

/. 7p

Figure 11.14

11.7 DIMENSIONLESS FORM OF TANGENT MODULUS CURVES

T h e m a t e r i a l s u s e d i n flight v e h i c l e c o n s t r u c t i o n a r e i m p r o v e d f r e q u e n t l y . D e s i g n -

e r s m u s t a d o p t n e w m a t e r i a l s a n d p r o c e s s e s w h i c h s a v e s t r u c t u r a l w e i g h t , e v e n

t h o u g h t h e n e w m a t e r i a l s a r e m o r e e x p e n s i v e t h a n s t a n d a r d i z e d o n e s . W h e n a

n e w o r i m p r o v e d m a t e r i a l i s i n t r o d u c e d , i t is d i f f i c u l t t o m a k e e x t e n s i v e c o l u m n

t e s t s a n d c r i p p l i n g t e s t s i n o r d e r t o e s t a b l i s h n e w d e s i g n a l l o w a b l e s t r e s s e s . I t

w o u l d b e m u c h b e t t e r t o o b t a i n s i m p l e c o m p r e s s i v e s t r e s s - s t r a i n c u r v e s f o r t h e

n e w m a t e r i a l a n d t o b a s e n e w c o l u m n a n d c r i p p l i n g a l l o w a b l e s t r e s s e s o n t h e s e

t e s t s t h a n t o t e s t n u m e r o u s b u i l t - u p c o l u m n s p e c i m e n s . T h e R o m b e r g - O s g o o d 1 9

e q u a t i o n f o r t h e s t r e s s - s t r a i n c u r v e , d i s c u s s e d i n S e c . 4 . 3 , p r o v i d e s t h e n e c e s s a r y

d a t a f o r c o m p a r i n g s i m i l a r m a t e r i a l s .

T h e R o m b e r g - O s g o o d e q u a t i o n o f t h e s t r e s s - s t r a i n c u r v e i s

+ ( 1 1 . 3 2 )

w h e r e 5 a n d e a r e d i m e n s i o n l e s s f u n c t i o n s o f t h e s t r e s s A a n d t h e s t r a i n e a n d t h e

m o d u l u s o f e l a s t i c i t y E :

e = — ( 1 1 . 3 3 )

( 1 1 . 3 4 )

T h e s t r e s s CTJ i s a p p r o x i m a t e l y e q u a l t o t h e y i e l d s t r e s s a t a p e r m a n e n t s t r a i n o f

0 . 0 0 2 , b u t i t m u s t b e d e f i n e d a s t h e s t r e s s a t a s e c a n t m o d u l u s o f e l a s t i c i t y o f 0 . 7 E ,

i n o r d e r f o r s t r e s s - s t r a i n c u r v e s w i t h e q u a l v a l u e s o f n t o b e g e o m e t r i c a l l y s i m i l a r .

T h e t a n g e n t m o d u l u s o f e l a s t i c i t y E, = dn/ik i s r e a d i l y o b t a i n e d f r o m E q s .

( 1 1 . 3 2 ) t o ( 1 1 . 3 4 ) :

E „ de di , _„_. „ , — = £ — = — = l + f / i f f " 1 ( 1 1 . 3 5 ) E, da da

^ = R — r URT®

S e e F i g / 1 1 . 1 5 . N o w t h e t a n g e n t m o d u l u s e q u a t i o n c a n b e w r i t t e n a s a s i n g l e

e x p r e s s i o n w h i c h i n c l u d e s b o t h t h e l o n g - a n d s h o r t - c o l u m n r a n g e s , s i n c c E q .

( 1 1 . 3 6 ) r e p r e s e n t s t h e m o d u l u s o f e l a s t i c i t y b e l o w a s w e l l a s a b o v e t h e e l a s t i c

l i m i t :

F o r l o w v a l u e s o f <r, t h e e x p r e s s i o n i n b r a c k e t s is a p p r o x i m a t e l y u n i t y , a n d E q .

( 1 1 . 3 7 ) c o r r e s p o n d s t o t h e E u l e r e q u a t i o n . T h e e x p r e s s i o n i n b r a c k e t s , c o r r e -

s p o n d i n g t o E J E o f E q . ( 1 1 . 3 6 ) , i s p l o t t e d i n F i g . 1 1 . 1 5 f o r v a r i o u s v a l u e s o f n.


A = —

Figure 11.15

I n o r d e r l o p l o t c o l u m n c u r v e s g i v e n b y E q . ( 1 1 . 3 7 ) i n a d i m c n s i o n l e s s f o r m

s i m i l a r t o t h a t s h o w n in F i g . 1 ! . 1 4 , t h e s r r e s s ax must b e u s e d r a t h e r t h a n ffc0 in E q s . ( 1 1 . 2 6 ) a n d ( 1 1 . 2 7 ) :

B = — ( 1 1 . 3 8 ) N J E J A i

A = — 7 = = n v

Figure 11.40

3 4 4 AIRCRAFT s t r u c t u r e s

R a = ~ ( 1 1 . 3 9 )

B y s u b s t i t u t i n g t h e s e v a l u e s f r o m E q s . ( 1 1 . 3 8 ) a n d ( 1 1 . 3 9 ) i n t o E q . ( 1 1 . 3 7 ) , t h e

f o l l o w i n g c o l u m n e q u a t i o n i s o b t a i n e d :

1 E, J ( U - 4 0 )

T h i s e q u a t i o n is p l o t t e d i n F i g . 1 1 . 1 6 f o r v a r i o u s v a l u e s o f n.

T h e f o r m o f d i m e n s i o n l e s s c o l u m n c u r v e g i v e n b y E q . ( 1 1 . 4 0 ) a n d F i g . 1 1 . 1 6

w a s p r o p o s e d b y C o z z o n e a n d M e l c o n 7 . T h e y a l s o u s e t h i s s a m e b a s i c d i a g r a m

f o r l o c a l c r i p p l i n g , i n i t i a l b u c k l i n g o f s h e e t i n c o m p r e s s i o n a n d s h e a r , a n d b u c k -

l i n g o f s h e e t b e t w e e n r i v e t s . T h e s e f u r t h e r a p p l i c a t i o n s a r e d i s c u s s e d l a t e r . T h e s e

c o l u m n c u r v e s h a v e a v e r y d i s t i n c t a d v a n t a g e w h e n s t r u c t u r e s o f n e w m a t e r i a l s

a r c a n a l y z e d . I t is n e c e s s a r y o n l y t o o b t a i n t h e b a s i c c o m p r e s s i o n s t r e s s - s t r a i n

d i a g r a m o f t h e m a t e r i a l . T h e s h a p e f a c t o r n a n d t h e s t r e s s a 1 c o r r e s p o n d i n g t o

t h e y i e l d s t r e s s s u p p l y a l l t h e n e c e s s a r y i n f o r m a t i o n o n t h e n e w m a t e r i a l . A l l t h e

i n f o r m a t i o n o b t a i n e d f r o m t e s t s o f c o l u m n s o f o n e m a t e r i a l a r e i m m e d i a t e l y

a p p l i c a b l e t o a n e w m a t e r i a l .

11.8 BUCKLING OF ISOTROPIC FLAT PLATES IN COMPRESSION

A f l a t p l a t e , i n w h i c h t h e t h i c k n c s s is s m a l l c o m p a r e d t o t h e o t h e r d i m e n s i o n s ,

d o e s n o t a c t a s a n u m b e r o f p a r a l l e l n a r r o w b e a m s w h e n r e s i s t i n g b e n d i n g

s t r e s s e s . T h e i n i t i a l l y f l a t p l a t e s h o w n i n F i g . 1 1 . 1 7 a m a y b e c o m p a r e d t o t h e

n a r r o w b e a m s h o w n i n F i g . 1 1 . 1 7 6 . T h e i n i t i a l l y r e c t a n g u l a r c r o s s s e c t i o n o f t h e

n a r r o w b e a m d i s t o r t s t o t h e t r a p e z o i d a l c r o s s s e c t i o n , b c c a u s e t h e c o m p r e s s i o n

s t r e s s e s o n t h e u p p e r f a c e o f t h e b e a m p r o d u c e a l a t e r a l e l o n g a t i o n , w h i l e t h e

t e n s i l e s t r e s s e s o n t h e l o w e r f a c e o f t h e b e a m p r o d u c e a l a t e r a l c o n t r a c t i o n . T h e

c r o s s s e c t i o n s o f t h e f l a t p l a t e , h o w e v e r , m u s t r e m a i n r e c t a n g u l a r .

I f t h e s h a d e d c l e m e n t o f F i g . 1 1 . 1 7 a , s h o w n t o a l a r g e r s c a l e in F i g . 1 1 . 1 7 c , is

c o n s i d e r e d , i t w i l l h a v e u n i t e l o n g a t i o n s e x a n d e y a s f o l l o w s :

a n d t 1 1 - 4 2 )

w h e r e v i s P o i s s o n ' s r a t i o . I n t h e p l a t e , t h e e l o n g a t i o n i n t h e y d i r e c t i o n m u s t b e

z e r o if t h e p l a t e i s a s s u m e d t o h a v e n o c u r v a t u r e i n t h e y d i r e c t i o n . S u b s t i t u t i n g

e y = 0 i n t o E q s . ( 1 1 . 4 1 ) a n d ( 1 1 . 4 2 ) y i e l d s

ff, = v f f x ( 1 1 . 4 3 )

= v 2 ) ( 1 1 . 4 4 )

BUCKLING DESIGN OF STRUCTURAL MEMBERS 345

T h u s , w h e n t h e f l a t p l a t e i s d c f l c e t e d w i t h s i n g l e c u r v a t u r e i n t h e x d i r e c t i o n , t h e

s t r e s s e s i n t h e y d i r e c t i o n a r e e q u a l t o P o i s s o n ' s r a t i o t i m e s t h e s t r e s s e s i n t h e JC

d i r e c t i o n . S i m i l a r l y , t h e u n i t e l o n g a t i o n s i n t h e x d i r e c t i o n h a v e t h e r a t i o o f

f — v 2 t o t h e c o r r e s p o n d i n g e l o n g a t i o n s i n a n a r r o w b e a m .

S i n c e a fiat p l a t e h a s s m a l l e r u n i t e l o n g a t i o n s t h a n t h e c o r r e s p o n d i n g n a r r o w

b e a m , t h e c u r v a t u r e r e s u l t i n g f r o m a n e q u i v a l e n t b e n d i n g m o m e n t w i l l b e s m a l l e r

b y t h e r a t i o 1 — v 2 , S i m i l a r l y , if t h e t e r m M/(EI) o f t h e g e n e r a l b e a m - d e f l e c t i o n

r e l a t i o n , E q . ( 1 1 . 6 ) , i s r e p l a c e d b y M ( l — v 2 ) / ( £ 7 ) , t h e E u l e r f o r m u l a . f o r a flat

p l a t e m a y b e o b t a i n e d a s f o l l o w s :

n EI (1 - v 3 ) L 2

T h i s e q u a t i o n a p p l i e s f o r t h e c o n d i t i o n s h o w n i n F i g . 1 1 . 1 8 w h e r e t h e u n l o a d e d

e d g e s a r e f r e e a n d t h e l o a d e d e d g e s a r e s i m p l y s u p p o r t e d , o r f r e e t o r o t a t e b u t

n o t f r e e t o d c f l c c t n o r m a l t o t h e p l a n e o f t h e p l a t e . S u b s t i t u t i n g / = bt3/12, L = 2 ,

Figure 11.18


a n d P c r = cr c r t b i n t o E q . ( 1 1 . 4 5 ) g i v e s

cr ( 1 1 . 4 6 )

F o r a p l a t e s i m p l y s u p p o r t e d o n a l l f o u r e d g e s , a s s h o w n i n F i g . 1 1 . 1 9 , t h e

b u c k l i n g c o m p r e s s i o n l o a d i s c o n s i d e r a b l y h i g h e r . A s t h e p l a t e d e f l e c t s , b o t h

v e r t i c a l a n d h o r i z o n t a l s t r i p s m u s t b e n d . T h e s u p p o r t i n g e f f e c t o f t h e h o r i z o n t a l

s t r i p s m a y b e s u f f i c i e n t t o c a u s e a v e r t i c a l s t r i p t o d e f l e c t i n t o t w o o r m o r e w a v e s ,

a s s h o w n i n F i g . 1 1 . 1 9 . I t c a n b e s h o w n t h a t t h e b u c k l i n g s t r e s s i s 1 5

w h e r e m i s t h e n u m b e r o f w a v e s i n t h e b u c k l e d s h e e t . T h e v a l u e o f v i s a p p r o x i -

m a t e l y 0 . 3 f o r a l l m e t a l s . S i n c e a l a r g e e r r o r i n v p r o d u c e s o n l y a s m a l l e r r o r i n

c r c r , i t i s s e l d o m n e c e s s a r y t o c o n s i d e r t h e v a r i a t i o n o f P o i s s o n ' s r a t i o . E q u a t i o n

( 1 1 . 4 7 ) m a y b e w r i t t e n a s f o l l o w s :

w h e r e K i s a f u n c t i o n o f a j b a n d i s p l o t t e d i n F i g . 1 1 . 2 0 f o r v = 0 . 3 . O n l y t h e

c u r v e o f F i g . 1 1 . 2 0 w h i c h g i v e s t h e m i n i m u m v a l u e o f K i s s i g n i f i c a n t , s i n c e t h e

s h e e t w i l l b u c k l e i n t o t h e n u m b e r o f w a v e s t h a t r e q u i r e s t h e s m a l l e s t l o a d . I t i s

s e e n f r o m F i g . 1 1 . 2 0 t h a t t h e w a v e l e n g t h o f t h e b u c k l e s i s a p p r o x i m a t e l y e q u a l t o

t h e w i d t h b, o r t h a t m = 1 f o r a j b = 1 , m = 2 f o r a j b = 2 , e t c . T h e r a t i o a j b a t

w h i c h t h e n u m b e r o f w a v e s c h a n g e s f r o m m t o m + 1 i s o b t a i n e d f r o m E q . ( 1 1 . 4 7 )

a s a j b = - J m { m + 1). T h e b u c k l i n g s t r e s s o b t a i n e d f r o m E q . ( 1 1 . 4 7 ) f o r a s q u a r e

p l a t e w i t h f o u r e d g e s s i m p l y s u p p o r t e d i s 4 t i m e s t h a t o b t a i n e d f r o m E q . ( 1 1 . 4 6 )

f o r t h e p l a t e w i t h s i d e s f r e e a n d e n d s s i m p l y s u p p o r t e d .

( 1 1 . 4 7 )

( 1 1 . 4 8 )

P = acib


BUC KLING DESIGN OF STRUCTURAL MEMBERS 3 4 7

A/H

Figure 11.20

. T h e b u c k l i n g l o a d s f o r r e c t a n g u l a r p l a t e s w i t h o t h e r e d g e c o n d i t i o n s a l s o c a n

b e f o u n d f r o m E q . ( 1 1 . 4 8 ) b y u s i n g t h e c o r r e c t v a l u e s o f K . V a l u e s o f K a r e

s h o w n i n F i g . 1 1 . 2 1 f o r v a r i o u s c o n d i t i o n s . T h e l o a d e d e d g e s a r e t e r m e d e n d s ,

a n d t h e u n l o a d e d e d g e s t e r m e d s i d e s , a s d e s i g n a t e d o n t h e c u r v e s . A f r e e e d g e

m a y r o t a t e o r d e f l e c t i n a d i r e c t i o n n o r m a l t o t h e p l a t e . A f i x e d e d g e , a s s h o w n i n

F i g . 1 1 . 2 2 o , i s p r e v e n t e d f r o m r o t a t i n g o r d e f l e c t i n g . T h e s i m p l y s u p p o r t e d e d g e

s h o w n i n F i g . 1 1 . 2 2 b i s f r e e t o r o t a t e , b u t n o t t o d e f l e c t n o r m a l t o t h e p l a n e o f t h e

p l a t e .

T h e t r u e e d g e - f i x i t y c o n d i t i o n s f o r f l a t p l a t e s i n a n a i r p l a n e s t r u c t u r e c a n n o t

b e c a l c u l a t e d i n m o s t c a s e s . I t i s n e c e s s a r y t o e s t i m a t e t h e e d g e f i x i t y a f t e r t h e

s u p p o r t i n g s t r u c t u r e h a s b e e n c o n s i d e r e d , i n a m a n n e r s i m i l a r t o t h a t f o r e s t i -

m a t i n g c o l u m n e n d - f i x i t y c o n d i t i o n s . T h e u p p e r s k i n o f a n a i r p l a n e w i n g , f o r

l/>) Figure 11.21


alb

' F igure 11.22

e x a m p l e , i s c o m p r e s s e d i n a s p a n w i s e d i r e c t i o n . I f t h e s t r i n g e r s a r e flexible t o r -

s i o n a l l y , t h e y w i l l r o t a t e a s t h e s h e e t b u c k l e s a n d w i l l a c t a l m o s t a s s i m p l e

s u p p o r t s f o r t h e s h e e t b e t w e e n t h e s t r i n g e r s , a s s h o w n i n F i g . 1 1 . 2 3 a . I f t h e

s t r i n g e r s h a v e c o n s i d e r a b l e t o r s i o n a l r i g i d i t y , a s d o t h e " h a t " s e c t i o n s a n d t h e

U'J Figure 11.23

BUCKUNG DESIGN OF STRUCTURAL MEMBERS 3 4 9

s p a r f l a n g e s h o w n i n F i g . 1 1 . 2 3 6 , t h e y w i l l r o t a t e o n l y s l i g h t l y a n d w i l l p r o v i d e

a l m o s t c l a m p e d e d g e c o n d i t i o n s . I n m o s t s t r u c t u r e s , i t i s n e c e s s a r y t o a s s u m e a

v a l u e f o r t h e t e r m K o f E q . ( 1 1 . 4 8 ) w h i c h w i l l r e p r e s e n t a c o n s e r v a t i v e m e a n

b e t w e e n s i m p l y s u p p o r t e d a n d c l a m p e d e d g e c o n d i t i o n s .

1 1 . 9 U L T I M A T E C O M P R E S S I V E S T R E N G T H O F I S O T R O P I C

F L A T S H E E T

T h e b u c k l i n g o f s h e e t s i n c o m p r e s s i o n d o c s n o t c a u s e t h e c o l l a p s e o f a s e m i -

m o n o c o q u c s t r u c t u r e , b c c a u s c t h e s t i f f e n i n g m e m b e r s u s u a l l y c a n r e s i s t s t r e s s e s

w h i c h a r e m u c h h i g h e r t h a n t h o s e a t w h i c h t h e i n i t i a l s h e e t b u c k l i n g o c c u r s . W e

s h o w e d t h a t a l o n g c o l u m n m a y r e s i s t a c o m p r e s s i o n l o a d w h e n i n t h e b u c k l e d

c o n d i t i o n a n d t h a t t h e l o a d i s t h e s a m e f o r a s m a l l l a t e r a l d e f l e c t i o n a s f o r a l a r g e

d c f i c c t i o n , p r o v i d e d t h a t t h e s t r e s s d o e s n o t e x c e e d t h e e l a s t i c l i m i t . T h e c o m -

p r e s s i o n l o a d r e s i s t e d b y a flat s h e e t w i t h t h e s i d e s f r e e a l s o r e m a i n s c o n s t a n t f o r

a n y l a t e r a l d c f l e c t i o n . If I h e s i d e s a r e s u p p o r t e d , h o w e v e r , t h e c o m p r e s s i o n l o a d

r e s i s t e d b y t h e s h e e t w i l l i n c r e a s e a s t h e l a t e r a l d e f l e c t i o n i n c r e a s e s , b e c a u s e t h e

s i d e s o f t h e s h e e t m u s t r e m a i n s t r a i g h t a n d c o n s e q u e n t l y m u s t b e s t r e s s e d i n

p r o p o r t i o n t o t h e s t r a i n i n t h e d i r e c t i o n o f l o a d i n g .

T h e p l a t e s h o w n i n F i g . 1 1 . 2 4 i s s i m p l y s u p p o r t e d a t a l l f o u r e d g e s a n d i s

l o a d e d b y a r i g i d b l o c k . T h e c o m p r e s s i o n s t r e s s e s a r e u n i f o r m l y d i s t r i b u t e d a s

s h o w n i n F i g . 1 1 . 1 9 if t h e l o a d is s m a l l e r t h a n t h e b u c k l i n g l o a d . T h e s t r e s s

d i s t r i b u t i o n o v e r t h e w i d t h o f t h e p l a t e i s i n d i c a t e d i n F i g . 1 1 . 2 5 b y l i n e s 1 a n d 2 ,

Figure 11.24

a. { Figure 11.25

w i t h l i n e 2 i n d i c a t i n g t h e . s t r e s s a t i n i t i a l b u c k l i n g . A s t h e l o a d i s i n c r e a s e d

b e y o n d t h e b u c k l i n g l o a d , t h e s t r e s s d i s t r i b u t i o n i s i n d i c a t e d b y l i n e s 3 , 4 , a n d 5 .

N e a r t h e m i d d l e o f t h e c r o s s s e c t i o n , t h e c o m p r e s s i v e s t r e s s r e m a i n s a p p r o x i -

m a t e l y e q u a l t o t h e b u c k l i n g s t r e s s , o r a v e r t i c a l s t r i p a c t s i n a s i m i l a r m a n n e r t o

a l o n g c o l u m n . A t t h e s i d e s o f t h e s h e e t , b u c k l i n g i s p r e v e n t e d , a n d t h e s t r e s s

i n c r e a s e s i n p r o p o r t i o n t o t h e v e r t i c a l m o t i o n o f t h e l o a d i n g b l o c k . T h e l o a d m a y

b e i n c r e a s e d u n t i l f a i l u r e occurs b y c r u s h i n g o f t h e s h e e t a t t h e sides, a l t h o u g h in c o m m o n a i r c r a f t s t r u c t u r e s t h e s t i f f e n i n g m e m b e r s s u p p o r t i n g t h e s h e e t u s u a l l y

f a i l b e f o r e t h e s h e e t f a i l s .

T h e c u r v e r e p r e s e n t i n g t h e d i s t r i b u t i o n o f c o m p r e s s i v e s t r e s s o v e r t h e w i d t h

o f a s h e e t i s d i f f i c u l t t o o b t a i n , a n d e v e n i f i t w e r e k n o w n , i t w o u l d b e t e d i o u s t o

u s e i n a n a l y s i s . I t is m o r e c o n v e n i e n t t o o b t a i n t h e t o t a l c o m p r e s s i o n l o a d c o r r e -

s p o n d i n g t o a g i v e n c o m p r e s s i o n s t r e s s a t t h e s i d e o f t h e s h e e t . I t i s c u s t o m a r y t o

w o r k w i t h e f f e c t i v e w i d t h s w , s h o w n i n F i g . 1 i . 2 6 , w h i c h a r e d e f i n e d i n s u c h a

w a y t h a t t h e c o n s t a n t s t r e s s e s <rc a c t i n g o v e r t h e e f f e c t i v e w i d t h s w i l l y i e l d t h e

t o t a l c o m p r e s s i o n l o a d . T h u s vv i s s e l e c t e d s o t h a t t h e a r e a u n d e r t h e t w o r e c -

t a n g l e s i n F i g . 1 1 . 2 6 a i s e q u a l t o t h e a r e a u n d e r t h e c u r v e o f t h e a c t u a l s t r e s s

d i s t r i b u t i o n . T h e t o t a l c o m p r e s s i o n l o a d P a n d t h e e d g e s t r e s s CC c a n b e f o u n d

e x p e r i m e n t a l l y , a n d t h e w i d t h s w m a y b e c a l c u l a t e d f r o m

A n a p p r o x i m a t e v a l u e o f vv m a y b e o b t a i n e d b y a s s u m i n g t h a t a l o n g s h e e t o f

t o t a l w i d t h 2 w w i l l h a v e a b u c k l i n g s t r e s s o f r r c . F r o m E q . ( 1 1 . 4 8 ) a n d F i g . 1 1 . 2 0 ,

2WCRC = P ( 1 1 . 4 9 )

BUCKLING DI-SIGN OF STRUCTURAL MI-MB1-RS 3 5 1

o r

T e s t r e s u l t s i n d i c a t e t h a t t h i s v a l u e i s t o o h i g h a n d t h a t i t i s m o r e a c c u r a t c t o u s e

( 1 1 . 5 0 )

I n o b t a i n i n g E q . ( 1 1 . 5 0 ) , w e a s s u m e t h a t t h e s h e e t i s f r e e t o r o t a t e a t a l l f o u r

e d g e s . I n a c t u a l s t r u c t u r e s s o m e d e g r e e o f r e s t r a i n t a l w a y s e x i s t s , a n d t h e e f f e c t i v e

w i d t h s m a y b e m u c h g r e a t e r i n m a n y c a s e s . T e s t s i n d i c a t e t h a t s t r i n g e r s p r o v i d e

c o n s i d e r a b l e e d g e fixity a t l o w s t r e s s e s , b u t d o n o t p r o v i d e m u c h r e s t r a i n t a t

s t r e s s e s a p p r o a c h i n g t h e u l t i m a t e s t r e n g t h o f t h e s t r i n g e r s . N u m e r o u s o t h e r e q u a -

t i o n s h a v e b e e n u s e d i n p l a c e o f E q . ( 1 1 . 5 0 ) ; n o e q u a t i o n s p r o v i d e a c c u r a t e c o r -

r e l a t i o n w i t h t e s t r e s u l t s u n d e r a l l c o n d i t i o n s . U n c e r t a i n t i e s r e g a r d i n g t h e e f -

f e c t s o f e d g e r e s t r a i n t s i n t h e a c t u a l s t r u c t u r e , a c c i d e n t a l e c c e n t r i c i t i e s i n t h e

s h e e t , a n d t h e e f f e c t s o f s t r e s s e s b e y o n d t h e e l a s t i c l i m i t f u r t h e r c o m p l i c a t e t h e

p r o b l e m . E q u a t i o n ( 1 1 . 5 0 ) y i e l d s a s m a l l e r e f f e c t i v e w i d t h t h a n d o m o s t o t h e r

e q u a t i o n s a n d i s c o n s e r v a t i v e f o r u s e i n d e s i g n . F o r n o r m a l a i r c r a f t s t r u c t u r e s i n

w h i c h t h e s h e e t is r e l a t i v e l y t h i n , t h e w e i g h t p e n a l t y i n t r o d u c e d b y u s i n g E q .

( 1 1 . 5 0 ) i s s m a l l ; f o r h i g h - s p e e d a i r c r a f t i n w h i c h t h e s k i n i s r e l a t i v e l y t h i c k , a m o r e

a c c u r a t e a n a l y s i s m a y b e j u s t i f i e d .

T h e b u c k l i n g s t r e s s f o r a f l a t s h e e t w i t h a l a r g e r a t i o o f l e n g t h t o w i d t h , w i t h

o n e s i d e s i m p l y s u p p o r t e d a n d t h e o t h e r f r e e , c a n b e o b t a i n e d f r o m F .q . ( 1 1 . 4 8 )

a n d F i g . 1 1 . 2 1 . F o r K = 0 . 3 8 5 , acr = 0 . 3 8 5 £ ( f / b ) 2 . T h e u l t i m a t e l o a d r e s i s t e d b y

s u c h a s h e e t w h e n t h e s u p p o r t e d s i d e i s s t r e s s e d b y a v a l u e a c i s f o u n d b y

c o n s i d e r i n g t h a t a n e f f e c t i v e w i d t h W[ r e s i s t s t h e s t r e s s cr,. a n d b y o b t a i n i n g vv t a s

i f r o m E q . ( 1 1 . 5 0 ) :

ac = 0.385

vi'j = 0 . 6 2 /

A m o r e c o n s e r v a t i v e v a l u e i s r e c o m m e n d e d :

0 . 6 0 , - 0 1 . 5 1 ) A.

T h e s e e f f e c t i v e s i i e e t w i d t h s a n d w l s h o w n i n F i g . 1 1 . 2 7 a r e o b t a i n e d f r o m E q s .

( 1 1 . 5 0 ) a n d ( 1 1 . 5 1 ) .

