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Shear lag analysis for thin-walled members by displacement method
Thin-Walled Structures 13 (1992) 337-354
Shear Lag Analysis for Thin-Walled Members by Displacement Method
K. K. Koo & X. S. Wu Department of Civil and Structural Engineering, Hong Kong Polytechnic, Hong Kong (Received 12 August 1990; received 4 May 1991)
ABSTRACT The classical theory of thin-walled members is unable to reflect the shear lag phenomenon since it is based on the assumption of no shear deformation. This paper presents a displacement method based on the potential variational principle and discusses the shear lag phenomenon. Compared with the mixed method, the displacement method is simpler in the analysis process and hence is easier to apply and to be combined with thefinite element method. Numerical examples show that quite good results can be obtained. The solution also includes the results from the classical theory.
NOTATION c
E G H Jd mc
Ms n Z
ex,G
Shear center of section Young's modulus (N/mm 2) Shear modulus (N/mm 2) Length of member (mm) St. Venant's torsional constant (mm 4) Line moment of the loading about c (N-mm/mm) St. Venant's torsional moment specified at ends of the member (N-mm) +lwherez=H;-lwherez=0 Line load in the x- and y-directions, respectively (N/mm)
337 Thin-Walled Structures 0263-8231/92/$05.00© 1992 Elsevier Science Publishers Ltd, England. Printed in Great Britain
338
Pz
Pz. S
t ta = /a(S) IJcx , V cy IJt
w(z,s) a
0 P ~,q
Es Y,s~
K. K. Koo, X S. W u
Intensity of the distributed load in z-direction per unit area of the wall (N/mm 2) Line load in z-direction at the n th node (joint or edge of walls) (N/mm) The curvilinear coordinate along the contour of the section with the positive direction arbitrarily assigned (mm) Equivalent thickness for shear continuum for the part of the wall with a band of openings and t = t. for the actual wall (mm) Actual thickness of wall at s, equal to zero for the part of the wall with a band of openings along the member (mm) x andy components of the shear center displacement vc(mm) Displacement along the tangent direction (mm) Longitudinal displacement (mm) Angle between x-axis and tangent at s (Radian) Angle of twist of section (Radian) Length of vector from point c to tangent at point s, positive when ~ × ds is along positive direction of z axis (mm) Normal stress and shear flow specified at ends of the member (N/mm 2, N/mm respectively) The whole cross-section Cross-section where the stress is specified
INTRODUCTION The thin-walled member has been widely used in engineering. The classical theory of thin-walled members is based on the assumption of no shear deformation and is unable to reflect the shear lag phenomenon, which is of increasing importance. A number of researchers have studied this problem and proposed some solution methods? Unfortunately, since these methods are based on an additional or individual judgment, they are short of generality and cannot be applied extensively in practice. The cited papers ~-6 present a mixed variational principle for thinwalled members and propose a mixed solution method. The mixed method is more general than the classical one. 7 The warping displacement distribution can be determined in a systematic manner. Satisfactory results can be obtained for both displacement and stresses. This paper presents a displacement method based on the potential variational principle and discusses the shear lag phenomenon. Compared with the mixed method the displacement method is simpler in the analysis process and hence is easier to apply, especially when it is combined with
Shear lag analysis for thin-walled members
339
the finite element method. Numerical examples show that quite good results can be obtained. The solution also includes the results from the classical theory.
POTENTIAL VARIATIONAL PRINCIPLE FOR THIN-WALLED MEMBERS For an arbitrary prismatic thin-walled member, shown in Fig. 1 and Fig. 2, the z axis is always taken to be parallel to the beam axis, while the principle axes of the cross-section coincide with the x and y axes. The present theory is based on the following two assumptions: (1) The cross-section can be regarded as rigid in its own plan. (2) The stress as in tangent direction s and or, in normal direction are much smaller than the axial stress crz and thus by Hooke's law 8 z ~ ~z/E.
H --
.,_.
y,
x/ Fig. 1. T h i n - w a l l e d m e m b e r a n d c o o r d i n a t e s x, y, z.
P
Fig.
L
7£
2. Cross-section o f a t h i n - w a l l e d m e m b e r .
340
K K Koo, X. S. Wu
For a thin-walled member the total potential energy II is given as
ri =
f t[fs ~ U z /
l(0w
to+~a ~s + Oz] t ds
J (dO)~ f + -~ OJa \-~zJ - J z f zwds - mcO
n
z=O.H
-
eyVcy
Pxv~x
-
so
+f ~cosadsvcxnz+f ~pdsO.nz+f ~towdsnz Sa
S~
S o,
+ M,Onz]
(1)
The potential variational principle states that of all admissible displacements which satisfy the prescribed boundary constraint condition the true one is that which makes the total potential energy a minimum. Hence, the following equation must exist: 6FI = 0
(2)
in which
617
f'tf ,~'Eoz [ owo,,w, (ow ovt Ozta+G O-ss+Oz]
= o
Os +
t ds + GJd dz
dz
s
-Py6vcy-Px6v~-ZPz"6w"}dz-Z [fz ~sinads6vcyn~ n
+
z =O,H
Ocosads6vcxnz +Ms6Onz + So
+fL.oqP ds6Onz]
so
d6wt,,dsnz So
(3a)
Shear lag analysis for thin-walled members
341
We now carry out the following integrations by parts:
fofz Ow O(6w) Oz Oz to ds dz -
-
=
fz Ow -dz tafW ds n:
E Z
OH
o2w t~6w ds -- fofz ~~-~z2
s N+
Oz) ~ t d s d z
=
Z
(3b)
s ~-~s + Oz) t6v'dsnz
Z=O.H
. (a2w a2.,) - flf~s-~ + Oz2j"v, ds ~-js + Oz )----O-~s t ds dz =
(3c)
6w -~s + Oz ) t
, m.x
_
)
Se
•~ ~SS
t ff~s + O z ] ) 6 w d s d z
Ov,I n
e
t.ex.~dz + Oz )) 6w ds dz nd(f0) dO dz dz
=
E
dO
fiO~-n:-
f~r~ d20
O~dz
.,. Os t as
(3d)
(3e)
Z=O.H
where E and E denote summation with respect to elements of the wall e
n
and nodes, respectively; Xne = + 1 when a positive direction of s in the wall e connected at node n points away from the node n; x.e = - 1 when positive s in the wall e points toward the node n; X.e = 0 when the wall e
K K. Koo, X S. Wu
342
5e,.Se,:ax
2tt
J
t
,I
l °s
part s+O
I
I
-J
pa~t s-O Fig. 3. Two parts of a cross-section.
