CE8502 STRUCTURAL ANALYSIS I UNITI STRAIN ENERGY …

CE8502 STRUCTURAL ANALYSIS I

UNITI STRAIN ENERGY METHOD

Determination of Static and Kinematic Indeterminacies – Analysis of continuous

beams, plane frames and indeterminate plane trusses by strain energy method (up to

two degree of redundancy).

UNITII SLOPE DEFLECTION METHOD

Slope deflection equations – Equilibrium conditions - Analysis of continuous beams

and rigid frames – Rigid frames with inclined members - Support settlements-

symmetric frames with symmetric and skew-symmetric loadings.

UNITIII MOMENT DISTRIBUTION METHOD

Stiffness and carry over factors – Distribution and carryover of moments - Analysis of

continuous Beams- Plane rigid frames with and without sway – Support settlement -

symmetric frames with symmetric and skew-symmetric loadings.

UNITIV FLEXIBLITY METHOD

Primary structures - Compatibility conditions – Formation flexibility matrices -

Analysis of indeterminate pin- jointed plane frames, continuous beams and rigid

jointed plane frames by direct flexibility approach.

UNITV STIFFNESS METHOD

Restrained structure –Formation of stiffness matrices - equilibrium condition -

Analysis of Continuous Beams, Pin-jointed plane frames and rigid frames by direct

stiffness method.

TOTAL: 45 PERIODS

TEXTBOOKS:

1. Bhavikatti, S.S,Structural Analysis,Vol.1,& 2, Vikas Publishing House

Pvt.Ltd.,NewDelhi-4, 2014.

2. Bhavikatti, S.S, Matrix Method of Structural Analysis, I. K. International

Publishing House Pvt.Ltd.,New Delhi-4, 2014.

3. Vazrani.V.N And Ratwani, M.M, Analysis of Structures, Vol.II, Khanna

Publishers, 2015.

4. Pandit G.S.andGupta S.P.,Structural Analysis–AMatrix Approach, Tata McGraw

Hill Publishing Company Ltd.,2006

REFERENCES:

1. Punmia. B.C, Ashok Kumar Jain & Arun Kumar Jain, Theory of structures, Laxmi

Publications, New Delhi, 2004.

2. William Weaver, Jrand James M.Gere, Matrix analysis of framed structures, CBS

Publishers & Distributors, Delhi,1995

3. Hibbeler, R.C.,Structural Analysis, VII Edition, Prentice Hall, 2012.

4. Reddy.C.S, “Basic Structural Analysis”,Tata McGraw Hill Publishing

Company,2005.

5. Rajasekaran. S, & G. Sankarasubramanian., “Computational Structural Mechanics”,

PHI Learning Pvt. Ltd, 2015

6. Negi L.S.and Jangid R.S.,Structural Analysis, Tata McGraw Hill Publishing

Co.Ltd.2004.

www.studymaterialz.in

UNIT 2 SLOPE DEFLECTION METHOD

(1). A beam ABC, 10m long, fixed at ends A and B is continuous over joint B and is loaded as shown in Fig. Using the slope deflection method, compute the end moments and plot the bending moment diagram. Also, sketch the deflected shape of the beam. The beam has constant EI for both the spans.

Solution.

(a) Fixed end moments

Treating each span as a fixed beam, the fixed end moments are as follows:

(b) Slope deflection equations

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The end rotations A and C are zero since the beam is fixed at A and C. hence there

is only o ne unknown, B. the ends do not settle and

hence for each span is zero. Let us assume B to be positive. The result will indicate

the correct sign. The slope deflection equations are as follows:

For span AB,

For span BC,

(c) Equilibrium equation

Since there is only one un known, i.e. B, one equilibrium equation is sufficient. For

the joint B, we have

MBA + MBC = 0




(0.8 EI B + 3.6) + (0.8 EI B + 5.0) = 0 1.6 EI B = 1.4

The plus sign indicates that B is positive (i.e. rotation of tangent at B is clockwise).

(d) Final moments

Substituting the values of EI B in Eqa. (1) to (4), we get




(2) A beam ABC, 10m long, hinged at ends A and B is continuous over joint B and is loaded as shown in Fig. Using the slope deflection method, compute the end moments and plot the bending moment diagram. Also, sketch the deflected shape of the beam. The beam has constant EI for both the spans.

SOLUTIONS


These are the same as calculated in the previous problem:

MFAB = -2.4 KN-m ; MFBA = +3.6 KN-m

MFBC = -5.0 KN-m ; MFCB = +5.0 KN-m

(b) Slope deflection equations.