E x a m p l e 1 1 . 1 T h e s h e e t s t r i n g e r p a n e l s h o w n i n F i g . 1 1 . 2 8 i s l o a d e d i n

c o m p r e s s i o n b y m e a n s o f r i g i d m e m b e r s . T h e s h e e t i s a s s u m e d t o b e s i m p l y

s u p p o r t e d a t t h e l o a d e d e n d s a n d a t t h e r i v e t l i n e s a n d t o b e f r e e a t t h e s i d e s .

^ J ^ (a)

P f i T j [ i l ^ j ^ ^

F igure 11.27

E a c h s t r i n g e r h a s a n a r e a o f 0 . 1 i n 2 . A s s u m e £ = 1 0 , 3 0 0 , 0 0 0 l b / i n 2 f o r t h e

s h e e t a n d s t r i n g e r s . F i n d t h e t o t a l c o m p r e s s i v e l o a d P :

(A) W h e n t h e s h e e t f i r s t b u c k l e s

(b) W h e n t h e s t r i n g e r s t r e s s <JC i s 1 0 , 0 0 0 l b / i n 2

(e ) W h e n t h e s t r i n g e r s t r e s s trc i s 3 0 , 0 0 0 l b / i n 2

S O L U T I O N (a) T h e s h e e t b e t w e e n t h e s t r i n g e r s i s s i m p l y s u p p o r t e d o n a l l f o u r

e d g e s a n d h a s d i m e n s i o n s o f a = 10 , b = 5 , a n d t = 0 . 0 4 0 i n . F r o m F i g . 1 1 . 2 0 ,

f o r ct/b = 2 . 0 t h e v a l u e K = 3 . 6 2 i s o b t a i n e d . T h e b u c k l i n g s t r e s s i s a c =

KE{tjb)2 = 3 . 6 2 x 1 0 , 3 0 0 , 0 0 0 ( 0 . 0 4 0 / 5 ) 2 = 2 3 9 0 l b / i n 2 . T h e e d g e o f t h e s h e e t

h a s d i m e n s i o n s o f a — 1 a n d b — 1 0 i n a n d is s i m p l y s u p p o r t e d o n t h r e e

_ L

T 0 . 0 4 0 in

to)

5 in * ' T " -

J h ^

e d g e s a n d f r e e o n t h e f o u r t h e d g e . F r o m F i g . 1 1 . 2 1 , K = 0 . 3 8 5 . T h e b u c k l i n g

s t r e s s i s <Jc = KE{t/b)2 = 0 . 3 8 5 x 1 0 , 3 0 0 , 0 0 0 ( 0 . 0 4 0 / 1 ) 2 = 6 2 0 0 l b / i n 2 .

T h e s h e e t t h e r e f o r e b u c k l e s i n i t i a l l y b e t w e e n t h e s t r i n g e r s . T h e t o t a l a r e a

o f t h e s h e e t i s a s s u m e d t o b e e f f e c t i v e b e f o r e b u c k l i n g o c c u r s . T h e b u c k l i n g o f

a flat s h e e t i n c o m p r e s s i o n i s a g r a d u a l p r o c e s s , a n d t h e l o a d d o e s n o t d r o p

a p p r e c i a b l y w h e n b u c k l i n g o c c u r s . T h e l o a d i s t h e r e f o r e c a l c u l a t e d a s f o l -

l o w s :

A = 3 x 0 . 1 + 1 2 x 0 . 0 4 0 = 0 . 7 8 i n 2

p = a c A = 2 3 9 0 x 0 . 7 8 = 1 8 6 5 l b

(b) T h e e f f e c t i v e s h e e t w i d t h s a r e o b t a i n e d f r o m E q s . ( 1 1 . 5 0 ) a n d ( 1 1 . 5 1 ) :

E / 1 0 , 3 0 0 , 0 0 0 «• = 0 . 8 5 - = 0 . 8 5 x 0 . 0 4 0 / ' ' = 1 . 0 9 i n

V ffe V 1 0 , 0 0 0

\ \ \ = 0 . 6 0 1 — = 0 . 7 7 i n

V Co

T h e e f f e c t i v e s h e e t a r e a i s

A , = (4vv + 2 w , ) f = ( 4 x 1 . 0 9 + 2 x 0 . 7 7 ) x 0 . 0 4 0 = 0 . 2 3 6 i n 2

T h e t o t a l c o m p r e s s i v e l o a d i s

p = a t A = 1 0 , 0 0 0 ( 0 . 3 -I- 0 . 2 3 6 ) = 5 3 6 0 l b

(c ) T h e s o l u t i o n i s s i m i l a r t o t h a t o f p a r t ( b ) :

E / 1 0 , 3 0 0 , 0 0 0 -u- = 0 . 8 5 / / — = 0 . 8 5 x 0 . 0 4 0 / ' ? = 0 . 6 3 i n

V a c V 3 0 , 0 0 0

[E u>, = 0 . 6 0 f / — = 0 . 4 4 i n

\ j < T c

A = 0 . 3 + (4 x 0 . 6 3 + 2 x 0 . 4 4 ) x 0 . 0 4 0 = 0 . 4 3 6 i n 2

P = a , A = 3 0 , 0 0 0 x 0 . 4 3 6 = 1 3 , 0 8 0 l b

11.10 PLASTIC BUCKLING OF FLAT SHEET

I n t h e d i s c u s s i o n o f b u c k l i n g o f s h e e t e l e m e n t s , w e a s s u m e t h a t t h e s t r e s s d o e s n o t

e x c e e d t h e p r o p o r t i o n a l e l a s t i c l i m i t f o r t h e m a t e r i a l . T h i s e l a s t i c b u c k l i n g a c t i o n

f o r flat s h e e t s i s s i m i l a r t o t h e e l a s t i c b u c k l i n g o f l o n g c o l u m n s i n t h a t t h e

m o d u l u s o f e l a s t i c i t y i s t h e o n l y s i g n i f i c a n t m a t e r i a l p r o p e r t y . E q u a t i o n ( 1 1 . 4 8 )

f o r s h e e t b u c k l i n g is s i m i l a r t o t h e E u l e r e q u a t i o n f o r c o l u m n s , a n d i n e a c h c a s e

t h e b u c k l i n g s t r e s s i s p r o p o r t i o n a l t o t h e m o d u l u s o f e l a s t i c i t y o f t h e m a t e r i a l .

I n t h e c a s e o f s h e e t e l e m e n t s f o r w h i c h t h e t h i c k n e s s i s g r e a t e r i n c o m p a r i s o n

t o t h e o t h e r d i m e n s i o n s , t h e c o m p r e s s i v e s t r e s s e s w i l l e x c e e d t h e e l a s t i c l i m i t

b e f o r e b u c k l i n g w i l l o c c u r , a s i s t h e c a s e f o r s h o r t c o l u m n s . E q u a t i o n ( 1 1 . 4 8 ) w i l l

b e v a l i d i n t h i s c a s e i f t h e t a n g e n t m o d u l u s o f e l a s t i c i t y E, i s s u b s t i t u t e d f o r t h e

m o d u l u s E:

r 2

= K F J - J ( 1 1 . 5 2 )

T h i s e q u a t i o n m a y b e w r i t t e n a s

KE, ( 1 3 . 5 3 )

" (b/t) E q u a t i o n ( 1 1 . 5 3 ) i s s i m i l a r t o t h e t a n g e n t m o d u l u s e q u a t i o n f o r s h o r t c o l u m n s :

T h e t a n g e n t m o d u l u s c u r v e a n d o t h e r c u r v e s f o r s h o r t c o l u m n s w e r e p l o t t e d w i t h

v a l u e s o f <rc a s o r d i n a l e s a n d v a l u e s o f Z / p a s a b s c i s s a s . V a l u e s o f <rcr a n d b / t

c o u l d b e s i m i l a r l y p l o t t e d f r o m E q . ( 1 1 . 5 3 ) f o r a k n o w n v a l u e o f K. I n f a c t , t h e

c o l u m n c u r v e s c a n b e u s e d f o r p l a s t i c s h e e t b u c k l i n g i f t h e v a l u e s o f b / t a r e

m u l t i p l i e d b y a c o n s t a n t w h i c h i s o b t a i n e d b y e q u a t i n g t h e r i g h t s i d e o f E q .

( 1 1 . 5 3 ) a n d t h e r i g h t s i d e o f E q . ( 1 1 . 5 4 ) a s f o l l o w s :

. £ 7 I B . E q u i v a l e n t — = — y = ~ ( 1 1 . 5 5 )

P V x t A t y p i c a l c o l u m n c u r v e f o r a n a l u m i n u m - a l l o y m a t e r i a l i s s h o w n i n F i g .

1 1 . 2 9 . T h e a l l o w a b l e c o l u m n s t r e s s i s o b t a i n e d f r o m t h e c u r v e f o r a k n o w n v a l u e

F igure 11.30

o f £ / / ) . I n t h e s h o r t - c o l u m n r a n g e , t h e t h e o r e t i c a l t a n g e n t m o d u l u s c u r v e m a y b e

r e p l a c e d b y t h e m o r e c o n s e r v a t i v e s t r a i g h t l i n e , i n o r d e r t o a c c o u n t f o r a c c i d e n t a l

e c c e n t r i c i t i e s o r o t h e r u n k n o w n c o n d i t i o n s . S i m i l a r l y , t h e c u r v e s i n F i g . 1 1 . 2 9

y i e l d t h e a l l o w a b l e b u c k l i n g s t r e s s f o r a flat p l a t e i n t h e p l a s t i c , o r s h o r t - c o l u m n ,

r a n g e . T h e v a l u e o f m a y b e o b t a i n e d f o r a n y k n o w n v a l u e o f

E i t h e r o f t h e s h o r t - c o l u m n c u r v e s m a y b e u s e d , d e p e n d i n g o n t h e p o s s i b l e i n i t i a l

e c c e n t r i c i t i e s o f t h e s h e e t e l e m e n t a n d t h e d e g r e e o f c o n s e r v a t i s m d e s i r e d . T h e

v a l u e o f K i s o b t a i n e d f r o m F i g . 1 1 . 2 0 o r 1 1 . 2 1 .

O n e c o m m o n a p p l i c a t i o n o f p l a s t i c b u c k l i n g i s t h e b u c k l i n g o f c o m p r e s s i v e

s k i n b e t w e e n r i v e t s a t t a c h i n g t h e s k i n t o t h e s t r i n g e r s o r s p a r c a p s . A s k i n

e l e m e n t o f t h i s t y p e . is s h o w n i n F i g . 1 1 . 3 0 . T h e r i v e t s h a v e a u n i f o r m s p a c i n g s

a l o n g t h e s t r i n g e r , a n d t h e r e s t r a i n t is s u c h t h a t t h e s k i n e l e m e n t o f l e n g t h s a n d

i n d e f i n i t e w i d t h h a s c l a m p e d e n d s a n d f r e e s i d e s . T h e e l e m e n t t h e r e f o r e r e s i s t s 4

t i m e s t h e l o a d o f a s i m i l a r e l e m e n t w i t h h i n g e d e n d s , w h i c h w a s a n a l y z e d b y E q .

( 1 1 . 4 6 ) . S u b s t i t u t i n g A = s a n d E - E, i n t o E q . ( 1 1 . 4 6 ) a n d m u l t i p l y i n g t h e r i g h t -

h a n d s i d e b y 4 ( t o a c c o u n t f o r t h e e n d fixity) y i e l d

TZE,

3 ( 1 - v 2 ) Vs.

o r , f o r v = 0 . 3 ,

3 . 6 2 £ ,

(s/tf < V = T - 7 ^ T ( 1 L 5 6 )

T h e t a n g e n t m o d u l u s s h o r t - c o l u m n c u r v e m a y b e u s e d i n s o l v i n g E q . ( 1 1 . 5 6 ) . A n

e q u i v a l e n t s l e n d e r n e s s r a t i o m a y b e o b t a i n e d b y e q u a t i n g t h e r i g h t - h a n d s i d e s o f

E q s . ( 1 1 . 5 4 ) a n d ( 1 1 . 5 5 ) :

E q u i v a l e n t — = — ? = = - = 1 . 6 5 - ( 1 1 . 5 7 ) P S/162 F T

E x a m p l e 1 1 . 2 F i n d t h e c o m p r e s s i o n b u c k l i n g s t r e s s f o r a s h e e t 4 b y 4 b y

0 . 1 2 5 i n w i t h a l l f o u r e d g e s s i m p l y s u p p o r t e d , a s s u m i n g t h a t t h e t a n g e n t

m o d u l u s c o l u m n c u r v e f o r t h e m a t e r i a l i s r e p r e s e n t e d b y F i g . 1 1 . 2 9 .

S O L U T I O N F o r t h i s s h e e t a = b = 4 a n d t = 0 . 1 2 5 . F r o m F i g . 1 1 . 2 1 f o r


a j b = 1 a n d s i m p l y s u p p o r t e d e d g e s , K = 3 . 6 2 . F r o m E q . ( 1 1 . 5 5 ) t h e e q u i v a -

l e n t L / p i s

71 b K 4 — 7 = ~ = , x = 5 2 . 8 s / K t y i 6 2 0 . 1 2 5

F r o m F i g . 1 1 . 2 9 , c c r = 2 8 , 0 0 0 l b / i n 2 . I f t h i s p o i n t h a d b e e n o n t h e r i g h t - h a n d

p o r t i o n o f F i g . 1 1 . 2 9 , c o r r e s p o n d i n g t o t h e l o n g - c o l u m n o r c l a s t i c r a n g e , t h e

b u c k l i n g s t r e s s w o u l d c o r r e s p o n d t o t h a t g i v e n b y E q . ( 1 1 . 4 8 ) .

E x a m p l e 1 1 . 3 T h e a n g l e e x t r u s i o n s h o w n in F i g . 1 1 . 3 1 i s l o a d e d i n c o m -

p r e s s i o n . E a c h l e g o f t h e a n g l e b u c k l e s a s a p l a t e s i m p l y s u p p o r t e d o n t h e

e n d s a n d o n o n e s i d e a n d f r e e o n t h e o t h e r s i d e . F i n d t h e s t r e s s a t w h i c h t h i s

b u c k l i n g o c c u r s . A s s u m e t h a t F i g . 1 1 . 2 9 r e p r e s e n t s p r o p e r t i e s o f t h i s m a t e r i a l .

S O L U T I O N F o r e a c h l e g , b = 1, a = 8 , a n d r = 0 . 0 7 2 . F o r a / b = 8 t h e v a l u e o f

K f r o m F i g . 1 1 . 2 1 i s a p p r o x i m a t e l y 0 . 3 8 5 . T h e e q u i v a l e n t ILjp i s

7C b TC 1

- 7 = - = , x = 7 0 . 4 t v / 0 3 8 5 0 . 0 7 2

F r o m F i g . 1 1 . 2 9 , o C a = 2 0 , 5 0 0 l b / i n 2 .

T h e t y p e o f f a i l u r e i n d i c a t e d f o r t h i s s e c t i o n is t y p i c a l o f c r i p p l i n g f a i l u r e s f o r

a l u m i n u m - a l l o y e x t r u s i o n s . T h e o r d i n a r y s h o r t - c o l u m n c u r v e s a p p l y o n l y t o

r o u n d t u b e s o r t o s t a b l e c r o s s . s e c t i o n s w h i c h d o n o t c r i p p l e l o c a l l y . S i n c e l i g h t

e x t r u s i o n s a r e u s e d e x t e n s i v e l y a s c o l u m n m e m b e r s i n a i r c r a f t s t r u c t u r e s , t h e

s u b j e c t o f c r i p p l i n g f a i l u r e is v e r y i m p o r t a n t a n d i s d i s c u s s e d i n d e t a i l l a t e r .

E x a m p l e 1 1 . 4 A n 0 . 0 4 0 s h e e t i s r i v e t e d t o a n e x t r u s i o n b y r i v e t s s p a c e d 1 i n

a p a r t . W h a t c o m p r e s s i o n s t r e s s i n t h e e x t r u s i o n w i l l p r o d u c e b u c k l i n g o f t h e

s h e e t b e t w e e n r i v e t s , a s s h o w n i n F i g . 1 1 . 3 0 , if t h e s h e e t h a s c o l u m n p r o p e r -

t i e s a s r e p r e s e n t e d b y F i g . 1 1 . 2 9 ?

S O L U T I O N F r o m E q . ( ! 1 . 5 7 ) t h e e q u i v a l e n t Z / p i s

1 . 6 5 - = 1 . 6 5 — — = 4 1 . 2 l 0 . 0 4 0

F r o m F i g . 1 1 . 2 9 , a c = 3 1 , 3 0 0 l b / i n 2 .

1 1 . 1 1 N O N D I M E N S I O N A L B U C K L I N G C U R V E S

T h e p l a s t i c b u c k l i n g s t r e s s e s in S e c . 1 1 . 1 0 a r e o b t a i n e d f r o m c o l u m n c u r v e s o f t h e

t y p e s h o w n i n F i g . 1 1 . 2 9 . A c o l u m n c u r v e o f t h i s t y p e i s a p p l i c a b l e t o o n l y o n e

m a t e r i a l , s i n c e t h e c o l u m n c u r v c is a f f e c t e d b y t h e s h a p e o f s t r e s s - s t r a i n c u r v e , t h e

m o d u l u s o f e l a s t i c i t y , a n d t h e y i e l d s t r e s s o f t h e m a t e r i a l . T h e r e a r e n u m e r o u s

a d v a n t a g e s t o p l o t t i n g c o l u m n c u r v e s i n d i m e n s i o n l e s s f o r m , a s s h o w n i n F i g .

1 1 . 1 6 a n d a s d i s c u s s e d i n S e c . 1 1 . 7 . W h e n s e v e r a l m a t e r i a l s h a v e s t r e s s - s t r a i n

c u r v e s o f t h e s a m e g e n e r a l s h a p e , a s i n d i c a t e d b y t h e v a l u e o f n , a s i n g l e - c o l u m n

c u r v e p r e s e n t s t h e d a t a f o r a l l t h e s e m a t e r i a l s . T e s t i n f o r m a t i o n f o r a n y o n e o f t h e

m a t e r i a l s i s t h e r e f o r e a p p l i c a b l e t o a l l o f t h e m .

C o 7 7 : o n e a n d M e l c o n p r o p o s e t h a t t h e n o n d i m c n s i o n a l c u r v e s o f F i g . 1 1 . 1 6

b e u s e d f o r a l l p r o b l e m s o f p l a s t i c s h e e t b u c k l i n g , i n t e r r i v e t b u c k l i n g , a n d l o c a l

c r i p p l i n g o f c o m p r e s s i o n m e m b e r s . T h e c u r v e s o f F i g . 1 1 . 1 6 a r e p r e s e n t e d b y

^ = ( 1 1 . 5 8 ) <?! E B V

w h e r e tr„ i s t h e a l l o w a b l e a v e r a g e s t r e s s f o r a c o l u m n f o r s h e e t b u c k l i n g , o r f o r

c r i p p l i n g , a n d r r , is t h e s e c a n t y i e l d s t r e s s c o r r e s p o n d i n g t o t h e s t r e s s a t t h e

i n t e r s e c t i o n b e t w e e n t h e s t r e s s - s t r a i n c u r v e a n d a l i n e t h r o u g h t h e o r i g i n h a v i n g

s l o p e 0 . 7 E .

F o r c o l u m n s , t h e t e r m B is d e f i n e d b y E q . ( 1 1 . 3 8 ) :

B = LJP ( 1 1 . 3 8 ) TC Y/ E/TR i

F o r p l a s t i c s h e e t b u c k l i n g , t h e v a l u e o f B i s o b t a i n e d f r o m E q s . ( 1 1 . 5 3 ) a n d

( 1 1 . 5 8 ) :

fl—^L= ( 1 1 . 5 9 ) JEKJIT,

F o r i n t e r r i v e t b u c k l i n g , t h e v a l u e o f B i s o b t a i n e d f r o m E q s . ( 1 1 . 5 6 ) a n d ( 1 1 . 5 8 ) :

B = ( 1 1 . 6 0 )


T h u s t h e v a l u e s o f B m a y b e c a l c u l a t e d f r o m E q s . ( 1 1 . 3 8 ) , ( 1 1 . 5 9 ) , o r ( 1 1 . 6 0 ) , a n d

t h e n t h e v a l u e o f a j a l m a y t h e n b e r e a d f r o m t h e p r o p e r c u r v e o f F i g . 1 1 . 1 6 .

11.12 COLUMNS SUBJECT TO LOCAL CRIPPLING FAILURE

T h e c o l u m n e q u a t i o n s p r e v i o u s l y d e r i v e d a r e a p p l i c a b l e t o c l o s e d t u b u l a r s e c -

t i o n s w i t h c o m p a r a t i v e l y ( h i c k w a l l s o r t o o t h e r c r o s s s e c t i o n s w h i c h a r e n o t

s u b j e c t t o l o c a l c r i p p l i n g f a i l u r e . M a n y o f t h e c o l u m n s u s e d i n s e m i m o n o c o q u e

( l i g h t v e h i c l e s t r u c t u r e s a r e m a d e o f e x t r u d e d s e c t i o n s o r o f b e n t s h e e t s e c t i o n s

a n d f a i l b y l o c a l c r i p p l i n g . T h e a s s u m e d c o l u m n c u r v e i s t h a t s h o w n b y l i n e A o f

F i g . 1 1 . 3 2 , w h e r e t h e a c c i s t h e c r i p p l i n g s t r e s s . T e s t s o f c o l u m n s o f e x t r u s i o n s o r

b e n t s h e e t w i t h t h i n w a l l s s u b j e c t t o l o c a l c r i p p l i n g y i e l d v a l u e s r e p r e s e n t e d b y

c u r v e B o f F i g . 1 1 . 3 2 a n d i n d i c a t e t h a t s e c t i o n s s u b j e c t t o c r i p p l i n g f a i l u r e s h o u l d

b e a n a l y z e d b y d i f f e r e n t c o l u m n e q u a t i o n s t h a n s t a b l e c r o s s s e c t i o n s o f t h e s a m e

m a t e r i a l .

U s u a l l y i t i s d e s i r a b l e t o m a k e t e s t s w h i c h w i l l c o v e r a r a n g e o f s l e n d e r n e s s

r a t i o s f o r e a c h t h i n - w a l l c d s e c t i o n w h i c h i s t o b e u s e d a s a c o l u m n . T h i s p r o -

c e d u r e is n o t a l w a y s p r a c t i c a l f o r p r e l i m i n a r y d e s i g n , s i n c e t h e d e s i g n e r , h a v i n g a

w i d e c h o i c e o f c r o s s s e c t i o n s , m u s t b e a b l e t o s e l e c t s o m e s e c t i o n s a n d p r e d i c t

t h e i r s t r e n g t h a t a n e a r l y s t a g e o f t h e d e s i g n . T e s t s o n a l u m i n u m - a l l o y c o l u m n s

s u b j e c t t o c r i p p l i n g f a i l u r e s s h o w t h a t t h e s h o r t - c o l u m n c u r v e c l o s c l y a p p r o x i -

m a t e s a s e c o n d - d e g r e e p a r a b o l a , a s r e p r e s e n t e d b y E q . ( 1 1 . 2 3 ) o r ( 1 1 . 2 9 ) . T h e

c r i p p l i n g s t r e s s n c c i s s u b s t i t u t e d f o r t h e s t r e s s a c 0 a s f o l l o w s :

' ( " . 6 1 ) 4N B.

A s i n t h e c a s e o f o t h e r s h o r t - c o l u m n c u r v e s , E q . ( 1 1 . 6 1 ) d o e s n o t a p p l y f o r v e r y

s h o r t c o l u m n s [IL/p < 12) b e c a u s e t h e e n d s u p p o r t s i n c r e a s e t h e c r i p p l i n g s t r e s s .

T h u s t h e c r i p p l i n g s t r e s s <r„ c a n b e o b t a i n e d b y t e s t i n g a c o l u m n w i t h a n L / p o f

a b o u t 12. A n a p p r o x i m a t e v a l u e o f t h e c r i p p l i n g s t r e s s m a y b e d e r i v e d b y f i n d i n g

t h e s u m o f t h e p l a s t i c b u c k l i n g s t r e n g t h s o f t h e r e c t a n g u l a r e l e m e n t s o f t h e c r o s s

s e c t i o n .

T h e c o l u m n c r o s s s e c t i o n s s h o w n i n F i g . 1 1 . 3 3 m a y b e c o n s i d e r e d a s m a d e o f

r e c t a n g u l a r p l a t e s o f w i d t h b, t h i c k n e s s f , a n d l e n g t h a , w h i c h is l a r g e in c o m p a r i -

s o n t o b. T h e p l a t e s w i t h w i d t h s d e s i g n a t e d b' a r e a s s u m e d t o b e s i m p l y s u p -

p o r t e d o n b o t h s i d e s , a n d t h o s e w i t h w i d t h s d e s i g n a t e d b a r e a s s u m e d t o b e f r e e

o n o n e s i d e a n d r e s t r a i n e d o n t h e o t h e r s i d e . I n t h e c a s e o f t h e a n g l e s s h o w n i n

F i g . 1 1 . 3 3 a a n d e , t h e p l a t e s a r e a s s u m e d t o b e s i m p l y s u p p o r t e d o n o n e s i d e ,

s i n c e t h e t w o p l a t e s b u c k l e a t t h e s a m e s t r e s s a n d n e i t h e r p l a t e s u p p l i e s a n y e d g e

r e s t r a i n t f o r t h e o t h e r . I n t h e c a s e o f t h e o t h e r c r o s s s e c t i o n s , h o w e v e r , t h e p l a t e s

w h i c h h a v e o n e s i d e f r e e h a v e e d g e c o n d i t i o n s b e t w e e n t h e c l a m p e d a n d s i m p l y

s u p p o r t e d c a s e s f o r t h e o t h e r s i d e . T h i s d i f f e r e n c e o n e d g e r e s t r a i n t i s s e e n b y

c o m p a r i n g t h e b u c k l e d f o r m o f t h e a n g l e s h o w n i n F i g . 1 1 . 3 1 t o t h e b u c k l e d f o r m

o f t h e c h a n n e l s h o w n i n F i g . 11 .34 . T h e l e g s o f t h e a n g l e b u c k l e i n o n e h a l f - w a v e

r e g a r d l e s s o f t h e l e n g t h o f t h e c o l u m n , a s i s t h e c a s e f o r a f l a t p l a t e w i t h o n e s i d e

f r e e a n d t h e o t h e r s i d e s i m p l y s u p p o r t e d . T h e l e g s o f t h e c h a n n e l b u c k l e i n t o t h e

s a m e n u m b e r o f h a l f - w a v e s a s t h e b a c k o f t h e c h a n n e l w h i c h b u c k l e s in a p p r o x i -

m a t e l y s q u a r e p a n e l s , a s s h o w n i n F i g . 1 1 . 3 4 .

T h e i n i t i a l b u c k l i n g s t r e s s o f t h e p l a t e s m a y b e s m a l l e r t h a n t h e s t r e s s a t

w h i c h c o l l a p s e o f t h e m e m b e r o c c u r s , s i n c e t h e c o r n e r r e s i s t s l o a d a f t e r t h e i n i t i a l

b u c k l i n g . T h i s e f f e c t i s c o n s i d e r e d e m p i r i c a l l y b y a s s u m i n g t h e e f f e c t i v e w i d t h b t o

b e l e s s t h a n t h e t o t a l w i d t h , a s s h o w n i n F i g . 1 1 . 3 3 e . T h e e x t r u s i o n s r e s i s t a

g r e a t e r l o a d a t t h e c o r n e r s t h a n t h e b e n t s h e e t s e c t i o n s , a s i n d i c a t e d b y t h e

K M h H

i" i (i) (</1

H

F i g u r e 11.40


Figure 11.34

w i d t h s b i n F i g . 1 1 . 3 3 . T h e e d g e c o n d i t i o n s f o r t h e p l a t e s i n a b u l b a n g l e e x -

t r u s i o n o f t h e t y p e s h o w n i n F i g . 1 1 . 3 3 / d e p e n d o n t h e b e n d i n g s t i f f n e s s o f t h e ^

b u l b , b u t u s u a l l y i t is a s s u m e d t h a t t h e b u l b s u p p o r t s t h e p l a t e a s i n d i c a t e d .