does not j o i n at n. se is the local c o o r d i n a t e of s in the wall (e) with the origin at one edge a n d Se = Se. maxat the other edge, as s h o w n in Fig. 3. Substituting eqns (3b)-(3e) into eqn (3a), collecting the terms w h i c h are c o n n e c t e d with the variations 6w, 6V~x,6Vcy a n d 60, respectively, a n d noting that vt = c o s a Vcx+ s i n a v~y + pO
(3f)
6vt = c o s a 6vc~ + sina 6Vcy + 060
(3g)
Zs = Zso + ZSu
(3h)
a n d 6w, 6vc~, 6 vcy, 60 are arbitrary, but equal to zero on Zs~ from eqn (2), we c a n o b t a i n the following five equations, each one being associated with o n e variable:
5w
" E ta ~ 2 +
Os t Os + Oz ] /] + Pz
=
6v,.~ •
, G ~ z -~s + O z j C O s a t d s + P x = 0
6v~y "
,G ~
60
"
o (Ow Ov,) Os + Oz ] sina t ds + Py
0
= 0
f ~2 ff~ ( OW OVt ] d20 sG -0-~ + Oz ]p t ds + G J d - ~ + mc = 0
Ow Ov,'X 6w. " Z G ~ +-~)..etnex.~ + Pz. = 0
(4)
(5)
(6)
(7)
(8)
e
At the two ends z = 0 a n d z = H, by the s a m e r e a s o n i n g we get
=
Oz on Zs~
(9)
Shear lag analysisfor thin-walled members
(owov,
cosa ds =
343
G -~s + Oz ] cosa t ds o n XSo
(10)
f~. { s i n a ds = f~. G ( O O Ws O+ I l t ~Oz ] s i n a t ds on Y~So Se Sa
(11)
Sa
So
L -~pds+M, so
=
L(owov, ao G -~s + O z ] p t d s + G J d - ~ o n ~ s o sa
(12)
It can be seen that eqns (4)-(8) a n d eqns (9)-(12) are exactly equilibrium equations a n d b o u n d a r y conditions on Zso, respectively, for a thin-walled member. METHOD OF ANALYSIS It will be a s s u m e d that the displacement w can be given by the sum of a n u m b e r of terms. I n matrix form, it is written as
w = leo(S), el(s), ¢2(s),.., ¢,(s)l [Wo(Z), w~(z), w2(z),...., w,(z)l r
= 1¢1 {w}
(13)
where functions $i(s) (i -- O, 1, 2 .... n) are chosen coordinate functions while wi(z) (i = O, 1, 2 .... n) are unknowns. Substitution o f e q n (13) forw a n d o f e q n (3f) for vt into eqn (1) gives
n = J0 f , ( \2 I _ E {w'} T Hl {w'} + ~1 Glw} r [BI {w} + Glw} T [Cl IvY}
I O,v:,T,O,,V:,+I-~GJa(O') 2)
+~
dz -
SO(
{Vc}T {P}
(fz ~ [hi {Vc}dsnz Z = O,H
n
sa
s°
(14a)
where the superscript ' represents d/dz, i.e. [w'} = d/dz{w}, {v~] = d/dz {vc} a n d 0' - dO/dz, [.4] = fzs [~l'r [~l t,, ds
(14b)
344
K. K. Koo, X. S. Wu
IBI =
f , d-s d [~]'r dss d [9]
lCl=
fz~ d i f i t In] tds
(14c)
tds
(14d)
[D] = Jz(s [r/]a" [r/]tds
(14e)
{ve] ={Vcx, Vcy, 017
(14f)
{P] = {Px, Pv, mcl r
(14g)
[1/] = [cosa, sina,o]
(14h)
{Pz]
= fz
(14i)
[¢]rpzds s
Performing the variation of II in eqn (14a) we get the following set of equations, which are similar to eqns (4)-(12). Along the member, [~w} x • E I A I { w " f - G [ B l l w l - G l C l l v ' }
+ ~ l ¢ ~ . ] x ? ~ . + IP, I = 0 n
05) (16)
{6vcl r • G l C l r l w ' l + GID'I {v"l + lPl = 0
and at the two ends z = 0 and z = H: ~,o~ I~] T ds - 610'1 tv;l - 6[C1 ~ {wl +
f:t ~
= 0onZso
f,~ ~ [~]Tt ads - EIA] {w'] = 0 on Y~so
(17)
(18)
SO
where [D'] = [D] + Ja [001]
(19)
From eqn (16), we get
[D']-' fl {P] dz + [D'] -I IF0] Z
G{v[I = - G[D'I-' [C]~r{wl -
(20)
Shear lag analysisfor thin-walled members
345
where IF01 = [Qx0, Qyo, M,0l r, Qx0, Oy0 and M,0 are shear forces in x, y direction, respectively, and the total torsional moment at the end z = 0. Substituting eqn (20) into eqn (15) we get the following equation:
EIAIIw"]- G [[BI- lCl [O'l-'lClrllw] + [Cl - lCl
[O'l-' f~ IPI dz
[D']-l{F0} + Z [0"la'Pz" + {Pz} = 0
(21)
n
Equation (21) is the governing differential equation for the analysis of a thin-walled member by the displacement method.
CHOICE OF FUNCTIONS ~Pi(s) The method of choosing the coordinate functions O;(s) (i = 0, 1, 2...) in eqn (13) has been introduced in references. 2-5 Here, for the sake of brevity, we derive it in a straightforward manner.
Open section member From eqn (13) it is obvious that the functions q~;(s) represent the distributions of the displacements along the cross-section. The better we choose the functions ~;, the higher the degree of accuracy in displacements we should reach. We are now to discuss this problem. From eqn (4) we get
v,33=
G Os t ffss + Oz ] )
(22)
- Eta Oz---7 - Pz
Noting that in fact Gt(dw/Os + dvt/dz) is the shear flow at the crosssection and eqn (22) denotes the longitudinal equilibrium. Integrating the above equation we get
G,
÷ Ozj = Ef. -
s_o ---~-2Z2ta Z
Pz.
ds-fpa~s_oPzdS (23)
n G parts - 0
where and in the latter derivation as shown in Fig. 3, 'parts- 0"and 'parts+ 0' denote the two parts of the cross-section divided by a cut at the points of
346
K. K. Koo, X. S. Wu
the cross-section, parts-0 denotes the part which includes a point - ~ > 0), while part~+ 0 denotes the other part.