(c) Equilibrium equations

Since end A is freely supported,




(d) Final moments : Substituting the values of EI A and EI B inEq. (2), we get

The bending moment diagram and the deflected shape of the beam are shown in the

Fig. Note. The beam is statically indeterminate to single degree only. This

problem has also been solved by the moment distribution method (example

10.2) treating the moment at B as unknown. However, in the4 slope- deflection

method, the slope or rotations are taken as unknowns, and due to this the

problem involves three unknown rotations A, B and C. hence the method of

slope deflection is not recommended for such a problem.




(3) A continuous beam ABCD consists of three spans and is loaded as shown in

fig. ends A and D are fixed. Determine the bending moments at the supports and

plot the bending moment diagram.

a)Fixed end moments

(b) Slope deflection equation

A and D are zero since ends A and D are fixed.





At join B, MBA + MBC = 0




At join C, MCB + MCD = 0

From (I) and (II), we get EI B = -2.03 kN-m and EI C = + 1.26kN-m

(d) Final moments

Substituting this values in Eqs. (1) to (6), we get

The bending moment diagram and the deflected shape are shown in Figure.

4) A continuous beam ABC is supported on an elastic column BD and is loaded

as shown in figure . Treating joint B as rigid, analyze the frame and plot the

bending moment diagram and the deflected shape of the structure.




(a)Fixed end moments

MFBD = MFDB = 0

(b)Slope deflection equations.

The slopes A and D are zero since ends A and D are fixed.

For span AB





At join B, MBA + MBC + MBD = 0




(d) Final moments

Substituting this values in Eqs. (1) to (6), we get

The bending moment diagram and the deflected shape are shown in Figure.

(5) Analyze the rigid frame shown in figure





(b) Slope deflection equations.


For the equilibrium joint B, MBA + MBD + MBC = 0

(2EIB + 2.67) + (EIB -2) + (-4) = 0

3EIB = 3.33

EIB = 1.11

(d) Final moments

Substituting this value of EIB in Eqs. (1) to (4), we get




The bending moment diagram and

The deflected shapes are shown in Figure.

(6) A portal frame ABCD is fixed at A and D, and has rigid joints at B and C.

The column AB is 3m long. The beam BC is 2m long, and is loaded with uniformly

distributed load of intensity 6 kN/m. The moment of inertia is 2.1 and that of BC

and CD is I (Fig). Plot B.M. diagram and sketch the deflected shape of the frame.


Let the joints B and C move horizontally by

(b)Slope deflection equations.




(c) Equilibrium equations. At joint B,

d )Shear equation




e) Final moments




(7) A portal frame ABCD is hinged at A and fixed at D and has stiff joints at B

and C. the loading is as shown in figure. Draw the bending moment diagram and

deflected shape of the frame.




Unit 3 - MOMENT DISTRIBUTION METHOD

INTRODUCTION AND BASIC PRINCIPLES

Introduction

(Method developed by Prof. Hardy Cross in 1932)

The method solves for the joint mo ments in continuous beams and rigid frames

by successive approxi mation

Statement of Basic Principles

Consider the continuous beam ABC D, subjected to the given loads,

as shown in Figure below. Assume that only rotation of joints occur

at B, C and D, and that no support d isplacements occur at B, C and

D. Due to the applied loads in spans AB, BC and CD, rotations occur at B, C and D

In order to solve the problem in a su ccessively approximating manner,

it can be visualized to be made up of a continued two-stage problems

viz., that of locking and releasing the joints in a continuous sequence.




The joints B, C and D are locked in position before any load is applied on the b

eam ABCD; then given loads are applied on the bea m. Since the joints of beam

ABCD are locke d in position, beams AB, BC and CD acts as ind ividual and

separate fixed beams, subjected to the applied loads; these loads develop fixed

ennd moments.

In beam AB

Fixed end moment at A = -wl2/12 = - (15)(8)(8)/12 = - 80 kN.m

Fixed end moment at B = +wl2/12 = +(15)(8)(8)/12 = + 80 kN.m

In beam BC

Fixed end moment at B = - (Pab2)/l2 = - (150)(3)(3)2/62

= -112.5 kN.m

Fixed end moment at C = + (Pab2)/l2 = + (150)(3)(3)2/62

= + 112.5

In beam AB

Fixed end moment at C = -wl2/12 = - (10)(8)(8)/12 = - 53.33 kN.m

Fixed end moment at D = +wl2/12 = +(10)(8)(8)/12 = + 53.33kN.m

Since the joints B, C and D were fix ed artificially (to compute the the fixed-end

moments), now the joints B, C and D are released and allowed to rotate. Due to

the joint release, the joints rotate maintaining the continuous nature of the beam.