A f t e r t h e p l a s t i c b u c k l i n g s t r e s s i s d e t e r m i n e d f o r e a c h e l e m e n t o f a r e a b y t h e

m e t h o d o f S e e s . 1 1 . 1 0 a n d 1 1 . 1 1 , t h e t o t a l c r i p p l i n g l o a d o n t h e c r o s s s e c t i o n i s

f o u n d a s t h e s u m o f t h e l o a d s o n t h e i n d i v i d u a l a r e a s . I f t h e a r e a s h a v e d i m e n -

s i o n s b 2 t 2 , a n d b3t3 a n d b u c k l i n g s t r e s s e s ff!, a 2 , a n d a 3 , t h e n t h e t o t a l

c r i p p l i n g s t r e s s i s

T h e d e n o m i n a t o r o f E q . ( 1 1 . 6 2 ) m a y n o t b e e q u a l t o t h e t o t a l a r e a b e c a u s e t h e

c o r n e r a r e a s a r e n o t i n c l u d e d . T h e c r i p p l i n g l o a d i s o b t a i n e d b y m u l t i p l y i n g t h e

s t r e s s a c c b y t h e t o t a l a r e a , a n d i t m a y b e g r e a t e r t h a n t h e n u m e r a t o r o f E q .

( 1 1 . 6 2 ) b e c a u s e o f t h e l o a d o n t h e c o r n e r s .

E x a m p l e 1 1 . 5 F i n d t h e e q u a t i o n o f t h e s h o r t - c o l u m n c u r v e f o r t h e e x t r u s i o n

s h o w n i n F i g . 1 1 . 3 5 g i v e n t h a t E = 1 0 , 7 0 0 , 0 0 0 , n = 1 0 , a n d = 3 7 , 0 0 0 .

S O L U T I O N A s s u m e t h e c o l u m n c u r v e f o r t h i s m a t e r i a l i s r e p r e s e n t e d b y F i g .

1 1 . 2 9 . F o r p l a s t i c b u c k l i n g , t h e e q u i v a l e n t L / p i s o b t a i n e d f r o m E q . ( 1 1 . 5 5 )

w i t h K = 3 . 6 2 f r o m F i g . 1 1 . 2 1 . F o r a r e a 1, b j t = 1 . 5 6 4 / 0 . 0 5 = 3 1 . 3 ; f o r a r e a 2 ,

b / t = 0 . 7 0 / 0 . 0 9 3 = 7 . 5 2 . F r o m E q . ( 1 1 . 5 5 ) ,

cc cr1fo it l 12 + a3 b313 Zabt

b1tl + b2t2 + b313 ~ Zbt (11.62)

E q u i v a l e n t — = t

BUCKLING DP-SIGN OF STRUCTURAL MEMBERS 3 6 1

F o r a r e a 1, t h e e q u i v a l e n t L/p i s 1 . 6 5 x 3 1 . 3 = 5 1 . 6 ; f o r a r e a 2 , i t i s

1 . 6 5 x 7 . 5 2 = 1 2 . 4 . F r o m F i g . 1 1 . 2 9 , cr = 2 9 , 0 0 0 f o r a r e a 1 a n d a = 4 5 , 0 0 0

l b / i n 2 f o r a r e a 2 . F r o m E q . ( 1 1 . 6 2 ) ,

Tain 2 9 , 0 0 0 x 1 . 5 6 4 x 0 . 0 5 + 4 5 , 0 0 0 x 0 . 7 0 x 0 . 0 9 3 x 2

Xbt 3 9 , 0 0 0 l b / i n 2

1 . 5 6 4 x 0 . 0 5 + 2 x 0 . 7 0 x 0 . 0 9 3

T h e c r i p p l i n g s t r e s s e s f o r t h e i n d i v i d u a l a r e a s a l s o c a n b e d e t e r m i n e d

f r o m t h e n o n d i m e n s i o n a l c u r v e o f F i g . 1 1 . 1 6 . F r o m E q . ( 1 1 . 5 9 ) ,

B = b/t = 0 . 0 3 0 8 b-V 1 0 , 7 0 0 , 0 0 0 x 3 . 6 2 / 3 7 , 0 0 0 f

F o r a r e a 1, B = 0 . 0 3 0 8 x 3 1 . 3 = 0 . 9 6 5 . F r o m F i g . 1 1 . 1 6 , CT/CT1 = 0 . 7 7 , o r

a = 2 8 , 5 0 0 l b / i n 2 .

F o r a r e a 2 , B = 0 . 0 3 0 8 x 7 . 5 2 = 0 . 2 3 2 . F r o m F i g . 1 1 . 1 6 , a/tr, = 1 .20 , o r

<7 = 4 5 , 0 0 0 l b / i n 2 . T h e s e c h e c k t h e v a l u e s o b t a i n e d f r o m F i g . 1 1 . 2 9 . T h e

s h o r t - c o l u m n c u r v e i s n o w o b t a i n e d f r o m E q . ( 1 1 . 6 1 ) :

A,- = c , , 1 -4 t t 2 £

= 3 9 , 0 0 0 1 -3 9 , 0 0 0 ( E / p ) 2

4 t i 2 1 0 , 7 0 0 , 0 0 0 _

3 9 , 0 0 0 [ 1 - 0 . 0 0 0 0 9 2 3 ( L / p ) 2 ]

E x a m p l e 11.-6 G i v e n t h e s e c t i o n . s h o w n i n F i g . 1 1 . 3 6 , a s s u m e t h a t n = 1 0 ,

E = 9 7 0 0 , a n d c , = 4 6 , 0 0 0 . F i n d t h e c r i p p l i n g s t r e s s f o r t h e c r o s s s e c t i o n .

S O L U T I O N T h e w e b i s a s s u m e d t o b e s i m p l y s u p p o r t e d o n b o t h s i d e s , w i t h

K = 3 . 6 2 , a n d b u c k l e s i n t o a p p r o x i m a t e l y r e c t a n g u l a r p a n e l s i n a m a n n e r

s i m i l a r t o t h e c h a n n e l s e c t i o n s h o w n i n F i g . 1 1 . 3 4 . T h e h a l f - w a v e s a r e a p -

p r o x i m a t e l y 1 . 1 2 i n l o n g ; t h e r e f o r e , t h e flanges m a y b e c o n s i d e r e d a s s i m p l y

s u p p o r t e d a t e n d s 1 . 1 2 i n a p a r t a n d o n o n e s i d e . F r o m F i g . 1 1 . 2 1 f o r

M 0 . 0 6 4 - -T

A = 0 . 1 3 6 i n 3

/ = O . O . ' O i n 4

- i r 4 « II'-

r •» 0 . 6 4 4

I

(H)

F i g u r e I I -36

a j b = 2 . 0 0 0 , K = 0 . 6 0 . F r o m E q . ( 1 1 . 5 9 ) ,

bjt B =

Y/EK/(T |

F o r t h e f l a n g e s ,

F o r t h e w e b ,

0 . 5 / 0 . 0 6 4 B = . ' = 0 . 6 9 3

7 9 , 7 0 0 , 0 0 0 x 0 . 6 / 4 6 , 0 0 0

1 . 1 2 / 0 . 0 6 4 B = ' = 0 . 6 3 3

- y / 9 , 7 0 0 , 0 0 0 x 3 . 6 2 / 4 6 , 0 0 0

F r o m F i g . 1 1 . 1 6 , ff/tr, = 0 . 8 8 , o r £7 = 4 0 , 5 0 0 l b / i n 2 , f o r t h e flanges, a n d

ff/ff, = 0 . 9 0 5 , o r <7 = 4 1 , 6 0 0 l b / i n 2 f o r t h e w e b .

F r o m E q . ( 1 1 . 6 2 ) ,

light Tab 2 x 4 0 , 5 0 0 x 0 . 5 - f 4 1 , 6 0 0 x 1 . 1 2

° c c ~T.bt~I.b~ 2 x 0.5 + 1.12 = 4 1 , 0 0 0 l b / i n 2

1 1 . 1 3 N E E D H A M A N D G E R A R D M E T H O D S F O R

D E T E R M I N I N G C R I P P L I N G S T R E S S E S

M o r e r e c e n t s e m i e m p i r i c a l m e t h o d s t h a n t h a t o f S e c . 1 1 . 1 2 f o r t h e d e t e r m i n a t i o n

o f c r i p p l i n g s t r e s s e s o f c o l u m n s w e r e d e v e l o p e d b y N e e d h a m 4 5 a n d G e r a r d . 4 6 - 4 9

I n t h e N e e d h a m m e t h o d , t h e s t r u c t u r a l m e m b e r s e c t i o n i s d i v i d e d i n t o a n g l e

e l e m e n t s , a s s h o w n i n F i g . 1 1 . 3 7 . T h e c r i p p l i n g s t r e n g t h o f t h e s e e l e m e n t s c a n b e

e s t a b l i s h e d b y t h e o r y a n d / o r t e s t s . T h e c r i p p l i n g f a i l u r e s t r e n g t h o f . t h e m e m b e r

s e c t i o n t h e n c a n b e d e t e r m i n e d b y s u m m i n g t h e c r i p p l i n g s t r e n g t h s o f e a c h a n g l e

e l e m e n t t h a t m a k e s u p t h e t o t a l s e c t i o n . T h r o u g h e x t e n s i v e t e s t s N e e d h a m a r -

r i v e d a t t h e f o l l o w i n g s e m i e m p i r i c a l e q u a t i o n f o r t h e c r i p p l i n g s t r e s s o f a n g l e

s e c t i o n s :

OC=- k E [ E c a X 5

( 6 ' / f ) 0 - 7 5 ( 1 1 . 6 3 )

C u t -

C u t I

C u t -


C u t

I C u t

U )

w h e r e t r c = c r i p p l i n g s t r e s s

E c = c o m p r e s s i v e m o d u l u s o f e l a s t i c i t y

rrL.r = m a t e r i a l c o m p r e s s i v e y i e l d s t r e s s

b' a + b T ~ 21

A,, = c o n s t a n t c o e f f i c i e n t w h o s e m a g n i t u d e d e p e n d s o n s u p p o r t c o n d i t i o n o f a n g l e e d g e s : k t , = 0 . 3 1 6 f o r t h e t w o e d g e s f r e e , 0 . 3 4 2 f o r o n e e d g e f r e e , a n d 0 . 3 6 6 w i t h n o e d g e f r e e

T h e c r i p p l i n g s t r e s s f o r t h e a c t u a l m e m b e r s e c t i o n i s o b t a i n e d b y u t i l i z i n g t h e

f o l l o w i n g e q u a t i o n :

'••-Tsr <1U4> w h e r e r r „ = m e m b e r s e c t i o n c r i p p l i n g s t r e s s

a d — / t h a n g l e c r i p p l i n g s t r e s s , c a l c u l a t e d f r o m E q . ( 1 1 . 6 3 )

A ; = / t h a n g l e c r o s s - s e c t i o n a l a r e a

G e r a r d ' s m e t h o d 4 " f o r c a l c u l a t i n g c r i p p l i n g s t r e s s e s i s a g e n e r a l i z a t i o n o f N e e d h a m ' s m e t h o d . H i s e x t e n s i v e i n v e s t i g a t i o n l e d t o t h e f o r m u l a t i o n o f t h r e e s e m i e m p i r i c a l e q u a t i o n s f o r d e t e r m i n i n g t h e c r i p p l i n g s t r e s s e s in v a r i o u s s h a p e s of s t r u c t u r a l m e m b e r s .

F o r s e c t i o n s w i t h s t r a i g h t u n l o a d e d e d g e s s u c h a s p l a t e s , t e e , c r u c i f o r m , a n d H s e c t i o n s , t h e f o l l o w i n g c r i p p l i n g s t r e s s e q u a t i o n a p p l i e s :

= 0 . 6 7 <xt.,. £ 1 A

i s \ i r

< W . (11.65,-;)

F o r s e c t i o n s w i t h d i s t o r t e d u n l o a d e d e d g e s s u c h a s t u b e s , V - g r o o v e p l a t e s , a n g l e s ,

s t i f f e n e d p a n e l s , a n d m u l t i c o r n e r s e c t i o n s , t h e f o l l o w i n g c r i p p l i n g e q u a t i o n

a p p l i e s :

= 0.56(7,., HL A

E^Y' A J _

( 1 1 . 6 5 6 )

F o r s e c t i o n s s u c h a s t w o c o r n e r s e c t i o n s , J , Z , a n d c h a n n e l s e c t i o n s , t h e f o l l o w i n g

e q u a t i o n a p p l i e s :

<7„ = 3 . 2 f f 0 . r j A

1/3 ( 1 1 . 6 5 c )

w h e r e A = s c c t i o n a r e a a n d tj = n u m b e r o f f l a n g e s w h i c h m a k e u p t h e s e c t i o n

p l u s t h e n u m b e r o f c u t s r e q u i r e d t o d i v i d e t h e s e c t i o n i n t o a n u m b e r o f flanges.

S e e F i g . 1 1 . 3 8 . T h e m a x i m u m c r i p p l i n g s t r e s s i7„ m u s t n o t e x c e e d t h o s e s p e c i f i e d

in T a b i c 1 l.'I u n l e s s it i s v e r i f i e d e x p e r i m e n t a l l y .

I (/'I

c m

( i ) (</) t o

Figure 11.38 [a) Basic angle sect ion; g = 2; (b) t ube ; g = 4 cu ts + 8 flanges = 12; (r) basic T sect ion; g = 3; (d) II sec l ion; g = 1 cut + 6 flanges = 7 ; (e) cruci form sect ion; g = 0 cut +- 4 flanges = 4.

T a b l e 1 1 . 1

Sect ion s h a p e M a x i m u m tr a

L 0.7<rc, T , + , H 0 . 8 ^ Z, J, u 0.9acy

• , mul t icorncr 0.8ffc ,

1 1 . 1 4 C U R V E D S H E E T I N C O M P R E S S I O N

A t h i n - w a l l e d c i r c u l a r c y l i n d e r l o a d e d i n c o m p r e s s i o n p a r a l l e l t o i t s a x i s m a y f a i l

b y l o c a l i n s t a b i l i t y o f t h e t h i n w a l l s . T h i s t y p e o f f a i l u r e i s s i m i l a r t o t h a t w h i c h

o c c u r s i n t h e c o m p r e s s i o n s k i n o f s e m i m o n o c o q u e w i n g a n d f u s e l a g e s t r u c t u r e s .

T h e c o m p r e s s i o n b u c k l i n g o f f l a t s h e e t s w a s c o n s i d e r e d , b u t m o s t a c t u a l s t r u c -

t u r e s a r e m a d e o f c u r v e d s h e e t s , a n d t h e c u r v a t u r e h a s a c o n s i d e r a b l e e f f e c t o n

t h e b u c k l i n g a n d u l t i m a t e s t r e n g t h s . A c y l i n d e r w h i c h i s l o a d e d i n c o m p r e s s i o n

w i l l a s s u m e a b u c k l c d f o r m s i m i l a r t o t h a t s h o w n i n F i g . 1 1 . 3 9 . T h e n u m b e r o f

c i r c u m f e r e n t i a l w a v e s d e p e n d s o n t h e r a t i o o f R / t , w h e r e R i s t h e r a d i u s a n d t i s

(A i Figure 11.39

b u c k l i n g d e s i g n o f s t r u c t u r a l m e m b e r s 3 6 5

the wall thickness of the cylinder. A large number of waves develop for a large value of R/t. The length of the longitudinal waves is the same magnitude as the length of the circumferential waves. Fo r the high ratios of R/t which are common in semimonocoque wing and fuselage skins, the wavelengths are so small that a sector of a cylinder, with simply supported edges as shown in Fig. 11.40a resists approximately the same buckling stress as the complete cylinder. This sector corresponds to the skin between adjacent stringers, as shown in Fig. 11.40b. Fo r smaller values of R/t, the length of the circumferential waves is greater, and the stringers or edge supports of Fig. 11.40b prevent the formation of the waves and thus increase the buckling stress.

The compression buckling stress for a thin-walled cylinder may be deter-mined theoretically in a manner similar to that used in obtaining the buckling stresses of flat plates. The classic analysis of cylinders, which is based on the assumption of small displacements, yields

if Poisson's ratio is 0.3. Test values, however, are much lower than those given by Eq. (11.66), and test results show considerable scatter. This is in contrast to the excellent correlation between theoretical and experimental values for the buckling stresses for flat sheet.

Von Karman, Dunn, and Tsien6 have shown that the assumptions made in the analysis by the classic theory arc in error, fn the case of buckling of a flat plate, a longitudinal strip of the plate is supported elastically by lateral strips which exert restraining forces in proport ion to their deflection. When the flat plate buckles, i t . m a y buckle in either direction, and the load after buckling remains equal to the buckling load, as in a Euler column. In the case of a compressed cylinder, however, the longitudinal strips are supported by circumfer-ential rings which exert restraining forces that are not proport ional to their radial deflection. The stillness of a circular ring increases as it is deflected outward and decreases as it is deflected inward. Thus the thin walls of a compressed cylinder buckle inward much more readily than they buckle outward. The buckling is accompanied by a sudden decrease in both the load and the length if there are no eccentricities of the walls. The buckling load is considerably reduced by small eccentricities of the walls. The buckling stress depends on the rigidity of the testing machine, since any testing machine has some elasticity, and the plates of

a c = 0 . 6 0 6 E — R t

(11.66)

Figure 11.40

the machine move together slightly as the resistance of the specimen decreases. The large effects of specimen eccentricity and or testing machine elasticity explain the large scatter of test results.

If the compression load on a cylinder P is plotted against the axial com-pressive deformation e, curves similar to those shown in Fig. 11.41 are obtained. Curve 1 represents a theoretical curve for an ideal cylinder in which the walls are perfectly cylindrical and homogeneous. The point A, corresponds to (he theo-retical buckling stress, obtained from Hq. (11.66), which cannot be determined experimentally by the most careful testing because the upper branch of the curve is so close to the lower branch. At a deformation e corresponding to point BL, the cylinder assumes a buckled form and the load drops. If the deformation has exceeded that corresponding to B1 before buckling occurs, the cylinder suddenly decreases in length when buckling occurs. Because of the elasticity of the testing machine, the plates of the machine move together and the cylinder decreases in length when the load drops, even for test specimens with small eccentricities, as represented by curve 2.

Buckling loads obtained experimentally are represented by points A2 and / l 3

of Fig. 11.41. In the case of unstiifened cylinders, these buckling loads represent the ultimate strength of the cylinder in compression. Several empirical equations have been derived from experimental results, and the various equations yield widely divergent values of buckling stress, as might be expected because of the scatter of test values. Kancmitsu and Noj ima propose the following equations:

where L is the length of the cylinder. This equation appears to give satisfactory agreement with test values within the ranges of 500 < R/t < 3000 and 0.1 < L/R < 2.5.

Another equation which yields reasonable values of the buckling stress for smaller values of R/t is obtained as approximately one-half of the value of Eq.

This equation yields results which are much higher than experimental values in

(11.67)

(11.66):

(11.68)

r

No L'Lxnitridty

/ Snwll ccccntricity

/ 1 /

v b u c k l i n g d 1 s i g n o l ' STRUCTURAL mi-mhlirs 3 6 7

cases where R/t is large. Perhaps it is reasonable to use Eq. (11.68) for values of R/t less than 500 and to use Eq. (11.67) for the range in which it applies.

In the case of curved sheet which is stiffened by longitudinal members, as is common in semimonocoque construction, the sheet would resist a buckling stress as given by Eq. (11.48) if there were no curvature and an additional stress, as given by Eq. (11.67), because of the curvature. While there is little theoretical justification for adding these buckling stresses, this procedure is substantiated reasonably well by tests.

The compression buckling stress for curved skin on the upper surface of a wing is increased considerably by the negative air pressure on the sheet. Since the curved sheet has a tendency to buckle inward, the aerodynamic forces reduce this tendency. Equation (11.67) is very conservative in this case. It is very important to prevent the buckling of the wing skin of high-speed aircraft because of the aerodynamic drag of the irregular airfoil section.

The ultimate strength of a stiffened, curved sheet panel may be found in a similar manner to that used in obtaining the ultimate strength of a flat sheet panel in Sec. 11.9. In addition to the compression load resisted by the stringers and by the effective widths of skin acting with the stringers, the sheet between stringers resists load because of its curvature, even though it has buckled. The load resisted by a buckled, curved sheet is indicated by the right-hand portion of the curves of Fig. 11.41. While this load depends on the elongation e cf the stringers, many other unknown factors are involved. The method for calculating this load is to assume that a skin width of h — 2W between stringers resists a stress of 0.25Et/R, as shown in Fig. 11.42. As an alternative method, this stress might be calculated by Eq. (11.67). Where this buckling stress for the curved sheet excceds the stringer stress a c , the entire sheet area is assumed to resist a stress a c .

Example 11.7 For the wing shown in Fig. 11.43, R = 50, f = 0.064, b = 6, and the rib spacing is L = 18 in. Find the compressive stress in the skin at which the buckling occurs if E = 107 lb/in2.

0 . 2 5 E'-

Figure 11.42 Figure 31.43

3 6 8 a i r c r a f t s t r u c t u r e s ,

SOLUTION The buckling stress is obtained as the sum of the buckling stress for a flat sheet simply supported on four sides, as obtained from Eq. (11.48), and the buckling stress for a cylinder, as obtained from Eq. (11.67). From Eq. (11.48),

<7Ccr = K S y j = 3.62 x 107 x = 4110 lb/in2

From Eq. (11.67),

50 } V 18

t7Ccr = 2130 + 1560 = 3690 lb/in2

The total buckling stress is the sum of these two values:

cCt[ = 4110 + 3690 = 7800 lb/in2

11.15 ELASTIC SHEAR BUCKLING OF FLAT PLATES

The buckling of rectangular plates which resist direct compression stresses is discusscd in Sec. 11.8. Other types of stresses, such as shear stresses and bending stresses, also may produce elastic buckling of thin plates. Only loads in the plane of the plate are discussed here, and components normal to the plane of the plate are assumed to be zero.

The elastic buckling stresses for thin rectangular plates in shear can be calcu-lated theoretically. The analysis is beyond the scope of this book, but the results may be expressed in the same form as Eq. (11.48) if Poisson's ratio v is assumed constant for all materials:

t „ = K E ^ 2 J 11.69)

The values of K are plotted in Fig. 11.44 for v = 0.3 and for the two conditions of all four edges clamped and all four edges simply supported. The term t is the plate thickness, and E is the elastic modulus of the plate material. For the compressed plate discussed in Sec. 11.8, the width b is perpendicular to the direction of loading and the length a is parallel to the loads; but since the plate in shear is loaded on all four sides, the dimension b is considered as the smaller of the two plate dimensions. The critical shearing stress Tcr is uniformly distributed along all four sides of the plate.

The rectangular plate which is loaded in pure shear has principal tension and compressive stresses at 45° to the edges. These principal stresses are equal to the shearing stresses. The diagonal compression stresses cause the sheet buckling, and

Figure 11.44

when buckling occurs, the wrinkles form at approximately 45° angles to the edges. The buckling shearing stresses t c r are considerably higher than the buck-ling compression stresses erCcr for plates with equal dimensions. This is a result of the restraining effect of the diagonal tension in the plate which is loaded by shearing forces.

The critical buckling stresses in a thin plate loaded in bending as shown in Fig. 11.45 also can be calculated theoretically and expressed in the same form as the equations for compression and shear buckling:

a*-70)

where <rbcr is the critical maximum bending stress shown in Fig. 11.45 and K is given by the curve of Fig. 11.45 if all four edges are simply supported.

In the case of buckling of thin plates under the combined action of two of the

3 7 0 AIRCRAFT STRUCTURES ,

1 1 1 1 ' IT

i K i K _ r v „ ; r . , 2

cr

i i i \ / Simp ly sui pork tl ClljSi s

\ . r — y A \ 0 0 . 2 0 . 4 0 . 6 0 . 8 1 . 0 1 . 2 1 . 4 1 . 6 1 . 8 2 . 0

a/ft

F igure 11.45

conditions of compression, shear or bending, the initial buckling stresses have been determined empirically by the method of stress ratios. The initial buckling occurs when one of the following equations is satisfied:

Compression and bending:

RL-15 + RC= 1 (11.71)

Compression and shear:

R'-5 + RC=1 (11.72)

Bending and shear:

R2b+Rt= 1 (11.73)

where RB, RS, and RE represent the ratios of the stresses in the plate to the critical buckling stresses <rk/aKr, crJaCcr, and T/T„ .

11.16 ELASTIC BUCKLING OF CURVED RECTANGULAR PLATES

A large part of the structure of a semimonocoque airplane consists of the outer shell, or skin. This skin usually is curved to provide the necessary aerodynamic shape, and it must resist tension, compression, shear, and bending stresses. In addition to the conditions of ultimate strength and yield strength, which must be considered in the design of flight vehicle structural members, often the skin must be designed so that it will not wrinkle under normal flight conditions. Skin wrinkles or other surface irregularities seriously affect the airflow in the case of high-speed aircraft, but may be permissible for slower-speed aircraft. Unfortu-nately, both the buckling strength and the ultimate strength of curved plates depend on many uncertain factors and are difficult to predict accurately. Initial


plate eccentricities, air pressure normal to the plate, and conditions of the sup-ports are difficult to evaluate; yet they may have a considerable effect on buck-ling loads.

The buckling stress for a curved plate in shear, such as shown in Fig. 11.46, is higher than the buckling stress for a flat plate with corresponding dimensions. The buckling stresses obtained experimentally usually are smaller than those calculated theoretically for an ideal plate with small deflections. A condition similar to that described in Sec. 11.14 for plates in compression exists for plates in shear; the theoretical buckling stresses for flat plates correspond closely with test results for practical plates, but theoretical buckling stresses- for curved plates usually arc higher than values obtained experimentally.

The theoretical shear buckling stresses for curved plates have been calculated by Batdorf, Stein, and Schildcrout.50 For a constant value of Poisson's ratio, v = 0.3, the shear buckling stress i c r may be expressed in the form of previous buckling equations:

t cc = K s E ( ~ J (11.74)

The term Ks is a function of the ratios ajb and b2/{rt) and is plotted in Figs. 11.46 and 11.47. When the circumfcrential length is greater than the axial length, Fig.

n

Figure 11.46


11.47 is applicable; when the axial length is greater, Fig. 11.46 must be used. The dimension b is smaller than the dimension a in either case. Both figures apply only to plates for which all four edges are simply supported. The points at the left side of the charts, for b2/(rt) - 0, correspond to the buckling stress coefficients for flat plates, as given in Fig. 11.44.

For design purposes, it is necessary to consider the effects of initial accidental eccentricities, which always cause the buckling stresses to be smaller than the theoretical values. Often the designer must use judgment in evaluating these effects for a particular structure. An empirical equation is proposed:

Tcr = X £ ( 0 ( 1 L 7 5 )

where the first term represents the buckling stress for a flat plate, as given by Fig. 11.44, and the last term represents the additional stress which can be resisted because of the curvature. The value •= 0.10 is recommended. By rewriting Eq. (1 1.75) aryd comparing it to Eq. (11.74), the following relations are obtained:

K+K>biHi b2

K ^ K + K i — ( 1 1 . 7 6 )

1000

20

1000

Figure 11.47


The values of Ks from Eq. (11.76) are plotted as the dotted lines in Figs. 11.46 and 11.47, assuming K t = 0.10. The values obtained from Eq. (11.75) are seen to represent conservative approximations for all values shown on the chart, except for the case of large values of ajb and b2/{rt) shown in Fig. 11.47. Except for this range, Eq. (11.75) approximates most of the available test information closely and conservatively and may be used in practical design. While the theoretical curves of Figs. 11.46 and 11.47 apply only to plates with simply supported edges, Eq. (11.75) may be used with Fig. 11.44 for plates with clamped edges or for other edge conditions, by interpolation of Fig. 11.44.

