Ugural - Stresses in Plates and Shells
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STRESSES IN PLATES AND SHELLS
STRESSES IN PLATES AND SHELLS
A. C. Ugural,
Ph.D.
Profe."sol' and Chairmall M l!l '/wllic:al Eugim"'ring DeplIrtmem Faide igh Dickillson University
McGraw-Hili Book CompanyNew York St. Lou is San Francisco Auckl~md Bogota Hamburg Johannes burg London Madrid Mexico Montreal N('w Delhi Panama Paris Sito Paulo Singapoft: Sydney Tokyo Toronto
This book was set in Times Roman. The editor was Frank J. Cerra; the production supervisor was Donna Piligra. Fairfield Graphics was printer and binder.
STRESSES IN PLATES AND SHELLS Copyright 1981 by McGraw~HiIl, Inc. All rights reserved. Printed ill th!! Ullitt!d States of America. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by means, electronic, mechanical, photocopying, recording, or otherwise, with om the prior written permission of the publisher. 1234567890FGFG8987654321
Library of Congress Cataloging in Publication Data
Ugura), A C Stresses in plates and shells.
Includes bibliographical references and index. I. Plates (Engineering) 2, Shells (Engineering) 3. Strains and stresses, l. Title. TA660.P6U39 624.1'776 80-13927 ISBN 0-07-065730-0
CONTENTS
Preface List of Symbols 1 Elements of Plate Bending Theory1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10 Introduction General Behavior of Plates Strain~Cllrvature Relations Stresses and Stress Resultants Variation of Stress within a Plate The Governing Equation for Denection of Plates Boundary Conditions Methods for Solution of Plate Deflections Strain Energy Methods Mechanical Properties and Behavior of Materials Problems
ix
xiii1 I
2
4610
12 14 16
20 222527 27
2 Circular Plates2.1 lot roduction
2.22.3 2.4
Basic Relations in Polar CoordinatesThe AsixYlTImetricai Bending Uniformly Loaded Circular Plates
2731 32 36
2.5 2.6 2.7 2.8 2.9 2.10 2.11 2.12
Effect of Shear on the Plate Deflection Circular Plates under a Concentrated Load at Its Center Annular Plates with Simply Supported Outer Edges Deflection and Stress by Superposition The Ritz Method Applied to Circular Plates on Elastic Foundation Asymmetrical Bending of Circular Plates Deflection by the Reciprocity Theorem Circular Plates of Variable Thickness under Nonuniform Load Problems
3740 41 42
444647 55v
\'i
CO~TI':--;
rs59
33.1 3.2 13 3.4 3.516 3.7
Rectangular PlatesIntroduction Navicr's Solution for Simply Supported Rectangular Plates Simply Supported Rectangular Plates under Varioll~ Loadings Levy's Solution for Rectangular Plates Levy's Method Applied to Nonuniformly Loaded Rectangular Pl c" and outer and inner edges
are determined from the following conditions at the
w=oM,= 0
111,=0
(r = a) (I' = b)
(e)
Upon substitution of the constants into Eqs. (d), the following expression for the plate deflection is obtainedIV =
5?e!"b 1(14D1
__ '~~) [--2 +-"-__ 2(1 + v)(/'
,b'(I- ---
b'
ln~]a r 2b' 1 + v b rl - In -- In a --I _b 2 1 -v a(2.29)
+- In - + a 2 a2 a
r'
It is observed that if the radius of the hole becomes infinitesimally small. b' In (bla) vanishes. and Eg_ (2.29), by letting Ql = Pl2nb. reduces to Eq_ (2..25), as expected.
2.8 DEFLECTION AND STRESS BY SUPERPOSITIONThe integration procedures discussed in the foregoing sections for determining the elastic deflection and stress of loaded plates are generally applicable to other cases of plates_ It is noted. however, that the solutions to numerous problems with simple loadings are already available_ For complex configurations of loads therefore, the method of superposition may be used to good advantage to simplify the analysis_ The method is illustrated for the case of the annular plate shown in Fig_ 2.6a. The plate is simply supported along its outer edge and is subjected to aPo
jUJIDHliiJJlli(aJ(b J
(c)
Figure 2.6
42
STRESsES 11'\ PLATES A:-":D SHELl.S
Table 2.3 Variously loaded annular platesUniform load: wma~ = k!(p~)aol/Et3), "m," = k 2(p(){//t l ) Concentrated load: ~~'m~~ = k 3 (Pa 2 /EtI ), a,nu = k 4 {Plt 2 )
A.
OtHer edge supported
alb1.5
k
,
k, 0.976 1.44
Fd---..-j.b U""i i---- u ~.B. Inner and outer t-'dges clamped
~U;
Pn
I
-JI'
2.03.0 4.0 5.0
0.414 0.664 0.824 0.830
1.882.08 2.190.273 0.71 1.54 2.23 2.80 k
0.8130.0062 0.0329 0.110 0.179 0.234
1.52.0 3.0 4.0 5.0
-'lUJ %'i IP1m.ti ~ . I .'c.Inner and oute-I"edges clamped
alb 1.5
k,0.0064 0.0237 0.062 0.092 0.114
,
1
FJ~I I
~
2.0 3.0 4.0 5.0
0.22 0.405 0.703 0.933 1.13
uniformly distributed load Po at its surface. Shown in Fig. 2.6b is a circular plate under a uniform load Po. Figure 2.6c is an annular plate carrying along its inner edge a shear force per unit circumferential length Po b/2 and a radial bending moment, defined by Eq. (2.18), 1'0(3 + v)(a' - b')j16. The solutions for each of the latter two cases are known from Sees. 2.4 and 2.7. Hence, the deflection and stress at allY point of the plate in Fig. 2.6a can be found by the supe~position of the results at that point for the cases indicated in Fig. 2.6b and 2.6c (Prob. 2.14). Employing similar procedures, annular plates with various load and edge conditions may be treated. Table 2.3 provides only the final results for several examples. 7 .' In all cases v. is taken as v = 0.3. Design calculations are often facilitated by this type of compilation.
2.9 THE RITZ METHOD APPLIED TO CIRCULAR PLATES ON ELASTIC FOUNDATIONIn the problems discussed thus far, support was provided at the plate edges, and the plate was assumed to undergo no deflection at these supports. We now consider the case of a plate supportedcontinllollsly along its bottom surface by a foundation, itself assumed to experience elastic deformation. The foundation reaction forces will be taken to be linearly proportional to the plate deflection at
CIRCULAR Pl.ATES
43
any point. i.e., 'vI.:.. I-Jere l\' is the plate dc1kctioll and k is a constant, termed as the modulus (~r the .!cJ/lI1dmiol1 or bedding COflstal1l or the foundation material, having the dimensions of force per unit surlacc area of plate per unit of deflection (e.g., Paim). The above assumption with respect to the nature of the support not only leads to equations amenable to solution, but approximates closely many real situations.' Examples of this type of plate include concrete slabs, bridge decks. floor structures, and airport runways. We shall apply the Ritz method (Sec. 1.9) to treat the bending of a circular plate of radius a resting freely on an elastic foundation and subjected to a center load P. In this case of axis ymmetrica I bending. the expression for strain energy given by Eq. (P2.1l) reduces toU = nD1
I. f(dd,." +-- - 2(1',.-l')dWd'W.l I'dI' -,- Id\l')2 - - -dr - , ' dr .0 dl'2 WI'
(2.30)
A solution can be assumed in the form of a series(a)
in which en are to be determined from the condition that the potential energy of the system in stable equilibrium is minimum. If we retain, for example, only the first two terms of Eq. (a)
n
(b)and the strain energy, from Eq. (2.30), is then[J, =
4clDrra 2 (1
+ v)
(e)
The strain energy owing to the de/ormation of the elastic foundation is determined as follows
o '0 The work done by the load is given by
U2 =
J
.21t .,"
I !kw r dr dll = trrk(c oa2
2
+ COC2a4 + !da 6 )
(d)
W=P'(w),"o=PeoThe potential enerlo'Y, IT = U,
(e)
+
[J 2 -
W, is thus
n
=
4clDrra 2 (1
+ 1') + j: (e5a' + c oc,a 4 + ~da6) - Pe o
Applying the minimizing condition, aIT/ilc, = 0, we find that= rr:a, [I
Co
+ (1/3) +P
3i~(l + V)/ka 4 j
(j)
44 STRESSES IN J>LATFS AND SHELLS
Theil by substituting Egs. center (,. = 0):>I'm" =
U)
into (b), we obtain the maximum del1ection at the
~1:a2 [I + (I/:3)+32~(l + v)/";;;; J
(2.31 )
An improved approximation results from retention of more terms of the series given by Eq. (a)
2,10 ASYMMETRICAL BENDING OF CIRCULAR PLATESIn the foregoing sections, our concern was with the circular plates loaded axisymmetrically. We now turn to asymmetrical bending. For analysis of deflection and stress we must obtain appropriate solutions of the governing differential equation (2.5). Consider the case of a clamped circular plate of radius a and subjected to a linearly varying or hydrostatic loading represented byI' = Po
+ 1'1 ~. cos 0a
,.
(a)
as shown in Fig. 2.7. The boundary conditions are
w=o
-
ow a,.
=0
(r = a)
(b)
where W = Hlp + W h . The particular solution corresponding to Po, referring to Sec. 2.4, is 4 w~ = l' o r /64D. For the linear portion of the loading,1/ Plr5'cos 0 w =A--_P a
\
Introduction of the above into Eg. (2.5) yields A = 1/192D. We thus haveWp
1'01'4 = 64D
1'1'" cos 0
+
192aD
(c)
or
"'~-o+~-
f a LFigure 2.7
cmCULAR Pl.ATES
45
It is noted that the general method of obtaining the particular solution or Eq. (2.5), given in Prob. 2.20, follows a procedure identical with that described in Sec. 3.4 (Levy's solution for rectangular plates). The homogeneous solution IV" will be symmetrical in e; thusr~ in Eq. (2.7) vanishes. Owing to the nature of p and wp' we take only the terms of series (2.8) containing the function /0 and /,. Tbe deflection w" and its derivatives (or slope, moment, and shear) must be finite at tbe center (r = 0). It follows that Bo = Do = B, = D, = 0 in expressions for f~ andr,. Hence,(d)
The conditions (b) combined with Eqs. (e) and (d) yield two equations in the four unknown constants
Since the term in each pair of parentheses is independent of cos 0, a solution exists for all 0 provided that
poa 2 64D- + Ao + Coa = 0
4
which upon solution, leads topoa AD = - ----64D4
2p,a C,= ---192D
PIa'A, = '192D
(e)
The deflection is theretore"' =
64D
Po (a2 _ r2)' +..J'.!.....!: (a 2 _ ,,2)2 cos 0I92Da
(2.32)
The center displacement is
(f)
46
STRESSES I~ PLATES A0:D SHELLS
We observe that. when the loading is uniform, 1', = 0 and Eq. (1.32) reduces to Eg. (2.14) as expected. The expressions for the bending and twisting moments are, h'mn Egs. (2.32) and (2.2)
M,M
=
Po 1', -;; 16 [a2( 1 + v) ' + \')) - 48 [1'3 (5 + v)- ar(3 -+ \') ] cos - 1'-(3l', [(l2(1 + v) - 1"(116'
iJ
=II
+ - - '. [,,3 (5v + 1) - ar(3v + 'lv)] 48 a
I)] cos 0
(2.33)
M,.!)