Due to the joint release, the fixed en d moments on either side of joints B, C and




D act in the opposite direction now, and cause a net un balanced moment to occur

at the joint.

These unbalanced moments act at th e joints and modify the joint moments at B,

C a nd D, according to their relative stiffnesses at the respective joints. The joint

moments are distributed t o either side of the joint B, C or D, according to their

reelative stiffnesses. These distributed moments al so modify the moments at the

opposite side of the beam span, viz., at joint A in span AB, at joints B and C in

span BC and at joints C and D in span CD. This modification is dependent on

the carry-over factor (which is equal to 0.5 in this case);

The carry-over moment becomes the unbalanced moment at the joints to

whic h they are carried over. Steps 3 and 4 are repeated t ill the carry-over

or distributed moment beco mes small.

Sum up all the moments at each of the joint to obtain the joint moments.

SOME BASIC DEFINITIONS

In order to understand the five steps mentioned in section 7.3, some words need

to be defined and relevant derivations made.

1Stiffness and Carry-over Factors

Stiffness = Resistance offered by m ember to a unit displacement or rotation at a

point, for given support constraint conditions




A clockwise moment MA is applied at A to produce a +ve bending in beam AB. Fin

d A and MB.

Using method of consistent defor mations

Considering moment MB,




MB + MA + RAL = 0

MB = MA/2= (1/2)MA

Carry - over Factor = 1/2

2 Distribution Factor

Distribution factor is the ratio according to which an externally applied

unbalanced moment M at a joint is apportioned to the various m embers mating

at the joint

M = MBA + MBC + MBD




Modified Stiffness Factor

The stiffness factor changes when t he far end of the beam is simply-supported.

As per earlier equations for deforma tion, given in Mechanics of Solids text-

books.

Solve the previously given proble m by the moment distribution method

Fixed end moments




Stiffness Factors (Unmodified Stifffness

Distribution Factors




Computation of Shear Forces




UNIT IV FLEXIBLITY METHOD

INTRODUCTION

These are the two basic methods by which an indeterminate skeletal structure

is analyzed. In these methods flexibility and stiffness properties of members are

employed. These methods have been developed in conventional and matrix forms.

Here conventional methods are discussed.

suitable number of releases. The number of releases required is equal to

staticalindeterminacy s. Introduction of releases results in displacement

discontinuities at these releases under the externally applied loads. Pairs of unknown

biactions (forces and moments) are applied at these releases in order to restore the

continuity or compatibility of structure.

The computation of these unknown biactions involves solution of? linear

simultaneous equations. The number of these equations is equal to

staticalindeterminacy s. After the unknown biactions are computed all the internal

forces can be computed in the entires tructure using equations of equilibrium and

free bodies of members. The required displacements can also be computed using

methods of displacement computation.

Inflexibility methods inceunknowns are forces at the releases the method is

also called force method. Since computation of displacement is also required at

releases for imposing conditions of compatibility the method is also called

compatibility method. In computation of displacements use is made of flexibility

properties, hence, the method is also called flexibility method.

EQUILIBRIUM and COMPATABILITY CONDITIONS

Thethreeconditionsofequilibriumarethesumofhorizontalforces,verticalforcesa

ndmom ents at anyjoint should beequal to zero.

i.e.?H=0;?V=0;?M=0

Forces should be in equilibrium

i.e.?FX=0;?FY=0;?FZ=0 i.e.?MX=0;?MY=0;?MZ=0

Displacement of a structure should be compatable

The compatibility conditions for the supports can be given as

1.Roller Support ?V=0 2.Hinged Support ?V=0, ?H=0




3.Fixed Support ?V=0, ?H=0, ??=0

DETERMINATE AND INDETERMINATE STRUCTURAL SYSTEMS

If skeletal structure is subjected to gradually increasing loads, without

distorting the initial geometry of structure, that is, causing small displacements, the

structure is said to be stable. Dynamic loads and buckling or instability of structural

system are not considered here. Ifforthestable structure it is possible to find the

internal forces in all the members constituting the structure and supporting reactions

at all the supports provided from statically equations of equilibrium only, the

structure is said to be determinate.