Let6(s-s.)Pz.
= [¢]t~{f.]
(24a) (24b)
p~ = [qqt.[f} where
6(s - sn) is a &function, a n d s,, is the coordinate s at node n.
{f} =
[(~]Xpzds = [AI-~IPzl
[O]Tl~lt. ds s
(24c)
s
Z {f"} = [ A ] - ~ Z [~"lTe~" n
(24d)
n
Substituting eqns (3t) a n d (24) into eqn (23) and integrating with respect to s yields w = Wo, + [Xo,,yo,,o9osl {v;I -
[x,y,o9] Iv'} +
7
arts-0
- [~]t~ds)ds(E {w",+ G({f] + Z {f.}) )
(25a)
n s
where os is the principal origin of coordinate s a n d o9 = fl O ds is the principal sectional coordinate. Let
Wo = wo~ + [Xos,Yos.Wos] {v'~]
(25b)
and w = [0]{w} = 1~o,~1,~2 ..... ][w}
(25c)
Equation (25a) can be written as
f l(fp
w = wo - [x,y, o91 {v'c} +
7
arts
-
0
-I 0, 1,02 .... ] ta ds) ds
n
It will be seen later that the first element in the last brackets in eqn (25d) will be equal to zero:
Shear lag analysisfor thin-walledmembers
Ewe' +f0 +
Zf,0
= 0
347
(26)
11
If this condition is met, the first term in the integration on the right-hand side of eqn (25d) will be cancelled. Comparing eqn (25d) with eqn (25c), we can choose 0;(i = 0,1,2,) as follows: t~0 =
04
=
05
=
1,01
= X,t~2 = Y,q)3 = 09
f~ 1(fp f~ 1(fp 7
arts - 0
7
arts - 0
(27a)
(-O,)tads)ds
(27b)
(-02)tads)ds
(27c)
(_Oi_3) tads)ds
(27d)
in general (~i =
f~ l (fp t-
arts-0
From the computational point of view, it would be more convenient to use orthogonal coordinate functions. They can be constructed from the function given by eqns (27a)-(27d) as follows: ~o =
1,t~i = X , ~ 2
= Y,~3
= 09
(28a)
k-I
~k ~- ~)k "~- Z 12ki~i
k >4
(28b)
i=0
Ctki = 0k
=
- f~.sOk~itads/ fxs ~2itads
f/l (fp 7
arts-0
( - ~ k 3- ) tads)ds
(28c) (28d)
The function [01 = [00, 01, ¢'2,....0/1] in eqn (13) can be replaced by [0] = [00, 01, 02,...0/1]. Since Oo -= 1, d~o/ds ----0, the first row in eqn (15) meets eqn (26). Closed section member As shown in Fig. 4, by introducing an imaginary longitudinal cut for each closed cell a closed section can be transformed into an open one. Apart from choosing a system of coordinates with its origin os it will also
K. K. Koo, X S. Wu
348
S
S 5. Os
S
S
Fig. 4. Open cross-section transformed from a closed section.
be necessary in this case to introduce an auxiliary curvilinear coordinate c; along each cell in the counterclockwise direction. Furthermore, the shear flows: (29)
{qcl = {q,.t,qc2 ...... ]x
must be i n t r o d u c e d into the cuts (qc; is positive w h e n it is in the same direction as positive c~) a n d be regarded as additional u n k n o w n variables. Define a matrix:
(30)
[~c] = [g,.~, ~,.2, V;,,3....... ] where tgc;(i = 1, 2 .... ) are functions o f s
I
ds
+ 1 c; a n d s are in same direction - 1 c; a n d s are in opposite direction c; a n d s do not coincide
dci
0
(31)
The shear strain o f a closed section m e m b e r can be expressed as the sum of the shear strain of the c o r r e s p o n d i n g o p e n section m e m b e r subjected to load a n d shear flows {qc}, i.e. for a closed section, e q n (23) would be written as
Gt (OW -~s + OY,'~ Oz / ~ - E
fp
02W ta ds - fp arts - 00Z2
-
E n G pans
p: ds arts - 0
P~"+ [~t¢]{q,,}
(32)
- 0
Substituting e q n (24) a n d w = [~] {wj into e q n (32), a n d writing
1~',1 = [
Jr~ar ts
(-[~d)tads
(33a)
- 0
where [Of] = [~l, ~2,-..], we get
Ow + c3v, Oz - Gt Iw,] ( E Iwi'l + be,} + ~ {L,I ) + lGitIw,.] Iq~l ?/
(33b)
Shear lag analysisfor thin-walled members
349
where Iwi} = [w,, w2..... ]T, bql = lf,,A..l T, be.,I = I L , . f . 2 .... l T and the condition, eqn (26), has been used. Integrating eqn (33b) with respect to each closed cell, which is equivalent to muliplying the eqn (33b) by [Vt~]T and integrating it with respect to Y.s, and noting that v, = Vcx cosa + V~y sina + 0p, we get [~t¢lr ~-s ds + v~, s
'
[~G]v sina ds + 0'
$
?_)
i
n
I
(34)
Since w must be single valued in each cell, the first integral on the lefthand side o f e q n (34) is equal to zero and it can be proved that the second and the third integrals are also equal to zero. As a result
{q,,} = - [B~c]-! [Bcl ] (E{w['} + {fl}+ Z {)cnl})+ [B~]-l {Dc]GO' H
(35) where [Bc~l = fz, [gc]T [V/~]dst
(36a)
[B~,I = fz, [V~AT lU,,I dst
(36b)
{Dcl = fzs [~'clrpds
(36c)
Substituting eqn (35) into eqn (33b), integrating eqn (33b) with respect to s from 0 to s to get an expression for w, comparing this expression with eqn (25c) and discussing it in the same way as eqns (27a)-(27d) were derived, we can finally get t~o =
1, t~t =
X,~2
= y,t~3
=
0.) - -
[gtc]
[Bcc]-'lDcl--ds t
(37a)
350
K. K. Koo, X. S. Wu
04 ----
0 (-- ~ 1 ) I a dS --[I//c]
[Bcc] -I
( I~,~1T Y~a~s-0 (37b)
in general
fpaas_o(-~Pi-3)tads) ~-) ds
(37c)
[q)] must be ta-orthogonalized by the Gram-Schmidt method as in the case of open section members; x and y are principal axes of the crosssection; and ~ 3 is the principal reduced sectorial coordinate with respect to the shear center for a closed section. SOLUTION OF GOVERNING EQUATION As mentioned above, analysis of the thin-walled member by the displacement method was reduced to solve the governing equation: EIA] {w"} - GIBI {w} + {P} = 0