11.17 PURE TENSION FIELD BEAMS51

The ultimate strength of thin webs in shear is much greater than the initial buckling strength. In the case of structural members which are not exposed to the airstream, such as wing spars, the shear webs may be permitted to wrinkle at a small fraction of their ultimate loads. To describe the manner in which loads are resisted by shear webs after buckling has occurred, it is convenient to consider a pure tension field beam in which the web buckles when the shearing forces are initially applied. Such a web never exists in practice, since even very thin webs have enough buckling resistance to affect the stress distribution appreciably.

The beam shown in Fig. 11.48a has concentrated flange areas which arc

— ^ j - 1 / c o s a

Figure 11.48

assumed to resist the entire beam bending moments. The beam web has thickness t and depth h between centroids of the flanges. The vertical stilfeners are spaced uniformly at a distance d along the span. The shear force V is constant for all cross sections. The shear flow at all points in the web is therefore equal to V/h, and the shear stress at all points T2 is Vj{th). If the web is shear-resistant, a web element at the neutral axis of the beam is stressed, as shown in Fig. 11.49. On the vertical and horizontal faces X and Y, the element resists only the shearing stresses t and no normal stresses. The principal stresses <r, and <rc occur on planes at 45" to the horizontal, as shown in Fig. 11.496. The magnitudes of the principal stresses are determined by* the Mohr circle construction of Fig. 11.49c, from which <r, = ac = r. If the beam web of Fig. 11.48a is assumed to be extremely flexible, it will not be capable of resisting the diagonal compressive stress. Then it will act as a group of parallel wires, inclined in the direction of the tension diagonal, or at an angle a. of approximately 45°, as shown. Such a group of wires cannot resist any of the beam bending moment, and it is customary to assume that every element in a tension field web resists the same stress as an element at the neutral axis. A web element for a pure tension field web is therefore stressed as shown in Fig. 11.50. The shearing stresses on the vertical and horizontal faces have the same values x = V/(th) as for the shear-resistant web. These planes, X and Y, also have tensile stresses ax and ay, respectively, which are obtained from the Mohr circle construction in Fig. 11.50c. From the geometry of the circle, the lengths of lines QX and PY are r/sin a, and the lengths of lines PX and QY are t/cos a. The following stresses are obtained:

a x = t cos a (11.77)

a , = T t a n a (11.78)

= — - E = (11.79) sin a cos a sin 2a

The relationships expressed by Eqs. (11.77) to (11.79) may be obtained with-out the use of the Mohr circle construction by referring to Fig. 11.486. The web of thickness r, which resists a maximum tensile stress a „ is assumed to be re-placed by wires a unit distance apart which resist forces of rr,. The vertical

Figure 11.49

b u c k l i n g d e s i g n o f STRUCTURAL m e m b e r s 3 7 5

(a)

F i g u r e 11.50

= 0

•

+ "r t

t Q a J

\\ M . ^ p 1

\ \

(/»

component of the wire tension is then a, sin a, and the horizontal component is cr, cos «. Along a horizontal line through the wires, the spacing is 1/sin a, corre-sponding to a horizontal web area of tj sin a. The web tension stress oy on the horizontal plane is obtained by dividing the vertical component of the wire tension <r, sin a by the web area tjsin a :

ay = <r, sin2 a (11.80)

The shearing stress x on a horizontal plane is found by dividing the horizontal component of the wire tension cr, cos a. by the web area £/sin a:

x = IT, sin a cos a (11.81)

Equation (11.81) corresponds to Eq. (11.79). Similarly, a vertical line through the wires gives spacing of 1/cos a corresponding to a web area of £/cos a. The hor-izontal web stress a x is obtained by dividing the horizontal component of the wire tension by the area:

crx = cr, cos2 a (11.82)

Equations (11.77) and (11.78) may be obtained from Eqs. (11.80) to (11.82). The tension field beam also differs from the shear-resistant beam in the

manner in which stresses are transferred to the stiffeners, beam flanges, and riveted connections. The vertical stiffeners in a shear-resistant beam resist no compression load; they only divide the web into smaller unsupported rectangles and thus increase the web buckling stress as calculated from Eq. (11.43). In a tension field beam, however, the vertical web tension stresses <jy tend to pull the beam flanges together, and this tendency must be resisted by compression forces in the stiffeners. Each stiffener must resist a compressive force P that is equal to the vertical tension force in the web for a length d equal to the stiffener spacing, as shown in Fig. 11.5In:

P = aytd • Vd

tan a (11.83)

The vertical web tension stresses also tend to bend the beam flanges inward. The flanges act as continuous beams supported by the stiffeners. If the ends of the

f T T i T ^ T ^ PI 2 P = a,.hl P P

= I - t a n a ( a )

M = a,!d~/24 M = ortd2l 12

C>) Figure 11.51

flanges are assumed to be fixed against rotation, the flange bending-moment diagram is as shown in Fig. 11.516. Al the stifleners the bending moment is

<rvtd2 Pd

M = ^ - = — (11.84)

Midway between the stifleners the flange bending moment is

The direction of the bending moment is such that it produces tension on the outside of the flange at the stifleners and on the inside of the flange between the stifleners.

The horizontal components of the web stresses a x tend to pull the end stifleners together with a force uxth = V cot a. This force is resisted equally by the two spar flanges, producing compression forces of K(cot a)/2, which must be superimposed on the forces M/h that result from beam bending.

The riveted connections for a shear-resistant web must be designed to resist a load g = t s£ per unit length. In a tension field web connection, the horizontal, riveted joints must resist shear flows q as well as tension forces of q tan a per unit length in a perpendicular direction. Hence all horizontal riveted joints must resist forces of q j 1 + tan2 a = q sec or per unit length. The vertical riveted joints at the ends of the beam or at web splices must resist shear flows of q = rst and tensile forces of axt + q cot a per unit length. Thus, the joints must be designed for a load of q^J 1 4- cot2 a. or q csc a. per unit length. This force does not apply for connections between the web and intermediate stifleners, since no appreciable load is transferred by this connection.

In earlier chapters, various shear-flow analyses are made in which the webs arc assumed to resist pure shear. These analyses remain valid even though the webs are in tension field, since the tension stresses on the X and Y planes may be superimposed on the shearing stresses without affecting the shear-flow analysis.

V = 6000 Hi

/ - 10 50 •

/ - 10 50 •

- t = 20

1

Figure 11.52

Example 11.8 The beam shown in Fig. 11.52 is assumed to have a pure tension field web. Draw free-body diagrams for the stiffeners and flanges, and plot the axial loads in the stiffeners and flanges. Assume a = 45°.,

SOLUTION The shearing stress on a horizontal or vertical plane of a web element is T = V/(th) = 6000/(0.020 x 20) = 15,000 lb/in2.. The running shear is q = T4. f = 300 lb/in. The tension stresses ax and <r on these planes and the tension loads pcr inch also equal Ts and q. The compression load on an intermediate stiffener, P = V<l/h, is 3000 ib. The stiffener at the left end has a compression load of P/2 and an additional compression force of 6000 lb applied at the lower end. Both beam flanges have compression loads of 7 /2 = 3000 lb at the left end. The beam flange loads vary linearly along the span. At the support, the flange loads from beam bending 'M/h are 6000 x f § = 15,000 lb. The compression flange resists a load —M/h—V/2 =

- 3 0 0 0 I b J

- 1 8 , 0 0 0 l b

3 0 0 0 j b

1 5 0 0 l b

3 0 0 I l i / i . n - 1 8 0 0

t r r t r r t r r r r t r r T - 1 5 0 0 ! b i 1 5 0 0 1 b 3 0 0 0 1b 1 3 0 0 0 1 b

~ V t 3 0 0 0 l b p q 3 0 0 0 1 b r L . 3 0 0 l b / i n

r . « M 0 . b ) n - 3 0 0 0 I b

3 0 0 0 1 b 1 5 0 0 1 b

- 7 5 0 0 l b

7 5 0 0 l b

r - 3 0 0 0 I b

3 0 0 0 I b

1 5 0 0 1 b 3 0 0 0 1 b 3 0 0 0 1 b 3 0 0 0 1 b 3 0 0 0 1 b 1 5 0 0 1 b

3 0 0 0 l b J J J U ^ J I J J I J J I j j 1 2 . 0 0 0 l b

3 0 0 l b / i n

- 3 0 0 0 l b { [

Figure 11.53

^ T T r r n r T r i T n T T T + 1 2 0 0 0 l b


- 1 8 , 0 0 0 lb, and the tension flange resists a load of M/h - V/2 = 12,000 lb. The free-body diagrams are shown in Fig. 11.53. All intermediate stiffeners resist the same loads.

11.18 ANGLE OF DIAGONAL TENSION IN WEB

This angle may be determined from the deflected geometry of the beam, framework, consisting of the flanges and stiffeners, has equal stiffness in resisting the horizontal tension o^'and the vertical tension cry, the two tension stresses will be equal and a will be 45°. In practical beams, the flanges are much more rigid in resisting compression loads than are the stiffeners. The stiffeners deform in com-pression and permit the flanges to move together, while the stiffeners remain approximately the same distance apart . The horizontal web stress a x is therefore greater than a y , and the diagonal tension stress uy has an angle less than 45°. This angle may be determined from the deflected geometry of the beam.

The beam shown in Fig. 11.54 is initially horizontal, and it has a shearing deformation y at all cross sections. Bending deflections are not considered here. The deformation y is caused by axial elongations of the stiffeners and flanges and by the elongation of the web diagonal resulting from the diagonal tension stress. A section of the beam of length h cot a and depth h is considered, and the value of a required to produce a minimum deformation y is determined. All elongations are assumed positive as tension in the derivation, although the stiffeners and flanges will always be in compression and have negative elongations. The defor-mations of the length, /i cot a, of the beam in Fig. 11.54 will be the same as those for the truss shown in Fig. 11.55, if the unit elongations in the horizontal, vertical, and diagonal directions are the same for the two structures.

The total elongation of a vertical stiffener is equal to the product of the unit elongation ey and and the length h. This elongation causes a shearing defor-mation Vi, as shown in Fig. 11.55a, which is obtained by dividing the total elongation by the radius h cot a :

her l'i = h cot a

= ey tan a ,(11.86)

T h e beam flanges have a unit elongation ex, or a total elongation ofeA./j cot

Figure 11.54


j-*-/? cot a -

J L _

7 1 ~ /; cot oc

fa)

Figure 11.55

R \ \ \ \ \ \ \

eJ t cot a

?2 c j l cot or 7 2 •sin a cos a

(O

a in the horizontal length considered. The angular deformation y2 is obtained by dividing this deformation by the radius h, as shown in Fig. 11.556:

y2 = ex cot a (11.87)

The diagonal strip of the web has a unit elongation e and a length /i/sin a. The angular deformation }>3 is obtained from the geometry of Fig. 11.55c:

ya sin a cos a (11.88)

The total shearing deformation for the beam is the algebraic sum of the three components :

V = - 7 l - >'2 +

Substituting from Eqs. (11.86) to (11.88) yields

— e., tan a — e , cot a + — > Ol sin a cos a (11.89)

The angle of the web diagonal tension a will be such that the deformation y is a minimum. Differentiating Eq. (11.89) and equating dy/da to zero yield

tan a • (11.90)

where e = aJE is the unit strain along the web diagonal, ex is the unit strain in the beam flanges resulting from the compression caused by the web tension a x , and ey is the unit strain in the vertical stiffeners caused by the compression load P. All strains are positive for tension and negative for compression. The bending of the beam flanges and the slip in the riveted joints at the flanges have the same effect as an elongation e} and may be included in the analysis.

The unit elongations used in Eq. (11.90) depend on the stresses, which in turn depend on the angle a. It is therefore necessary to solve this equation simul-taneously with other equations obtained from the web stress conditions. For normal beam proport ions, the flanges do not compress appreciably as a result of


the tension field stress, and ex may be assumed zero. The web diagonal strain c is uJE, or 2r, csc 2a/E, from Eq. (11.79). The unit strain ey is obtained as — tstd tan a/AcE, where A,. is the effective area of a vertical"stiflcncr. Substituting these values into Eq. (11.90) gives

The cfTcctive stiffener area is equal to the true stiffener area A if the stiffener consists of two members symmetrically attached on opposite sides of the web. Where a single stiffener is attached to only one side of the web, it is loaded in bending as well as compression. The compression load P has an eccentricity e measured from the ccnter of the web to the ccntroid of the stiffener area. The combined bending and compression stress at a distance e from the neutral axis is

where p is the radius of gyration of the stiffener cross-sectional area and

The differentiation of Eq. (11.89) may appear questionable, since the elonga-tions e, ex, and ey are treated as constant with respect to a. Equation (11.91) also may be obtained by substituting values for the strains as functions of a into Eq. (11.89) before differentiation, which is a more rigorous mathematical procedure but yields the same angle a. Equation (11.90), however, is a general expression for the angle of the principal planes at a point in any structure with two-dimensional stress conditions when the strains ex, ey, and e are known. Lahde and Wagner55

first applied this equation to the analysis of tension field webs. Langhaar67 ex-pressed the strain e in terms of a known distortion y and equated de/da to zero in order to find the angle a for the maximum or principal strain. The angle a thus obtained is equal to that yielded by Eq. (11.90). Langhaar also expressed the total strain energy as a function of a and equated the derivative of the strain energy to zero in order to find the angle a, which yielded a minimum of the total 'strain energy. This also gave the same result as that obtained from Eq. (11.90). If flange bending and other deformations are considered, the value of ey is greater than that used in obtaining Eq. (11.91).

In Sec. 11.18 wc assume that the beam web is perfectly flexible and is not capable of resisting any diagonal compressive stress. In practical beams, the webs resist some diagonal compressive stress after buckling, and thus they act in an inter-mediate range between shear-resistant webs and pure tension field webs. Such beams are termed semitension field beams, partial-tension field beams, or incom-

(11.91)

11.19 SEMITENSION FIELD BEAMS


pletely developed diagonal tension field beams. The pure tension field theory is conservative for the design of all parts of a practical beam, but may yield stiffener loads or flange bending moments which are as much as 5 times the true values. Hence a more accurate theory is necessary for design purposes.

The theory of pure tension field beams was first published by Wagner in 1929. Since then, many investigators have studied the problem of semitension field beams. Lahde and Wagner55 published empirical data in 1936 which were based on strain measurements of buckled rectangular sheets. These data provided information for the practical design of beams, but the test points had consider-able scatter because of the difficulty in making strain measurements of buckled sheet. Many aircraft manufacturers conducted tests and developed empirical design formulas, but usually, one particular type of beam has been tested, and the equations must be used with caution in designing beams of different materials or proportions different from those on which the tests were conducted. The most extensive test program has been conducted by the NACA under the direction of Paul Kuhn. Kuhn et al.51 and Peterson5 2 measured strains in the vertical stiff-eners of a large number of beams and derived empirical equations from these measurements. The stiffener stresses supply information required for the stiffener design and for the design of the flanges to resist secondary bending. A theoretical analysis of the stresses in a buckled rectangular sheet has been made by Levy et a l . 5 3 , 5 4 While some simplifying assumptions are made in the analysis, it provides valuable information regarding the stress distribution in the web. The analysis of Levy et al. shows that the stress conditions vary considerably at different points in the web and that any practical analysis in which the same stress conditions are assumed at all points will have some discrepancies with observed test conditions.

In the analysis of semitension field beams by Kuhn et al., it is assumed that part of the shear load kV is resisted by pure tension field action and that the remaining load (1 — k)V is resisted by the beam acting as a shear-resistant beam. All points of the web are assumed to have the same stress distribution, except for the web adjacent to the vertical stiffeners. This portion of the web is riveted to the stiffeners and does not wrinkle. The stresses in the web may be found by multi-plying the values shown for a shear-resistant web in Fig. 11.49 by 1 — k and those shown for a pure tension field web in Fig. 11.50 by k and then superimposing them. The values shown in Fig. 11.56c represent the total stresses on horizontal and vertical planes and arc found by superimposing the conditions shown in Fig. 11.56a and b. The angle a is obtained from Eq. (11.90) with sufficient accuracy. There is a conservative error involved in using Eq. (11.90) for partial tension field webs, since this equation yields the angle of principal stress. The principal tension stress for the element of Fig. 11.56c has an angle which is between the 45° angle for the principal stress of the element of Fig. 11.56a and the angle a for the element of Fig. 11.56b.

The diagonal tension factor k is given by the empirical equation of Kuhn et al.:

k = tanh ^0.5 log1 0 (11.93)

(I - k)as (I - k)a,

which is plotted in Fig. 11.57. The angle a is obtained from Eq. (11.90) after functions of k, a, and Ac are substituted for the strains. Values of tan a are plotted as functions of k and td/Ae in Fig. 11.58. Since Eq. (11.90) must be solved by trial for tan a, it is much more convenient to use Fig. 11.58 than to solve the equation for each particular case.

The stress conditions for any web clement are known after k and a arc found and are as shown in Fig. 11,56c. The stiffener compression forces and the flange bending moments are proportional to the vertical component of the web tensile stress <7j., which is shown in Fig. 11.56:

<7, = kr tan a (11.94)

1.0

0.8

0.6 K

0 . 4

0.2

0 1 2 4 6 1 0 2 0 4 0 1 0 0 2 0 0 4 0 0 1 0 0 0

a J a K , F i g u r e 1 1 . 5 7

-

s

/

/ /

/ /

The stifTener compression load P is obtained as follows:

P = aytd = kztd tan a (11.95)

The flange bending moments are obtained from the stress cry, as in Eqs. (11.84) and (11.85), for pure tension field webs, but the values ofcr, and P are smaller for semitension field webs:

tjf V y t d 2 Pd '' 1 2

M = ay td2 Pd

24 24

at stifleners (11.84)

between stifFeners (11.85)

The compression load on a vertical stifiener is resisted by the stifTener and by the effective web that is riveted to the stiffener: If sufficient rivets are provided that the web wrinkles do not extend through the riveted joint at the stiffener, as in the customary construction, the web must have the same vertical compression strain and approximately the same compression stress as the stiffener at the rivet line. An effective width of web equal to 0.5(1 — k)d is assumed by Kuhn et al. to act with the stiffener, and the stiffener compression stress then has the following value:

(11.96) + 0.5(1 - k)td

Since the values of k are obtained empirically from measurements of ac, an approximate expression for the effective width of web will yield an accurate value of <rc if the same effective width is assumed in calculating k from experimentally determined values of a c . It seems probable that the true effective widths of the web are less than those assumed and that the empirical values of k thus yield conservative values for cry and the flange bending moments.


The riveted joints between the webs and the beam flanges must be designed for the resultant of the shearing stress z and the tension stress cry or for a running load of yfx1 + (ay)2 t. Similarly, vertical web splices' must be designed for a run-ning load of s/z2 + (ax)2 t. The rivets connecting the vertical stiffeners to the beam flanges should be designed to transfer the load Pu = Ae(7c, according to the above theory. In actual beams, frequently it is impractical to provide this strength. The theoretical analysis by Levy ct al. and many tests indicate that the. stiffener load decreases near the end of the stiffener and that as much as half of this load is transferred to tlje web near the end of the stiffener rather than to the beam flange. In practical beams, it is customary to provide a total strength in the rivets connecting the stiffener to Ihe flange and the rivets connecting the end of the stiffener to the web to resist the load Pu. The stilfener-web rivets which are assumed to transfer part of this load Pu are spaced as close to the end of the stiffener as possible.

The allowable strength for beam webs has sometimes been obtained by equating the calculated diagonal tension stresses to the allowable tension stress for the web material, with empirical corrections for rivet holes and various stress concentration factors. More accurate web strength predictions can probably be obtained by equating the total web shearing stress t to an allowable stress asv, obtained from tests of beams of common proportions. These allowable stresses are plotted in Fig. 11.59 as functions of r /x c r . The curves of Fig. 11.59 were obtained by analyzing data from extensive tests of semitension field beams. The tests were conducted under the supervision of S. A. Cordon of the Glcon L. Martin Company.

3 4

32

3 0

« 2 8

cL 12

i 2 6

24

2~>

20

1 2 4 b S 1 0 2 0 4 0 6 0 1 0 0

Figure 11.59

R U C K U N G DESIGN OF STRUCTURAL MEMBERS 3 8 5

It is assumed that the wcb-flange rivet spacing is in the normal range of 3 to 5 times the rivet diameter. A smaller rivet spacing reduces the net area exces-sively, while a very large spacing may permit the web wrinkles to extend through the rivet lines. The allowable web stresses often are plotted as functions of the sheet thickness, with the heavier sheet gages resisting higher stresses. This prac-tice is permissible because stiffener spacings usually are kept to a maximum of about 8 in, even for very heavy sheets, and the thicker webs therefore have lower ratios of T/Tci.. For geometrically similar, webs, however, the allowable stress would be independent of the web thickness. Where beam flanges and web stiff-eners are attached symmetrically to both sides of a web, the edge of the web is better supported and resists higher stresses than those given in Fig. 11.59 for flanges attached to only one side of the web.

Example 11.9 Determine the margin of safety for the web intermediate stiff-eners, and riveted joints of the beam shown in Fig. 11.60. The web is, 2024S-T Alclad sheet, and the flanges and stiffeners are 2024S-T extrusions.

SOLUTION The buckling stress for the web in shear is obtained from Eq. (11.69) and Fig. 11.44. The web dimensions for computing buckling stresses are measured between rivet lines, as a = 15 and b = 8 in. By entering Fig. 11.44 with a/b = 1.875, the values K = 10.3 for clamped edges and K = 5.9 for simply supported edges are obtained. An average value, K — 8.1, is used because of the restraining effects of the flanges and stiffeners. It is given that E = 9,700.000 lb/in2. Substituting in Eq. (11.69) yields

x„r = KE[ - J = 8.1 x 9,700,000 x 0.032

1260 lb/in2

The shear stress is V 10,000 /if _ 16 x 0.032

= 19,500 lb/in2

1 = 0 . 0 3 2

/<= it.

- « = 8 -

0 . 0 5 1 I 0 . 7 5 0

^ F ^ j - f — v

V \ \ A N 4 3 0 A D 4 f . ' t i n

I ' = 1 0 , 0 0 0 l b A N 4 3 0 A D 5 ( 2 rivets)

A = 0 . 1 1 0 i n 2

/ . , = 0 . 0 I S . i n "

0.-10

Figure 11.60


The web depth h is always the distance between centroids of the flange areas, rather than the distance between rivet lines, when used in shear stress or shear-flow calculations.

The effective stiffener area is computed from Eq. (11.92), since the stiff-eners are attached to only one side of the web and are loaded eccentrically. The stiffener radius of gyration is /i = ^/'7/Vt = ^/O.Ol 5/0.110 = 0.37 in. The stiffener eccentricity is e = 0.40 + f/2 = 0.416 in. The effective stiffener area is obtained from Eq. (11.92):

- ^^l+(00416/0.37)2 = a ° 4 9 i n 2

The ratio of web area to effective stiffener area is

td 8 x 0.032 — = = 5.22 Ae 0.049

This ratio and xjxcr determine the stress distribution in the beam:

_r_ _ 19^500 _ • T c r 1 2 6 0

From Fig. 11.57, k = 0.53, and from Fig. 11.58, tan a = 0.79. From Fig. 11.59, the allowable web stress is rlv = 20,400 lb/in2. The margin of safety for the web in shear is

Web MS = ^ - 1 = 0 . 0 4 t 19,500

The vertical component of the web tension a y is obtained as follows:

ay = xk tan a = 19,500 x 0.53 x 0.79 = 8160 lb/in2

The stiffener compression load is obtained from Eq. (11.93):

P = cytd = 8160 x 0.032 x 8 = 2090 lb

The load per inch in the web-flange rivets is

q, = y/x2 + o-2 t = v/19,5002 + 81602 x 0.032 = 660 lb/in

The allowable load for one i-in A17S-T rivet is 375 lb shear and 477 lb bearing on a 0.032 gage sheet, as obtained from ANC-5. For the | - in spacing, the allowable rivet load is 750 lb/in, as determined from the shear strength:

Rivet MS = sfo — 1 = 0.14

The maximum compression stress in the stiffener, resulting from the eccentric compression load, is obtained from Eq. (11.96):

2 0 9 0 = 19,200 lb/in2

0.049 + 0.5(1 - 0.53) x 0.032 x 8

This stress exists in the leg attached to the web and decreases to zero or a tension stress in the outstanding leg. The allowable stress is the compression crippling stress for a leg with b = 0.70, t = 0,051, or b/t = 13.7, as computed by the methods of Sec. 11.9. The allowable crippling stress is approximately 22,000 lb/in2.

22,900

The allowable compression stress in the stiffener depends on many fac-tors and is only roughly approximated here. The attached leg of a stiffener usually is made at least one gage thicker than the web in order to prevent a forced crippling from the web wrinkles. Since the outstanding leg is not highly stressed in compression, it supplies torsional rigidity, and the attached leg probably can be assumed to have one side clamped and the other side free in computing the crippling stress.

The two rivets connecting the stiffener to the flange are ^ - i n A17S-T rivets with a single shear strength of 596 lb each. They must transfer the compression load in the stiffener:

' pu = ac Ae = 19,200 x 0.049 = 940 lb

The remaining-part of the force P is resisted by compression in the effective sheet and is not transferred by the rivets. The margin of safety of the two stifTener-flange rivets is

MS = ^ ? - 1 = 0.27 940

PROBLEMS

i l . l A long co lumn has an inilial curva ture defined by the equa t ion y0 = sin (TIX/L). Der ive an equa t ion for the add i t iona l def lect ion y by in tegra t ing Eq. (11.6). Show t h a t the center deflection <5 is defined by

(5 + A = -i - P / r a

where P c r is defined by Eq. (11.11). C o m p a r e values of + a f rom this equa t ion to those o b t a i n e d f rom Eq. (11.14) for / 7 P c r = 0.2,0.4, 0.6, 0.8, 0.9, 0.95, and 1.00.

11.2 F r o m the express ions Ra = 1 - KB and R„ = \/B2 derive Eqs. (11.24) and (11.31) by e q u a t i n g Ihe slopes of the two curves al their point of tangency. Find the coord ina te s of the point of tangency.

11.3 F r o m the express ions R„ = I - KB2 and R„ = 1 ,'B2 derive Eqs . (11.23) and (11.29) by e q u a t i n g lhc slopes of the two curves at their point of tangency. F ind the coord ina te s of the point of tangency.

11.4 Find the co lumn loads which may be resisted by round steel tubes hea t - t rea ted to an u l t ima te tensile s t rength of 180,000 lb / in 2 , with the ends welded before heat t r ea tmen t . T h e dimensions a r e :

Tube size L

1 x 0.05 20 2 l | x 0.049 20

x 0.065 40 1 l i x 0.05 30 2

11.5 Repeat Prob. ! 1.4 for steel heat-treated to an ultimate tensile strength of 150,000 lb/in2 .

11.6 Repeal Prob. i 1.4 for steel heat-treated to an ult imate tensile strength of 125,000 lb/in2 .

11.7 The skin of the upper side of an airplane wing is made of 2024-T6 Alclad. The stringer spacing is 5 in. and the rih spacing is 20 inT Assuming the edges to be simply supported, find the compression buckling stress for skin gages of(d) 0.020, (b) 0.032, (c) 0.040, and (d) 0.004 in.

11.8 Repeat Prob. 11.7, assuming the values of K to be the average of values-for simply supported edges and clamped edges.

11.9 Calculate points on the curve for m = 1 or Fig. 11.20 for values of ajb or 0.25, 0.33, 0.5, 1, 2, 3, and 4. Calculate points on the curve for m = 2 for values of ajb of 0.50, 0.66, 1, 2, 4, 6, and 8. Note the similarity between the two curves, and devise a system of coordinates which would show all the curves of Fig. 11.20 as a single curve.

11.10 Calculate the compression buckling stress for a sheet with a = S in. h = 4 in, and t = 0.156 in. The tangent modulus column curve for the material is shown in Fig. 11.29.

(«) Assume all four edges are simply supported. (h) Assume all four edges are clamped. (c) Assume the ends are simply supported and the sides arc free.