If it is possible to determine all the support reactions from equations of

equilibrium alone the structure is said to be externally determinateelse externally

indeterminate. If structure is externally determinatebutitis not possible to determine

all internal forces then structure is said to be internally indeterminate. There

foreastructural system may be:

(1)Externally indeterminate but internally determinate

(2)Externally determinate but internally indeterminate

(3)Externally and internally indeterminate

(4)Externally and internally determinate

DETERMINATEVs INDETERMINATESTRUCTURES.

Determinate structures can be solving using conditions of equilibrium alone

(?H=0;?V=0 ;?M=0). No other conditions are required.

Indeterminate structures cannot be solved using conditions of equilibrium

because (?H?0; ?V?0;?M?0). Additional conditions are required for solving such

structures. Usually matrix methods are adopted.

INDETERMINACYOF STRUCTURAL SYSTEM

The indeterminacy of a structure is measured as statically (?s) or kinematical

(?k) Indeterminacy.




?s= P (M - N + 1) - r = PR- r ?k= P (N - 1) + r - s+?k= PM -c P = 6 for space frames

subjected to general loading

P = 3 for plane frames subjected to inplane or normal to plane loading. N = Numberof

nodes in structural system.

M=Numberofmembersofcompletelystiffstructurewhichincludesfoundationas singly

connected system of members.

Incompletely stiff structure thereisnorelease present. In singly connected

system of rigid foundation members there is only one route between any two

points in which tracks are not retraced. The system is considered comprising of

closed rings or loops.

R = Number of loops or rings in completely stiff structure. r = Number of releases

in the system.

c = Number of constraints in the system. R = (M - N + 1)

For plane and space trusses ?sr educes to:?s=M- (NDOF)N+ P

M= Number of members in completely stifftruss.

P = 6 and 3 for space and plane truss respectively

N= Number of nodes in truss.

NDOF = Degrees of freedomat node which is 2 for plane truss and 3 for space truss.

For space truss?s=M- 3N+ 6

For plane truss?s= M- 2 N+ 3

Test for static indeterminacy of structural system

If ?s> 0 Structure is statically indeterminate

If ?s= 0 Structure is statically determinate

and if?s<0 Structure is a mechanism.

It may be noted that structure may be mechanism even if ?s >0 if thereleases

are present in such away so as to cause collapse as mechanism. The situation of

mechanism is unacceptable.

Statically Indeterminacy




It is difference of the unknown forces (internal forces plusexternal reactions)

and the equations of equilibrium.

Kinematic Indeterminacy

It is the number of possible relative displacement softhenodes in the directions

of stress resultants.

PRIMARY STRUCTURE

A structure formed by the removing the excess or redundant restraints from an

indeterminate structure making it statically determinate is called primary structure.

This is required for solving indeterminate structures by flexibility matrix method.

Indeterminate structure Primary Structure

ANALYSIS OF INDETERMINATE STRUCTURES :BEAMS

1Introduction

Solve statically indeterminate beams of degree more than one.

To solve the problem in matrix notation.

To compute reactions at all the supports.

To compute internal resisting bending moment at any section of the

continuous beam.

Beams which are statically indeterminate to first degree, were considered. If the

structure is statically indeterminate to a degree more than one, then the approach

presented in the force method is suitable.




Problem 1.1

Calculate the support reactions in the continuous beam ABC due to loading as shown

in Fig.1.1 Assume EI to be constant throughout.

Select two reactions vise, at B(R1 ) and C(R2 ) as redundant, since the given beam

is statically indeterminate to second degree. In this case the primary structure is a

cantilever beam AC. The primary structure with a given loading is shown in Fig. 1.2

In the present case, the deflections (? L)1 and (? L) 2 of the released structure at B

and C can be readily calculated by moment-area method. Thus

(? L) 1 = ? 819.16 / EI

(? L) 2 = ? 2311.875/ EI (1)

For the present problem the flexibility matrix is,

a11= 125/3EI ,a21= 625/6EI

a12= 625/6EI , a22 = 1000/3EI (2)

In the actual problem the displacements at B and Care zero. Thus the

compatibility conditions for the problem may be written as, a11 R1+ a12 R2 + (? L)

1 = 0

a21 R1+ a22 R2+ (? L) 2 = 0(3)




Substituting the value of E and I in the above equation,

R1 = 10.609 KN and R2 = 3.620 KN

Using equations of static equilibrium, R3 = 0.771 KN m and R4 = ?0.755KN m

Problem 1.2

A Fixed beam AB of constant flexural rigidity is shown in Fig.1.3 The beam is

subjected to auniform distributed load of w moment M=wL2 kN.m. Draw Shear force

and bending moment diagrams by force method.