= _I~V p, ra( I
-
~;: ) sin (I
The case of the simply supported plate under hydrostatic loading can be treated in a similar way,
2.11 DEFLECTION BY THE RECIPROCITY THEOREMPresented in this section is a practical approach for computation of the cenler deflection of a circular plate with symmetrical edge conditions under asymmetricalor nonuniform loading. The method utilizes the reciprocity theorem together with expressions for deflection ofaxisymmetricaUy bent plates. Consider, for example, the forces P, and P 2 acting at the center and at r (any 0) of a circular plate with simply supported edge (Fig. 2.8). According to the reciprocity theorem,' due to E. Betti and Lord Rayleigh, we may write:(a)
That is, the work done by P, owing to displacement w" due to 1'" is equal to the work done by P, owing to displacement IVI2 due to 1',. For the sake of simplicity let P, = 1, IV" = w" and IV" = "'(r). The deflection at the center IV, of a circular plate with a nonuniform loading p(r, IJ) but symmetric boundary conditions may therefore be determined through application of Eg. (a) as followsw, =
I I p(r, (1)",(1')1' dr diJ'0 '0
,211:
,0
(2.34)
Clearly, ",(r) is the deflection at r due to a unit force at the center. In the cases of fixed and simply supported plates, Iv(r) is given by the expressions obtained by setting l' = 1 in Eqs. (2.22) and (2.25), respectively.
,
...........
-
... --)
C'IRClJl.AR PLA1TS
47
To illustrate the application of the approach. reconsider the bending of the plate described in Sec. 2.10. Upon substituting 1'(1'. Ii) = Po + PI (ria) cos () and Eg. (2.22) into Eg. (2.34), setting I' = 1 and integrating. we havelV,
=
'1~61Drr
.0
("
0
((Po + p, ': cos 0)(2/'2 In " a
I'
+ ,,'
_,,2)1' d" dO =
poa~64D
The above is identical with the value given by Eg. (I) of the preceding section.
2.12 CIRCULAR PLATES OF VARIABLE THICKNESS UNDER NONUNIFORM LOADIn this section we discuss an approximate method for computing stresses and deflections in solid or anllllal circular plates of variable thickness, subjected to arbitrary lateral loading 9 Except for the requirement ofaxisymmetry, no special restrictions are placed on the manner in which either the thickness or the lateral loading vary with the radial coordinate. Several applications immediately come to mind: turbine disks, clutches, and pistons of reciprocating machinery. Consider a circular plate (Fig. 2.9a), and the division of the plate into small (finite) ring segments, as in Fig. 2.9b. Note that radial lengths of the segments need not be equal, but that the thickness is taken as constant for each. For each element defined as in Fig. 2.9b. the development of Sec. 23 applies. In order to accommodate the substitution of a series of constant-thickness elements for the original structure of varying thickness, it is necessary to match slopes and moments at the boundary between adjacent segments. The boundary conditions are handled in the usual manner. As the method treats the plate as a collection of constant-thickness disks, it is unnecessary to determine an analytical expression for thickness as a function of radius. Prior to illustrating the technique by means of a numerical example, the general calculation procedure is outlined. A knowledge of the derivation of the basic relationships [Sec. 2.12(b)) is not essential in applying the method.
r I ---,--,-_.J~(a)
--r,
_(b)
Figure 2.9
48
STRESSES IN PLATES AND SHELLS
(11) Calculation ProcedureThe expressions developed in Sec. 2.12(b) may be so arranged as to facilitate the calculation process. Consider, with this end in mind, a plate subdivided into a number of annular elements with the applied lateral loading on each element denoted Q, the average load on that element (Fig. 2.9). We shan apply thenotation(J.
=t(1 -- fI')
p, =
~
i+v
In p -
i-v s;r- (I -
(r)
.,
Et'D= 12(! _ 1'2)
1- p' ),' = -------
4(1
+
p2 + '---- In p v) 2(1 - v)
(2.35)
fi. =
- - In p
I+v
4"
+ - - (I _ p')8"p2
I-v
,"Q =
I - p2 I + p' ---g;- + -g;;- In p
2, = 4(1 + -~) -2(T::::~) In p
1- p2
In addition, for each joint between adjacent elements,n'/ _.
tn+.! ~ I 3
3
tn
(2.36)
The notation thus introduced is next applied in the determination of the following quantities. Bending moments The change in bending moment in proceeding from the inner edge to the outer edge of any element may be ascertained by rearranging Eqs. (2.44) as follows:
(2.37) Ll.M, = -aIM"~ - M,,) + PoQ At the inner edge, the moments are either given or assumed. The outer moments acting on an element are then (2.38) M,o = M" + Ll.Mo Similarly, the moment increments corresponding to the interface between adjacent elements arcLl.M, = 0(2.39)It now follows that the moments at the inner dege of the next element are found fromMrQ=Mrl+/lM,
Ll.M, = a(M" - M,,) + PJJ
(M"),,+l = (M,,),
(M o,)" + 1 = (M o,)"
+ Ll.M,
(2.40)
CIRCULAR PLATES
49
Table 2.4Assumed values at inner boundary of plate (A and 8 are any arbitrary values B of 0) Given inner houndary conditions Particular solutionHOJnogl!1l0US
solution M;r.=OAI~lI = B M~u= B
AIr"Clamped Solid plate
A-1;,,,
= AI",
M~'r= A
iH;" = A M;J,,"'" I'M:" M~" = A M~" = M;Q
Al~" = I'M;"M~([
M;,,=B = M;,.
When Egs. (2.37) to (2.40) are applied successively, beginning with the innermost element, the moments at any intermediate edge may be found in terms of the moment at the inner boundary of the disk. Boundary conditions The following steps are taken to satisfy the boundary conditions at the inner and outer edges of the disk:Step 1. Apply Egs. (2.37) to (2.40) to obtain a particular solution (denoted by a single prime). Begin the calculations with the appropriate inner boundary values specified in Table 2.4. Step 2. Repeat step 1 with Q = 0 to obtain a homogeneous solution (denoted by a double prime). Seep 3. Superimpose the values found in steps 1 and 2 to obtain the general solution:Mr=M~+kM~
M!I= M~
+ kM~
(2.41 )
The constant k is calculated as indicated in Table 2.5 from the given boundary conditions at the outer edge of the disk.
Table 2.5Given outer boundary conditions
Formula for kin Eq. (2.41)
M'bClamped
k k
~=', 0) cos ,,0 dO n- .. r.
1 .'
(II
~
O. 1. ... )
R,,('") ~ is expressed in the general form: 6
1 ."
1t._~
I
p(t, 0) sin "U dO
(II
~
1,2 ....)
I\'JI =
Fo(r)
+ L [Fn(r) cos nO + Gn(r) sin nOn""J
(Pl.20)
Here /-'0(1'), F,,(r), /2, and to satisfy Eg. (1.18). We proceed with the description of the method by assuming that two opposite sides of the rectangular plate at x = 0 and x = a are simply supported as shown in Fig. 3.4. In this case, Eg. (b) appears as;>
IVII
=
, ,m (}') t.... {,m 0:= 1
sin - ... :II
m1TY
(3.13)
Eg. p.13) satisfies the conditions (1.26b) along the x edges. To complete the solution, we must now apply to \V the boundary conditions on the fl.\'O arbitrary sides at y = 1>/2. Substituting Eg. (3.13) into Eg. (1.18) yields . -"
In order that this eguation be valid for any .\,14' ')m _ 2()2d2{'~' + (".!~) 41m = 0 1I1n ..c ..
dy4
"dy-.
II
68
snO:::SSES I!\" PLATES A~D SHELLS
Tile gelleral solution of the above is (see Proll. 3.9)
(3.14)or by employing trigonometric identities.fm.=
Am Sinh
.
mITy'-0-
+ Bm cosh '--::,--' + (mY smhl'
mrcy,.
mITy a
+ Dm.V cosh --.::... a
mIT\'
(3.15)
The homogeneous solution is thereforeH'h
=
III~ 1.
L:
.x:' (
. mrry Am smh - -
a
+ Bm
nm.y cosh _ . .......(/
+C
III.
. mn:v v smh --' a
+D
/U
I1m y ). nmx J' cosh - - Sill ... ~. a' a
(3.16)
where .,.1 n" Bm ) CII!> and Dill are constants, to be determined later' for specified cases. It is observed that (he boundary conditions (1.26/ along the edges x = 0 and x = a are satisfied if the particular solution is expressed by the following single Fourier seriesIV p =
m...:;l
mnx l: k",(y) Sill -a. That is
where
(/Similarly, for plate 2, setting Po = 0, the deflection expressed in terms of the coordinates X2 and y, and for a different set of constants isW2
=
L:m=t.3, ..
00
[Em sinh
A.mX2
+ Fill cosh
AmX2
The boundary conditions for plate I and plate 2 are represented as follows, respectively:
w, =0
2 wl --2 l
8 OX
=0wm~1.3,
(Xl = 0)(e)
w,andW2
=0
-D--;- =
a2 }VIilx,
L
Mm sin }'mY..
(x, = a)
=0= 0
-=0ax~
82 w2
(X2
= a)(d)
IV2
-D-= ax~
a2 H.'2
",m= 1. 3 ....
L
/1..1 m sin Am Y
(X2
= 0)
We thus have eight equations (e) and (d) containing nine unknowns Am . ... ,H." Mm. The required additional equation is obtained by expressing the condi-
tion that the slopes must be the same for each panel at the middle support. This continuity requirement is expressed
RECTANGl'LAR PL.'!'TI:S
85
Introducing Eqs. (3.44) and (3.45) into the above we have
Application of the edge conditions (e) and (d) to Eqs. (3.44) and (3.45) leads to values for the constants as follows:1 coth - - _.. 4 = -- Ia - - . }'m" ItV!m m ." '1' f) I sll1h I'ma .... 1./11
(3.47)
-
-- ...
csch Ama [Mm -- .. _D 2}.m~-.