11.U Solve Prob. 11.10, using a dimensionless buckling curve for n= 10. Assume E = 10,700,000 and ff, = 37,000 lb/in2 .

U . t 2 Find the buckling stress for a column with («) both ends fixed. (b) both ends free, (c) one end fixed and one end free.

11.13 Use the Rayleigh-Ritz method to find the buckling load Tor a column with both ends pinned. 11.14 The skin on a fuselage is supported by stringers which are spaced at 5 in and by rings spaced at 20 in. Assume E = 107 lb/in2 and an average between simply supported and clamped-edge condi-tions. Find the shear buckling stresses for the flat sheet if (a) t = 0.020, (b) L = 0.032, (c) t = 0.040, and (</) f = 0.064 in.

11.15 Plot the axial loads in Ihe flanges and stifleners of a pure tension field beam similar to that shown in Fig. 11.46 with h = 10 in, d - 10 in, and V = 10,000 lb. Compute the flange bending moments and the load per inch on all rivets. Assume a = 45°.

11.16 Solve Example 11.5 by the Ncedham method and the Gerard method. Compare your results.

11.17 Solve Example 11.6 by the Ncedham method and the Gerard method. Compare your results.

C H A P T E R

TWELVE JOINTS A N D FITTINGS


A flight vehicle structure is manufactured from many parts. These parts are made from sheets, extruded sections, forgings, castings, tubes, or machined shapes, which must be joined to form subassemblies. The subassemblies must then be joined to form larger assemblies and then finally assembled into a complete flight vehicle. Many parts of the completed vehicle must be arranged so that they can be disassembled for shipping, inspection, repair, or replacement and are usually joined by bolts.

In order to facilitate assembly and disassembly, it is desirable for such con-nections to contain as few bolts as possible. For example, a semimonocoque metal wing usually resists bending stresses in numerous stringers and sheet el-ements distributed around the periphery of the wing cross section. The wing cannot be made as one continuous riveted assembly from tip to tip, but usually must be spliced at two or more cross sections. Often these splices are designed so that four bolts transfer all the loads across the splice. These bolts connect mem-bers called fittings, which are designed to resist the high concentrated loads and to transfer them to the spars, from which the loads are distributed to the sheet and stringers. The entire structure for transferring the distributed loads from the sheet and stringers outboard of the splice to a concentrated load at the fitting and then distributing this load to the sheet and stringers inboard of the splice is considerably heavier than the continuous structure which would be required if there were no splice.

389

Many uncertainties exist concerning the stress distribution in fittings. Manu-facturing tolerances are such that bolts never fit the holes perfectly, and small variations in dimensions may affect the stress distribution. An additional margin af safety, of 1 a i r p l a n e s and 20jwrcenTfoT^mTairplaneffis used in the design of fittings. A common procedure is to multiply the design loads by a "fitting factor of 1.15 or 1.20 before the stresses are calculated. This fitting factor must be used in designing the entire fitting, including the riveted, bolted, or welded joint attaching the fittirfg to the structural members. The fitting factor need not be used in designing a continuous riveted joint, although the stress distribution in such a joint-is also indeterminate.

The allowable stresses for rivets are rather conservative to account for such uncertainty.

12.2 BOLTED OR RIVETED JOINTS

Bolted or riveted joints must be investigated for four types of failure: bolt or rivet shear, as shown in Fig. 12.1; bearing, as shown in Fig. 12.2; tear-out, as shown in Fig. 12.3; and tension, as shown in Fig. 12.4. The true stress distribution is rather complex and is discussed later. It is customary to assume a simple uniform or

n s r r

< o

Figure 12.2 Figure 12.1

JOINTS A N D FITTINGS 3 9 1

Figure 12.3

average stress distribution in all cases, and the allowable stresses which are used in design are also average stresses which have been obtained from tests of similar joints:

JJ.1S J ^ r e f o r c pojsiWeJo predict. the. strengdijp^a jo in t with an accuracy of a few percent, a l t h o u j h j h e j r u e maximum stresses may be 3 or 4 times as much as the average'stresses. The average stress for any of the four types of "failure"'is ~ "

<r = 4 f 1 2 - 1 )

where cr is the average stress, P is the load, and A is the area of the cross section on which failure may occur. The margin of safety (MS) is found from

MS = — - 1 cr

(12.2)

where oa is the allowable stress and the stress cr is obtained from the load P, which includes the safety factor of 1.5 and usually the fitting factor of 1.15 or 1.2 as well. If this fitting factor is included in the stress cr, the margin of safety should be zero or a small positive value. Some designers may not include the fitting factor in the stress a, and thus they must show a minimum margin of safety of 0.15 or 0.20 from Eq. (12.2). In any analysis, it should be clearly stated whether the fitting factor is included in the mafgitr"of~s1iTety.,_The symbol ca always representTan"attd\vahTe~stress,"TriBThe symbol a represents a calculated stress. A subscript is used to designate the type of stress; that is, za and t are shearing stresses, <r„hr and <rhr are bearing stresses, aut and a, are tensile stresses, cac and crc

are compression stresses, and oab and cr,, are bending stresses.


For investigating the shear strength of a bolt or rivet, the area to be used in Eq. (12.1) is the area of the bolt or rivet cross section, ox A = nd2/4, where d is the diameter of the bolt or rivet. See Fig. 12.1. The shearing stress is then obtained from Eq. (12.1):

4 P r = - 2 (12.3)

In Figs. 12.1 through 12.4, the bolt is shown to be in single shear and one plate is assumed to be rigid in bending, so that the forces on the thin plate are in static equilibrium. Hence the bolt would resist a bending moment Ptf2 at the cross section subjected to shear. It is shown later that this bending moment on the bolt does not exist in most actual single-shear connections, and it is customary to disregard this bolt bending moment when the two plates are clamped together by the bolt. When a washer or a filler plate is used between the two stressed plates, the bolt bending must be considered.

The_ bearing failure of a riveted or bolted joint usually consists of an_elonga-tion of the ho l e jn the plate, as shown in Fig. 12.2a. The aUowableTiearing stress usually depends on the permissible elongation of the hole. For riveted joints, the allowable bearing stress is determined by arbitrarily specifying a hole elongation equal to a certain percentage of the rivet diameter. The bearing failure is some-what similar to the tear-out failure shown in Fig. 12.3, and the allowable bearing stress for rivets is reduced when the rivets are too close to the edge of the sheet. The bearing stress is assumed to be uniformly distributed over an area A = td, as shown in Fig. 12.2. By' substituting this area into Eq. (12.1), the equation for the assumed average bearing stress is obtained:

= £ ' ( 1 2 - 4 )

Bolt h o l e s j d w a y s j ^ t J ^ j h ^ M y J a r g e r than, t h e ^ b ^ t ^ i a m ^ t c r j f the joint is subjecfecTto shock or vibrational loading, as in a landing-gear member, there is a much greater tendency for a bolt hole to elongate than when the joint resists only static loading. Similarly, when relative rotation of the two parts occurs, the bolt hole is more likely to become enlarged. In such cases, the bearing stress rfiust be low in order to prevent frequent replacement of the bolt or the hole bushing-JThe licensing agencies therefore specify that a bearing factor of 2.0 or more be usccfin obtaining the bearing s£ress_when ajbolted joint is subject to relative rotation under design loads or to shock or vibration loads. This bearing factor is used in place of the fitting factor, not in addition to the fitting factor.

A tear-out failure of a bolt or rivet hole is shown in Fig. 12.3. The plate materia} fails in shear on the areas A = 2xt, and the tear-out stress is found from

The distance .x is obtained as length ab in Fig. 12.3, but it is conservative to use

JOINTS AND FITTINGS 3 9 3

length cd, which is easier to calculate. It is seldom necessary , to calculate the tear-out stresses for riveted joints in a sheet of the type shown in Fig. 12.4. Etflra practical considerations, it is desirable tojeeep the distance f rom th e c e nter.of the rivets to the edgc~orTrie~shcet ecfuaTto at least two.diameters of the rivet, and Tlvcreis :no danger of tear-out with this edge distance.

A riveted or bolted joint must be investigated for a possible tension failure through the bolt or rivet holes, as shown in Fig. 12.4. The tension stress is assumed to be uniformly distributed over the area A =(a> — d)t for the bolted fitting shown in Fig. 12.4a.

(12.6) (w - d)t

For the riveted joint shown in Fig. 12.4/;, the tension stress is

(s - d)t (12.7)

where P = load pcr rivet 5 = rivet spacing d = rivet diameter t = sheet thickness

Example 12.1 T h e fitting shown in Fig. 12.5 is made of a 1014 aluminum forging, for which = 65,000, za = 39,000, and <rabr = 98,000 lb/in2. The bolt and bushing arc made of steel for which aM = 125,000, za = 75,000, and <rabr = 175,000 lb/in2 . The fitting resists limit or applied loads of 15,000-lb compression-and 12,000-lb tension. A fitting factor of 1.2 and a bearing factor of 2.0 are used. Find the margins of safety for the fitting for various types of failure.

0.5(05 r r ^ T i

15,0001b | compression ^ i

1 ?..000 lb ! !

1

! i

tcu.Mun | i

. U j U

_ J

1.40

0 . 7 0 R

0.125 Figure 12.5

394 AIRCRAFT STRUCTURES ,

SOLUTION The design or ultimate fitting loads are obtained by multiplying the loads given by a safety factor of 1.5 and a fitting factor of 1.2:

Design fitting loads:

15,000 x 1.5 x 1.2 = 27,000 lb compression

12,000 x 1.5 x 1.2 = 21,600 lb tension

The bearing of the bolt on the bushing is investigated by using the bearing factor of 2.0 in place of the fitting factor of 1.2: '

Design bearing loads:

15,000 x 1.5 x 2.0 = 45,000 lb compression

12,000 x 1.5 x 2.0 = 36,000 lb tension

The bolt is in double shear; therefore one-half of the 27,000-lb load must be resisted by each cross section of the bolt in shear. From Eqs. (12.3) and (12.2):

4 x 13,500 „ 0 „ 2 T = - ^ n 6 8 ' 6 0 0 1 b / i n

and MS = — 1 = 0.09 includes fitting factor 68,600

The bearing stress also is calculated from the larger of the loads for tension and compression. From Eqs. (12.4) and (12.2), the bearing of the bolt on the bushing is investigated:

MS = _ i _ o,09 includes bearing factor 160,000

For bearing of the bushing on the forging, one need only use the fitting factor, because the bushing fits tightly in the hole:

g f c r = 27,000 = 7 6 8 0 Q l b / i n 2 br 0.5625 x 0.625 ' '

MS = 9 S ' ° 0 0 - 1 = 0.29 includes fitting factor 76,800

The tear-out of the bolt hole is investigated first by assuming that the length x shown in Fig. 12.3 is equal to cd rather than ab:

cd = 0.70 + 0.125 - 0.3125 = 0.5125 in

The tension load must be used in calculating the tear-out stress, since the compression load produces no stress on this cross section. From Eqs. (12.5)


i n 2 4 0 ° - ~ cos 4 0 °

4 0 °

F i g u r e 12.6

and (12.2),

and

21,600 2 x 0.5125 x 0.5625

39,000

= 37,400 lb/in2

includes fitting factor

A more accurate value of the distance x may be calculated from the equation given in Fig. 12.6. The term in brackets may be plotted for various values of r/R in order to reduce the labor of the calculations, where it is necessary to repeat such calculations frequently. For"/? = 0.7, r = 0.3125, B = 0.125, and x = 0.562,

21,600

and MS =

2 x 0.562 x 0.5625

39,000

= 34,000 ib/in2

0.14 includes fitting factor 34,200

The tension stress through the bolt hole is obtained from Eq. (12.6):

21,600

MS =

(1.4 - 0.625)0.5625

65,000 49,600

- 1 =*= 0.13

= 49,600 lb/in2

includes fitting factor

12.3 ACCURACY OF FITTING ANALYSIS

The ultimate strength of a lilting usually may be calculatcd accurately by the methods previously described. True stress distribution at stresses below the elas-tic limit often is much different from the assumed distribution. Before the ultimate strength of the fitting is reached, however, the material yields and the stresses are redistributed so that they usually approach the assumed stress dis-tribution. Because of this plastic yielding of the material and because the allow-able shear and bearing stresses are obtained from tests on specimens similar to those in the actual structure, it is possible to achieve accurate calculated strengths by means of inaccurate assumptions. While the conventional methods are satis-


i m f f f t f r d

J

F i g u r e 12 .7

factory for making design calculations for fittings, the designer must keep in mind the true stress distribution and must avoid conditions of high local stress wher-ever possible.

One very common case of stress concentration, shown in Fig. 12.7, is that of a tension plate containing a circular hole. For small loads, the tensile stress at the side of the hole is 3 times the average tensile stress in the plate, as indicated by line 1. As the loads increase, the stress at tixe_sidejojLihe ii.olexxceeisJie_elastic limit, and lorar^Tastfc yieTdirfg of the material j jccurs j jear the ho.lg._The stresses near ihe "Hole remain almost constant at the yield point, while the stresses at a distance from the hole increase with the load, as indicated by line 2. Before failure occurs, yielding has progressed over the entire width of the plate, and the stress is constant over the net section, as shown by curve 3. Thus the customary assump-tion that failure occurs at a load equal to the product of the ultimate tensile stress and the net area is accurate for ductile materials. Brittle materials, which fail suddenly with no plastic elongation, should never be used for aircraft structural members.

Stress concentrations are much more serious in engine parts on which the loads are repeated millions of times than in airframe parts on which the maxi-mum loads occur only a few times during the life of the airplane. In airframe design, usually it is safe to consider only average stresses and to neglcct stress concentrations, although certain unfavorable conditions, such as radial cracking of sheet around holes when the holes are press-countersunk, may lead to service failures from stress concentrations.

The double-shear connection shown in Fig. 12.8 is assumed to resist one-half

I 1

the load by shear on each bolt cross scction. Manufacturing tolerances may permit the hole in the lower lug to be slightly to the left of the hole in the upper lug, as shown. For small loads, then, the entire load is resisted by shear on the upper cross section. As the load is increased, the parts deflect so that the lower end of the bolt is also bearing on the lug, but the upper lug continues to resist more than one-half the load. The fitting factor is intended to account for such eccentric loading conditions; in this case, the use of a fitting factor of 1.2 is equivalent to the assumption that one side of the fitting may resist 60 percent of the total ultimate load.

Most of the bolted and riveted joints in aircraft structures are single-shear joints. For the joints shown in Figs. 12.1 and 12.2, we assume that one member is rigid, and only the forces acting on the other member are considered. For this assumed loading, the bolt resists a bending moment of Pt/2 and the heavy member resists a larger bending moment. The usual single-shear joint has both members of comparable size. At first it might appear that each of the members shown in Fig. 12.9 could be treated in the same manner as the upper member of Fig. 12.2.'?. In fact, many textbooks show the forces as in Fig. 12.9, and this assumed stress distribution is customary and satisfactory for design. The forces shown in Fig. 12.9 cannot be in equilibrium, however, because there is an unbal-anced moment Pt on the plates in Fig. 12.9a and a similar unbalanced moment on the pin in Fig. 12,9b. The correct stress distribution must be as shown in Fig. 12.10. For the forces P to balance, they must act on the same line, as shown in Fig. 12.10a. The stresses in the plate are no longer P/A, but must also include stresses from the bending moment Pt/2. If the plate width is b, the plate stress is P/A ± My/I:

bt~ 2 bt2 bt~ bt

At the inside faces of the plates, the tensile stress from Eq. (12.8) is AP/A, and at the outside faces the compressive stress is 2P/A, as shown.

In order for the pin to be in equilibrium under the bearing stresses, it must bear on opposite corners of the hole, as shown in Fig. 12.106. The most opti-mistic assumption of bearing stresses is the straight-line assumption shown in Fig. 12.10c, which yields maximum bearing stresses 4P/(tb) at the inside corner and 2P/(tb) at the outside corners. If the pin does not fit tight in the hole, the

, I I N ;

u 1 '

J </>) Figure 12.9


P

F i g u r e 12.10

bearing stresses must be higher than those assumed. Thus, for the single-shear pin joint between plates of equal thickness, the maximum plate tension stresses and the bearing stresses are both 4 times the values assumed in Figs. 12.2 and 12.9.

The bending moment in the pin of Fig. 12.10 is zero at the cross section of maximum shear, and the maximum pin bending moment is £Pt at a cross section a distance tj3 from the inside of the plates.

The ultimate strength of conventional riveted and bolted joints approaches that assumed in the original simple analysis, because of the clamping action of the rivet heads or bolt heads. For a riveted joint between two sheets in tension, the bending and tension stresses in the sheets exceed the elastic limit, and the sheets deform as shown in Fig. 12.11. The two forces P are almost in the center plane of the sheets, as shown in Fig. 12.11a, as the ultimate strength is ap-proached. The moment Pt of the bearing forces on the rivet is balanced by the moment of clamping forces under the head of the rivet, as shown in Fig. 12.116. These forces on the rivet head have a moment arm D which is slightly less than the diameter of the rivet head. The bending moment in the rivet shank varies from Pf/2 at each end of the shank to 0 at the plane of rivet shear. After plastic

p

p Figure 12.11


yielding has progressed in the sheet, the bending stresses shown in Fig. 12.10a are eliminated, and the sheet is in almost uniform tension at all points. The angular change in the sheet is arctan (f/D), as shown in Fig. 12.11c, and the force exerted by the rivet head on the sheet is just sufficient to keep the resultant tension in the center plane of the sheet. The force triangle at the bend in the sheet is represented by Fig. 12.1 W. The angular change in the sheet is exaggerated in Fig. 12.11 For a rivet shank diameter of 4 times the sheet thickness and a rivet head diameter D of twice the shank diameter, the angle is arctan (tjD) or arctan •§•.

Where a tension joint has two lines of rivets, the deformation is as shown in Fig. 12.12. If the tension stresses in the sheet were uniform at all points, the sheet would deform as shown in Fig. 12.12u. Between the rivet lines, however, the sheets have only one-half the average tensile stress that they have at the ends and therefore may resist the bending deformation and assume the deformed shape shown in Fig. 12.126. The forces on the rivets will remain approximately as shown in Fig. 12.116, since the clamping forces on the rivet head must balance the moment of the bearing forces.

Any riveted or bolted single-shear joint will have stress conditions which vary between the extreme conditions of Fig. 12.10 and 12.11. At low loads, the sheet must resist bending stresses, as shown in Fig. 12.10; but as local yielding occurs, the stresses arc redistributed so that they approach the conditions of Fig. 12.11. The ultimate strength is predicted accurately from an assumed average tension stress in the sheet and an average bearing stress on the bolt or rivet. Many types of "b l ind" rivets or of countersunk rivets do not provide a sufficient amount of clamping action by the rivet head, and strength calculations based on simple stress distributions must be verified by tests.

It is interesting to compare the action of aircraft rivets with the action of hot-driven steel rivets, such as those used in bridges, buildings, boilers, and other steel structures. The steel rivet is upset when red-hot and cools and con-tracts in place. The contraction makes the rivet slightly smaller in diameter than the hole and provides a residual tension stress in the rivet approximately equal to the yield stress of the rivet material. The rivet tension clamps the plates so tightly that small loads are resisted by friction between the plates, and the rivet shank bears on the hole only at higher loads. This tension does not exist in aircraft rivets, which are driven at room temperatures.

It is common aircraft practice to assume the same allowable bearing stresses

^ c g ^ T

Figure 12.12


for single-shear joints and the double-shear joints of the type shown in Fig. 12.5. The common practice in bridge or structural steel design is to use higher allow-able bearing stresses for joints in double shear. This practice is logical, since the eccentric distributions, as shown in Fig. 12.10, are eliminated in double-shear joints.

When several similar rivets or bolts act together in a joint, it is customary to assume that each rivet or bolt carries a proportionate share of the load. This assumption is very inacccurate when the joint is not highly stressed; but is more accurate as the loads approach the ultimate strength of the joint and local plastic yielding and rivet "slip" "have occurred. In Fig. 12.13 the deformations of ths various rivets in a double-shear joint are exaggerated in order to show the relative motion between the plates, although the actual plates would be in close contact and the actual deformation would consist of hole elongations as well as rivet shear deformations. It is assumed that the two outside plates have the same total area A as the inside plate and that the average stress in all plates is p = P/A. If each of the five rivets transfers one-fifth to the total load, as commonly as-sumed, stresses in the various plates between rivets will be 0.2p, 0.4p, 0.6p, and 0.8p, as shown in Fig. 12.13. Between rivets 1 and 2, the outside plates resist tensile stresses of 0.8p, and the inside plate resists a tensile stress of 0.2p; there-fore, the outside plates must elongate 4 times as much as the inside plate. Thus, rivet 1 must be deformed much more than rivet 2 and must resist a higher shear. Between rivet 2 and rivet 3, the outside plates have 1.5 times the stress and deformation of the inside plate; therefore, rivet 2 must resist more load than rivet 3. Study of other deformations shows that the end rivets 1 and 5 are equally stressed and must resist much higher shears than the other rivets. Rivets 2 and 4 are equally stressed and resist higher shears than rivet 3.

In the case of a longer line of bolts or rivets than that shown in Fig. 12.13, the end bolts or rivets are still more highly stressed relative to the bolts and rivets near the center of the line. As the load is applied gradually, first the two end rivets must resist most of the load, until they slip or yield in shearing and bearing. Then the load is transferred to the next rivets in the line, until they also slip and transfer load to other rivets. The ultimate strength of the joint is accurately predicted as the sum of the strengths of the individual rivets, provided th'cre is enough ductility to permit each rivet to slip considerably and yet still retain its maximum strength after slipping. It is desirable, however, to vary the plate areas in order to obtain approximately constant tension stresses in the plates and thus distribute small loads more equally to all the rivets or bolts. Bolted or riveted joints in brittle materials arc undesirable, since the end bolts may fail before they

0.8/' 0.6/" 0.-1/' o.:/' 1 1 2 i 3 1 4 | 5 l l l i l l l i l

T 0.2P 0 , 4 V 0 . 0 / ' O . R / ' r F i g u r e 1 2 . 1 3


I R F i g u r e 12.14

deform enough to redistribute the load, and then each bolt in the line will fail in turn. Some spot welds do not have enough ductility to be satisfactory for this type of loading, and occasional rivets are used in most lines of spot welds so that a progressive failure will not extend past the rivet.

12.4 E C C E N T R I C A L L Y L O A D E D C O N N E C T I O N S

In many connections, tiic resultant forcc docs not act through the ccnter of the bolt or rivet group. In such cases, it is usually convenient to superimpose the eflccts of an equal parallel force acting at the center of the rivet group and a moment about the center which is equal to the product of the force and its distance from the centcr. The rivet forces in the typical connection shown in Fig. 12.14 may be obtained by superimposing the forces for the concentric loading of Fig. 12.15a and for the moment Re, shown in Fig. 12.15b.

First we assume that all the rivets are critical in single shear and that all

/>.= AV.I \

Figure 12.15


plates are rigid. For the concentric load shown in Fig. 12.15a, the shearing stresses on all rivets are assumed equal. The force Pc on any rivet resulting from the concentric load is

. „ RA

where A is the area of the rivet cross section and J.A is the total cross-sectional area of all the rivets in the group. For a rivet group of n rivets of equal area, Eq. (12.9) reduces to the following form:

Pc = * (12.10)

The resultant of the forces on the individual rivets passes through the centroid of the areas of the rivet cross sections; hence this point must be used as the center of moments for the rivet group.

When the rivets resist a moment, the shearing stresses are assumed to be proportional to the distance r from the centroid of the rivet areas. The force Pc

on any rivet of area A resulting from this moment is

Pc = KrA (12.11)

The constant K is obtained by equating the sum of the moments of the individual rivet forces to the external moment:

M = XPcr = KT.r2A (12.12)

The constant K may be eliminated from Eq. (12.11) and (12.12) and the force Pc

obtained:

Mr A = (12-13)

Equation (12.13) is similar in form to the common equations for bending or torsion.

The resultant force P on any rivet can be determined now from the com-ponent forces P c and P„, as shown in Fig. 12.15c. When an algebraic solution is desired, usually it is more convenient to obtain the horizontal and vertical com-ponents of the rivet forces. The distance r does not need to be calculated if the coordinates x and y are used. From Fig. 12.16 and Eq. (12.13), the following equations for the components Prx and Pey are obtained:

-My A MxA " Zx2A + Zy2A cr Zx'A + Zy'A [ A )

The method of analysis for an eccentric conncction, like methods of analysis for several other types of fittings, must be considered only as a rough approxi-mation. Where bearing stresses are critical in the design of the bolts or rivets, it is customary to substitute Pa, the allowable bearing load for each bolt or rivet, into


Eqs. (12.13) and (12.14) in place of the shear area A. Tn some cases, it may be assumed that the loads on all bolts or rivets approach their ultimate strengths, rather than that the loads are proportional to r or that bolts or rivets near the center are not highly stressed. Frequently bolts or rivets are attached to members which are more rigid in one direction than another. If the supporting structure is rigid horizontally but flexible vertically, for example, it may be assumed that an applied moment is resisted by horizontal rivet or bolt forces, rather than by forces perpendicular to the radial line. Where both standard bolts and rivets are used in the same connection, it is neccssary to design the connection so that either the rivets alone or the bolts alone can resist the total load. Rivets fill the holes completely, but bolts must be slightly smaller than the holes; consequently, bolts resist no loads until rivets slip enough to be permanently, damaged. Close-tolerance, drive-fit bolts are occasionally used with rivets, and each may be assumed to resist a proportionate share of the load.

Example 12.2 Find the resultant force on each rivet of the connection shown in Fig. 12.17. Also find the margin of safety of the most highly stressed rivet. All rivets arc ^ - i n - O D aluminum alloy in single shear, and the sheet is 0.051 gage 1024 aluminum Alclad.

SOLUTION The rivet loads are calculated in Table 12.1. The centroid is deter-mined by inspection, and values of x, x2, y, and y2 are tabulated in columns 2 to 5, respectively. The rivet forces Pcx and Pcy are obtained by dividing the loads of 1800 and 300 lb by 6, since these loads are resisted equally by each

Table 12.1 Analysis of riveted connection

Rivel .V i' x2 P. , P, P.y r, P

(1) (2) (3) (4) (5) (6) (7) (8) (9) (10) ( ID (12)

1 - 1 1.5 1 2.25 300 - 1 2 0 180 50 - 8 0 - 3 0 183 2 1 1.5 1 2.25 300 - 1 2 0 180 50 80 130 221

3 - 1 0 1 0 300 0 300 50 - 8 0 - 3 0 302 4 1 0 I 0 30(1 0 300 50 80 130 327 5 - I - 1 . 5 I 2.25 A . j 120 420 50 - 8 0 - 3 0 422

6 1 - 1 . 5 1 2.25 300 120 420 50 80 130 440

Z 6 9.00

n


of the six rivets. The values of Pex and Pcy are found from Eqs. (12.14). Since A is the same for all rivets, the values of A may be omitted from Eqs. (12.14). The moment M is 1200 in • lb. The values Px and Py are each obtained as the sum of the terms in the two preceding columns, with care being taken with regard to the algebraic signs. The resultant rivet forces P arc found as the square root of the sum of the squares of the rectangular components Px

and Py. The allowable load for the rivet is obtained from MILHDBK-5 as

593 lb. Rivet 6 resists the greatest load, 440 lb. The margin of safety is obtained from these loads :

593 MS = — - 1 = 0.35

440

It has been assumed that the loads of 1800 and 300 lb were design fitting loads, or that they were obtained by multiplying the applied or limit loads by the safety factor of 1.5 and the fitting factor of 1.2 or 1.15.