Fig 1.3 Fixed Beam with R1 and R2 as Redundant

Select vertical reaction (R1)and the support moment(R2) at B as the redundant.

The primary structure in this case is acantilever beam which could be obtained by

releasing the redundant R1 andR2.

The R1 is assumed to positive in the upward direction and R2 is assumed to be

positive in the counterclockwise direction. Now, calculate deflection at B duetoonly

applied loading. Let ( L ) be the transverse deflection at1 B and( L 2 bethe slope




at B due to external loading. The positive directions of the selected redundant are

shown in Fig.8.3b.




The deflection(? L1)and(? L2)of the released structure can be evaluated from unit

load method. Thus,

(? L1) =wL4/8EI - 3wL4/8EI = ?wL4/2EI - (1)

(? L2) = wL3/6EI - wL3 /2EI = ? 2wL3/3EI --- -(2)

The negative sign indicates that ( L )is downwards and rotation( is 1 L2) clockwise.

Problem 1.3.

A continuous beam ABC is carrying a uniformly distributed load of 1 kN/m in

addition to a concentrated load of 10kN as shown in Fig.7.5a, Draw bending moment

and shear force diagram. Assume EI to be constant for all members.




It is observed that the continuous beam is statically indeterminate to first degree.

Choose the reaction at B,RBy as the redundant. The primary structure is a simply

supported beam as shown in Fig.1.11. Now, compute the deflection at B, in the

released structure due to uniformly distributed load and concentrated load. This is

accomplished by unit load method. Thus,

In thenextstep, apply a unit load at B in the direction of

RBy(upwards)and

Calculate the deflection at B of the following structure. Thus(seeFig.7.5c),




Now, deflection at B in the primary structure due to redundant RB is,

In the actual structure, the deflection at B is zero. Hence, the compatibility equation

may be written as

L+ B=0(4)

The other two reactions are calculated by static equilibrium equations (videFig.

1.13)

RA =7.8125kN

RB =2.8125kN




UNIT V STIFFNESS METHOD

INTRODUCTION

The given indeterminate structure is first made kinematic ally determinate by

introducing constraints atthenodes. The required number of constraints is equal to

degrees of freedom at the nodes that is kinematic indeterminacy ?k. The kinematic

ally determinate structure comprises of fixed ended members, hence, all nodal

displacements are zero. These results in stress resultant discontinuities at these

nodes under the action of applied loads or in other words the clamped joints are not

in equilibrium.

Inorder to restore the equilibrium of stress resultants at the nodes the nodes

are imparted suitable unknown displacements. The number of simultaneous

equations represen ting joint equilibrium of forces is equal to kinematic

indeterminacy ?k. Solution of these equations gives unknown nodal displacements.

Using stiffness properties of members the memberend forces are computed and

hence the internal forces through out the structure.

Since nodal displacements are unknowns, the method is also called

displacement method. Since equilibrium conditions are applied at the joints the

method is also called equilibrium method. Since stiffness properties of members

are used the method is also called stiffness method.

In the displacement method of analysis the equilibrium equations are written

by expressing the unknown joint displacements in terms of loads by using load-

displacement relations. The unknown joint displacements (the degrees of freedom

of the structure) are calculated by solving equilibrium equations. The slope -

deflection and moment - distribution methods were extensively used before the high

speed computing era. After the revolution in computer industry, only direct stiffness

method is used.

PROPERTIES OFTHESTIFFNESS MATRIX

The properties of the stiffness matrix are:

It is asymmetric matrix

The sum of elements in any column must be equal to zero.

It is an unstable element therefore the determinant is equal to zero.




ELEMENT AND GLOBAL STIFFNESS MATRICES

Local co ordinates

In the analysis for convenience we fix the element coordinates coincident with

the member axis called element (or) local coordinates (coordinates defined along the

individual member axis )

Global co ordinates

It is normally necessary to define a coordinate system dealing with the entire

structure is called system on global coordinates (Common coordinate system dealing

with the entire structure)

Transformation matrix

The connectivity matrix which relates the internal forces Q and the external

forces R is known as the force transformation matrix. Writing it in a matrix form,

{Q} =[b]{R}

Where Q=member force matrix/vector, b=force transformation matrix R =

external force/load matrix/ vector

ANALYSIS OF CONTINUOUS BEAMS




ANALYSIS OF PIN JOINTED PLANE FRAMES

An introduction to the stiffness method was given in the previous Page. The

basic principles involved in the analysis of beams, trusses were discussed. The

problems were solved with hand computation by the direct application of the basic

principles. The procedure discussed in the session (page)

though enlighteningarenot suitable for computer programming. It is

necessary to keesphand computation to a minimum while implementing this

procedure on the computer.