+ 21'0 (-1 + cosh Ima) X41;
'J'm
and
Mm a E = - --'~ (1 "- coth 2 ) ~ "In
2)'IJI D
I'm
=
0 (3.48)
H = - .-- ... cothIII
A1m 2)'m D
I
,"III
a
Having Eqs. (3.47) and (3.48) available we obtain. from Eq. (3.46). the moment coefficients M",. Equations (3.44) and (3.45) then give the dellection of the continuous plate from which moments and stresses can also be computed. The foregoing approach may be extended to include the case of long rectangular plates with many supports. subjected to loading which is symmetric in y. In so doing, an equation similar to that of the three-moment equation of continuous beams is obtained.' It is noted that there are situations where the intermediate beams arc relatively flexible compared with the flexibility of the plate. The deflections and rotations of the plate arc not then taken as zero along the supports, but are functions of the hending and torsional stitTnesses of the supporting beams'" The design methods used in connect ion with continuous plates ut ilize the solutions derived in the foregoing sections and a number of approximations 1i
86
STRESSES IN PLATES A~D SHELLS
3.10 RECTANGUl.AR PLATES SUPPORTED BY INTER MEDIA TE COLUMNSIn this section we consider the bending of a thin continuous plate over many columns. To attain a simplified expression for the lateral deflection it is assumed that: the plate is suhjected to a uniform load Po. the column cross sections are so small that their reactions on the plate are regarded as point loads, the columns arc equally spaced in mutually perpendicular directions, and the dimensions of the plate are large as compared with the column spacing. The loregoing set of assumptions enables one to assume that the bending in all panels. away !i'om the boundary of the plate, is the same. We can therelore restrict our attention to the bending of one panel alone, and consider it as a unifomlly loaded rectangular plate (a x b) supported at the corners by the columns. The origin of coordinates is placecl at the center of a panel shown by the shaded area in Fig. 3.12a. Clearly, the maximum dellection occurs at the center of the panel. A solution can be obtained utilizing Levy's approach (Sec. 3.4). Accordingly, the deflection may be expressed as a combination of that associated with a strip with uniform load and fixed ends y = b/2 and that associated with a rectangular plate. That is,IV
=
wp
+ w" = 3~4D +
P b* (
1w
4V2)2
bi(
+ Bo1nny mny"---
L:111""'2,4. ...
Bm cosh_- + Em _ . sinh a a
mn y )
a
COS - - -
mn.\:
a
(3.49)
I>
-...J
'. 1--.2c
.
lal
(h) Figure 3.12
RECTA!'..:Gl"L,\R PI.ATf'S
87
where Bu , Bm, and Em are constants of WI!' yet to he determined, The aboH" satisfies the boundary conditions for the rectangular panel along the x edges,-- = 0
thv
ax
(a)
and Eqs. (1.17) and (1.18). For purposes of simplifying the analysis, the support forces are regarded as acting over short (infinitesimal) line segments, between x = a(2 - e and x = 1/2 + c or 2c (Fig. 3.12,,). The plate loading is transmitted to the columns by the vertical shear forces. From the conditions of symmetry of the bent panel we are led to conclude that the slope in the direction of the normal to the boundary and the vertical shear force vanish everywhere on the edges of the panel with the exception of the corner points. Thus,
Qy=O
(o. = - (I - v)
.
8Dk a2h 2 X)' .
At the center (x
= 0, y = 0), the moments arelvl x = 4Dk ( a2
I\' ) + b2 .(d)\' +-, ) a-
1 lvl .Y = 4Dk ( -, h-
At the ends of the minor and major axes, we have
My8vDk lvl = - - - . x b2
~
8vDk -- (;2 SDk
(x= a,y=O)(4.12)(x =0,)' =
lvl = - - -2 y
b
b)
where k is given in Eg. (e), It is observed from Egs, (4,12) that the maximum moment occurs at the eXlremity t?f the minor axis. For a = b the solution derived in this section reduces to the results obtained for a circular plate with clampcd edge (Sec, 2.4). In the extreme case, a b b, Egs, (4.10) and the second of (d) appear asW
= -----------mox
Po(2b )4 384D
(e)
Equations (e) correspond to the deflection and moment at the center of a fixed-end strip of span 2b under uniformly distributed load.
102 STI~ESSES IX PLATES A;\1D SHELL";
Table 4.1(//hW .
1.0
1.51.26 0.2220.321
2.0
3.0
4.0
A-(~ . (1/p,.. b2 )
(Et 3 /P ob-l)
M,. (l1J"h')
0.70 0.206 0.206
1.580.210 0.379
1.88 O.lSg0.433
2.020.184 0.465
Uniformly loaded elliptic plate with simply supported edge The boundary conditions are now given by\1.1
=
0
(for
a' +
X'
v' i,2
= 1
)
where, as before, 11 is a normal to the plate boundary. After routine hut somewhat more lengthy calculations than in the case of a fixed plate, expressions for deflection and hending moments may be ohtained. Table 4.1 provides, for \. = 0.3, some final numerical results for the deflection and bending moments at the center of the plate."
4.5 STRESS CONCENTRATION AROUND HOLES IN A PLATEThe bending theory of plates applies only to plates of constant cross-sectional area. If the cross section changes gradually, reasonahly accurate results can also be expected. On the other hand, where abrupt variation in the cross section takes place, ordinary bending theory cannot predict the high values of stress which actually exist. The condition referred to occurs in plates with holes, notches, or fillets. In some instances, the stresses in these regions can be analyzed by applying special theories which take into account the shear deformation. The solutions more often involve considerable difficulty. However, it is more usual to rely upon experimental methods. The finite element approach (Sec. 5.7) is also very efficient for this purpose. The ratio k, of the actual maximum stress to the nominal stress in the minimum section is defined as the stress concentration factor. Hence,
(4.13)where 0'" is obtained by applying the familiar flexure formula for beams. The technical literature contains extensive information on stress concentration factors presented in the form of graphs and tables 7 14 It is convenient to examine the stress concentration at the edge of a small hole in a very large plate (Example 4.1). The results obtained for this case have been proven applicable to plates of any shape. Example 4.1 A large, thin plate containing a circular hole of radius a and width 2a, small as compared with the overall dimensions, is subjected to
PLATES OF VARJO[}S m;O\lETRIC'AL FOR~fS
103
J
--'-~r
,'}
-,--
~2.6
\
\
\\ - \
~:~- ~' \':,
L0;=J! ..,rI!
{JmJ~
x
8
\
\ , Mo0 ,...................
__ ?,' )/ -? Z!
,,,,+=) MO
2.0 1.81.7
..
.......,Ie, ---'::
-------
/kb
,o
,2
,
-2a
3
4
5
6
7
Figure 4.4
pure bending ililx = Mo and My = 0 (Fig. 4.4). Determine the stress distribution around the hole.SOLUTION
We place the origin of coordinates at the center of the hole. In the absence of any hole, one has, from polar form of Eq. (1.30)\V!
M ,2 = - ._."--,-) [1 4D(1 I'
- \' + (1 + v) Gas 20]
(a)
At r = a the radial moment and effective transverse force owing to the M 0, obtained by means of Eqs. (a), (22), and (2.4), are:
M" =I",
1M 0(1 + cos 20)I
(b) cos 20
= M o
a
When a hole of radius a is drilled through the plate, we superimpose on the original state of stress an additional state of stress so that: the combined radial moments and forces equal to zero at /' = a; the superimposed stresses ollly are zero at /' = ro. Guided by the expression for 11'" we assume the additional deflection of the form'!2 2
11',
-
= - ~o~ [ A In /' M a
21)
+
(
B - C"' Q
/'
)
cos 2() J
(e)
104
,sTRESSES IN PLATES AND SHELLS
thus satislying both conditions stated above and Oq. (2.6). The moment andcflcctivc transverse force corresponding to Eq. (c) at r = a are determined as
follows
M" = -M,,{(1 - viA + [4vB - 6(1 - v)C] cos 2e}
v"
= 1 Mo[2(3 - v)B
a
+ 6(1 - v)C] cos 20
(d)
The conditions along the boundary of the hole are represented by
v,.1 +
~~2 =
0
(,. =
0)
(e)
Substitution of Eqs. (b) and (d) into Eqs. (e) leads to the values of the constants. The final deflection expression (w, + IV,) is then obtained. It follows that the tangential moment and shear force at the periphery of the circle (r = a) areM, = M0 [I -
2~1 : \~) cos 20 ](4.14)
4Mo . Qo = . sm 20 (3 + v)a
From Eqs. (4.14), it is evident that the maximum values of stress resultants occur at e = 1[/2 and 0 = 1[/4, respectively:M 0. mal( -M 0 3 + v-
5 + 31'
(4.15)
The stress concentration factor is thereforek _~+ 31' b - 3 + \'
(4.16)
where reduction in the cross-sectional area due to small hole has been neglected. For I' = 1/3, Eq. (4.16) yields kh = 1.80. Upon introducing Eqs. (4.15) into Eqs. (1.12) and (1.20) and setting v = 1/3, the maximum stresses are found to be O'm" = IO,8M,,/t' and T",,, = L8Af 0 fat. The ratio of these stresses is
(f)When, for example, the hole width is equal to the plate thickness the above quotient is 1/3, For 2a < t, the ratio '",,,/O'm,, is larger. The intluence of shear upon the plate deformation is thus pronounced. In Fig. 4.4, the dashed line represents a plot of the true stress concentration factor k" obtained from Reisner's theory which takes the shear deformation into account." In both cases \' is taken to be 1/3. It is observed from the graphs that the stress concentration factor determined using the thin plate theory is quite crude,
PLATES OF VARIOL'S GEOMETRICAL f-ORMS
105
The case of plates \vitb noncircular holes may also be treated hy thL' usc of the bending theory, as illustrated in the example. For the reasons cited above, the solutions offer unacceptably poor accuracy.
PROBLEMSSl.'t.'S. 4.1 to 4.3
4.1 A simply supported wing panel in the form of an isosceles right triangle is subjected to a uniform load of intensity Prj (Fig. 4.1). Determine, by retaining only the first term of the serics solution, at point .4(x "" a/4, y = aI4): (a) the deflection; (h) the bending moments. 4.2 Consider an equilateral triangular plate with simply supported edges under a uniform moment Mo along its boundary (Fig. 4.2). Taking v = 1/3. find: (a) the twisting moment on the side AC; (b) Ihe concentrated reactions at the corners. Sees. 4.4 10 4.54.3 A manhole cover consists of a cast-iron elliptical plate 1.0 m long, 0.5 m wide, and 0.02 m thick.