12.5 WELDED JOINTS

Welding is used extensively for steel-tube truss structures, such as engine mounts and fuselages, and for steel landing gears and fittings. The most common'type of welding consists of heating the parts to be joined by means of an oxyacetylene torch and then using them together with a suitable welding rod. The grain structure of the material at the weld becomes similar to that of cast metal, and it is more brittle and less able to resist shocfi and vibration loading than is the original material. Aircraft tube walls, are thin and more difficult to weld than other machine and structural members. All aircraft welding was previously torch welding, but electric arc welding has been developed so that it is also satisfactory for the thin aircraft members. In arc welding, the welding rod forms an electrode from which current passes in an arc to the parts being joined. The-electric arc simultaneously heats the parts and deposits the weld metal from the electrode. The heating" is much more localized than in torch welding, and the strength of heat-treated parts is not impaired as much by arc welding as by torch welding. Design specifications normally require that the same allowable stresses be used for arc welding and for torch welding.


30" or less

J

l i F i g u r e 12.18

The strength of welded joints depends greatly on the skill of the welder. Often the stress conditions arc uncertain, and it is customary to design welded joints with liberal margins of safety. It is preferable to design joints so that the weld is in shear or compression rather than tension, but frequently it is necessary to have welds in tension. Steel tubes in lension are usually spliced by "fish mouth" joints, as shown in Fig. 12.18«, which are designed so that most of the weld is in shear and the local heating of the tube at the weld is not confined to one cross section. Where a bult weld must be used, as shown in Fig. 12.186, the weld is not perpendicular to the centerline of the tube.

Fuselage truss members often arc welded as shown in Fig. 12.19a. Only the horizontal member is highly stressed, and usually the size of the other members is determined as a minimum tube size, because they resist small loads. When these members are highly stressed, it is neccssary to insert gusset plates, as shown in Fig. 12.196. Steel tubes often have walls as thin as 0.035 in and the welder must control the temperature to keep from overheating the thin walls and burning holes in them, fl is extremely difficult to weld a thin member to a heavy one,

i / j i Figure 12.19

bccause more heat is required for the heavy member. The thickness ratio of parts being welded should always be less than 3 :1 and preferably less than 2 :1 .

The allowable load on the weld metal in welded seams is specified in MIL-HDBK-5 by the following equations:

' P = 32,000Li (low-carbon steel) (12.15)

P = 0.48Lr<r,„ (chrome-molybdenum steel)

where P — allowable load, lb L = length of welded seam, in r = thickness of thinnest material joined by weld in lap welds between

two steel plates or between plates and tubes, in t = average thickness of weld metal in tube assemblies (cannot be as-

sumed greater than 1.25 times thickness of welded stock), in <7,u = 90,000 lb/in2 for material not heat-treated after welding a,u = ultimate tensile stress of material heat-treated after welding, (heat-

treatable welding rod must be used), but not to exceed 15.0,000 lb/in2

The local healing during welding also reduces the allowable tension or bend-ing stress in the material near the weld. For normalized tubing with no heat treatment after welding, the allowable tensile stress is 90,000 lb/in2 near the weld for tapered welds making an angle of 30° or less with the axis of the tube and 80,000 lb/in2 for other welds. For tubing which is heat-treated after welding, the allowable tensile stress is <T,u .

Example 12.3 The l j - by 0.065-in chrome-molybdenum steel tube shown in Fig. 12.20 resists a limit or applied tension load of 15,000 lb. Find the margin of safety of the weld and of the tube near the weld if Lx = 2.5, L2 = 3, and ^ = 0.20 in; the tube area is A = 0.293 in2. (a) Assume that the ultimate tensile stress a tu is 100,000 lb/in2 before welding

and that there is no subsequent heat treatment. (b) Assume that the tube assembly is heat-treated to a <jlu of 180,000 lb/in2

after welding and that the limit load is 22,000 lb.

SOLUTION (a) The weld on the curved end of the tube is neglected, since loads transmitted to this portion of the tube tend only to straighten and flatten the end of the tube and do not increase the tensile strength of the weld appreci-ably. Because the load P is applied at the center of the tube, one-half of this load is resisted by the weld on each side of the tube. Thus the two welds of length L j must resist the load of Pj2. The ultimate or design load is obtained by multiplying the applied load by the safely factor or 1.5:

P = 15,000 x 1.5 = 22,500 lb

The allowable load P„ is obtained from Eq. (12.15). The tube thickness t — 0.065 is critical since the forging thickness is more than twice the tube wall thickness. The length Lx is welded lo the tube on both sides of the

j o i n t s a n d f i t t i n g s 4 0 7

to)

Figure 12.20

forging; therefore a. length L = 5 in must resist half of the load. The allow-able load is

y = 0.48Ltfflu = 0.48 x 5 x 0.065 x 90,000

or Pa = 28,000 lb

The fitting factor of 1.20 must be included in the calculation of the margin of safety:

P.. . 28,000 MS = 1.20P

1 = 1.2 x 22,500

1 = 0.04

This margin appears small for a weld, but was calculated conservatively. The ultimate tensile stress in the tube near the weld is 90,000 lb/in2 since for a slotted tube, the weld makes an angle of 0° with the tube axis. The allowable tension in the tube near the weld is

l \ - 90,000 x 0.293 = 26,400 lb

It is not necessary to use a fitting factor here, since the tube itself, rather than the fitting, is being investigated:

26,400 MS = 1 =0.17

22,500

(/>) The allowable load is computed in the same manner as for part (a), but now a,,, = 150,000 lb/in2:

= 0.48 x 5 x 0.065 x 150,000

408 a i r c r a f t s t r u c t u r e s ,

or •

pa = 46,400 lb

The design load is .

P = 22,000 x 1.5 = 33,000 lb

The fitting factor is included in calculating the margin of safety:

46,400 MS = ' 1 . 2 - 1 =0 .18

33,000

The allowable tensile stress in the tube near the weld is c,u, and

pa = a,u A = 180,000 x 0.293 = 52,700 lb

= 1 = 0 . 6 0 33,000

PROBLEMS

I2.J An end filling similar to those shown in Figs. 12.5 and 12.6 is m a d e of steel wilh a n u l t imate tensile s t rength o,„ of 180,000 lb / in 2 a n d has n o bushing. It has a ^ - i n steel bolt in doub le shear , a thickness of 0.5 in, and d imens ions R = 0.5 and e = 0.05 ill, as shown in Fig. 12.6. Find the m a x i m u m limit loads in tension a n d in compress ion if the fining f ac to r is 1.2 and the bear ing fac tor is 1.0.

.Obta in a l lowable stresses f rom M I L H D B K - 5 .

12.2 Design a fi t t ing to resist a limit tension load of 15,000 and a limit compress ion load of 20,000 lb. Assume the mate r ia l s and uni t stresses t o b e the s ame as those used in P r o b . 12.1. 1 2 3 Design an end fitting of steel with an ul t imate tensile s t rength of 125,000 lb / in 2 . T h e applied o r limit loads are 15,000-lb tension and 20,000-ib compress ion . Use a fitting f a c t o r of 1.2 and a bear ing fac tor or 2.0.

12.4 Find the margin of safety for the jo in t shown in Fig. 12.17 if Rx = 3000 Ib.R,. = 200 lb. and the rivets are m a d e of f - i n -d i ame te r 2017-T.j in single shear . T h e p la te is 0.072 gage 2024-T 3 clad a luminum.

12.5 Assume the tube of Fig. 12.20 to be 2 by 0.083 in with i , = 0.2, L , = 3.0, and L 2 = 4.0 in. F ind the al lowable load P if (a) the tube has an al lowable stress <r,„ of 95,000 lb / i n 2 and (6) the assembly is heat- t reated a f t e r welding to a tensile s t rength <r,u of 150,000 lb/ in2 .

MOMENTS OF INERTIA,

A P P E N D I X

A MOHR'S CIRCLE

A.I CENTROIDS

The force of gravity acting on any body is the resultant of a group of parallel forces acting on all elements of the body. The magnitude of the resultant of several parallel forces is equal to the algebraic sum of the forces, and the position of the resultant is such that it has a moment about any axis equal to the sum of the moments of the component forces. The resultant gravity force on a body is its weight W, which must be equal to the sum'of the weights \vt of all elements of the body. If the forces of gravity act parallel to the z axis as shown in Fig. A.l, the moments of all forces about the x and y axes must be equal to the moment of the resultant:

If the body and the axes are rotated so that the forces are parallel to one of the other axes, a third moment equation can be used:

W'\Y = Xjiv, + x2w2 4- • • • = X.-cw or x dW (A.L)

Wy = j',H', + y2w2 + • • - = Zyw or y dW (A.2)

Wz — Zj H'j + z2 w2 + - - • = 2zvv or z dW (A.3)

409


C c n t c r o f \ N y f f$ C c n t c r o f \

Y i T ' g r a v i t y 1

l i ^ T -

1 » : 2 -r It' r It'

V

Figure A. l

The three coordinates x, y, and z of the center of gravity may be obtained from Eqs. (A.l) to (A.3): .

(A.4) ~Lxw

or J x dW

W

Zyw or

J y dW or

W

Xzvv or

J z dW ~w

or W

3' = "ITT or 77/ (A-5)

(A.6)

The summations or integrals for Eqs. (A.4) to (A.6) must include all elements of the body. In many engineering problems, the weights and coordinates of the various items are known, and the center of gravity is obtained by a summation procedure, rather than by an integration procedure.

In the case of a plate of uniform thickness and density which lies in the xy plane, as shown in Fig. A.2, the coordinates of the center of gravity are

I A- dW W

i'J x dA wA

J x dA

A (A.7)

w = ii-.-l

n - n J i - r o f J11' = ii- dA / g r a v i t y

1 / - Z . y 7 > s

Figure A.14

m o m e n t s o f i n e r t i a , m o h r ' s c i r c l e 4 1 1

_ | y dW wf y dA J y dA } ' ~ W ~ w A ~ A (A.8)

where A is the area of the plate and w is the weight per unit area. I t is seen that the coordinates x and y will be the same regardless of the thickness or weight of the plate. In many engineering problems, the properties of areas are important, and the point in the area having coordinates x and y as defined by Eqs. (A.7) and (A.8) is called the centroid of the area.

A.2 MOMENT OF INERTIA

In considering inertia forces on rotating masses, it was found that the inertia forces on the elements of mass had a moment about the axis of rotation of

Lt = a r2 dM

as shown in Fig. A.3. The term under the integral sign is defined as the moment of inertia of the mass about the z axis:

r2 dM (A.9)

Since (he x and y coordinates of the elements are easier to tabulate than the radius, frequently it is convenient to use the relation

r 2 = x 2 + r

h = x2 dM + y2 dM (A.10)

An area has no mass, and consequently no inertia, but it is customary to designate the following properties of an area as the moments of inertia of the area since they are similar to the moments of inertia of masses:

I r = y2 dA

x2 dA

(A. 11)

(A.12)

\ ret ,IM

Figure A.3

412 AIRCRAFT s t r u c t u r e s ,

T h e c o o r d i n a t e s a r e s h o w n in Fig . A.4. T h e polar moment of inertia of a n a r e a is defined as

(A.I 3)

F r o m Ihe r e l a t i o n s h i p u sed in E q . (A. 10),

r2 = x2 + y2

.x dA + >-2 dA = /,, + Jx (A. 14)

It is frequently necessary to find the moment of inertia of an area about an axis when the moment of inertia about a parallel axis is known. The moment of inertia about the y axis shown in Fig. A.5 is defined as follows:

Iy = I x2 dA

S u b s t i t u t i n g t h e r e l a t i o n x = d + x' in E q . (A.1S) y ie lds

(d + x')2 dA

(A. 15)

= d2 dA 4- 2d x' dA + x'J dA

or

I,. = Ad2 + 2x' Ad + (A. 16)

where x' represents the distance of the ccntroid of the area from the / axis, as defined in Eq. (A.7), !'y represents the moment of inertia of the area about the y' axis, and A represents the total area. Equation (A. 16) is simplified when the y' axis is through the centroid of the area, as shown in Fig. A.6:

I}. = Ad2 + Ic (A. 17)

The term represents the moment of inertia of the area about a centroidai axis. The moment of inertia of a mass may be transferred to a parallel axis by a


Figure A.5

similar procedure to that used for the moment of inertia of an area. For the mass shown in Fig. A.7, the following relations apply, where the centroidal axis C lies in the xz plane:

1. = v2 dM (.x2 + y ) dM

[id + x'f + j'2] dM

(d2 + 2dx' + .x'2 + y2) dM

and substituting r2 = x'2 + y2 gives r

I . = d2 dM + 2d I* dM + r2 dM

Since x' is measured from the ccntroidal axis, the second integral is zero. The last integral represents the moment of inertia about the centroidal axis:

!z = Md2 4- Ic (A. 18)


Equations (A. 17) and (A. 18) can be used to find either the moment of inertia about any axis when the moment of inertia about a parallel axis through the ccntroid is known or the moment of inertia about the centroidal axis when the moment of inertia about any other parallel axis is known. In transferring mo-ments of inertia between two axes, neither of which is through the centroid, it is necessary first to find the moment of inertia about the centroidal axis, then to transfer this to the desired axis, by using Eq. (A. 17) or (A. 18) twice. It is seen that the moment of inertia is always a positive quantity and that the moment of inertia about a centroidal .axis is always smaller than that about any other parallel axis. If Eq. (A.16) is used, it is necessary to use the proper sign for the term x'. All terms in Eqs. (A. 17) and (A. 18) are always positive.

The radius of gyration p of a body is the distance from the inertia axis over which the entire mass would be concentrated in order to give the same moment of inertia. Equating the moment of inertia of the concentrated mass to that for the body yields

p2M = I

o r

(A. 19)'

It is seen that the point where the mass is assumed lo be concentrated is not the same as the center of gravity, except for the case where lc in Eq. (A. 18) is zero. The point at which the mass is assumed to' be concentrated is also different for each inertia axis chosen.

The radius of gyration of an area is defined as the dislancc from the inertia axis lo the point where the area would be concentrated in order to produce the same moment of inertia:

p2A = I

P = (A.20)

- The moment of inertia of an area is obtained as the product of an area-and the square of a distance and usually is expressed in units of inches to the fourth power. The moments of inertia for the common areas shown in Fig. A.8 should

I -I C I J _

= M / ' / l 2

M)

Figure A.14

m o m e n t s o f i n e r t i a , m o h r ' s c i r c l e 415

Figure A.9

be memorized, sincc I hey are used frequently. Moments of inertia for other areas may be found by integration or from engineering handbooks.

Example A.l Find the center of gravity of the airplane shown in Fig. A.9 a. The various items of weight and the coordinates of their individual centers of gravity are shown in Table A.l. It is customary to take reference axes in the directions shown in Fig. A.9b with the x axis parallel to the thrust line and the z axis vertical. While the z axis is at the wing leading edge in this example, it may be taken through the propeller or through some other con-venient reference point.

SOLUTION The y coordinate of the center of gravity is in the plane of sym-metry of the airplane. The coordinates 3c and z are obtained from Eqs. (A.4) and (A.6). The terms W, Sxvv, and Zz>v are obtained by totaling columns 3, 5, and 7 of Table A.l.

Zwx 50,723 W 4243

26,109

4243

= 12.16

= 6.2

Table A.l

N o . I tem Weight w X wx z wz

(1) (2) (3) (4) (5) (6) P)

1 W i n g g r o u p 697 22.6 + 15,781 40.9 + 2 8 , 5 7 4

2 Tai l g r o u p 156 198.0 30,904 33.1 5,171

3 Fuselage g r o u p 792 49.8 39,430 3.9 3,092

4 L a n d i n g gear (up) 380 19.2 7,297 - 1 1 . 7 - 4 , 4 2 9

5 Engine sect ion g r o u p 160 - 3 8 . 6 - 6 , 1 7 9 - 7 . 1 - 1 , 1 3 8

6 P o w e r p l a n t 1,302 - 4 8 . 8 - 6 3 , 6 7 4 - 6 . 0 - 7 , 7 8 2

7 Fixed equ ipmen t 756 35.9 27,164 3.5 2,621

T o t a l weight e m p t y 4,243 50,723 26,109


- 2 in-*

in

| /

©

t

2 h 1 © 1

p 6 in

Example A.2 Find the centroid and the moment of inertia about a horizontal axis through the centroid of the area shown in Fig. A. 10. Find the radii of gyration about axis xx and about axis cc.

SOLUTION The area is divided into rectangles and triangles, as shown. The areas of the individual parts are tabulated in column 2 of Table A.2. The y coordinates of the centroids of the elements are tabulated in column 3, and the moments of the areas Ay are tabulated in column 4. The centroid of the total area is now obtained by dividing the summation of the terms in column 4 by the summation of the terms in column 2:

XAy 79.5 y = = -zz- = 2.94 in

A 27

The moment of inertia of the total area about the x axis will be obtained as the sum of the moments of inertia of the elements about this axis. In finding the moment of inertia of any element about the x axis, Eq. (A. 17) may be written as

Ix = Ay2 + T0

where Ix is the moment of inertia of the element of area A about the x axis, y is the distance from the centroid of the element to the x axis, and 70 is the moment of inertia of the element about its own centroid. The terms Ay2 for

Table A.2

Element A y Ay Af h (1) (2) (3) (4) (5) (6)

1 12 1 12 12 4.0

2 1.5 2.5 3.75 9.4 0.2

3 1.5 2.5 3.75 9.4 0.2

4 12 5 60 300 36.0

Total 27.0 79.5 330.8 40.4


all the elements are obtained in column 5 as the product of terms in columns 3 and 4. The values of I 0 are found from the equations shown in Fig. A.8. The moment of inertia of the entire area about the x axis is equal to the sum of ail terms in columns 5 and 6:

l x = 330.8 + 40.4 = 371.2 in4

This moment of inertia may be transferred to the centroid of the entire area by using Eq. (A.17J as follows:

Ic = / , - (TA)y2

where HA represents the total area and y represents the distance from the x axis to the centroid of the total area. Thus

Ic = 371.2 - 27.0(2.942) = 138.0 in4

The radii of gyration may now be obtained as defined in Eq. (A.20);

Example A.3 In a metal stressed-skin airplane wing, the sheet-metal covering acts with the supporting spanwise spars and stringers to form a beam which resists the wing bending. Figure A.l la shows a cross section of a typical wing which has a vertical web and extruded angle sections riveted to the spar web and to the skin. The stringers are extruded Z sections which are riveted to the skin. The upper surface of the wing is in compression, and the sheet-metal skin buckles between the stringers and is ineffective in carrying load. The

-i •! 1

4 s

b i

• r

s

b i

t

6 11 s 7

t

6

Figure /V.11


skin is riveted to the stringers at frequent intervals, and a narrow strip of skin adjacent to each stringer is prevented from buckling and acts with the string-er in carrying compressive load. The effective width of skin acting with each stringer is usually about 30 times the skin thickness. On the underside of the wing, the entire width of the skin is effective in resisting tension. It is usually sufficiently accurate to assume the area of each stringer and its effective skin to be concentrated at the centroid of its area in computing the moment of inertia of the area. The wing cross section would then be represented by the nine elements of area shown in Fig. A.l lb . The moment of inertia of each clement about its own "centroid is neglected. In this particular wing, the skin and stringers to the right of the spar are very light and are assumed to be nonstructural.

The moment of inertia of the area shown in Fig. A.l lb is obtained about horizontal and vertical axes through the centroid of the total area. The areas and coordinates of the elements are given in columns 2, 3, and 6 of Table A.3.

SOLUTION This example is solved by the method used for Example A.2, except that the column for I 0 is omitted. Table A.3 shows the calculations for the moments of inertia about both the horizontal and vertical axes.

Then -82.4

- 1 1 . 5 6 7.135

I , c = 2310 - 7.135(11.562) = 1358 in4

- 4.72 „ ^ z = — = 0 . 6 6

Ixc = 331.6 - 7.135(0.662) = 328 in 4

Table A.3

Element A •X Ax Ax2 r Az Az2

(1) (2) (3) (4) (5) (6) (7) (8)

I 0.358 - 3 4 . 5 - 1 2 . 3 4 426 + 8.6 3.08 26.5

2 0.204 - 2 8 . 1 - 5 . 7 3 16! + 9 . 6 1.96 18.8

3 0.395 - 1 9 . 9 - 7 . 8 5 156 + 10.0 3.95 39.5

4 0.204 - 1 0 . 1 - 2 . 0 6 21 + 9.6 1.96 18.8

5 1.615 + 0.5 + .81 0 8.8 14.21 125.2

6 1.931 + 0.5 + .97 I — 5.V - 1 1 . 0 2 62.8

7 0.752 - 1 0 . 1 - 7 . 6 0 77 - 5 . 2 - 3 . 9 1 20.4

8 0.7S4 - 2 2 . 4 - 1 7 . 6 5 394 - 4 . 3 - 3 . 3 7 14.5

9 0.892 - 3 4 . 7 - 3 0 . 9 2 1074 - 2 . 4 - 2 . 1 4 5.1

To ta l 7.135 - 8 2 . 4 0 2310 4.72 331.6

m o m e n t s o f i n e r t i a , m o h r ' s c i r c l e 419

A.3 MOMENTS OF TNERTIA ABOUT INCLINED AXES

The moment of inertia of an area about any inclined axis may be obtained from the properties of the area with respect to the horizontal and vertical axes. For the area shown in Fig. A.12, a relationship between the moments of inertia about the inclined axes x' and y' and the axes x and y may be found. The moment of inertia about the x' axis is

ya dA (A.21)

The coordinate / of any point is

/ = y cos 4> — x sin (j> (A.22)

If this value of / is substituted in Eq. (A.21), the following value is obtained:

Ix, = (cos2 <j>) y2 dA - (2 sin </> cos cp) xy dA + (sin2 cj>) x2 dA (A.23)

The integrals must extend over the entire area. The angle <j> is the same regardless of the clement of area considered and is therefore a constant with respect to the integrals. The first and 'last integrals of Eq. (A.23) represent the moments of inertia of the area about the .x and y axes. The second integral represents a term which is called the product of inertia Ixy :

^ xy xy dA

Equation (A.23) may now be written as

Ix, = Ix cos2 (p - IX). sin 2(p + Iy sin2 <j)

(A.24)

(A.25)

A similar expression may be derived for the moment of inertia about the / axis:

Iy, = l x sin2 (j> + Ixy sin 2<p + Iy cos2 (A. 2 6)

If Eqs. (A.25) and (A.26) arc added, die following relationship is obtained:

Ix + Iy=Ix, + Iy,

From Eq. (A. 14), the sum of the moments of inertia about any two perpendicular axes is seen to be equal to the polar moment of inertia, which is the same regardless of the angle <p of the axes.

The product of inertia about the .v' and / axes is defined as follows:

K A - ' / dA (A.27)

By subs t i t u t i ng the re la t ions

.x' = x cos (j) + y sin

a n d

/ = y co s (j) — x sin tp

into Eq. (A.27), the fo l lowing value of I , is o b t a i n e d :

Kiy< — (cos2 <p) xv dA — (s in2 <j>) xy dA

+ (sin $ cos <p) y1 dA — (sin $ cos (p)

or lx,v, = !xy(cos2 <p — s in 2 (•/)) + (Ix — /j,) sin tp cos <p

x2 dA

( A . 2 8 )

A.4 P R I N C I P A L A X E S

The moment of inertia of any area about an inclined axis is a function of the angle (p, as given in Eqs. (A.25) and (A.26). The angle <p at which the moment of inertia Ix, is a maximum or minimum is obtained from the derivative of Eq. (A.25) with respect to (p:

'Ipr = — 21 % cos <p sin <p — 27,.. cos 2(p + 21. sin (p cos tp dip

This derivative is zero when Ix, is a maximum or minimum. Equating the deriva-tive to zero and simplifying yield

(Iy - Ix) sin 2<p = 21 xy cos 2<p

or tan 2<p = j ^ f - (A.29) * y * x

Since there are two angles under 360° which have the same tangent, Eq. (A.29) defines two values of the angle 2rp, which will be at 180° intervals. The two corresponding values of the angle cp will be at 90° intervals. It can be shown that the value of Ix, will be a maximum about one of these axes and a minimum about

the other. These two perpendicular axes about which the moment of inertia is a maximum or minimum are called the principal axes.

The product of inertia about the inclined axes may be expressed in terms of the angle 2$, by making use of the trigonometric relations

sin 2<j) = 2 sin <j> cos (j> (A.30)

and cos 2e/> = cos2 <fi — sin2 <j) (A.31)

Substituting these values in Eq. (A.28) yields

lx,y, = Ixy cos 2$ + / j ~ I f sin 2<f> (A.32)

An important relation is obtained Tor the angle at which Ix,y, is zero. Substituting Ix,y, = 0 in Eq. (A.32) produces

tan 2(b = 2I*>' 1, - >*

This is identical to the expression defining the principal axes in Eq. (A.29). The product of inertia about the principal axes is therefore zero.

The moments of inertia about the principal axes may be obtained by substi-tuting the value of 4> obtained from Eq. (A.29) into Eq. (A.25):

I p = ^ ± l 2 + l I 2 y + ( L H J x ) (A.33)

and j ^ l ^ . J ^ + ^ - J z J (A.34)

where I p represents the maximum value of Ix, and Iq represents the minimum value of l x , . These values are moments of inertia about perpendicular axes de-fined by Eq. (A.29).

A.5 PRODUCT OF INERTIA

The product of inertia of an area is evaluated by methods similar to those used in evaluating the moment of inertia. Products of inertia for various elements of the area usually are evaluated separately and then added to find the product of inertia for the entire area. When both x and y are positive or negative, the product of inertia is positive; but when one coordinate is positive and the other negative, the product of inertia is negative. In the case of an area which is symmetrical with respect to the x axis, as shown in Fig. A. 13, each element of area dA in the first quadrant will have a corresponding area in the fourth quad-rant with the same x coordinate but with the y coordinate changed in sign. The sum of the products of inertia for the two elements will be zero, and the integral


y(-): dA Figure A. 13

of these terms for the entire area will be zero:

xy dA = 0

The same relation is true if the area is symmetrical with respect to the y axis. Therefore, when either axis is an axis of symmetry, the product of inertia is zero and the axes are principal axes.

When the product of inertia of an area about one set of coordinate axes is known, the product of inertia about a set of parallel axes can be found. For the area shown in Fig. A.14, the product of inertia about the x and y axes is defined as

xy dA

By substituting the values

x = h + u

y = k + v

the transfer theorem is obtained:

lx>. = J"(A + u){k + v) dA

hk dA + h v dA + k

/,„ = hkA + hvA + kuA + L

u dA + UP dA

(A.35)

where u and v are the coordinates of the centroid of the area and /„„ is Ihe

Figure A.14


product of inertia of the area with respect to the u and v axes. If the u and v axes are through the centroid of the area, Eq. (A.35) becomes

Ixy =-- hkA + /„„ • (A.36)

With u and v axes also principal axes of the area, Iuv = 0 and Eq. (A.36) becomes

Ixy = hkA ( A . 3 7 )

For an area composed of several symmetrical elements, the product of inertia may be obtained as the sum of the values found by using Eq. (A.37) for each element.

Example A.4 Find the product of inertia for the area shown in Fig. A.15.

SOLUI 'ION The total area is divided into the three rectangular elements A, B, and C. Rectangle A is symmetrical about both the x and y axes; hence the product of inertia is zero. Rectangle B is symmetrical about axes through its centroid; therefore the product of inertia may be found from Eq. (A.37):

, Ixy = hkA = - 3 x 5 x 8 = - 1 2 0 in 4

For rectangle C,

For the total area,

Example A.5- Find the product of inertia about horizontal and vertical axes through the centroid of the area shown in Fig. A.16.

SOLUTION The total area is divided into the two rectangular elements A and B as shown. The x and y reference axes are chosen through the centroids of

Ixy=hkA = 3 x - 5 x 8 = - 1 2 0 in4

/ „ = 0 - 120 - 120 = - 2 4 0 in4

- 4 i n

4 i n

6 i n

t i n

- S i n -

Fignre A. 15 Figure A.16


these rectangles. Since rectangle A is symmetrical abou t the y axis and rec-tangle B is symmetrical about the x axis, Ixy — 0. The centroid of the area is obtained as follows:

4 x 2.5 '

x = — = 1 . 0 - 6 * 2 - s , «

— I O T - 1 - 5

The product of inertia about the centroidal axes may now be derived from Eq. (A.36):

Ixy = xyA + IXcyc

0 = 1.0 x 1.5 x 10 + IXcyc

or IX c f c = —15 in 4

Example A.6 Find the product of inertia about horizontal and vertical axes through the centroid of the area shown in Fig. A . l l . The areas and coordi-nates of the elements are given in Table A.3 and are repeated in columns 2, 3, and 4 of Table A.4.