(38a)
where 181 =
[ B I - [CI [D'] -j [CI T
IP]
[C] [D']-' Ii {P} dz - [C] [D'I-'
(38b)
{ro} + Z
[q),IT p~, + [p~}
~o n
(38c) The first row in eqn (38a) reduces to condition eqn (26). The remainder of eqn (38a) leads to the solution of the following form: {wll = [a] [sh,~zla{cl + [a]
EG~Ai
"],i =
[al
=
[ch,~zla{d]+ {wl,[
i = 1, 2, 3,....n
[{all, {a2}....... la,,}l
(39)
(40) (41)
Shear lag analysis for thin-walled members
351
n is the number of terms included, and Ai and {ai} are the eigenvalues and eigenvectors of the matrix equation: EA~ [AI]
{ai} - GIBd {a;} =
0
(42)
where [AI] is reduced from [A] by cancelling the first row and first column and [Bi] from [BI in the same way, and
[ f(/l'z)f('~2z) "f(,~,z)J]
[f(A'Z)]d =
(43)
{Wlp} is a particular solution of {w~} from eqn (38a). It can be determined from the applied load. [c} and {d} are constants to be determined by the boundary condition, andf(gt;z) is a function of A.;z. Substituting eqn (39) into eqn (20), we can get [vc}. The boundary conditions are generally as follows: (1) {w} = 0, {vc} = 0 at the fixed end; (2) {w'] = 0, [D'] {v~} + [C] r {w} = 0 at the free end; and (3) {w'} = 0, = 0 at the simply supported end. Based on the method mentioned above, a computer program (named ATWAM) for analysis of the thin-walled member has been worked out.
{Vc}
EXAMPLES
Example 1 A thin-walled tube, shown in Fig. 5, is subjected to a horizontal load 2p at the free end is analyzed by the ATWAM program. The stress distribution in the bottom cross-section is shown in Fig. 6. It is obvious that the difference between the stresses computed by the proposed method and by the classical theory is quite significant, i.e. the effect of shear lag must Z
B
,r
2p A'
H'Sh
~-
X
i
!I
y (s)
I__
!
b
b/h-Z
L
(-lib)
b
_1
--7
Fig. 5. Cantilever beam subject to point load. (a) Cantilever beam; (b) cross-section.
352
K.K.
Koo, X S. Wu
~37.79> [32.78] •
21.4 <37.79> [32.781
<16 21> [18 "001
14 ."' ~
<"116- ' -4 1 > t J ".0 8 ]
"
~
21.4
<13.89>
[15,771
~, [
[ ] ....
N
< > ....
Two-term
solution
Three-term Htxed
^
three-term
.....
solution
me~hod
solution
Classical theo~/
Fig. 6. N o r m a l stress at lower end.
be considered in engineering design. It is also noted that the result computed by the displacement method is in good agreement with the result by the mixed method. Example 2
Consider another thin-walled tube, shown in Fig. 7. There are four bands of openings adjacent to four corners over the full length of this tube. The well-known shear continuum technique is applied, which assumes that the lintel beams between adjacent openings in a band can be replaced by an equivalent shear continuum of thickness to. The normal stresses crz/ (PHh/Ix) in the lower end of the cantilever are shown in Fig. 8. It can be seen that due to the existence of the openings, the shear lag phenomenon is significant. 2P
o.lt
/ t
Y
h
E/G-2.3
O.Zh I I
11
2h
~
ah
llo.2h
17
Fig. 7. C r o s s - s e c t i o n o f c a n t i l e v e r b e a m with o p e n i n g s .
x
Shear lag analysis for thin-walled members (1.21)
<1.02>
(0.67)
[ 1.15]
~L"~"
353 (0.49)
<0.66>
<0.71>
I0.601
10.571
(2.66)
<3.09> 13.391
! (
) ....
< > [ A
] t
Two-t.erm solution Three-tet~
"/~,
solution
Four-Lem solution Nixed
method
llatee- tern .....
c
Classical
solution theory
xx
Fig. 8. Normal stress at lower end.
CONCLUSIONS Based on the potential variational principle, a displacement method for the analysis of thin-walled members is proposed. This method is more general than the classical one and can be applied to open and closed sections. In particular, it is very suitable for the analysis of shear lag phenomena. Compared with the mixed method, the displacement method is simpler in the analysis process and hence is easier to apply. Numerical examples have shown that quite good results can be obtained. Numerical examples have shown that the effect of shear lag on stress distribution is quite significant in some cases and must be considered in engineering design.
REFERENCES 1. Reissner, E., Analysis of shear lag in box-beams by the principle of potential energy. Q. Appl. Math., 4, (1946) 268-78. 2. Koo, K. K., Analysis of thin walled structures in tall buildings. PhD Thesis, Hong Kong University, May 1986. 3. Koo, K. K. & Cheung, Y. K., Mixed variational formulation for thin-walled beams with shear lag, J. Eng. Mech., ASCE, 115(10) (1989) 2271-86. 4. Cheung, Y. K. & Koo, K. K., Analytical method for thin-walled members in general bending and torsion. Thin-Walled Struct., 6 (1988) 355-69. 5. Koo, K. K. & Cheung, Y. IC, Analytical method for thin-walled members of closed sections. Thin-Walled Struct., 6 (1988) 419-32. 6. Koo, K. K. & Cheung, Y. IC, Shear lag analysis of core of variable crosssections. Proc. Fourth Int. Conf. on Tall Buildings, Hong Kong and ShanghaL
354
K.K. Koo, X S. Wu
Hong Kong University and Hong Kong Institute of Engineers, 1988, pp. 483-9. 7. Vlasov, V. Z., Thin-walled Elastic Beams, 2nd edn, Program for Scientific Translation, Jerusalem, 1961.