Determine the maximum uniform loading the plate can carry when the allowable stress is limited to 20 MPa. Usc \' = 0.3. Assume that (a) the plate is simply supported at the edge and (0) the plate is built in at the edge.4.4 Consider two plates, one having a circular shape of radius c, the other an elliptical shape with semiaxes a = 2c and 0 = c, both clamped CIt the edge and under a uniform load of intensity Po.
Determine the ratio of the maximum deflection and the maximum bending moment for the elliptical plate to those for the circular plate. Take v = 0.3. 4.5 An elliptical plate built in at the edge is subjected to a linearly varying pressure Jl = Pox (Fig. 4.3). Given un expression for the deflection of the formw
~ E-~ 36
(b)
Figure- 5.5
Example 5.4 Consider the case of a uniformly loaded rectangular plate with two adjacent edges simply supported, the third edge free, and the fourth edge built in (Fig. 5.5a). Use the finite difference method with h = a/4 to obtain w at various points.SOLUTION The domain is divided into twelve squares of sides a/4 (Fig. 5.5b). At the nodes of the simply supported and the built-in edges, we have w = O. The finite difference formula corresponding to Eq. (1.17) is applied to all interior nodes and to nodes 16 and 24 on the free edge, where the detlections are to be found. This application gives eight equations involving thirty-six nodes, indicated in the mesh on the figure. The boundary conditions for each edge, replaced by the finite difference forms, are listed in Table 5.2. It is seen
Table 5.2Edge(s) A. Clamped atConditionsw = 0: Ws = W6 = ~\'7 = "'s = \\'9 = 0 (lw!cs = 0: WI~ = WI' WI'; = w2 , \\'15 = IV 3 W l6 = W 4 Continuity of zero curvature (B2wj(JX 1 = 0) along x axis: Ws= --w10=0w
y=oB. Simply supported at x = 0 and y = 3a/4
= 0: "'12 = "'20 = "'n =
W 2 1'(W:lI
iPw/iJ... 2 =0:(!2 W /
oy 2 = 0: "'21 = -W.l3' -W34' W 2 3 = ~ W3~' W~ ... = -W36 Continuity of zero curvature (c 2 w/ax2 = 0) along edge y = 3a/4: w 30 = ~W32 = 0Vx = 0, at nodes 16 and 24, respectively:WJ8 "'26
"'
13
= ~WII'
= \~'29 =
W:H1
=
W 31
= 0
= \\'22 =
~W19
c.
Free at x=a
~ W l4
+ (\.
~ 2)(1\'8 -
W25
+
W 23 -
\VIO)
+ (6 -I- (6 -
2v)(wl~
At x\VIS
= 0, at nodes+W + H
~
W 22
+ (v 2)(W I5W 2 .;)
-I-
W3$
~
W 17 -
\I'n)
- W 17 ) = 0 2l')(W23 ~ W2S) = 0
16 and 24, respectively;~ 2(1
+ W l7 + \I(W9 +v(w l6
W.u
+ "'.~l) -
2(1
+ V)W I6 = 0 + v}w 14 = 0
PLAT!' BF~DI;-';(j BY NUl\1LQUCAI. \I[TI-IODS
'19
from the tahle that all edge conditions are represented by a total of tH'eJ/(Yeiilh,. expressions. We thus have a total or thirty-six independent equations in terms of the thirty-six values of the deflections corresponding to each node. Writing the finite difference expressions for Eg. (U7) by applying the operator V' of Table 5.[ at nodes 13, 14. 15, 16,21,22,23, and 24, and inserting the lV'S given in cases A and B of Table 5.2, we have eight simultaneous equations containing twelve unknown values of 11' at nodal points 13 through 18 and 21 through 26. Case C of Table 5.2 provides four more equations in terms of the same unknown values of w. The resuiting twelve independent expressions are represented in the following matrix form (for \' = 0.3):20
-821
1
0
0
0
-H2
-81 0
-821
0
-81 2
-821
-s0 2
1
0 0
2 .. R
02
11
0()
0
02
0
00 18
-80 0 02
t0 0 00
-82[)
0 0 0 0 0
-8 2 0 -I
0 02
-8()
2 0 -8 19
-82
-8
1
-8
82 5.4
-81
t9-8-1.7 5.40
0 00
-1.7 1 0
8 0 0 -2.6 0.3
-5.4 1.7 1 0
1 00
0
00 0
0
0 -I 0 0 10
1
0 0 0 0 t 0 -8 1 19 -8 1.7 0 0 0 - 5.4 0.) 0 0 --2.6 1 0
0 2 0 0
0
j"'W"15W 17W IS
Ir 14
wI(,
\= pl>h~D
Ii
1
W21(
t1 0 0 0 0
\\.'24
""}\\123W 25
"'26
The above equations are solvedIV13 W I6
yield\V15
= O.25819N= 0.51951N
IVI4
= 0.38943N= 0.79839N
= 0.45037N= 1.07433N
W I7W22
"'IS11'23
"''' = 0.30383N"'24
= 0.46598N
= 0.63989N
IV" = 0.96226N
= 0.54558N IV 26 = 2.27753N
where N = Po 114/D and h = a/4. From these values, we determine the bending moment at any node in the plate as illustrated in the previous examples.
5.4 PLATES WITH CURVED BOUNDARIESWe now treat the bending problems of simply supported plates having curved or irregular boundaries. Dividing a portion of such a boundary into a square mesh (Fig.5.6a) shows that the V 2 operator (Table 5.1) does not apply to point 0 because of the unequal lengths of arms 01. 02, 03, and 04. When at least one of the arms is not equal to 11, the operator pattern is referred to as an irregular star. There are available various approaches to such situations. The one that will be discussed assumes that it is required to develop an irregular star using the actual boundary points, rather than those falling" outside," associated with the continued regular mesh.
120
STHESSES Ii\" PLATES A:\"D SHELLS
..--Boundary
D
2 ,:j' 0.1'1' . OX}' . _zm' {Q \, = i Q~n ~ = IQ. Q. Q. QZ)' Q . QBJ')' Q QOXm' QOym)\ (5 ,~38)
" i
I
132 STRESSES
f!\! PLATES AND SHULS
3
'i',~"~~
J -3030 -15 0 -300
0. 15
HI(l
t5
--15Il
0 0 0 0 0 0 0\)
60 30 0 -30 15 030
20
Symmetric
6 -84 --6
0
,--60 846
-2
"(l (l
0
0()
-150IS
[k.d60 -30 0 -60- 30
=
-6 -84
-8-6(l
0(I
200 30 10\}
60()
-,"6
6 84-6 -6 -84
6 6030
15I)
lO\}
" 00
0
2000
0
0
0
0
"6-6where
2\}
"0
-6\l
-66
, 84 2 " " -, " -84 " -R\1(J
Symmdr:c .
6
,)
.. 6I)
6
:';4(,.. (1
-
-6 6
-,
Il
,"
IR)
=
[[') [ll) 10) 10]1 (0) I') [0] (0) [0] 10) ['I [OJ[OJ [0) [OJ
['I =
[I 0]0 0 ; 0 00(/
['I
:;;
'"
136
STRESSFS [.....; I'LAT~S A1\;D SHELLS
-II.
~
t--:I\
il
,/
Sliding edges
te) Nodal force and
(a) Actual plate
(b) Substitutt! plate
dis plact'lll ents
Figure 5.15
Example 5.8 Consider a square plate of sides II with two opposite edges and x = a simply supported and the remaining edges clamped x= (Fig. 5.15(/). Compute maximum value of 1\' if the plate is subjected to uniformly distributed load of intensity Po. Take a = 2 m and \' = 0.3. SOLUTION Symmetry in deflection dictates that only one quarter-plate need be analyzed. provided that slic/illy-edge conditions (1.28) are introduced along the lines of symmetry. The substitute plate is shown ill Fig. 5.151>, For the sake of simplicity in calculations, we employ only one element per quarter-plate. A concentrated load Po(1 x 1)/4 = 1'0/4 is assigned to node I. The boundary constraints permit only a lateral displacement \VI at node 1 and a rotation Ox2 at node 2 (Fig. 5.15c). Nodal force and the displacement matrices are
{QL = {Po/4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, O}{I},= {Wi' 0, 0, 0,
0_'2,0,0,0,0,0,0, OJ
The forcedisplacement relationship (5.26), together with the values of stiffness coefficients (Table 5.3), is then readily reduced to the form IPo!41
I
1
r'
=
lso7i-=o.9)(60 [
+ 60 + OJ
x 30 + 0.35 x 84) (0 - 30 + 0,35 x 6)
-H'J
(0-30+0-0.35 x 6))111'11 (0 + 20 + 0 + 035 x 8) 10x,I
From the above, we obtain= H'max= O.3617~E 3 .1
Po
The "exact" solution of thi' problem (see Example 3.7) is 0.3355J'oEi'. When the quarter-plate is divided into 4, 16, and 25 elements, the resultsart!, respectively:Wn1ax =
O.3512Po/Et.) O.3378Po/Et 3
H'max = 0.3397Po/(3Wn1llX =
It is apparent that the accuracy of the solution increases as the mesh is refined.
PROBLEMSSecs. 5.1 to 5.3 S.1 Verify thaI the effective (Fig. 5.1b), in the form~hear
o
forces arc represented,
tiS
finite difference approximations at point
Vx = - ;;-J3 [W9 -
.,
j)
11'"
-
2(3 -
V){W j
--
1\'3)
+ (2
- 1')(W, --
\\'(> -
\\'7
+ WIl)](PS.I)
Referring to Table 5.1. check the correctness orille reslllt for
I-~.CIt
5.2 Determine a finite difTercncc expression corresponding to \7 4 1\' = "ID (Fig. 5.1h), for a rectanglliar mesh. Take.1 __ = "and L1y = k. 5.3 Determine the /inite difference cquivalent of M ... rectangular mesh with .1.x = II and Ay == k.M,I' ,
a nodal point ()II
and Vx at a nodal point 0 (Fig. 5.lh) for
5.4 Calculate the maximum stress al the nodal point 22 for the plate shown in Fig. 5.5. 5.5 Swept wing of an aircraft is approximated by a simply supported skew plate subjected to uniform loading Po (Fig. P5.5). Determine the deflection w at the nodal points 1 through 5.
\,
,,
'Ia_:0.
r~--------"Figure P5.5 Figure P5.6
Free
-_.
II
~-
3
5.6 Consider a uniformly loaded plate with two opposite sides ~imply supported. the third side clamped, and the fourth side free (Fig. P5.6). Determine: (tI) the displacement II at the nodttl point~ 1 through 9; (IJ) the bending moments al nodal point 9.
138
STRESSES I~ PLr\TE:S AND SHELLS
"/
-Ita
"
I
T" J, .. ,
II"" _.
a
th,II
4'
Ii, ""
11
6
-
5
Figure PS.7
Figure PS.S
a/3
1.L,6
a/3
r
'.