SOLUTION The product of inertia about the x and z axes is obtained as the summation of the terms Axz, in column 5 of Table A.4. The centroidal axes were found in Example A.3 to have coordinates x = — 11.56, z = 0.66. F r o m Eq. (A.36),

/ „ = xzA + L xc Z,

- 6 8 . 2 = - 1 1 . 5 6 x 0.66 x 7.135 + /

13 .8 '

*c 2C

Table A.4

Element A X z Axz

(1) (2) (3) (4) (5)

t 0.358 - 3 4 . 5 + 8^6 - 1 0 6 . 2

2 0.204 - 2 8 . 1 + 9 . 6 - 5 5 . 0 3 0.395 - 1 9 . 9 + 10.0 - 7 8 . 5 4 0.204 - 1 0 . 1 + 9.6 - 1 9 . 8 5 1.615 + 0.5 + 8.8 7.1 6 1.931 + 0.5 - 5 . 7 - 5 . 5 7 0.752 - 1 0 . 1 - 5 . 2 39.5 8 0.784 - 2 2 . 4 - 4 . 3 75.9 9 0.892 - 3 4 . 7 - 2 . 4 74.3

To ta l 7.135 - 6 8 . 2

A.6 MOIIR'S CIRCLE FOR MOMENTS OF INERTIA

The equations for the moments and products of inertia about inclined axes are difficult to remember. It is often convenient to use a semigraphic solution, which is easier to remember and which is an aid in visualizing the relationship between the moments of inertia about various axes. If the values of Ixy, from Eq. (A.32) are plotted against values of Ix, obtained from Eq. (A.25) for corresponding values of <j>, the points all fall on the circle shown in Fig. A. 17. The maximum and mini-mum values of the moments of inertia are represented by points P and Q.

In order to prove that the circle shown in Fig. A.17 represents the values given by Eqs. (A.25) and (A.32), the values of l x , and Ixy/ are expressed in terms of the moments of inertia about the principal axes I p and l q and the angle 0 from the x' axis to the principal axis. If the x and y axes are principal axes, a substitu-tion of the values Ixy = 0, Ix = / , , ly = Iq, and cj> = B into Eqs. (A.25) and (A.32) yields

Jx, = Ip cos2 6 + sin2 6 (A.38)

= L l = J i s i n 20 (A.39)

The following trigonometric relations for double angles are used:

sin2 9 = i - i cos 29

cos2 0 = i + I cos 20

Substituting these values in Eq. (A.38) yields

l x , = i ^ 2 + l-JL~* cos 26 (A.40)

Equations (A.39) and (A.40) correspond to the coordinates Ixy , and Ix, which are computed from the geometry of the circle shown in Fig. A.17. The angle of inclination of the .x' axis from the principal axis is one-half the angle 28 measured between the corresponding points on the circle. If Ixy, is measured as positive


upward on the circle, a counterclockwise rotation of the x' axis corresponds to a counterclockwise rotation around the circle. Points at opposite ends of the diam-eter of the circle correspond to perpendicular inertia axes. The products of inertia about perpendicular axes are always equal numerically but opposite in sign, since rotation of the axes through 90° interchanges the numerical values of the coordi-nates x' and y' for any element of area and changes the sign of one coordinate.

Example A,7 Find the'moments of inertia about principal axes through the centroid of the area shown in Fig. A. 18. Find the moments and product of inertia about axes Xj and yx and axes x2 and y2.

SOLUTION The moments of inertia about the x and y axes are

' 2 x 123 ' U x 23

12 • + 21

12 + 8 x 52 = 693.3 in4

h = + 2f—7-— + 8 x 32 I = 173.3 in4

1 2

From Example A.4,

12

Ix y = - 2 4 0 in4

Mohr's circle for the moments and products of inertia about all inclined axes may now be plotted from these three values. The products of inertia Ixy, are plotted against the moments of inertia l'x, as shown in Fig. A.19. Point x in Fig. A.19 has coordinates 693.3 and -240 .0 , as shown. If the x' axis is rotated through 90°, it will coincide with the y axis, and the coordinates of point Y will be Ix, = 173.3, Ixy, = 240. The positive sign for Ixy, results from the fact that after the x' axis is rotated through 90° from the x axis, the coordinate x' is positive up and the coordinate / is positive to the left. The

Figure A.14


value of Ixy, is numerically equal but opposite in sign to the value of I x y . Points X and Y are at opposite ends of a diameter of the circle, since a rotation of the axes of 90° corresponds to an angle of 180° on the circle.

The center of the circle is a distance -j(693.3 + 173.3), or 433.3, from the origin. Point X is a distance 260 horizontally and 240 vertically from the center of the circle; therefore

Radius = v/ 2 4 0 2 + 2602 = 353.8

20 = arctan = 42.7°

or 0=21.35"

The principal axes are represented by points P and Q on the circle. The moments of inertia have maximum and minimum values, and the product of inertia is zero for these axes. The principal moments of inertia are equal to the distance from the origin to the center of the circle plus or minus the radius of the circle:

I p = 433.3 + 353.8 = 787.1 in4

/ , = 433.3 - 353.8 = 79.5 in4

The P axis is counterclockwise from the x axis, at an angle 9 = 21.35°. Similarly, since .point Q on the circle is counterclockwise from point Y, the Q axis is countcrclockwisc from the y axis.

The moments and product of inertia about the Xj. and y\ axes are ob-tained from the coordinates of points on the circle. Since the and axes are 15° clockwise from the x and y axes, the points X1 and VL on the circle will be 30° clockwise from points X and 7 . The coordinates of points Xl and Yi may be obtained from the geometry of the circle as follows:

/ M = 433.3 + 353.8 cos 72.7° = 538 in4

I y i = 433.3 - 353.8 cos 72.7° = 328 in4

I x i n = - 3 5 3 . 8 sin 72.7° = - 3 3 8 in4

428 a i r c r a f t STRUCTURES

0 = 7.97°

L.v ' r<2

Figure A.20

The moments and product of inertia about the x2 and y2 axes are also derived from the geometry of the circle:

IX2 = 433.3 - 353.8 cos 77.3° = 355 in4

l n = 433.3 + 353.8 cos 77.3° = 511 in4

/ „ „ = - 3 5 3 . 8 sin 77.3° = - 3 4 5 in4

Example A.8 Find the principal axes through the centroid and the moments of inertia about these axes for the area shown in Fig. A.20. The x and z axes are through the centroid, and the moments and product of inertia have the values I x = 320, l z = 1160, and / „ = - 120.

SOLUTION The coordinates of point X on the circle of Fig. A.21 are 320 and — 120. The coordinates of point Z are 1160 and +120. These points are at opposite ends of the diameter and thus determine the circle. It is important to show l x , with the correct sign at point X so that the direction of the principal axes may be obtained correctly. The distance of the center of the circle from the origin of coordinates is i (320 + 1160) = 740. From the geom-etry of the circle,

Radius = . / 4 2 0 2 + 1202 = 437

tan 20 = 20 = 15.94°

or 0 = 7.97°

The principal moments of inertia are represented by points P and Q on the circle,"at which the moments of inertia have maximum and minimum values

z r

J o H

l i p

(1.161 r l i2i

V" tfi'

160 20

1,177 0

Figure A.21


and the product of inertia is zero:

l p = 740 + 437 = 1177 in4

Iq = 740 - 437 = 303 in 4

Since point P is 15.94" clockwise from point Z on the circle, the P axis will be half this angle, or 7.97° clockwise from the z axis, as shown in Fig. A.20.

A.7 M O H R ' S C I R C L E F O R C O M B I N E D S T R E S S E S

The relationship between normal stresses and shearing stresses on planes at various angles of inclination is similar to the relationship between moments and products of inertia about inclined axes. Most structural members are subjected simultaneously to normal and shearing stresses, and it is necessary to consider the combined effect of the stresses in order to design the members. The landing-gear strut shown in Fig. A.22, for example, is subjected to bending stresses which produce tension in the direction of the strut, internal oil pressure which produces a circumferential tension, and torsion which produces shearing stresses on the horizontal and vertical planes. The maximum tensile stress does not occur on either the horizontal or the vertical plane, but on a plane inclined at some angle to them. It can be .shown that there are always two perpendicular planes on which the shearing stresses are-zero. These planes are called principal planes, and the stresses on these planes arc called principal stresses.

Any condition of two-dimensional stresses can be represented as shown in Fig. A.23, in which the principal stresses ap and crq act on the perpendicular principal planes. The orientation of these planes depends on the condition of stress and will be'Tound from known stress conditions. The normal and shearing stresses a„ and i can be found on a plane at an angle 8 to the principal planes, from the equations of statics. The stresses are in pounds per square inch and must always be multiplied by the area in square inchcs in order to obtain the force. In Fig. A.24, a small triangular element is. shown, with principal planes forming two sides of the element and the inclined plane a third side. If the inclined plane has

Figure A.24 Figure A.23

an area A, the sections of the principal planes have areas A cos 6 and A sin 0, as shown. From a summation of forces along the n and s axes, which are perpen-dicular and parallel to the inclined plane, the following equations are obtained:

£ = CnA - c„A cos2 9 - <rqA sin2 0 = 0 (A.41)

£ = xA - ar A cos 0 sin 0 + aq A sin 0 cos 0 = 0 (A.42)

Using the trigonometric relations for functions of double angles yields

cos2 0 = j + % cos 20

sin2 0 = i - i cos 29

sin 0 cos 0 = j sin 20

Dividing Eqs. (A.41) and (A.42) by A gives

= + (A.43)

T = sin 20 (AM)

If values of t and a„ are plotted for different values of the angle Q, as shown in Fig. A.25, all the points will lie oh the circle. This construction, first used by Mohr, is similar to that used for finding moments of inertia about inclined axes.

Normal stresses are considered positive when tensive and negative when compressive. Compressive stresses are therefore shown to the left of the origin on Mohr's circle. In the following examples, shearing stresses are considered positive when they tend to rotate the element clockwise and negative when they tend to rotate the element counterclockwise. Thus, on any two perpendicular planes, the shearing stress on one plane would tend to rotate the element clockwise and would be measured upward on the circle, and the shearing stress on the other

MOMENTS OE INERTIA, MOHR'S CIRCLE 4 3 1

T, n + "r =5 + i c o s 20 1-

0

'r Figure A.25

plane would be equal numerically but opposite in sign and would be measured downward. If this sign convention for shearing stresses is followed, a clockwise rotation of the planes of stress corresponds to a clockwise rotation on the circle. In some books, the opposite sign convention for shearing stresses is used, and a clockwise rotation or the planes corresponds to a counterclockwise rotation on the circle.

Example A.9 The small clement shown in Fig. A.26 represents the conditions of two-dimensional stress at a point in a structure. Find the normal and shearing stre'sses on planes inclined at an angle 0 with the vertical plane for

. values of 0 at 30° intervals. Find the principal planes and principal stresses. Find the planes of maximum shear and the stresses on these planes.

SOLUTION The values of the normal stress a„ and the shearing stress x on the horizontal and vertical planes are plotted as shown in Fig. A.27. The vertical plane has a normal stress of 10,000 and a shearing stress of 4500 lb/in2, and these coordinates are shown for point A on Mohr's circle. The stresses on the horizontal plane are represented by point B on the circle, with coordinates of — 2000 and +4500. The circle is now drawn with line AB as a diameter. The

2000 Ill/in-

200!) Ji./in- Figure A.26


/? a

distance OC is i ( - 2 0 0 0 + 10,000) = 4000. Points A and B have a horizontal distance of 6000 and a vertical distance of 4500 from the center of the circle:

Radius = . / 6 0 0 0 2 + 45002 = 7500

tan 20 = _ 0.75 6000

20 = 36.86° and 0 = 18.43°

The principal stresses are represented on the circle by points P and Q. The coordinates of these points are obtained by adding and subtracting the radius from distance OC:

a p = 4000 + 7500 = 11,500 lb/in2

aq = 4000 - 7500 = - 3 5 0 0 lb/in2

The point P on the circle is at an angle 20 counterclockwise around the circle from point A. The principal plane P is therefore at the angle 0 = 18.43° counterclockwise from the vertical plane A, as shown in Fig. A.28a. Similarly, point Q is counterclockwise from the horizontal plane B, or perpendicular to plane P.

The planes of maximum shearing stress are always at 45° to the principal planes, regardless of the stress conditions. Points R and S at extremities of the vertical diameter of Mohr ' s circle represent the maximum shearing stresses. On the circle, these points are always 90° f rom the points at the extremities of the horizontal diameter, which represent the princip. stresses. The normal stresses on the two planes of maximum shear are always equal, since they are both equal to the distance OC on the circle diagram. The maximum shearing stresses are shown on an element in Fig. A.28b. Plane S is

MOMENTS Of iNERT i a , MOHR'S CIRCLE 4 3 3

7 4 5 0 lb / ' in :

F i g u r e A . 2 8

26.57° ciockwisc from the vertical, sincc point S is twice this angle clockwise from point A on the circle. Plane S is also 45° clockwise from plane P, and plane R is 45° counterclockwise from plane P. The shearing stress on plane R is positive, tending to rotate the element clockwise. The shearing stress on plane S is negative, lending to rotate the element counterclockwise.

Planes E and F are 30° counterclockwise from planes A and B. Points E and F on the circle-must be 60° counterclockwise from points A and B, respectively. The stresses on plane E are obtained by calculating the coordi-nates of point E on the circle:

The stresses on planes £ and F arc shown in Fig. A.28c in the correct directions. -

The stresses on plane G, which is 60° counterclockwise from the vertical plane A, arc

r = 7500 sin 83.14" = 7450 lb/in2

an = 4000 + 7500 cos 83.14° = 4900 lb/in2

The stresses on planes H, which are perpendicular to plane G, are

r = - 7 5 0 0 sin S3.14° = - 7 4 5 0 lb/in2

<r„ = 4000 - 7500 cos 83.14° = 3100 lb/in2

t = 7500 sin 23.14" = 2950 lb/in2

a„ = 4000 + 7500 cos 23.14° = !0,900 lb/in 2

These stresses are shown in Fig. A.28d.

A P P E N D I X

B MATRIX ALGEBRA

B.l MATRIX DEFINITIONS

A matrix is defined as a rectangular array of elements (symbolic or numerical) arranged in rows and columns as follows:

* I 2 " -

m = k21 k22 •• k2J

ki2 •

(i= 1, 2, 3 , . . . ; ; = 1 , 2 , 3 , . . . ) (B.l)

where an clement has the subscripts i and j to indicate the element location in the ith row and j th column, respectively. For instance, element k2l is located in row 2 and column 1.

For i not equal to j, as in Eq. (B.l), the matrix is called a rectangular matrix of order i x j. For i =j = n, the matrix is defined as a square matrix of order n x n:

" f e n k i 2 - ••• ku-£K1 = k21 • k2n

(B.l)

K 2

Elements kllt k22, ••• > k„„ form the main diagonal of a square matrix. Elements kij (i T^ j) are referred to as the off-diagonal elements.

For i = 1 and j > 1, the matrix is said to be a row matrix of order 1 x j and is written as

[ K ] = [ * u ( j = 1 , 2 , 3 , . . . ) (B.3)

For / > 1 and j = 1, the matrix is defined as a column matrix of order i x 1

434

MATRIX ALGEBRA 4 3 5

and is written as

k21 ( ' • = 1 , 2 , 3 , . . . ) (B.4)

Diagonal , Identi ty and Nul l Mat r ices

A diagonal matrix [ K ] is one whose oIT-diagonal elements ku = 0 (£ # j ) and main-diagonal elements ki} ^ 0 (/ = j):

~ / c n 0 . . . 0

0 k22 . . . 0 [A'] =

0 0 fc.ij

(B-5)

An identity, or unit, matrix [ K ] is a diagonal matrix whose main-diagonal el-ements k;j — 1 (i = j):

iKi = m =

1 0 . . 0 0 1 . 0 (B.6)

_ 0 0 1 _

A null matrix [ K ] is one whose every element k;j- = 0:

0 0 . . . 0

[ K ] = (B.7)

Symmet r i c and Transposed Mat r ices

A square matrix [A'] is said to be symmetric if the following holds true:

kij = ( i j i j )

A numerical example of a 3 x 3 symmetric matrix is

(B.8)

[K] = 15x - 2 7

- 2 21 7 21 " 1 6 J

Notice that the main diagonal (dashed line) is the line of symmetry. The transpose matrix [ K ] T of a matrix [ K ] is obtained by interchanging rows

and corresponding columns. For example, if a matrix [ K ] is given by

k i

[K] k n k 1 2

k 2 \ k 2 2

(B.9)


then its transpose [A']T can be easily written as

" t n k [ K ] r = k i 2 k

k

Determinants

Determinants are defined for square matrices only. Thus, the determinant |K[ of a square matrix [ K ] is defined as the quantity which results upon performing the following arithmetic operation on [/v]:

m s

£ kjjCij ( j = any integer between 1 and m) (B.II) i = 1

m or = £ kjjCij (i = any integer between 1 and m) (B.12)

J=I Elements k;] in Eqs. (B.ll) and (B.12) appear in the;'th column and the ith row, respectively, of matrix Likewise, cu are the cofactors which correspond to the j th column of matrix [A'] in Eq. (B.ll) or to the ith row of matrix [/Tj in Eq. (B.12).

The cofactors c - of a matrix [ K ] are determined by

Cij = ( - i y + J M i j (B.13)

where M y are the minors of matrix [ K ] and are defined as the determinant of the matrix which results after deleting the /th row and y'th column of matrix [K].

Let us illustrate the calculation of a determinant by considering the following 3 x 3 matrix:

[K] = 2 1 6

3 5 4 . 2 2 1 .

Using Eq. (B.l 1) and choosing,/' = 3 arbitrarily, we gel

|K|= Y, fei3cf3 = kuc13 + k23c23 + k33c33 i=i

where, from Eq. (a),

^13 = 6, k23 = 4, k33 = 1

For j = 3, Eq. (B.13) becomes

c i 3 = ( - ! ) ' • * 3 M a - / = 1 , 2 , 3

MATRIX ALGHiRA 4 3 7

Ilcnce

C t 3 = / H 13 C2 3 = - M 2 3 C33 = M 3 3

As defined previously, the minor A-f,,, or the given matrix in Eq. (a) is the determinant of the matrix resulting from deleting the first row and the third column, corresponding to subscripts 1 and 3 on M i 3 . Or,

Similarly,

A f » =

M „ =

M,, =

3 5 2 2]

1 2 1

2 2

2 1

3 5

= 3 x 2 — 5 x 2 = — 4

= 2 x 2 - 1 x 2 = 2

= 2 x 5 - 1 x 3 = 7

rherefore,

= - 4 c2i = - 2 c 3 3 = 7

and the determinant is

|K| = 6( —4) + 4( — 2) + 1(7) = - 2 5

Propert ies of De te rminan t s

1. Interchanging any two rows or columns of a determinant changes the sign of the determinant:

2 5 1 10 1 10 2 5

5 2 10 1

15

2. If two rows or two columns of a determinant are identical, then the value of the determinant is zero.

3. If a determinant has dependent rows or columns, the value of the determinant will be zero:

4 1

4

6 5

10 = 0

Column 3 is dependent on columns I and 2:

Column 3 = column 1 + column 2

4. If rows and corresponding columns of a determinant are interchanged, the value of the determinant is not changed:

= 20 5 10 j 5 1 1 61 ~ 10 6


5. Multiplying the determinant by a scalar quantity a, is equivalent to multi-plying the elements of one row or one column by that quantity:

fcn k12 aktl k12 aku ok12

ki i k22 ak21 k22 k2i k22

6. If a row or a column of a determinant is changed by adding to or subtracting from its elements the corresponding elements (or corresponding elements multiplied by a common factor) of any other row or column, then the value of the determinant is not changed:

2 1 6 7 3 1 2 2 4 4 4 6 4 6 10 6 4 8 4 6

row 2 column 2 half of half of added added to column 1 row 2 to row 1 column 1 added to added to

column 2 row 1

Singular Matrices The matrix [K] is said to be singular if the value of its determinant is zero.

4 2 8 4

If 1X1 =

Then | K | =

Hence, matrix [K] is singular.

4 2 8 4

= 1 6 - 1 6 = 0

B.2 MATRIX ALGEBRA

Matrix Addition Two matrices [B] and [C] of order m x n can be added by adding each element bij of matrix [B] to the corresponding element c1J of matrix [C] :

bu+cn b12 + c12

l_b2i + c 2 1 b22 + c22_

or, in general, [A]"*" = [5] m X " + [C]""x" (B.14)

where aa = b,7 + ctJ i = 1 , 2 , . . . , m;j = 1, 2 , . . . , n) (B.15)

" 6 x i bl2~ +

Cll Cl2 " 6 x i + Cll Cl2

=

A : b22_ +

_C21

Matrix Subtraction

Matrix subtraction is similar to matrix addition except elements by and c,j are subtracted instead of being added:

[S] M * N = [B]"" 1" — [ C ] m X " (B.16)


where bi} - cu (i = 1, 2 , . . . , m;j = 1, 2 , . . . , n) (B.17)

Matrix Multiplication

The multiplication of two matrices [B]m X"and [CJ"'"' is performed by multi-plying each element of a row i of matrix [B] by the corresponding element in a column j of matrix [C] and summing the products:

[P]"""- = [/?]"• x " [C]" x r (B.18)

where Pij = Z Kc,i (' = 1. 2 , . . . , m;j = 1, 2, r) ( B . 1 9 )

In expanded form, Eq. (B.19) bccomes

piJ = bnclj + bi2c2J+ ••• +bi„c„j (B.20)

The two matrices [B] and [C] can be multiplied only if the number of columns in [B] is equal to the number of rows in [C].

Let us consider the following example:

15J = [2 1 4 ] l x 3

c n = 2 1 3 x 2

4 2 _3 5.

[ / > ] ! * * = L / ? J > X ' [ C ] 3 X 2

" 2 r

4 2 L3 5 J

= [2 1 4] = [20 24]

or Pij = I! bik ctJ = buCij + ba c2J + bi3 c3J

For i = 1 and j = 1,

F o r i = 1 a n d j = 2 ,

Pu =bt lCI j + ij, 2 1 + b13c31

= 2 x 2 + 1 x 4 + 4 x 3 = 20

P12 = b n C 1 2 + b 12 ^22 + ^ 1 3 ^ 3 2

= 2 x 1 + 1 x 2 + 4 x 5 = 24

Propert ies of M a t r i x Multiplication

1. D i s t r i b u t i v e l a w :

[ / I ] X M + [ C ] ) - [ / l ] [ B ] + [ / i ] [ C ]


2. Commutative law: :

M M * IBM 3. Associative law:

M x ( [5] x [C]) = ([/!] x [B]) x [C]

4. If a matrix is multiplied or divided by a scalar quantity, this is equivalent to multiplying or dividing each element of the matrix by the scalar quantity:

or

A l l ban ba12

« 2 2 _ _ba21 ba22

~an a\i~ b b

a2l a22

i_ b b _

M a t r i x Inversion

The inverse of a square nonsinguiar matrix [iC], denoted by can be defined as an operation in matrix algebra analogous to division in ordinary algebra. Among the many techniques used for inverting a matrix, only two are presented here.

To invert a matrix [ K ] which must be square and nonsinguiar, do the following:

1. Form a matrix of cofactors [C] whose elements are found by utilizing Eq. (B.13):

cu = (-l),+JMy " ( i = l , 2 , ...,n;j = 1 , 2 , . . . , n) (B.13)

The new matrix of cofactors [CJ is referred to as the adjoint of [K] :

A d j [ K ] = [C]

2. Transpose the adjoint matrix:

Transpose [C] = [ C ] r

3. Find the determinant of [ i f ] by using Eq. (B.l 1) or (B.12):

1*1 = £ k(JCij ( / = 1 , 2 , . . . . or in) i = l

or 1*1 = I V y (/ = 1, 2, . . . , or in)

( B . l i )

(B.12)

4. Divide Adj [ / f ] T , or [C]T , by |/C| to obtain the inverse:

A d j [ * ] T [C ] T

1 * 1 1 * 1 (B.21)


To illustrate, consider the following 2 x 2 matrix:

~5 2 VQ =

First, find [C], the adjoint of [K'j:

Q , = ( 1 )*'"*"A/f/

or f u = M n c i 2 = - M l 2

where Mu = 1 A/12 = 2 M2I

Therefore, [C] = ^

Second , t r a n s p o s e [ C ] :

2 1

c21 = —M2l

2 A122 = 5

[C]7 1

- 2

Third, find the determinant of [K] :

|Kj = 5 x 1 - 2 x 2 = 1

Fourth, divide [C"| r by |K|:

or

Check the results:

or

- 2

5.

5 2 I - 2 " "l 0" 2 1_ . - 2 5. .0 1.

C22 = M 2 i

In matrix inversion by the Choleski method, any square nonsingular matrix can be expressed as the product of an upper triangular matrix [ [ / ] and a lower triangular matrix [ / J . Then the inversion of the original matrix is reduced lo inverting two triangular matrices [ t / ] and [L], which is quite simple. The de-tailed development of this method can be found in Refs. 10 and 24. Restricting ourselves to symmetrical matrices only, we have

but

Therefore Eq. (B.22) becomes

[ A ] = [ L ] [ t / ]

[W] = M r

[ K ] = [ L ] [ L y

(B.22)

(B.23)

(B.24)

The inverse can be wrilten as

I K T 1 = l L T y l i L T l

[xr 1 =

The elements of [L] and [."?] arc from Rcf. 10:

or

where

1 \ l / 2 '2

- E 'i

V i - i ; I . ij I 'im'mj hj

Su - mj

bj

0

' > J

' < y

>>j

i<j

If the inverse of [ K ] is denoted by [ / / ] :

[ / / ] = [ X ] " 1 = [S ] T [S ] = [S][S]

then the elements of [H] can be determined as follows:

E ^im mj

S„„ = 0 ( i > in)

s,„, = 0 ( / < m )

Let us consider the following 2 x 2 symmetric matrix:

"2 3~ [K] =

3 5

(B.25)

(B.26)

(B.27)

(B.28)

(B.29)

(B.30)

From Eq. (B.28),

/ „ = ( W J = Y 2


2 - 1 \ l / 2

From Eq. (B.29),

= ( 5 - f ) " 2 = V5

1 1

hi J l

S22 = 7 - = V ^ '22

E2 s 2 m ' m l " 1

m= I ' 11

S 2 2 / 2 1 _ - V 2 ( 3 / V 2 )

in N/2

- 3

.v,2 = 0

Or

[S] =

[S]T = [S] =

7 5

• f i V2

0 J i

From Eq. (B.30),

h 1 1 = Z Si».•'». 1 m - 1

: -S11 i-S I I + S 1 2 S 2 l

1 1


' ' 1 2 = ZSlm\,2

1 - 3 r 75 75

' ' 2 2 — X

— •'] i S j 2 + -^12 5 2 2

= - 3 = h2l

— s 2 I -f I 2 + •''2 2 s 2 2

= 0(0) + 7 2 ( V 2 ) = 2

Hcncc [K] " ' = [ / / ] =

Properties of the Inverse of a Matrix

5 - 3

- 3 2

1. The inverse of the product of two square matrices [R] and [K] is equal to the inverse of [A'] multiplied by the inverse of [/?]:

[ [ W ] ] - ' = [ * ] - ' [ * ] (B.31)

2. The inverse of a symmetric matrix is also symmetric.

Matrix Partitioning

A matrix K can be partitioned (divided) into smaller submatrices by drawing vertical and corresponding horizontal dotted lines as indicated:

[K] =

O

©

p n k ^ I k13 k 14 1 fcis" k21 k22 | fe23 k 24 | k 25

k31 k32 r -

k3 A | k3S

k4l k42 J _ * 4 3 K 4 J k,,

kS2 1 k5i ^ 5 4 I

where [ X w ] =

[ K I 2 ] =

[ £ » ] =

[ * u ] [ K i J

[ * 2 l ] [ K « ] [ * 2 3 ]

[ * 3 l ] [ * 3 J LK3i] k12

-kll ^ 2 2 .