Shear Lag Analysis for Thin-Walled Members by Displacement Method
K. K. Koo & X. S. Wu Department of Civil and Structural Engineering, Hong Kong Polytechnic, Hong Kong (Received 12 August 1990; received 4 May 1991)
ABSTRACT The classical theory of thin-walled members is unable to reflect the shear lag phenomenon since it is based on the assumption of no shear deformation. This paper presents a displacement method based on the potential variational principle and discusses the shear lag phenomenon. Compared with the mixed method, the displacement method is simpler in the analysis process and hence is easier to apply and to be combined with thefinite element method. Numerical examples show that quite good results can be obtained. The solution also includes the results from the classical theory.
NOTATION c
E G H Jd mc
Ms n Z
ex,G
Shear center of section Young's modulus (N/mm 2) Shear modulus (N/mm 2) Length of member (mm) St. Venant's torsional constant (mm 4) Line moment of the loading about c (N-mm/mm) St. Venant's torsional moment specified at ends of the member (N-mm) +lwherez=H;-lwherez=0 Line load in the x- and y-directions, respectively (N/mm)
337 Thin-Walled Structures 0263-8231/92/$05.00© 1992 Elsevier Science Publishers Ltd, England. Printed in Great Britain
338
Pz
Pz. S
t ta = /a(S) IJcx , V cy IJt
w(z,s) a
0 P ~,q
Es Y,s~
K. K. Koo, X S. W u
Intensity of the distributed load in z-direction per unit area of the wall (N/mm 2) Line load in z-direction at the n th node (joint or edge of walls) (N/mm) The curvilinear coordinate along the contour of the section with the positive direction arbitrarily assigned (mm) Equivalent thickness for shear continuum for the part of the wall with a band of openings and t = t. for the actual wall (mm) Actual thickness of wall at s, equal to zero for the part of the wall with a band of openings along the member (mm) x andy components of the shear center displacement vc(mm) Displacement along the tangent direction (mm) Longitudinal displacement (mm) Angle between x-axis and tangent at s (Radian) Angle of twist of section (Radian) Length of vector from point c to tangent at point s, positive when ~ × ds is along positive direction of z axis (mm) Normal stress and shear flow specified at ends of the member (N/mm 2, N/mm respectively) The whole cross-section Cross-section where the stress is specified
INTRODUCTION The thin-walled member has been widely used in engineering. The classical theory of thin-walled members is based on the assumption of no shear deformation and is unable to reflect the shear lag phenomenon, which is of increasing importance. A number of researchers have studied this problem and proposed some solution methods? Unfortunately, since these methods are based on an additional or individual judgment, they are short of generality and cannot be applied extensively in practice. The cited papers ~-6 present a mixed variational principle for thinwalled members and propose a mixed solution method. The mixed method is more general than the classical one. 7 The warping displacement distribution can be determined in a systematic manner. Satisfactory results can be obtained for both displacement and stresses. This paper presents a displacement method based on the potential variational principle and discusses the shear lag phenomenon. Compared with the mixed method the displacement method is simpler in the analysis process and hence is easier to apply, especially when it is combined with
Shear lag analysis for thin-walled members
339
the finite element method. Numerical examples show that quite good results can be obtained. The solution also includes the results from the classical theory.
POTENTIAL VARIATIONAL PRINCIPLE FOR THIN-WALLED MEMBERS For an arbitrary prismatic thin-walled member, shown in Fig. 1 and Fig. 2, the z axis is always taken to be parallel to the beam axis, while the principle axes of the cross-section coincide with the x and y axes. The present theory is based on the following two assumptions: (1) The cross-section can be regarded as rigid in its own plan. (2) The stress as in tangent direction s and or, in normal direction are much smaller than the axial stress crz and thus by Hooke's law 8 z ~ ~z/E.
H --
.,_.
y,
x/ Fig. 1. T h i n - w a l l e d m e m b e r a n d c o o r d i n a t e s x, y, z.
P
Fig.
L
7£
2. Cross-section o f a t h i n - w a l l e d m e m b e r .
340
K K Koo, X. S. Wu
For a thin-walled member the total potential energy II is given as
ri =
f t[fs ~ U z /
l(0w
to+~a ~s + Oz] t ds
J (dO)~ f + -~ OJa \-~zJ - J z f zwds - mcO
n
z=O.H
-
eyVcy
Pxv~x
-
so
+f ~cosadsvcxnz+f ~pdsO.nz+f ~towdsnz Sa
S~
S o,
+ M,Onz]
(1)
The potential variational principle states that of all admissible displacements which satisfy the prescribed boundary constraint condition the true one is that which makes the total potential energy a minimum. Hence, the following equation must exist: 6FI = 0
(2)
in which
617
f'tf ,~'Eoz [ owo,,w, (ow ovt Ozta+G O-ss+Oz]
= o
Os +
t ds + GJd dz
dz
s
-Py6vcy-Px6v~-ZPz"6w"}dz-Z [fz ~sinads6vcyn~ n
+
z =O,H
Ocosads6vcxnz +Ms6Onz + So
+fL.oqP ds6Onz]
so
d6wt,,dsnz So
(3a)
Shear lag analysis for thin-walled members
341
We now carry out the following integrations by parts:
fofz Ow O(6w) Oz Oz to ds dz -
-
=
fz Ow -dz tafW ds n:
E Z
OH
o2w t~6w ds -- fofz ~~-~z2
s N+
Oz) ~ t d s d z
=
Z
(3b)
s ~-~s + Oz) t6v'dsnz
Z=O.H
. (a2w a2.,) - flf~s-~ + Oz2j"v, ds ~-js + Oz )----O-~s t ds dz =
(3c)
6w -~s + Oz ) t
, m.x
_
)
Se
•~ ~SS
t ff~s + O z ] ) 6 w d s d z
Ov,I n
e
t.ex.~dz + Oz )) 6w ds dz nd(f0) dO dz dz
=
E
dO
fiO~-n:-
f~r~ d20
O~dz
.,. Os t as
(3d)
(3e)
Z=O.H
where E and E denote summation with respect to elements of the wall e
n
and nodes, respectively; Xne = + 1 when a positive direction of s in the wall e connected at node n points away from the node n; x.e = - 1 when positive s in the wall e points toward the node n; X.e = 0 when the wall e
K K. Koo, X S. Wu
342
5e,.Se,:ax
2tt
J
t
,I
l °s
part s+O
I
I
-J
pa~t s-O Fig. 3. Two parts of a cross-section.