~, _ _ _ = . t = J.i....-~ a ~-~_.
a ___ -1- 1
,-~
Figure PS.9
Figure P5.1 0
Figure PS.ll
Figure PS.12
PLATE HE:\Dl~G BY -"";t;MF:JUCAL METHODS
J39
Sees. 5.4
(0
5.6
5.7 through 5.12 Each of the \ariously shaped plaks ~howll in thL' figurcs is simply supported at ~
EI,
.~)"
C: Torsional rigidity of one ribI: Moment of inertia about neutral axis of a Tscction of width. s (shown as shaded) G~,.: Torsional tigidity of the plating E; Elastic modulus of thc plating
n.
Corrugated plate
Dx
=_.-...... _.... .,{
s
EI~
l20 _
v~)
D)"
~-"
1. H = .... _._ ......-
).
Et J
(/12.(1
+ I')
Dx ,'
""
0
where
I = O.Sh I 1 - .. _ _ .... _ ... 0.81 1 + 2.5(h/2s)J
'I
J
144
SntESSES 1;-.; PLATES A:-':D SHELLS
When it is /lot possihle. however. to determine the constants of (6.10) experimentally, one resorts to approximations deri\'ed using analytical techniques. The lalter approaches consist of constructing an orthotropic plate with elastic properties equal to the average properties of components of the original plate. 21 Such a plate is lern1(~d an eql/ivale1ll or t.ransformed orthotropic plate. For example, in the case of a plate reinforced by ribs, the bending stiffness of the ribs and the plating are combined and taken to be uniform across the replacement model. Subsequently, the constants of (6.10) are approximated with Eqs. (6.11) giving the rigidities. For reference purposes, Table 6.1 presents the rigidities for some commonly encountered cases. ll . 22 It is noted that when E~ = E~ = E (and hence l'x = l'y = 1'), Eqs. (6.3) become
Ex = )' =Consequently, D =D =_.Et' .... __x "
~l~~2~v
(6.12)
12(1
~
,,')
G
X)~
~
.-..-. __......
(' 24(1 + v)
H=--~- =/)
Et 3 12(1 - v')
(6.13 )
It follows that Eq. (6.9), as expected. reduces to that of an isotropic plate [Eq, (1.17)].
6.4 RECTANGULAR ORTHOTROPIC PLATESThe general procedure for the detemlination of rectangular orthotropic plates is identical with plates, We now apply Navier's method (Sec, 3.2) supported rectangular orthotropic plate under (Fig. 3.1). Introduction of Egs. (3.1) into (6.9) yields the deflection and stress in t hat employed for isotropic to treat the case of a simply a nonuniform load p(x, y)
m=II1"'1
I I I laoc00
("m+1[4mll
--'-4-
a
Dx + 2H ---~-2b2 + ~b'-4 Dy I
m21l 2n 4
li4n+)
I .
Innx .
Pmlr{SUl - - - Sin
a
-b = 0
111ry
Inasmuch as the above must be valid for all "and y. it follows that the terms in the brackets must be zero, leading toa
"'"
= . -...... -.~-.-.-.. -p"'''-.--.
(1I/ 4,,4/a4)Dx + 2H(m'n2,,4(a' b2) + (n""4/b 4)D,,
.
(6.14)
The expression of the platedeflection surface, by substitution of Eqs. (6.14) and (3.3) into (3.1b), is thereforeIV =
ai; m~, "~,
4
oc
00
I"
/."
'0'"
(;;"4,,4/a")D:'+ 2lf(m 2n 2,,4/a 2b') + 'F;4,,4/h4)D~xSill - - 8111--
p(x. y) sin (mn,,/a) sin (n"y/b) dx dy
. 1nnx . nnya
h
(6.15)
ORTfIOTROPIC PLATES
145
For an isotropic material, from Eqs. (6.[2) and (6.1 J), D, ~ D,. = H = D, and the above coincides with Eg. (3.5). In the particular case of a rectangular plate under a uniformly distributed load Po, referring to Sec. 3.3 we can readily obtain from Eq. (6.15):161'0W
=
"Teo'
~ ~ ;;'~,[(m4Ia4)D, + 2H(m'n';a'b'f+(;;"lb4jDy]
w
~
sin (mITx!a) sin (IlITyl")
(6.16)
When the material is isotropic the above reduces to Eq. (3.6). For example. if the plate is made of a reillforced COllacte, from Table 6.1, we have H = J DxD~. Based upon the notation
(a)Eq. (6.16) becomes\\' =
16po ~ ~ sin (lIl"xla) sin (IInYlb)
DIT 6
-;-
'i;'
mll(m' /ai + /1 2lhn'
(/1/,
II
= 1, 3, ... )
(b)
which is of the same form as Eq. (3,6). We are led to conclude that the center deflection of the reinforced concrete plate (a x h) having rigidities D." D,. is equal to that of an isotropic plate (lI, x 0,) of rigidity D. Having the expression for the deflection of the plate available, we can obtain the bending moments from Eqs. (6.5) and the stresses by applying the relationships of Sec. 1.4. Deflection of orthotropie rectangular plates is also determined by following the same basic procedures that were prescribed in Chaps. 3 and 5 for the Levy's solution and for the finite difference approach. Referring to Table 5.1 the pertinent coefficient pattern for finite difference expression of Eq. (6.9) of the orthotropic plate may readily be obtained. It is given in Fig. 6.1 for the case of evenly distributed nodes, i.e" for Ax = Ay = h.
211 r----t-.:-~4~(/~f;+~D~,~,1J---{ 211
2H )-----f-:-:;4;Z(I1-:-:,+~D[i)~.10----12Ff
Figure 6.1
146
STREssES
[1'\
PLATES AND SHELLS
, ,,:>;'.
.~ ".2I
,
.till = a.'
',
13
;
!
t
4
I IFignre 6.2
Example 6.1 A square orthotropic plate is subjected to a uniform loading of intensity Po. Assume the plate edges are clamped and parallel with the principal directions of orthotropy. Determine the defiection, using the finite differencc approach by dividing the domain into equal nets with h = a/4. Take D, = Do, D,. = O.5Do, If = 1.248D o , and ''x = v, = 0.3.SOLUTION For this case, the governing cxpression for del1ectioll, Eq. (6.9), appears
Po Do
(e)
Considerations of symmetry indicate that only one-quarter (shaded portion) of the plate need by analyzed (Fig. 6.2). The conditions that the slopes vanish at the edges are satisfied by numbering the nodes located outside the plate surface as shown in the figure (See Example 5.2). The values of ware zero on the boundary. Applying Fig. 6.1 at the nodes I, 2, 3, and 4, we obtain18.718 -8.859 .. 6.859 2.429 -17.718 20.718 4.859 -6.859 - 13.718 4.859 19.718 -8.859 9.718]I -13.718 J w2 i -17.718 ii' 11'3 ( 21.718W
II
=
E )
PLATFS l"':\~DFtt COMJlI:\'EJ) LATERAL AND DIRJ:CT LOADS
ISS
Ina,much '"
/i and fi' are sm,dl. sin /i '" /i " hl'ex and sin /1' " Ii', and hellce , cjJ alV ,,211' II "II +- dx = cc- +,-- dx [!X (IX ('Xl
Neglecting higher-order terms, Eg_ (b) is therefore -N,
dJ'~\~ -+ ex
(N x + ?~x dX)dY(~.\1' + ~~~ dX) ex ox('X
The z components of the shear forces N" on the x edges of the element are determined as follows. The slope of the deflection surface in the y direction on the x edges eguals ""'jay and awlc')' -+ (,,2 wlex ?y) dx_ The: directed component of the shear forces is thenN"'----:;-- dx dv . - rx oy .12 ( W
-+ ---'---- dx d)' (Ix cy
ONxOH>
An expression identical to the above is found for the z projection of shear forces N yx acting on the y edges: N)'x
ax oy -
;y2\-\> eN }>x i1W I ---d,dy+-------,xdy fly ax
For the forces in Figs_ 7.1 and 1.5, from
I
F, = 0, we thus have
iJQx -l};',
oQ) r2 w tJ2 w (32\-1.' + -CV -+ p + N \ --- + N } .-- + 1 N xYax a.v --,,-2 ox 8y2 ....
+ (ON x + i1N,,,)~\v + (o..l'jXl' + a~)')8W =ex0)'
ox
Ox
0)'
c7y
0
(c)
It is observed from Egs. (7.1) and (7.2) that the terms within the parentheses in the above expression vanish. As the direct forces do not result in any moment along the edges of the element, Egs. (b) and (e) of Sec. 1.5, and hence Egs. (1.l6), are unchanged. Introduction of Eqs. (1.16) into Eg. (e) yields02W + 2NXJ' -:--- ) 8x iJy
(7 .3)
Expressions (7.1), (7.2), and (7.3) are the governing differential equations for a thin plate, subjected to combined lateral and direct forces. It is observed that Eg. (1.17) is now replaced by Eg. (7.3) to determine the deflection surface of the plate. Either Navier's or Levy's method may be applied to obtain a solution.
Example 7.1 A rectangular plate with simply supported edges is subject to the action of combined uniform lateral load Po and uniform tension N (Fig. 7.2). Derive the eguation of the deflection surface.
156
STRESSES 1:-': PtA rES A:,\:D SHELLS
----
!-> -, ----(/------,.1~
_
~) j...--
! :::,i~'
13..~' ,\, ... ~I
'I .. - .. -J'-
.\"
I
:
t "
Figure 7.2
SOLUTION In this particular case, N x = N = constant and N,. = N x.\" = 0, and hence Eqs. (7.1) and (7.2) are identically satisfied. The lateral load 1'0 can be represented by (Sec. 3.3):I' = - i "TC
16po
I Im I Sill -_. .':Y)
I1lnx
.
mr)!
-._. S1l1 ._",
III
/I
11111
a
b
(111,
,,= I, 3, ... ). n11tx .mt)'
Inserting the above in Eq. (7.3), we have4 Dw
a,:;;r + - 3x 2 {!J~2 + t3y4 - 15 2~'\.2 =
..,
c4 w
a4 w
N alw
16poTC2D
~~
00
00
1m; SIn -~- SIn
1
b
(7.4)
The conditions at the simply supported edges, expressed by Eqs. (a) of Sec. 3.2, are satisfied by assuming a deflection of the form given by Eq. (3.lh). When this is introduced into Eq. (7.4), we obtain
am"
= _._6
""~ .!6.P--"-~'____1 N( na n Dmn r(111' +- +""a' b' D
n')'
m )'
(m,
11
=
1. 3, ... )
The deflection is thusW
= 16po6
nD
f f . .."Lfl ("",x/a) sin (:,,,Y/I>2.I'll/I
mll[(m~+!.~_~_)' +,,' b'
Upon comparison of Eqs. (3.6) and (7.5), we are led to conclude that the presence of a tensile (compressive) force decreases (increases) the plate deflection.