In this session a formal approach has been discussed which may be

readily programmed on a computer. In this less on the direct stiffness method as

applied to planar truss structure is discussed.

Planetrusses are made up of short thin members inter connected a thin gesto form

triangulated patterns. Ahinge connection can only transmit forces from one member

to another member but not the moment. For analysis purpose, the truss is loaded at

the joints. Hence, a truss member is subjected to only axial forces and the forces

remain constant along the length of the member. The forces in the member at its two

ends must be of the same magnitude but actin the opposite directions for equilibrium

as shown in Fig.2.8




STIFFNESS MATRIX METHOD

1. What are the basic unknowns in stiffness matrix method?

In the stiffness matrix method nodal displacements are treated as the basic

unknowns for the solution of indeterminate structures.

2. Define stiffness coefficient kij.

Stiffness coefficient 'kij' is defined as the force developed at joint 'i' due to

unit displacement at joint 'j'while all other joints are fixed.

3. What is the basic aim of the stiffness method?

The aim of the stiffness method is to evaluate the values of generalized coor

dinates 'r' knowing the structure stiffness matrix 'k' and nodal loads 'R' through

the structure equilibrium equation.

{R} =[K]{r}




4. What is the displacement transformation matrix?

The connectivity matrix which relates the internal displacement 'q' and the

external displacement 'r' is known as the displacement transformation matrix 'a'.

{q} =[a]{r}

5. How are the basic equations of stiffness matrix obtained?

The basic equations of stiffness matrix are obtained as:

Equilibrium forces

Compatibility of displacements

Force displacement relationships

6. What is the equilibrium condition used in the stiffness method?

The external loads and the internal member forces must be in equilibrium at the

nodal points.

7. What is meant by generalized coordinates?

For specifying a configuration of a system, a certain minimum no of indepen dent

coordinates are necessary. The least no of independent coordinates that are

needed to specify the configuration is known as generalized coordinates.

8. What is the compatibility condition used in the flexibility method?

The deformed elements fit together at nodal points.

9. Write about the force displacement relationship.

The relationship of each element must satisfy the stress-strain relationship of the

element material.

10. Write the element stiffness for a truss element.

The element stiffness matrix for a truss element is given by

11. Write the element stiffness matrix for a beam element.

The element stiffness matrix for a beam element is given by




12. Compare flexibility method and stiffness method.

Flexibility matrix method

The redundant forces are treated as basic unknowns.

The number of equations involved is equal to the degree of static indete

rminacy of the structure.

The method is the generalization of consistent deformation

method. Different procedures are used for determinate and indeterminate

structures

Stiffness matrix method

The joint displacements are treated as basic unknowns

The number of displacements involved is equal to the no of degrees of

freedom of the structure

The method is the generalization of the slope deflection method.

The same procedure is used for both determinate and indeterminate

structures.

13. Is it possible to develop the flexibility matrix for an unstable structure?

In order to develop the flexibility matrix for a structure, it has to be stable and

determinate.

14. What is the relation between flexibility and stiffness matrix?

The element stiffness matrix 'k' is the inverse of the element flexibility

matrix 'f' and is given by f=1/k or k =1/f.

15. What are the type of structtures that can be solved using stiffness matrix

method?

Structures such as simply supported, fixed beams and portal frames can be solved

using stiffness matrix method.

16. Give the formula for the size of the Global stiffness matrix.




The size of the global stiffness matrix (GSM) =No: of nodes x Degrees of free

dom per node.

17. List the properties of the stiffness matrix

The properties of the stiffness matrix are:

It is a symmetric matrix

The sum of elements in any column must be equal to zero.

It is an unstable element there fore the determinant is equal to zero.

18Why is the stiffness matrix method also called equilibrium method or

displacement method?

Stiffness method is based on the superposition of displacements and hence is also

known as the dispalcement method. And since it leads to the equilibrium

equations the method is also known as equilibrium method.

19 If the flexibility matrix is given as

20 Write the n stiffness matrix for a 2D beam element. The stiffness matrix for a 2

D beam element is given by




CE8502 STRUCTURAL ANALYSIS I UNITI STRAIN ENERGY …

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