does not j o i n at n. se is the local c o o r d i n a t e of s in the wall (e) with the origin at one edge a n d Se = Se. maxat the other edge, as s h o w n in Fig. 3. Substituting eqns (3b)-(3e) into eqn (3a), collecting the terms w h i c h are c o n n e c t e d with the variations 6w, 6V~x,6Vcy a n d 60, respectively, a n d noting that vt = c o s a Vcx+ s i n a v~y + pO
(3f)
6vt = c o s a 6vc~ + sina 6Vcy + 060
(3g)
Zs = Zso + ZSu
(3h)
a n d 6w, 6vc~, 6 vcy, 60 are arbitrary, but equal to zero on Zs~ from eqn (2), we c a n o b t a i n the following five equations, each one being associated with o n e variable:
5w
" E ta ~ 2 +
Os t Os + Oz ] /] + Pz
=
6v,.~ •
, G ~ z -~s + O z j C O s a t d s + P x = 0
6v~y "
,G ~
60
"
o (Ow Ov,) Os + Oz ] sina t ds + Py
0
= 0
f ~2 ff~ ( OW OVt ] d20 sG -0-~ + Oz ]p t ds + G J d - ~ + mc = 0
Ow Ov,'X 6w. " Z G ~ +-~)..etnex.~ + Pz. = 0
(4)
(5)
(6)
(7)
(8)
e
At the two ends z = 0 a n d z = H, by the s a m e r e a s o n i n g we get
=
Oz on Zs~
(9)
Shear lag analysisfor thin-walled members
(owov,
cosa ds =
343
G -~s + Oz ] cosa t ds o n XSo
(10)
f~. { s i n a ds = f~. G ( O O Ws O+ I l t ~Oz ] s i n a t ds on Y~So Se Sa
(11)
Sa
So
L -~pds+M, so
=
L(owov, ao G -~s + O z ] p t d s + G J d - ~ o n ~ s o sa
(12)
It can be seen that eqns (4)-(8) a n d eqns (9)-(12) are exactly equilibrium equations a n d b o u n d a r y conditions on Zso, respectively, for a thin-walled member. METHOD OF ANALYSIS It will be a s s u m e d that the displacement w can be given by the sum of a n u m b e r of terms. I n matrix form, it is written as
w = leo(S), el(s), ¢2(s),.., ¢,(s)l [Wo(Z), w~(z), w2(z),...., w,(z)l r
= 1¢1 {w}
(13)
where functions $i(s) (i -- O, 1, 2 .... n) are chosen coordinate functions while wi(z) (i = O, 1, 2 .... n) are unknowns. Substitution o f e q n (13) forw a n d o f e q n (3f) for vt into eqn (1) gives
n = J0 f , ( \2 I _ E {w'} T Hl {w'} + ~1 Glw} r [BI {w} + Glw} T [Cl IvY}
I O,v:,T,O,,V:,+I-~GJa(O') 2)
+~
dz -
SO(
{Vc}T {P}
(fz ~ [hi {Vc}dsnz Z = O,H
n
sa
s°
(14a)
where the superscript ' represents d/dz, i.e. [w'} = d/dz{w}, {v~] = d/dz {vc} a n d 0' - dO/dz, [.4] = fzs [~l'r [~l t,, ds
(14b)
344
K. K. Koo, X. S. Wu
IBI =
f , d-s d [~]'r dss d [9]
lCl=
fz~ d i f i t In] tds
(14c)
tds
(14d)
[D] = Jz(s [r/]a" [r/]tds
(14e)
{ve] ={Vcx, Vcy, 017
(14f)
{P] = {Px, Pv, mcl r
(14g)
[1/] = [cosa, sina,o]
(14h)
{Pz]
= fz
(14i)
[¢]rpzds s
Performing the variation of II in eqn (14a) we get the following set of equations, which are similar to eqns (4)-(12). Along the member, [~w} x • E I A I { w " f - G [ B l l w l - G l C l l v ' }
+ ~ l ¢ ~ . ] x ? ~ . + IP, I = 0 n
05) (16)
{6vcl r • G l C l r l w ' l + GID'I {v"l + lPl = 0
and at the two ends z = 0 and z = H: ~,o~ I~] T ds - 610'1 tv;l - 6[C1 ~ {wl +
f:t ~
= 0onZso
f,~ ~ [~]Tt ads - EIA] {w'] = 0 on Y~so
(17)
(18)
SO
where [D'] = [D] + Ja [001]
(19)
From eqn (16), we get
[D']-' fl {P] dz + [D'] -I IF0] Z
G{v[I = - G[D'I-' [C]~r{wl -
(20)
Shear lag analysisfor thin-walled members
345
where IF01 = [Qx0, Qyo, M,0l r, Qx0, Oy0 and M,0 are shear forces in x, y direction, respectively, and the total torsional moment at the end z = 0. Substituting eqn (20) into eqn (15) we get the following equation:
EIAIIw"]- G [[BI- lCl [O'l-'lClrllw] + [Cl - lCl
[O'l-' f~ IPI dz
[D']-l{F0} + Z [0"la'Pz" + {Pz} = 0
(21)
n
Equation (21) is the governing differential equation for the analysis of a thin-walled member by the displacement method.
CHOICE OF FUNCTIONS ~Pi(s) The method of choosing the coordinate functions O;(s) (i = 0, 1, 2...) in eqn (13) has been introduced in references. 2-5 Here, for the sake of brevity, we derive it in a straightforward manner.
Open section member From eqn (13) it is obvious that the functions q~;(s) represent the distributions of the displacements along the cross-section. The better we choose the functions ~;, the higher the degree of accuracy in displacements we should reach. We are now to discuss this problem. From eqn (4) we get
v,33=
G Os t ffss + Oz ] )
(22)
- Eta Oz---7 - Pz
Noting that in fact Gt(dw/Os + dvt/dz) is the shear flow at the crosssection and eqn (22) denotes the longitudinal equilibrium. Integrating the above equation we get
G,
÷ Ozj = Ef. -
s_o ---~-2Z2ta Z
Pz.
ds-fpa~s_oPzdS (23)
n G parts - 0
where and in the latter derivation as shown in Fig. 3, 'parts- 0"and 'parts+ 0' denote the two parts of the cross-section divided by a cut at the points of
346
K. K. Koo, X. S. Wu
the cross-section, parts-0 denotes the part which includes a point - ~ > 0), while part~+ 0 denotes the other part.