. N(-"-'-)' 1D na
(7.5)
7.3 COMPRESSION OF PLATES. BUCKLINGWhen a plate is compressed in its midplane, it becomes unstable and begins tohuckle at a certain critical value of the in-plane force. Buckling of plates is
qualitatively similar to column buckling.' However, a buckling analysis of the fonner case is not performed as readily as for the latter. Plate-buckling solutions using Eq. (7.3) usually involve considerable difficulty and subtlety,24 and the conditions that result in the lowest eigenvalue, or the actual buck lillY load, are not
PLATES Ui'I'd increases with increasing N. When N reaches the critical value given by Eq. (7.8), the denominator of Eq. (d) vanishes and w, grows without limit. This means that the plate buckles, as descrihed in the alternate manner in Sec. 7.3. To write the equation for a plate subjected to uniform tensile forces (Fig. 7.2), it is required only to change rite sign of N in the equations of the foregoing example. By following an approach similar to that described above, the deflection of an initially curved plate under simultaneous action of in-plane forces N" N)., and N", may also be readily obtained.
7.7 BENDING TO A CYLINDRICAL SURFACEAssume that a /ollg rectangular plate of uniform thickness t is bent into a cylindrical sUI}ace with its generating line parallel to the}' axis. For this case, \\' = w(x). and the goverl1il1g equatiol1for deflection, Eg. (7.3), reduces to
(7,31)When axial force N x is zero, the foregoing agrees with Eq. (c) of Example 1.1. Observe that the calculation of the plate deflection simplifies to the solution of Eq. (7.31), which is of the same fonn as the differential equation for deflection of beams under the action of lateral and axial forces. If plate edges are 110t Fee to move horizontally, a tension in the plate is
PLATES l!~OER COMBll'\'ED LATERAL AND DIRECT LOADS
169
\,-~ .... ~uTTL~->-c _-x .-/,-.. ,. // '\x
r--.~>__
{J(,
__ . ___ ,_--"
i Po ,. t------~- L -----~~t} poLIzFigure 7.9
produced depe/UliJlf} "1'0/1 the magnitude of lateral deflection w. The problem then becomes complicated. The tensile forces in the plate curry part of the lateral loading through membrane action. A typical case in which bending to a cylindrical surface occurs is illustrated below. Example 7.6 A rectangular plate, the length of which is large in comparison with its width, is subjected to a uniform loading of intensity Po. The longitudinal edges of the plate afe free to rotate but otherwise immovable. Determine the lateral deflection and stresses in the plate. SOLUTION A strip of unit width removed from a plate of this type will be in the same cond ition as a laterally and axially loaded beam or so-caJJed tie-rod (Fig. 7.9). The value of the axial tensile forces N x = N is such that horizontal movement of the edges is prevented. The bending moment at any section x of the strip is described by Af(x) = tPoLx ~ 1PoX2 ~ Nw In terms of this moment Eq. (7.31) may be writtenD ses at the faces of the plate, upon introduction of Eqs, (9.20) and (921b) into (9.8), are as follows:
a, =
(T,
= 2 (=:--,;'j (T, -
~E
T,J
(e)
The resultant stress in the original plate is obtained by addition of the stresses given by Egs, (b) and (e):
This is the result expected, Equation (92Ia) leads to the relationship
"'",,, =2'1 (7;
rxa 2
- 12)
(922)
for the maximum deflection of the original plate,
9.8 AXISYMMETRICALLY HEATED CIRCULAR PJ"ATESConsider the bending of an axisymmetrically heated circular plate having simply supported or clamped edge conditions and subjected to temperatures varying with the /' and z coordinates, T(/', z), such that the equivalent transverse load
194
STRESSES IN PLATES AND SHELLS
1'* = p*(r). The expressions for moments, Eqs. (:~.9). for the situation described hecome
M,.=
_D(d2~ + \~'.':) _d .. .. ' = --..... ------ ..-----...... ------- ......... ------ .. ------.----- ---
(a l
_
b I )l
_ 4.111[>2
In 2 (a/h)-
(/\~.~IJf({/l _
b2 ) tn (a/b)] + w~(Ifb2
{/2 -
211 z In (a/h)J
c.l
~"' --------~-(;:;i-~i,if-~-4~i-PIr1(~ibT---------
stress resultant at ,. = u. Given a temperature distribution T(r, z), the deflection and moment in a solid or annular plate with simplY supported or built-in edges can thus be obtained from Eqs. (9.26) and (9.23). Plates with other edge conditions can similarly be treated. Example 9.2 Redo the problem discussed in Example 9.1 using the relationships developed in this section.SOLUTION
As p* = 0, we have, from Eqs. (9.29), to row B of Table 9.3,=
w:;') =
0 for alln. Referring6111'= 1 3 '
2(1-=,;2)15
M:
where I'v1: is replaced by M*, because the thermal loading does not vary with radius. Equation (9.26) then yields))' =
~~1" 31
(a' _ ,.2)
This expression with Eg. (9.20) results in a solution of w, identical to that found in Example 9.1. Upon inserting Eqs. (9.21a) and (9.20) into Eqs. (9.23), we find that At, = M. = O. Applying Eqs. (9.8), we again obtain Or = a" = o.
196
STRESSES 1;-; PLATES AND SHELLS
(0 )
(b)
Figure 9.2
Example 9.3 Determine the dellcctions and stresses in a circular plate with clamped edge for the following cases: (a) the plate is stress-free at temperature and uniformly heated to a temperature (Fig. 9.2a); (b) the plate experiences a steady-state, linear temperature variation between its faces (Fig. 9.2b). Assume that the plate is free of stress at 0"(' and that T1 > T2 .
To
1.
SOLlTTION (a)
have p'
=
The uniform temperature differential is I1T = T, - To. We 0 and thus W~' = 0 for all /l. Row A of Table 9.3 then givesC1
=
C2
=
C3
=
('4-
=
0
Hence, Eq. (9.26) yields\\,=0
(a)
The thermal stress resultant isAi"
= a(I1T)
I
./.:2
0
dz
=0
. -/:2
Expressions (9.23) then lead to At, = M" = O. The magnitude of the bending stresses at the plate surfaces, from Eqs. (9.8), is therefore
a =r
(fu
0:(111') = --_. __ .1_ v
(9.30)
(1)) We now have I1T= (1; -1;)/2 (Fig. 9.2b). The displacement >1'= 0 asbefore. The thermal stress resultant At" is defined by Eq. (9.20) and the bending moments are found from .Eqs. (9.23):
M, = Mil = -12([~~ (T1 - 7~)By means of Eqs. (9.8), we haverJ,.
aEt 2
E~ = cr, = E~(I1T) = ..... __. - (T1 - T2) -.-------------
I-v
2(1-1')
(9.31 )
for the magnitude of the bending stress at the upper and the lower faces of the plate. It is noted that inasmuch as p* = 0, solutions (9.30) and (9.31) may readily be found without any calculation by the use of analogy between heated and isothermal clamped plates.
THER\IAL STRESSES '" PLATES
197
PHom.EMSSees. 9.1 109.7 9.1 An aluminum airplane wing panel is assumed to be stress~free at 20C. Arter a time al cruising speed. the temperatures on the heated (upper) ;md the cooled (lower) surfaces are 54 and 24"('. respectivdy. Compute the thermal slI'ess resultants for a linear temperature transition. Assume that the panel edges arc clamped. Use E =- 70 GPa. ~ = 23.2 X 10""6 per c, \' =- 0.3. and t = 6 mm. 9.2 Derive expressions for the bending moments in a simply supponed square- plate of sides temperature distribution linearly varying over the thickness olily.(I
with
9.3 Determinc the maximum deflection II' and th\.! maximum moment Alx in a simply supported square plate or sides (I. subject to a temperature field Az J , where A is a conManl. Retain only the first term of the series solutions. 9.4 Consider a builtjll edge, square plate of sides a, under an equivalent tralls\'ers~ tbermalloading p*(\:. y) (Fig. 3,13). Employ the analogy between isothermal and thermal problems and apply the Ritz method. taking m = /I = 1 to obtain an t'xpression for [he deflection surface It,.". 9.5 A square plate is simply supported at .'(=0 and.\' = (/. The remaining edges at y = a/2 arc clamped (Fig.3.9a). The plate is subjected to a temperature distribution A.\'2.:\ where A is a constant. Derive an approach to the evaluation of the center displacement II', Use the analogy with the is{)tiJermaJ solutions given in Sees. 3.6 and 3.7. Sec. 9.8 9.6 Verify there~uJts
given in Cases A and R of Table 9.3.
9.7 Determine the deflection wand the stress aT at the center of a simply supported ell'cular plate experiencing a temperature lleld A;:r 3 , where A is a constant 9.8 Redo Prob. 9.7 for a plate clamped on all edges. 9.9 Determine the displacement II' at r = 2a of a hollow plate having inner (r = (I) and outer (r = 30) edges built in. Assume a femperature distribution 8Z,/"2, where B is a constant.