~kl3

> ' 2 3 k24_

-k2s_

Properties of Partitioned Matrices

1. Two matrices can be added or subtracted in terms of their partitioned sub-matrices if the partitioning is done on both matrices in the same manner.

2. Two matrices can be multiplied in terms of their partitioned submatrices if the rule of matrix multiplication applies.

3. The inverse of a matrix can be found in terms of its partitioned submatrices.

B.3 S I M U L T A N E O U S LINEAR ALGEBRAIC E Q U A T I O N S

The most important application of matrix algebra is in the solution of a large set of linear algebraic equations. For instance, the following is a set of n simul-taneous linear algcbraic equations with unknowns qt (i = 1, 2, . . . , n) expressed in terms of known quantities Q; (' = 1, 2, . . . , n) and the coefficients K;j (i, j = 1, 2 , . . . , « ) :

fciltfl + /fl2<Zl + 1" kln1n = Qi

' k2lqt + k22q2 + ••• + k2nqn = Q2

k„i</i + kn2 q2 + • • • + k„„q„ = Q„

or, in matrix algebra notation,

ku fcl2 ' • kin q i Q k2i k22 • • k2n <?2

= Q

A . K1 • fc„„_ _Q

In shorthand matrix notation, Eq. (B.32) can be written as

m{q} = { f i }

Upon inverting [iC], the unknown column vector {q} can be found:

UJ-cxrMe} To illustrate, consider the following simultaneous equations:

2qx + 3 q2 = 14

In matrix form, (a) becomes

+ 5 q2 = 22

"2 3~j h ~14~ _3 5J Ai- _22.

'/i '2 3" -1 "14

J?2- J 5_ _22_

(B.32)

(B.3 3)

(a)

00

(c)

The inverse of the matrix is

Thus

or

5 - 3 14 4 12„ — 3 2_ _22_ 2

<h = 4 and 92 = 2

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NACA Tech. No te 1348, 1947. 51. Kuhn, P., et a!.: A Summary of Diagonal Tension, pt. 1, vol. 1, NACA Tech. Note 2661, 1952. 52. Kuhn, P., and J. P. Peterson: Strength Analysis of Stiffened Beam Webs, NACA Tech. Note

1364, 1947. 53. Levy. S. , et al.: Analysis of Square Shear Web above Buckling Load, NACA Tech. Note 962,

1945. 54. Levy, S., et al.: Analysis of Deep Rectangular Shear Web above Buckling Load, NACA Tech.

Note 1009, 1946. 55. Lalide, R., and H. Wagner, Test for the Determination of the Stress Concentration in Tension

Fields, N A C A Tech. Mem. 809, 1936. 56. Stowell, E. Z. : Compressive Strength of Flanges, N A C A Tech. No te 2020, 1950, 57. Gatewood, B. F..: Thermal Stresses, McGraw-Hill , New York, 1957. 58. Nowacki , W.: Thermoelasticity, Addison-Wesley, Reading, Mass., 1962.


59. Burgreen, David: Elements of Thermal Stress Analysis, C. P. Press, Jamaica, N.Y., 1971. 60. Calcote, Lee R.: The Analysis of Laminated Composite Structures, Van Nostrand, New York,

1969. 61. Plantema. F. J.: Sandwich Construction. Wiley. New York, 1966. 6 1 Zicnkiewicz, O. C., and G. S. 1 lolistcr: Stress Analysis, Wiley, New York, 1965. 63. Argyris, J. H„ and S. Dclscy: Modern Fuselage Analysis and the Elastic Aircraft. Hutterworth,

London, 1963. 64. Gallagher, R. II.: A Correlation Study of Methods of Matrix Structural Analysis, Pergamon, New

York, 1964. 65. HolT, N. J.: The Analysis of Structures, Wiley, New York, 1960. 66. Kuhn, P.: Stresses in Aircraft arid Structures, McGraw-Hill, New York, 1956. 67. Langhaar, H. L.: Energy Methods in Applied Mechanics, Wiley, New York, 1960. 68. Lee, C. W.: "Thermoelastic Stresses in Thick-Walled Cylinders under Axial Temperature Gradi-

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p . 4 i 8 .

70. Janak, J. F.: "Thermal Expansion in a Constrained Elastic Cylinder," IBM J. Res. Develop., May 1969.

71. Miller, R. E.: Theory of the N onhomogeneous Anisotropic Elastic Shells Subjected to Arbitrary Temperature Distribution, University of Illinois Bulletin 458.

72. Sokolowski, M.: "The Axially Symmetric Thermoelastic Problem of the Infinite Cylinder," Arch. Mech. Stos., vol. 10, 1958, p. 811.

73. Strub, R. A.: "Distribution of Mechanical and Thermal Stresses in Multilayer Cylinders," AS,ME Trans., Jan. 1953, p. 73.

74. Sun, C. T„ and K. C. Valanis: " O n the Axially Symmetric Deformation of a Cylinder of Finite Length," Engineering Research Institute, Iowa Stale. University Rep. 48,1966.

75. Gallagher, R. H.: "Computerized Structural Analysis and Design—The Next Twenty Years," Computers and Structures, vol. 7. 1977, pp. 495-501.

76. Norris, C. H., J. B. Wilbur, and S. Utku: Elementary Structural Analysis, 3d ed., McGraw-Hill, 1976.

77. Livesley, R. K.: Matrix Methods of Structural Analysis, 2d cd„ Pergamon, New York, 1976. 78. Schmit, L. A. and B. Farshi: "Minimum Weight Design of Stress Limited Trusses," Proc. ASCE,

' J. Struct. Div., vol. 100, no. ST1, Jan. 1974, pp. 97-107. 79. Glockner, P. G. : "Symmetry in Structural Mechanics," Proc. ASCE, J. Struct. Div.. vol. 99, no.

STi , Jan. 1973, pp. 71-89. 80. Fox, L.: An Introduction m Numerical Linear Algebra, Oxford University Press, New York,

1965. 81. Meyer, C.: "Solution of Linear liquations—State of the Art," Proc ASCE, J. Struct. Dir., vol.

99, no. ST7, July 1973, pp. 1507-1526. 82. Meyer, C.: "Special Problems Related to Linear Equation Solvers," Proc. ASCE, J. Struct. Div.,

vol. 101, no. ST4, Apr. 1975, pp. 869-890. 83. 1 licks, J. G.: Welded Joint Design, Halsted Press, N.J., 1979. 84. Kirby, P. A., and D. A. Nethercot: Design for Structural Stability, Halsted, N.J., 1979. 85. Donnell, L. H.: Beams, Plates, and Shells, McGraw-Hill , New York, 1976. 86. Jones, R. M.: Mechanics of Composite Materials, McGraw-Hill, New York, 1975. 87. Tauchert, T. R.: Energy Principles in Structural Mechanics, McGraw-Hill , New York. 1974. 88. Zienkiewicz, O. C.: The Finite Element Method in Engineering Science, 3d ed., McGraw-Hill,

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Engineers, Wiley, New York, 1974. 90. Brick, R. M.: Structure and Properties of Engineering Materials, 4th ed., McGraw-Hill, New

York, 1977. 91. Brush, D. O., and B. O. Almrnth: Duckling of Bars, Plates and Shells, McGraw-Hill, New York,

1975. 92. Rockey, K. C.: The Finite Element Method: A Basic Introduction, Wiley, New York, 1975.

RIMRKNCES 4 4 9

93. Brooks, C. R.: Heat Treatment of Ferrous Alloys, McGraw-Hill, New York, 1979. 94. Cherepanov, G. P.: Mechanics of Brittle Fracture, McGraw-Hill, New York, 1979. 95. Cordon. VV. A.: Properties, Evaluation, Control of Engineering Materials, McGraw-Hill, New

York, 1979. 96. Dieter, G. E.: Mechanical Metallurgy. 2d ed.. McGraw-Hil l , New York, 1976. 97. Palmer, A. C.: Structural Mechanics, Clarendon Press, Oxford, 1976. 98. Burgreen, A. C.: Design Methods for Power Plant Structures, Isl ed.. C. P. Press. Jamaica, N.Y.,

1975. 99. Fertis, D. G.: Dynamics and Vibration of Structures, Wiley, New York, 1973.

100. Lyon, R. H.: Statistical Energy Analysis of Dynamical Systems: Theory and Applications, M.I.T. Press, Cambridge, Mass., 1975.

101. Clough, R. W.: Dynamics of Structures, McGraw-Hill , New York, 1975. 102. Budynas, R. G . : Advanced Strength and Applied Stress Analysis, Rochester Inst, of Technology,

Rochester, N.Y., 1977. 103. Shames, I. H. : Introduction to Solid Mechanics. Prcnticc-Hall. Englewood ClilTs, N.J., 1975. 104. Harrison, H. B.: Structural Analysis and Design: Some Minicomputer Applications, Pergamon,

New York, 1979.

INDEX

Aerodynamic bonds , 31 —33 Airy s tress func t ion . 1 2 6 - 1 2 7

Four ie r ser ies solut ion. 1 2 9 - 1 3 0 polynomial solut ion, 127—129

Algebraic equat ions , s imul taneous linear, 4 4 5 - 4 4 6

Applied loads, 27 Argyris , J . H . , 145 Associa t ive law for matr ix multiplication, 440 Axial loads, 3 Axial rods, 1

Batdorf , S.. 371 Beam-delliction equat ions , 328 — 330 Beam e lements , global st i l lness matr ices for,

2 1 4 - 2 2 6 Beam shear in multicell boxes . 2 7 2 - 2 7 5 Beams:

curved , bending s t resses in. 307 — 3 10 normal s t resses in, 99 — 101 shear s t resses in. 101 — 104

alternate solut ions for , 104— 107 tapered, shea r flow in, 118—121 unres t ra ined, thermal s t resses in, 289 — 291 with var iable str inger a reas . 122—126

Bending modulus of rupture , 301 —302 Bending momen t s , 3 Bending s t resses :

cons tant , 3 0 2 - 3 0 4 in curved b e a m s , 307 —310 trapezoidal dis t r ibut ion of, 304 — 307

Body loads, 2

Bolted joints . 3 9 0 - 3 9 5 Bolts, eccentr ical ly loaded. 401—404 Boundary condi t ions . 70

in thermal s t r ess problems, 284 Box beams:

c losed-sect ion: shea r flow in. 113—118 torsion of, 112-113

• multicell: beam shear in. 272 — 275 torsion of. 2 7 0 - 2 7 2

redundancy of , 269 — 270 torsion of, 2 6 1 — 2 6 3 warping of, 266 — 269

Buckling design, 3 2 8 - 3 8 7 fo r crippling fa i lure of columns, 358 — 362 for curved rec tangular plates, elastic

buckling. 3 7 0 - 3 7 3 for curved shee t s , 3 6 4 - 3 6 8 for eccentr ical ly loaded columns. 33 1—332 for fixed end co lumns , 334 — 339 fo r fiat plates, elast ic shear buckling,

3 6 8 - 3 7 0 for isotropic flat p la tes in compression,

3 4 4 - 3 4 9 for isotropic flat shee t s , 349 — 353 for long co lumns , 3 3 0 - 3 3 1 nondimension:: i buckl ing curves, 357 — 358 plastic buckling of flat sheet, 3 5 3 - 3 5 7 of pure tension field beams, 373 — 377 for short co lumns . 3 3 2 - 3 3 4 , 3 3 9 - 3 4 1

Built-up s t ruc tures , thermal stresses in, 2 9 1 - 2 9 5

Bulkheads , 238 (See utsu Fuse lage bulkheads)

450

I N D E X 4 5 1

Cast igl iano's first t h e o r e m . 144 Cast igl iano's second theorem. 146

use in deflect ion analysis of s t ruc tures . 1 5 0 - 1 5 6

Centroids , 4 0 9 - 4 1 1 Cholcski method . 441 Circular shaf t s , tors ion of, 310 —311 (."loscd-scclinn box beams:

shear flow in, 1 1 3 - 1 1 8 torsion of, 112—113

Column end fixity. 334 — 339 Column failures, local crippling, 358 —362 Column mat! ices. 4 3 5 Column yield s t r e s s , 339 Co lumns (see Eccentr ical ly loaded columns,

buckling of : Long columns, buckling of; Short co lumns)

Combined s t resses . 3 2 0 - 3 2 5 M o h r ' s circle for , 4 2 9 — 4 3 3

Commuta t ive law for matr ix multiplication. 4 4 0

Compar i son of s imilar materials , dimensionless column curves Tor. 342 — 344

Compatibil i ty equa t ions for thermal s t resses . 283. 284 — 285

Complemen ta ry s train energies and strain. 140 Complementa ry w o r k and work . 140 Compress ion :

of curvcd shee ts . 3 6 4 — 368 of isotropic flat plates , buckling. 344 - 3-19

Compress ion tes ts , 81 Compress ive s t rength of isotropic flat sheet .

3 4 9 - 3 5 3 Concent ra ted loads . 2 Cons tan t bending s t resses , 302 — 304 Convers ion f a c t o r s . S i /met r i c , inside f ront

cover Coordina te sys tem. 192. 194 Cor ten . H . J. . 86 C o z z o n e . F. P., 304 . 305. 343. 357 Crippling s t resses :

of co lumns , 358 — 362 Gera rd method for calculat ing. 362 — 363 N e e d h a m m e t h o d for calculating. 362—363

Curved beams , bending s t resses in. 307 — 310 Curved rectangular plates, elastic buckling of.

3 7 0 - 3 7 3 Curved sheets :

buckling of. 3 6 4 - 3 6 8 compress ion of, 364 — 368

Cu tou t s in s e m i m o n o c o q u e s t ructures . 2 5 1 - 2 6 0

Deflection analysis of s t ructural systems. 1 3 9 - 1 8 6

Design loads. 27 Dete rminan ts . 4 3 6 — 437

proper t ies of, 437 — 438 Diagonal matr ices , 4 3 5

Diagonal web tens ion , angle of, 377—380 Dimens ionless co lumn curves for compar ison

of materials , 3 4 2 — 3 4 4 Displacement b o u n d a r y condit ions, 70 Disp lacements , sign convent ion for, 194 Distr ibuted loads. 2 Distr ibutive law for matr ix multiplication. 439 Dolan. .1. J . . 86 Drag, 31, 32 Dunn , L. G . , 365 Dynamic loads. 2

Eccentrically loaded co lumns , buckling of, 3 3 1 - 3 3 2

Eccentrically loaded connect ions . 4 0 1 — 4 0 4 Elastic axes of wings, 263—266 Elastic buckling of curved rectangular pla tes ,

3 7 0 - 3 7 3 Elastic limit, tors ional s t resses above,

3 1 7 - 3 2 0 Elastic modulus of elasticity, 148 Elastic nonl inear behavior , 71, 72 Elastic shear buckl ing of fiat plates, 368 — 370 Elasticity of s t ruc tures , 62—77 Llcment discret izat ion, 191 Element formula t ions , t ransformat ions of , to

sys tem formula t ion , 211 —214 Element shape func t ions . 207 — 209

polynomial me thods . 207 — 209 Element st iffness matr ices , 209—211 Element s t ructural relat ionships:

formulation p rocedures , 202 — 203 direct me thod , 2 0 3 - 2 0 5 energy me thods , 205 — 207

End restraint of tors ion members , 316 — 317 Engesser equat ion , 333 — 334 Equilibrium condi t ions for thermal s t resses ,

281, 2 8 4 - 2 8 5 Equivalent load solut ion for thermal s t ress

problems, 2 8 6 - 2 8 9 Euler, L., 331 Euler column, 365 Euler curve , 339. 3 4 0 litilcr equat ion . 331 . 3 3 9 - 3 4 2 . 345, 353 Euler load, 331, 332

Fac to r of safe ty , 27 Failure theories in s t ructura l design,

3 2 5 - 3 2 7 Fatigue. 8 4 - 8 9 Fatigue failure, 84 Fatigue-life predic t ion, 84 — 87 Fatigue tests, S-N cu rves , 85, 8 7 - 8 9 Finite difference me thod , 158—165 Finite e lement matr ix methods , appl icat ions of,

192 Finite element s t i f fness method, 190 — 226 First central d i f fe rence approximation, 159

452 I N D E X

Filtings, 389 strength of. accuracy of analysis of, 395

Fixed supports . 5 Flat plates:

elastic shear buckling of. 368 — 370 isotropic, buckling of, 344—349

Flat sheets, buckling of, 349 — 357 Flight loading conditions, 27 — 31 Flight-vehicle aerodynamic loads, 3 1 —33 Flight-vehicle imposed loads, 26 — 58 Flight-vehicle inertia loads, 34 — 37 Force-stress relationships, 97 — 98 Forces, sign convention for, 194, Free contraction of solids. 279 Free expansion of solids. 279 Full monocoque structures, 233 Fuselage bulkheads, loads on. 2 3 8 - 2 4 3

Gassner , E.. 86 Gaussian coordinate systems. 97 Gerard method, 362 — 363 Global stillness matrices for special beam

elements , 214 — 226 Gordon . S. A.. 384 Gust load factors, 43—46

Hinge-roller supports , 5 Hinge supports , 4 — 5 Honeycomb cores, 92 Hooke 's law, 148

Identity matrices, 435 Inclined axes, moments of inertia about,

4 1 9 - 4 2 0 Incompletely developed diagonal tension field

beams, 3 8 0 - 3 8 1 Inelastic nonlinear behavior, 71, 72 Inertia loads, 34 — 37 Isotropic bodies, 72

• Joints, 3 8 9 - 3 9 5

Kanemitsu, S., 366 Kelsey .S . , 145 Kowalewski, J., 86 Kuhn, P.. 3 8 1 , 3 8 3

Lahde, R„ 381 Langhaar, H. L „ 380 Levy, S., 3 8 1 , 3 8 4 Lift. 3 1 , 3 2 Limit-load factor, 27 Limit loads, 27 Linear clastic behavior, 71, 72

Linear elastic structural systems, 147— 150 Load factors:

gust, 43 — 46 for translational acceleration, 37 —41 V-ti diagrams, 41 —43

Loads: applied. 27 classification of, 2 —3 design. 27 distribution to this webs, 2 3 3 - 2 3 7 on fuselage bulkheads. 2 3 8 - 2 4 3 imposed on flight vehicles, 2 6 - 5 8 limit, 27 ultimate. 27

Long columns, buckling of. 3 3 0 - 3 3 1

Main diagonal of a square matrix, 434 Materials:

behavior and evaluation of, 78 — 95 strength-weight comparisons of, 89 — 92 testing of. 7 9 - 8 2 typical design data for, 94 — 95

Mathematical models of structures, 191 Matrices. 6. 4 3 5 - 4 3 6 Matrix addition, 438 Matrix algebra, 438—445 Matrix definitions. 434 — 438 Matr ix inversion, 440 — 444 Matrix multiplication. 439

properties of, 439 — 440 Matrix partitioning, 444

properties of. 445 Matrix subtraction. 438 Melcon, M. A., 343, 357 Membranes , 1 Method of joints . 1 0 - 12 Method of sections, 12 -Metric conversion factors, endpaper Minimum guaranteed values, 94 — 95 Minimum weight, 89 Modulus of elasticity, 80 Mohr ' s circle. 320, 322, 374

for combined stresses, 429—433 for moments of inertia, 425—429

Moments of inertia, 411—418 about inclined axes, 419 — 420 Mohr ' s circle for. 4 2 5 - 4 2 9

Monocoque structures, 233 Multicell box beams:

beam shear in, 272 —275 torsion of. 2 7 0 - 2 7 2

Multiphase materials. 79

Needham method, 362 — 363 Negat ive high angle of attack ( N H A A ) ,

2 9 - 3 0 Negat ive low angle of attack (NLAA) . 30 Noj ima, H „ 366

I N D E X 4 5 3

Nonci rcular shaf t , tors ion of. 311 - 3 16. Nondimens ional buckling curves , 357 — 358 Normal strains, 67 — 68 Null matr ices , 435

Off-diagonal e lements of a square matrix. 434 O r t h o t o p i c bodies . 72 Osgood . W. R.. 83. 342

I 'a l tngren-Miner theory . 85 Partial-tension field b e a m s . 3 8 0 - 3 8 1 Partit ioned matr ices . 4 4 4

propert ies of, 445 Peterson, J. P., 381 Pitching moment , 32 Plane motion, 35 Plane-strain problems, compatibi l i ty equat ions

for. 6 9 - 7 0 P l a n c - . s t r e s s p r o b l e m s , c o m p a t i b i l i t y e q u a t i o n s

for . 6 9 - 7 0 Plastic bending. 3 0 0 - 3 0 2 l'iustic buckling of flat shcc ls . 353 — 357 Plastic range of m e m b e r s . 300 Plate e lements . ! Poisson ' s ratio, 73, 148, 3 4 4 - 3 4 6 . 368. 371 Polar moment of iner t ia , 412 Positive high angle of a t t ack ( P H A A ) . 2 7 - 2 9 Positive low angle of a t t ack (1 ' l .AA), 29 Principal axes, 4 2 0 - 4 2 1 Principal planes, 4 2 9 Principal s t resses . 4 2 9 Product of inert ia. 4 1 9 . 4 2 1 - 4 2 4 Proport ional limit.. 80 Pure tension field b e a m s , buckling of.

3 7 3 - 3 7 7

Radius of gyrat ion, 4 1 4 Rayleigh-Ritz me thod . 143. 1 5 6 - 1 5 8 React ions . 4 — 5 Rectangular mat r ices . 4 3 4 Reduced modulus equa t ion . 334 Redundancy of box b e a m s . 269 — 270 Redundant s t ruc tures , 6 — 7

multiple r e d u n d a n c y . 173—181 single r edundancy , 167—173 and the unit-load m e t h o d . 165— 166

Riveted jo in ts , 390 — 395 Rivets , eccentr ical ly loaded . 401 - 4 0 4 Roiling. 35 Romberg . W „ 83 Romberg-Osgood equa t i on , 342 Row matr ices . 4 3 4 Ryder , E. I., 321 , 325

Sandwich cons t ruc t ion . 92 — 94 Schildcrotit , M.. 37 I

Semimonocoque s t ruc tures : analysis of m e m b e r s of, 233 — 275 cutouts in, 2 5 1 — 2 6 0 shearing de fo rma t ions of. 260 — 261

Semitension field b e a m s , 380 — 387 Shaf ts :

circular, tors ion of, 310 — 311 noncircular . torsion of. 311 —316

Shanley. E. R.. 321, 3 2 5 ' Shear ccn tc rs . 109 - I 12

of wings, 263 — 266 Shear (low, 106

in c losed-sect ion box beams . 1 13 —1 18 in tapered b e a m s . 118—121 in la; , - i webs . 2 4 6 - 2 5 1 in ttim webs , 1 0 7 - 109

Shear lag. 1 8 1 - 1 8 6 Shear lqads, 3 Shear modulus , 73 Shear modulus of elasticity, 82 Shear panels . 2 Shear s t resses in beams , 10 I — 104

al ternate solut ions for , 104— 107 Shear tests . 81 - 82 Shearing d e f o r m a t i o n s in semimonocoquc

s t ruc tures . 2 6 0 - 2 6 1 Shearing s t ra ins . 67 •Shells, 2 Short co lumns:

buckling of. 3 3 2 - 3 3 4 empirical f o inmlas for buckling design of.

3 3 9 - 3 4 1 Sign convent ion for forccs and displacements ,

194 S imul taneous l inear algebraic equat ions,

4 4 5 - 4 4 6 Single-phase mater ia l s , 79 Singular mat r ices , 6 S-N cu rves . 85 . 8 7 - 8 9 Spanwise taper eli'ect, 1 1 8 - 1 2 1 Square mat r ices , 4 3 4 , 436 Stable s tat ic equi l ibr ium, 5 Static analysis of s t ruc tures , 1—21 Static equi l ibr ium, equat ions of, 5 —6 Static loads, 2 Statically d e t e r m i n a t e s t ructures , 6 - 7 Statically inde te rmina te s t ructures , 6 — 7 Stein, M., 371 Stiffness m e t h o d concep t , 197 — 202 Strain and c o m p l e m e n t a r y strain energies .

140 S t ra in -d i sp lacement relat ionships, 68 — 69, 2 8 2 Strains, 6 6 - 6 7

(See also S t r e s ses and strains, t r ans fo rmat ion of)

S t rength-weight compar i sons of materials, 8 9 - 9 2

Stress analys is , 9 7 — 1 3 0 Stress equi l ibr ium equat ions , nonuni form

s t ress field, 65 — 66

454 I N D E X !

-. • »

Stress function solution for thermal stress problems. 286 • .

Stress ratios, 321—325 •• Stress-strain curve idealization, equations for,

' 8 2 - 8 4 Stress-strain relationships, 71—73 Stresses, 62—64

normal, in beams, 99 — 101 Stresses and strains, t ransformation of. 73 — Stringer areas in beams, 122—126 Structural systems, defined. I Structures, static analysis of, 1 —21 Supports, 3—5 j. . Surface loads, 2 •• Symmetric matrices! 435—436 • System formulations, t ransformed f rom

element formulations,. .211—214

Tangent modulus curves , dimensionless form of, 3 4 2 - 3 4 4 . . ' - .-

Tangent modulus pf elasticity, 333 ^ . Tangent modulus equat ion, 333 — 334 Tapered webs, shear (low in, 246—251 Taylor series, 143, 146 Tensile tests, 7 9 - 8 1 Tension members, 2 9 9 - 3 0 0 Thermal coefficient of expansion, 280 Thermal equilibrium, 2 8 4 - 2 8 5

equations for, 28 1-. Thermal loads, 2 1

Thermal stresses, 2 7 9 - 2 9 5 boundary condit ions, 284 in built-up s tructures , 2£1 —295 formulation of equat ions for analysis of,

2 8 1 - 2 8 4 . solution methods for problems in,

2 8 4 - 2 8 9 in unrestrained beams, 289 — 291.

Thermoelast ic problems, solution methods for 2 8 4 - 2 8 9

Thermoelastic strain-stress relat ionships. ' 2 8 2 - 2 8 3 '

Thin webs: load distribution to, 233 —237 • shear flow in, 1 0 7 - 1 0 9

Torques , 3 Torsional stresses:

of box beams, 261—263 of circular shafts , 3 10 —311 ,

Torsional-stresses (Con/.): i of closed-section box tjcams. 112—113

: above the elastic limit, 317—320 end res-tfaint of member?,' 316 — 317 of multicell box beams, 2 7 0 - 2 7 2 _ ..' of mmeirculiir shaf ts . 31 I —3 16

Translational acceleration, load factors for. 3 7 - 4 1 „ -

Transpose matrices, 436 Trapezoidal distribution of bending stresses.

3 0 4 - 3 0 7 Trusses . 1 Tsicn, H. S.. 365 Two-force members , I

Ultimate compressive strength of isotropic flat sheet, 3 4 9 - 3 5 3

Ultimate-load factor, 27 Ultimate loads. 27 Ultimate strength, 81 Unit-load method. 145

and redundant s tructures, 165—166 Unit matrices, 435

Velocity-load-factor (J-'-H) diagrams. 41 - 4 3 Virtual displacements, principle of, 140—144

^ Virtual forces, 144— 145 Virtual work, 140 Von Karman, T. L.. 365

Wagner . H.. 380, 381 Warping of beam cross sections, 266 — 269 Webs , diagonal tension in, angle of. 377 —3 SO

(SF£" also Tapered webs, shear How in: Thin webs)

Welded joints, 4 0 4 - 4 0 8 Wing ribs, analysis of, 243 — 246 Wings:

elastic axes of. 263 — 266 shear centers of, 263 — 266

Work and complementary work, 140

Yawing, 35 Yield point, 80 Yield stress, 80 Y'oung's modulus, 73

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