Let6(s-s.)Pz.
= [¢]t~{f.]
(24a) (24b)
p~ = [qqt.[f} where
6(s - sn) is a &function, a n d s,, is the coordinate s at node n.
{f} =
[(~]Xpzds = [AI-~IPzl
[O]Tl~lt. ds s
(24c)
s
Z {f"} = [ A ] - ~ Z [~"lTe~" n
(24d)
n
Substituting eqns (3t) a n d (24) into eqn (23) and integrating with respect to s yields w = Wo, + [Xo,,yo,,o9osl {v;I -
[x,y,o9] Iv'} +
7
arts-0
- [~]t~ds)ds(E {w",+ G({f] + Z {f.}) )
(25a)
n s
where os is the principal origin of coordinate s a n d o9 = fl O ds is the principal sectional coordinate. Let
Wo = wo~ + [Xos,Yos.Wos] {v'~]
(25b)
and w = [0]{w} = 1~o,~1,~2 ..... ][w}
(25c)
Equation (25a) can be written as
f l(fp
w = wo - [x,y, o91 {v'c} +
7
arts
-
0
-I 0, 1,02 .... ] ta ds) ds
n
It will be seen later that the first element in the last brackets in eqn (25d) will be equal to zero:
Shear lag analysisfor thin-walledmembers
Ewe' +f0 +
Zf,0
= 0
347
(26)
11
If this condition is met, the first term in the integration on the right-hand side of eqn (25d) will be cancelled. Comparing eqn (25d) with eqn (25c), we can choose 0;(i = 0,1,2,) as follows: t~0 =
04
=
05
=
1,01
= X,t~2 = Y,q)3 = 09
f~ 1(fp f~ 1(fp 7
arts - 0
7
arts - 0
(27a)
(-O,)tads)ds
(27b)
(-02)tads)ds
(27c)
(_Oi_3) tads)ds
(27d)
in general (~i =
f~ l (fp t-
arts-0
From the computational point of view, it would be more convenient to use orthogonal coordinate functions. They can be constructed from the function given by eqns (27a)-(27d) as follows: ~o =
1,t~i = X , ~ 2
= Y,~3
= 09
(28a)
k-I
~k ~- ~)k "~- Z 12ki~i
k >4
(28b)
i=0
Ctki = 0k
=
- f~.sOk~itads/ fxs ~2itads
f/l (fp 7
arts-0
( - ~ k 3- ) tads)ds
(28c) (28d)
The function [01 = [00, 01, ¢'2,....0/1] in eqn (13) can be replaced by [0] = [00, 01, 02,...0/1]. Since Oo -= 1, d~o/ds ----0, the first row in eqn (15) meets eqn (26). Closed section member As shown in Fig. 4, by introducing an imaginary longitudinal cut for each closed cell a closed section can be transformed into an open one. Apart from choosing a system of coordinates with its origin os it will also
K. K. Koo, X S. Wu
348
S
S 5. Os
S
S
Fig. 4. Open cross-section transformed from a closed section.
be necessary in this case to introduce an auxiliary curvilinear coordinate c; along each cell in the counterclockwise direction. Furthermore, the shear flows: (29)
{qcl = {q,.t,qc2 ...... ]x
must be i n t r o d u c e d into the cuts (qc; is positive w h e n it is in the same direction as positive c~) a n d be regarded as additional u n k n o w n variables. Define a matrix:
(30)
[~c] = [g,.~, ~,.2, V;,,3....... ] where tgc;(i = 1, 2 .... ) are functions o f s
I
ds
+ 1 c; a n d s are in same direction - 1 c; a n d s are in opposite direction c; a n d s do not coincide
dci
0
(31)
The shear strain o f a closed section m e m b e r can be expressed as the sum of the shear strain of the c o r r e s p o n d i n g o p e n section m e m b e r subjected to load a n d shear flows {qc}, i.e. for a closed section, e q n (23) would be written as
Gt (OW -~s + OY,'~ Oz / ~ - E
fp
02W ta ds - fp arts - 00Z2
-
E n G pans
p: ds arts - 0
P~"+ [~t¢]{q,,}
(32)
- 0
Substituting e q n (24) a n d w = [~] {wj into e q n (32), a n d writing
1~',1 = [
Jr~ar ts
(-[~d)tads
(33a)
- 0
where [Of] = [~l, ~2,-..], we get
Ow + c3v, Oz - Gt Iw,] ( E Iwi'l + be,} + ~ {L,I ) + lGitIw,.] Iq~l ?/
(33b)
Shear lag analysisfor thin-walled members
349
where Iwi} = [w,, w2..... ]T, bql = lf,,A..l T, be.,I = I L , . f . 2 .... l T and the condition, eqn (26), has been used. Integrating eqn (33b) with respect to each closed cell, which is equivalent to muliplying the eqn (33b) by [Vt~]T and integrating it with respect to Y.s, and noting that v, = Vcx cosa + V~y sina + 0p, we get [~t¢lr ~-s ds + v~, s
'
[~G]v sina ds + 0'
$
?_)
i
n
I
(34)
Since w must be single valued in each cell, the first integral on the lefthand side o f e q n (34) is equal to zero and it can be proved that the second and the third integrals are also equal to zero. As a result
{q,,} = - [B~c]-! [Bcl ] (E{w['} + {fl}+ Z {)cnl})+ [B~]-l {Dc]GO' H
(35) where [Bc~l = fz, [gc]T [V/~]dst
(36a)
[B~,I = fz, [V~AT lU,,I dst
(36b)
{Dcl = fzs [~'clrpds
(36c)
Substituting eqn (35) into eqn (33b), integrating eqn (33b) with respect to s from 0 to s to get an expression for w, comparing this expression with eqn (25c) and discussing it in the same way as eqns (27a)-(27d) were derived, we can finally get t~o =
1, t~t =
X,~2
= y,t~3
=
0.) - -
[gtc]
[Bcc]-'lDcl--ds t
(37a)
350
K. K. Koo, X. S. Wu
04 ----
0 (-- ~ 1 ) I a dS --[I//c]
[Bcc] -I
( I~,~1T Y~a~s-0 (37b)
in general
fpaas_o(-~Pi-3)tads) ~-) ds
(37c)
[q)] must be ta-orthogonalized by the Gram-Schmidt method as in the case of open section members; x and y are principal axes of the crosssection; and ~ 3 is the principal reduced sectorial coordinate with respect to the shear center for a closed section. SOLUTION OF GOVERNING EQUATION As mentioned above, analysis of the thin-walled member by the displacement method was reduced to solve the governing equation: EIA] {w"} - GIBI {w} + {P} = 0