CHAPTER
TENMEMBRANE STRESSES IN SHELLS
10.1 GENERAL BEHAVIOR AND COMMON THEORIES OF SHELLSUntil now, our concern was with the analysis of thin flat plates. We now extend the discussion to curved sw:face stmcll/res termed thill shells. Examples of shells include pressure vessels, airplane wings, pipes. the exterior of rockets, missiles, automobile tires, incandescent lamps, caps, roof domes, factory or car sheds. and a variety of containers. Each of these has walls that arc curved. Inasmuch as a curved plate can be viewed as a portion of a shell, the general equations for thin shells are also applicable to curved plates. We shall limit our treatment (except Sees. 11.11 and 11.12) to shells of cqpstant .thickness.smallincomparison with the other two dimensions. As in the treatment of plates. the plane bisec'iIng the shell thickness is called the midsur/ace. To describe the shape of a shell, we need only know the geometry of the midsurface and the thickness of the shell at each point. Shells of technical significance are often defined as thil1 when the ratio of thickness t to radius curvature r is equal to or less than 1/20. For thin shells of practical importance. this ratio may be 1/1000 or smaller. The analysis of shell structures often embraces two distinct. commonly applied theories. The tirst of these, the membralletheol'Y, usually applies to a rather large part of the entire shell. A' membrane. either flat or curved~ is identified as a body of the same shape as a plate or shell, but incapable of conveymg moments or shear forces. In other words, a membrane is a two198
MEMBRANE STRf:SSES 1:-': SHELLS
199
dimensional analog of a llcxible string with the exception that it can rcsi~l compression. Tile second, the bel/dillg theory or gel/era! theory. includes the etTects of bending. Thus it penn its the treatment of discontinuities in the stress distribution taking place in a limited region in the vicinity of a load or structural discontinuity. However, information relative to shell membrane stresses is usually of much greater practical signHicance than the knowledge oftbe bending stresses. The former are also far simpier to calculate. For thin shells having no abrupt changes in thickness, slope, or curvature. the meridional stresses are uniform throughout the wall thickness. The bending theory generally comprises a membrane solution. corrected in those areas in which discontinuity etTects are pronounced. The goal is thus not the improvement of the membrane solution, but rather the analysis of stresses and strains owing to the edge forces or concentrated loadings. which cannot be accomplished by the membrane theory only. It is important to note that membrane forces are independent of bending' and are completely defined by the conditions of static equilibrium. As no material properties are used in the derivation of these forces, the membrane theory applies to all shells made of any material (e.g., metaL fabric. reinforced concrete, sandwich shell, soap film, gridwork shell, and plywood). Various relationships developed for bending theory in Chaps. 11 to 13 arc, however. restricted to homogeneous, elastic, isotropic shells. The basic kinematic assumptions associated with the deformation of a thin shell as used in small-detlection analysis are as follows:
(1) The ratio of shell thickness to radius of curvature of the midsurface is small in compari.on with unity. (2) Deflections are small compared with shell thickness. (3) Plane sections through a shell taken normal to the midsurface. remain plane and become normal to the de{imlled midsurface after the shell is subjected to bending. This hypothesis implies that strains I" and ';'" are negligible. Nonnal strain, B" owing to transverse loading may also be omitted. (4) The z-directed normal stress. rI" is neglible.In this chapter, we consider only shells and loadings for which bending stresses are negligibly small. Applications are also presented of the governing equations of the membrane theory to specific practical cases.
10.2 LOAD RESISTANCE ACTION OF A SHELLThe common defonnational behavior of beams, thin plates, and thin shells is illustrated by the unified set of assumptions (Secs. 1.2 and 10.1). The loadcarrying mechanisms of these members do not resemble one another, however. That the Ioad,resisting action of a shell differs from that of other structural forms is underscored by noting the extraordinary capacity of an eggshell or an
200
STRESSES IN PLATES A:-.:n SHELLS
FigUrl' 10.1
electric light bulb to withstand normal forces. this despite their thinness and fragility. (A hen's egg has a radius along the axis of revolution I' = 20 mm and a thickness t = 0.4 mm; thus tfr = 1/50.) The above behavior contrasts markedly with similar materials in plate or beam configurations under lateral loading. A shell, being curved, can develop in-plane forces (thrusts) to form the primary resistance action in addition to those forces and moments existing in a plate or beam. To describe the phenomenon, consider a part of a spherical shell of radius I' and thickness t, subjected to a uniform pressure of intensity p (Fig. 10.1). The condition that the sum of vertical forces be zero is expressed: 2m' o N sin ' The governing equations for these forces are derived fronl two equilibrium con ditions. Figures 10.30 and 1O.3h show two different views of the element .4BCD cut from the shell of Fig. 10.2. Prescribed by the condition of symmetry, the membrane forces and the loading display no variation with O. The externally applied forces per unit surface area are represented by the components I>,. and 1', in the y and z directions, respectively. Description of the z equilibrium requires that the z components of the loading as well as of the forces acting on each edge of the element be considered. The z-directed distributed load carried on the surface area of the element is
p,l'or, d8 dThe force acting on the top edge of the element equals N. 1'0 dll. Neglecting higher terms, the force on the bottom edge is also N. 1'0 dO. The z-directed component at each edge is then N~l'o dO sin (d!2). This force is nearly equal to N ~ /'0 dO d/2, yielding the following magnitude of the resultant for both edges:N~ro
dO d
As the cross-sectional area along each of the two sides of the element is 1', ii, the force on these areas is Nor, d. The resultant in the direction of the radius of the parallel plane for both such forces is Nol', d . dO, producing the iollowing component in the z direction Nor, d dl! sin For the forces considered above, from
L F, =
0, we have(a)
N.ro + Nor, sin + 1',"01', = 0
F
(a)
(b)
(Cl
Figure 10,3
204
STRESSES IN PLATES AND SHELLS
This c~pressioll may he converted to simpler form by.' dividing hy fOr] and replacing I'(l b.y f2 sin 1;. By so doing. Olle of the basic rdations lor the axisymmetrically loaded shell is found as followsN,.I' j
Nfl +-.=/'2
-p,
(10.111)
The equilibrium of forces in the direction of the meridional tangent. that is. in the .I' direction is expressed
,iip (N"I',,) dq, dO -
d
N"r j dq, dli cos cf' + p,.rj dq,
1'0
dO = 0
(b)
The first term represents the sum of normal forces acting on edges !lC and RD, while the third term is the loading component. The second term ofEq. (b) is the component in the y direction of the radial resultant force Nol'j (/q, dO acting on faces .1R and CD. Dividing Eq. (b) by dO dq, the equation of equilibrium orthe y d ireeted forces is now
d iJ;(N"l'o) - N,l'j cos q,
=
-p,.r, ro
(102)
It is noted that an equation of equilibrium which can be used instead of Eq. (10.2) follows readily by isolating part of the shell intercepted by angle q, (Fig. 1O.3c). Here force F represents the res"lram of all externallvadillY applied to this free body. Recall that from conditions of symmetry, lorces N,; are constant around the edge. Equilibrium of the vertical forces is therefore described by 2"1',, N", sin + F = 0 and it follows thatN~=
F ----.--sm q,
2"1',,
(10.11
We verify below that Eq. (lO.1b) is an alternative form orEq. (10,2). Substitution of Nfl from Eq. (IO.1a) into (10.2) and multiplication of the resulting expression by sin q, leads Lo
,:q, (l'oN,,) sin q, + "oN,; cos q,
= -I'II',p,COS
(psin q, -
I' j
l'21',sin' q,
Clearly, the lell-hand side of the foregoing equation may be written
-ri1> ("oN. sm q, =
d
')
d . , i (1', N. sm-
"
.!.)
and force N", determined through integration: N" = - --..t1'2S1112.
:~-- ...... [1'1'
1
r2(p, cos q, + p sin q,) sin q, drj! + cJY
(e)
Here constant c represents the effects of the loads which may be applied to a shell element (Fig. 10.3c). Thus, introduction of 2nc = F, p, = p, = 0, and = 1'2 sin q, into Eq. (e), results in the value of N", defined by Eq. (lO.1b).
1'0
~IE~.fllHA;-.r~ STRESSES l~ ~l-IELLS
205
Equations (10.1) arc sufficient to determine the so-called hoopji)l'c(' N'I and the meridional jc.Jrce N", from which the stresses arc readily determined. Ncgativl.> algebraic results indicate compressive stresses. Because of their freedom of motion in the: direction, for the axisymmetrically loaded shells of revolution considered, strains are produced such as to assure consistency with the field of stress and compatibility with one another." The action cited demonstrates the basic difference between the problem of a shell membrane and one of plane stress, In the latter case, a compatibility equation is required. However. it is clear that when a shell is subject to the actioll of concentrated surface loadings or is constrained at its boundaries, membrane theory cannot everywhere fulfill the conditions of deformation. The complete solution is obtained only by application of bending theory.
10-5 SOME TYPICAL CASES OF SHELLS OF REVOLUTIONThe membrane stresses in any particular axisymmetrically loaded shell in the form of a surface of revolution may be determined from the governing expressions of equilibrium developed in the preceding section. Treated in the following paragraphs are several common structural members. Spherical shell For spherical shells one can set the mean radius a = 1', = 1',. Then Eqs. (10.1) appear in the form
F N = - - - -... 2na sin'
(IOJ)
1>
The simplest case is that of a spherical shell subjected to constant internal gas press"re 1', like a balloon. We now have I' = -1'" 1> = 90, and F = - na'p. Inasmuch as any section through the center results in the identical freebody, N. = No = N. The stress, from Eqs. (10.3), is therefore(T
=-
N
t
I'a 21
(10.4)
where t is the thickness of the shell. The expansion of the sphere, applying Hook's law, is thena pa' b = - (N - vN) = --- (1 - ,,) , Et 2EI
(10.5)
Conical shell (Fig. 10.4) In this typical case, angle 1> is a constant (r, = (0) and can no longer serve as a coordinate on the meridian. Instead we introduce
206
STRESSFS 1;-"; PLATES Al'1J SHELLS
Figure 10...
coordinate s, the distance of a point of the midsurface, usually meaSllred Fan! the !'erlex, along the generator. Accordingly, the length of a meridional element ds = 1'1 d, Hence-~.
- r -_. d - 1 ds
d
d
(a)
Also,1'0
=
seas
1',
=
scot
(b)
These relationships, when introduced into Egs. (to.2) and (10.1,,), lead to1S
-I (N,s) - No = -p,.S
d
(lO,6a, h).
N n = "'p.s cot =
PJo --.-sm
where 1'" is the mean radius at the base. Clearly, load components p,. and p. are in the s and radial directions, respectively. The sum of Egs. (10.6) yields .
d'S (N,s)
d
= -
(Py + p, cot )s
The meridional force, upon integration of the above expression, isN, = I s.I' (p,. + p, cot )s lis
(lO,7)
An alternate form of Eg. (I0.6a) may be obtained from Egs. (to,lb) and (b). The membrane forces are thenNs.
= - .....--~ ...:-...--.2"1'0
F
sm
(lO,8)
-
sin
..
p, ro~.-'-
It is observed that gIven an external load distribution, hoop and meridional
stresses can be computed illdepellliently.
~IE\IRRA.;-','E STRESSES I'\; SHELLS
207
Circular cylindrical shell To ohtain the stress resultants in a circular cylindricalshell, one can begin with the cOile equations. setting radius a = 1'" = constant. Hence Eqs. (10.8) become
q) =
rr/2, p=
= p,. and mean
N.=N.= _._oS
x
F 2na
(10.9)
in which x is measured in the axial direction.For a closed-end cylindrical vessel under constal1[ illtel'llal pressure, p
and F =stresses:
-no'".
= - Pr Equations (10.9) then yield the following axial and hoop
a=x
po
2t
(III
= .--
"a t
(10.10)
From Hooke's law, the extension of the radius of the cylinder under the action of the stresses given above is (10.11 ) Solutions of various other cases of practical significance may be obtained by employing a procedure similar to that described in the foregoing paragraphs, as demonstrated in Examples 10.1 through 10.6.