(38a)
where 181 =
[ B I - [CI [D'] -j [CI T
IP]
[C] [D']-' Ii {P} dz - [C] [D'I-'
(38b)
{ro} + Z
[q),IT p~, + [p~}
~o n
(38c) The first row in eqn (38a) reduces to condition eqn (26). The remainder of eqn (38a) leads to the solution of the following form: {wll = [a] [sh,~zla{cl + [a]
EG~Ai
"],i =
[al
=
[ch,~zla{d]+ {wl,[
i = 1, 2, 3,....n
[{all, {a2}....... la,,}l
(39)
(40) (41)
Shear lag analysis for thin-walled members
351
n is the number of terms included, and Ai and {ai} are the eigenvalues and eigenvectors of the matrix equation: EA~ [AI]
{ai} - GIBd {a;} =
0
(42)
where [AI] is reduced from [A] by cancelling the first row and first column and [Bi] from [BI in the same way, and
[ f(/l'z)f('~2z) "f(,~,z)J]
[f(A'Z)]d =
(43)
{Wlp} is a particular solution of {w~} from eqn (38a). It can be determined from the applied load. [c} and {d} are constants to be determined by the boundary condition, andf(gt;z) is a function of A.;z. Substituting eqn (39) into eqn (20), we can get [vc}. The boundary conditions are generally as follows: (1) {w} = 0, {vc} = 0 at the fixed end; (2) {w'] = 0, [D'] {v~} + [C] r {w} = 0 at the free end; and (3) {w'} = 0, = 0 at the simply supported end. Based on the method mentioned above, a computer program (named ATWAM) for analysis of the thin-walled member has been worked out.
{Vc}
EXAMPLES
Example 1 A thin-walled tube, shown in Fig. 5, is subjected to a horizontal load 2p at the free end is analyzed by the ATWAM program. The stress distribution in the bottom cross-section is shown in Fig. 6. It is obvious that the difference between the stresses computed by the proposed method and by the classical theory is quite significant, i.e. the effect of shear lag must Z
B
,r
2p A'
H'Sh
~-
X
i
!I
y (s)
I__
!
b
b/h-Z
L
(-lib)
b
_1
--7
Fig. 5. Cantilever beam subject to point load. (a) Cantilever beam; (b) cross-section.
352
K.K.
Koo, X S. Wu
~37.79> [32.78] •
21.4 <37.79> [32.781
<16 21> [18 "001
14 ."' ~
<"116- ' -4 1 > t J ".0 8 ]
"
~
21.4
<13.89>
[15,771
~, [
[ ] ....
N
< > ....
Two-term
solution
Three-term Htxed
^
three-term
.....
solution
me~hod
solution
Classical theo~/
Fig. 6. N o r m a l stress at lower end.
be considered in engineering design. It is also noted that the result computed by the displacement method is in good agreement with the result by the mixed method. Example 2
Consider another thin-walled tube, shown in Fig. 7. There are four bands of openings adjacent to four corners over the full length of this tube. The well-known shear continuum technique is applied, which assumes that the lintel beams between adjacent openings in a band can be replaced by an equivalent shear continuum of thickness to. The normal stresses crz/ (PHh/Ix) in the lower end of the cantilever are shown in Fig. 8. It can be seen that due to the existence of the openings, the shear lag phenomenon is significant. 2P
o.lt
/ t
Y
h
E/G-2.3
O.Zh I I
11
2h
~
ah
llo.2h
17
Fig. 7. C r o s s - s e c t i o n o f c a n t i l e v e r b e a m with o p e n i n g s .
x
Shear lag analysis for thin-walled members (1.21)
<1.02>
(0.67)
[ 1.15]
~L"~"
353 (0.49)
<0.66>
<0.71>
I0.601
10.571
(2.66)
<3.09> 13.391
! (
) ....
< > [ A
] t
Two-t.erm solution Three-tet~
"/~,
solution
Four-Lem solution Nixed
method
llatee- tern .....
c
Classical
solution theory
xx
Fig. 8. Normal stress at lower end.
CONCLUSIONS Based on the potential variational principle, a displacement method for the analysis of thin-walled members is proposed. This method is more general than the classical one and can be applied to open and closed sections. In particular, it is very suitable for the analysis of shear lag phenomena. Compared with the mixed method, the displacement method is simpler in the analysis process and hence is easier to apply. Numerical examples have shown that quite good results can be obtained. Numerical examples have shown that the effect of shear lag on stress distribution is quite significant in some cases and must be considered in engineering design.
REFERENCES 1. Reissner, E., Analysis of shear lag in box-beams by the principle of potential energy. Q. Appl. Math., 4, (1946) 268-78. 2. Koo, K. K., Analysis of thin walled structures in tall buildings. PhD Thesis, Hong Kong University, May 1986. 3. Koo, K. K. & Cheung, Y. K., Mixed variational formulation for thin-walled beams with shear lag, J. Eng. Mech., ASCE, 115(10) (1989) 2271-86. 4. Cheung, Y. K. & Koo, K. K., Analytical method for thin-walled members in general bending and torsion. Thin-Walled Struct., 6 (1988) 355-69. 5. Koo, K. K. & Cheung, Y. IC, Analytical method for thin-walled members of closed sections. Thin-Walled Struct., 6 (1988) 419-32. 6. Koo, K. K. & Cheung, Y. IC, Shear lag analysis of core of variable crosssections. Proc. Fourth Int. Conf. on Tall Buildings, Hong Kong and ShanghaL
354
K.K. Koo, X S. Wu
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