Example 10.1 Consider a simply supported covered lIlal'ket dome of radius a and thickness t, carrying only its own weight p per unit area. (a) Determine the stresses, for a dome of half-spherical geometry (Fig. 1O.5a). (b) Assume that the hemispherical dome is constructed of 70-mm-thick concrete of unit weight 23 kN/m' and span 2a = 56 m. Apply the maximum principal stress
u
.l!.~ +--~-,--------
a,
0+---poI-"'::~
-)f'------,--++
5150'
90
___
21
(a)
Ib)
Figure 10.5
208
STRESSES Il\! PLATES AND SHELLS
theory to evaluate the shell's ability to re~isi failure by fracture. The ultimate compressive strength or crushing strength or concrete 0u' = 21 MPa, and
E = 20 GPa. Also check the possibility of local buckling. (e) Determine the stresses in a dome which is a truncated half sphere (Fig. IO.5b).SOLUTIO)\;
The components o/the dome weight are
I,x = 0
P.I'
=
p sin
(p
[1, =
I' cos r/>
(10.12)
(a) Referring to Fig. 10.5a. the weight of that part of the dome subtended by q, is
._,
F=
I p. 211a sin q, . a dq, ='0
211a'p(1 - cos q,)
(c)
Introduction into Eqs. (10.3) of p, and F given by Eqs. (10.12) and (e), and division of the results by I yields the membrane stresses:up
- - t( I + cos 1
At the equator (base of the shell) 1', = 1>' /a and /', = a and Eqs. (10.20) appear as(j
pa = ._--
2t
It is observed that the hoop stress 17" becomes compressive for a2 > 2b'. Clearly, the meridian stresses 17" are always tensile. A ratio a/b = I, the case of a sphere, yields the lowest stress.
Example 10.5 Analyze the membrane stresses in a thin metal cOl1tailwJ' of conical shape, supported from the top. Consider two specific case,: (a) the shell is subjected to an internal pressure p; (b) the shell is filled with a liquid of specific weight I (Fig, 10,9).SOLUTION
(a) For this case, p, = - 1', = (n/2) + CI. and F = - pm'!.. Expre"ions (10.8), after dividing by the thickne" t, then become(J~
.
=
pro . 21 cos CI.
(J
pro.-~~-
(J -
t
cos
IX
(10.21)
\lHIB\{.-\~,f-: STRESSES 10.: SHELLS
213
Figure 10.9
(f,) According to the familiar laws of hydrostatics, the pressure at any point in the shell equals the weight coIUIn11 u11it cross-sectional area oJ the liquid at that point. At any arbitrary level y. the pressure is therefore:
or a
or
I' = --1', = r(h - y)
(g)
Employing becomes
I'D =
Y tan
IX,
the second of Eqs. (10.8), after division by t,;'(i! - y)y tan ~ - . - ... t cos~
(1
a-
-
."~'--
(1O.22a)
Differentiating with respect to y and equating (0 zero reveals that the maximum value of the above stress occurs at y = "/2 and is given by
a fI, max
yh 1 tan 1".1. = . 4t- cos~
The load is equal to the weight of the liquid of volume acOdb. That is,
F = -rc;')"(h - Y + 1v) tan'
IX
introducing this value into the first of Egs. (10.8), and dividing the resulting expression by t leads to(J
= -'-.. _.. _ . - - s
,,(h . 2y/3) tan 2t cos
C(
Cf.
(1O.22b)Ct,
The maximum value of this stress. (J.~, mil" = 3h 2 }' tan rJ.j16t cos at y = 3h/4.
occurs
Examl.)e 10.6 Delemline the membrane forces in a spherical storage tank filled with liquid of specific weight I', and supported on a cylindrical pipe (Fig. 10.10).SOLUTION
The loading is expressed
211
sn~r:ssl:S I!'-' PLATES AND SHELLS
. .1
b
i ,Figure 10.10
Owing to this pressure, the resultant force F for the portion intercepted by is:F = -2",,'=3
I )'a(1 -- cos 1>} sin cos 1> d'0
.oP
-2"a ){t -
t cos' (1 - I cos 1>)],'al ('
Inserting the above into Eqs. (10.3).N.; =
6 sin' [1
~'a2
- cos
2
1>(3 - 2 cos 1>)) = -6' 1 - 1 + cos 1>
2 cos
2
rJ> )
Nn=
6-
).a' (
2 cos' 1> ) 5 - 6 cos + T+~o;;
(10.23)
Equations (10.23) are valid for 1> > 0' In determining F for > 1>", the sum of the vertical support reactions 4)'"a'/3, must also be taken into account in addition to the internal pressure loading. That is F=
-'!na'), .- 21ra')H -} cos'
(1 - i
cos 1]
Equations (10.3) now yield
i'a' N,,=- ( 5+ 2 cos' 1> .. )' 6 I-cos2 - -... No ="a' ( 1 - 6 cos - -cos' ) '6, I-cos
(10.24)
From Eqs. (10.23) and (10.24) it is observed that both forces N,p and N" change values abruptly at the support ( = 1>0). A discontinuity in No means a discontinuity of the deformation of the parallel circles on the immediate sides of the '111. Thus, the deformation associated with the membrane solution is no/ compatible with the continuity of the structure at support !l".
.\IE/lfIIRA;-. +cjsin e/> . SOle/>
(I)l'
The constant of integration c is determined Irom a boundary condition. Once has been found, one can readily obtain \l' from Eq. (e).
Example 10,7 Determine the displacements of the spherical roof dome supporting its own weight (Fig. IO.Sa).
For the half sphere under consideration 1", = 1", = a and stresses IT. and lTv are given by Egs. (10.13). Equation (10.25) is thereforeSOLUTION
d4)
dr
-1'
cot
a'p(1 + 1')( e/> = ~- cos 4) -j+
cos =f(e/
2).
Inserting this expression into Eq. (I), we obtain sin e/> . l' = a' p(1 + ") [SIO In (I + cos e/ - ..... _-_.. j + CSill e/> - . EI I + cos e/> necessary to choose c such that v = 0 at e/> = a (Fig. to.Sa). It c = {/'p(l ~.vll___L.~ -In (1 EI I + cos CI.(g)
It is that
follows
+ cos
all
(h)
MEMBRASE STRF.SSfS IN SHEI.l.S
217
Upon substituting this yalw: of c into Eq, (y), uel"leclioll r is obtained and Eq. (c) then yields II". It js noted that if the slIpport displacement \\' is to be determined, one need not employ Eq, (q), as l' = 0 tilere: the second of Eqs. (e) and Eg, (e) directly give the solution,
10.7 ASYMMETRICALLY LOADED SHELLS OF REVOLUTIONIn the bending of a shell of revolution under unsymmetrical loading, not only do normal forces NfjJ and No act on the sides of an element. but shearing forces NrN ancl No as well (Fig. 10.12). The moment equilibrium requircs that Ny. = N q." as is always the ('(Ise for a thil1 shell (Sec. 1I .2). The surface load, referred (0 the unit area of the midsurface, has components Px, p,., and p,. The x-directed forces are as follows. The force(a)
is owing to the variation of N". Horizontal components of the forces Nil . 1', d acting on the faces AB and CD of the element make an angle dO, and thus have the following resultant in the.\' direction:Nil, . r , d . cos . d(l
(b)
The difference of the shearing forces acting on faces AC and ED of the element is expressed(e)
Figure 10.12
218
STRESSES IN PL/\TES A}.;[) SJ-IELLS
The
compon~nt
of the external rore!.? isl)xfOl'l
dO (4)
(d)
Thc x equilibrium condition thus reads:
(I0.26n)To the expression governing the v equilibrium of the symmetrically loaded case (Sec. lOA), we must add the force
"';;0'" o
tNo~1",
dO def.>
produced by the difference in the shearing forces acting on the faces AB and CD of the element. Inasmuch as the projection of the shearing forces on the z axis vanishes, Eq. (IO.la) remains valid for the present case as well. The equilibrium of y- and z-directed forces is therefore satisfied by the expressions:
-
c (N
"",
I" ) '" 0
+ ..... --'-~ I"
oN ao'
-
N
I"
0"
cos
(h
P
+p
1",1""
=
)'
0
(10.261 (10.26(")
Equations (10.26) permit determination of the membrane forces in a shell of revolution with nonsymmetrical loading that may, in general. vary with 0 and $. Such a case is discussed in the next section. We note that the governing equations of equilibrium for the spherical, conical, and cylindrical shells may readily be deduced from Eqs. (10.26) upon following a procedure identical with that described in Sec. 10.5.
10,8 SHELLS OF REVOLUTION UNDER WIND LOADINGIt is usual to represent dynamic loading such as wind and earthquake effects by statically equivalent or pseudoslatic loading adequate for purposes of design. The wind load on shells is composed of pressure on lhe wind side and suction of the leeward side. Only the load compollent acting perpendie,dar to the l1lidsw:face p, is considered important. Components p" and p,. are due to friction forces and are of negligible magnitude. Assuming for the sake of simplicity that the wind acts in the direction of the meridian plane 0 = 0, the components of wind pressure are as follows: px = 01',. = 0p, = p sin
K y
,
1\x,\'
lJl
U. =
to
JJ [(X., + 1.).)2.A
2(1 - ")(ZxXy - X;y)] dx dy
(11.7)
where A represents the surface area of the shell. The membrane energy is associated with mid surface stretching produced by the in, plane forces and is given by
Urn
=
~
JJ (Nx"x" + Ny")'o + Nx)')'x)'o) dx dyA
(11.8a)
Introduction of Eqs. (11.3) into the above expression leads to the following form involving the strains and clastic constants:
EtUno= 2(1 _
vll.1.i [(e xo + B,O)' - 2(1 A
..
V)(B'OByo'"
!-;;,o)] dx dy (1I.Sb)
Expressions (11.7) and (11.8) permit the energy to be evaluated readily for a number of commonly encountered shells of regular shape and regular loading. The strain energy plays an important role in treating the bending and buckling problems of shells (Chap. 13).
11.6 AXISYMMETRICALLY LOADED CIRCULAR CYLINDRICAL SHELLSPipes, tanks, boilers, and various other vessels under internal pressure exemplify the axisymmctrically loaded cylindrical shell. Owing to symmetry, an element cut from a cylinder of radius a will have acting on it only the stress resultants shown in Fig. 11.3: N", M ", N" and Qx' Furthermore, the circumferential force and moment, No and M", do not vary with O. The circumferential displacement l' thus vanishes and we need consider only the .\ and z displacements, /I and w. Subject to the foregoing simplifications, only three of the six equilibrium
HEND1NG STltFSSES IN SHELLS
237
/xadOf
\z~ M,I
Q + -_.': dxx
dQ
dx
I II
I . I
I IN,
No
/
/
dO -11/
V
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