Exact solution for general variable cross-section members

m43-7949~1S3.00 + 0.00 l’tqmon I’maspk

Cmtprtlrs& Stitis Vol.41,No. 4, pp.765-772, 1991 Printed in Greati3riti.

EXACT SOLUTION FOR GENERAL VARIABLE CROSS-SECTION MEMBERS M. EISENBERGBR Department of Civil Engineering, Technion-Israe~~~itute

(Received

14 September

of Te~lmology, Technion City, Haifa 32oo0,

MO)

paper concerns the formulation of a new element-based method for the solution of beams with variable cross-section. Using only one element it is possible to derive the exact stifkss matrix and equivalent loads (up to the aocuracy of the computer), for any polynomiaf variation of axial, torsional, Abetsact-This

and bending stiffnesses, and load along the beam. Examples are given for the accuracy and efficiencyof the method.

INTRODUCTION

The use of variable cross-section members can help the designer reduce the weight of structures; improved strength and stability can be achieved too. The cost of fabricating such members is relatively high, and offsets their advantges. However, when weight and performance are the most important requirements, members with variable cross-section are the most suitable choice. One such application might be for the construction of large structures in space. Many authors have suggested exact and approximate solutions for tapered members. The recent publications by Banerjee and Williams [1,2] gave exact solutions for few cases of such members. Their solutions are good only for linear taper of the depth, and the cases of I-section and box section with constant width are not included. The variation of the cross-sectional area, polar moment of inertia, and the moment of inertia of the area are not inde~ndent, and their variation is dependent on the choice of only two parameters. In the dynamic stiffness matrix the axial force cannot bc accounted for, so that dynamic stability calculations are not possible. Karabalis and Beskos [3] gave exact stiffness matrices for linear depth variation and constant width. For dynamics and stability they got good approximate solutions. Eisenbcrger [4] gave exact static stiffness matrix for beams in bending with linear and parabolic depth variation and constant width, and for linearly varying width with constant depth. Eisenberger and Reich fq presented an approximate method for general polynomial variation of width and/or depth for beams, All the above, and many other earlier publications, do not give exact solution for any general polynomial variation of width and depth along the member. The need for exact solutions is obvious: modeling for tapered members will be simplified, and the solution time cut substantially. Another issue deals with optimization: if one tries to optimize the member

dimensions using the currently approximate or exact (limited cases) solutions, the problem becomes large and requires very long solution time, yielding only a local minimum of the objective function. Better solutions will enable much faster optimization with convergence to the global minimum. The word ‘exact’ in the title means exact up to the accuracy of the computer. It means also that if we solve a particular case using one element, and solve it again using two or more elements, we are going to get the same numerical solution. The same is true for regular prismatic beam elements, where we get the exact solution using any number of elements. If we use the finite element method we get better results as we refine the mesh. In this paper, a new method [6] is used for variable cross-section members, yielding exact results, and avoiding the need for mesh refinements and error estimates.

The differential equations that govern the displacement (axial, twist and bending) of a tapered member should be solved in order to obtain the required stiffnesses. A typical member is shown in Fig. 1 with the coordinate system and the numbered degrees of freedom. The stiffness matrix is of size 12 x 12, for the following degrees of freedom at the two ends of the member: three translations and three rotations.

5

11

2

8 L

7

10

* x

Fig. 1. Typical member and degrees of freedom.

766

M. Ersarmaacza

These are governed by the following equations: (a) For axial displacements

differential

A(x)=

Q(x) = where A(x) is the cross-section area along the beam, E is Young’s modulus of elasticity, u is the axial displacement, and Q(x) is the distributed axial load along the member. (b) For torsional rotations -

A,?

(5)

,go

(6)

i i-0

Qix’

where 1,m are integers representing the number of terms in each series. This presentation is very general, and many functions can be represented in this way, exactly or up to any desired accuracy. If we introduce a new local variable c

-$ GJ(x) 2 =T(x), [

(7)

1

where J(x) is the polar second moment of inertia, G is the shear modulus, B is the angle of twist, and T(x) is the distributed torsional moment along the member. (c) For bending displacement (translations and rotations) in the two planes of bending

we have for eqns (l), (S), and (6)

=4(C)(8)

a(T) = f: EA,L’~’ = i i-0

q(C) =

a,(’

(9)

i-0

i$o QiLi+25i =i

i-0

qrt’.

(10)

Now we choose the solution a(&) as the following infinite power series where Z,(x) and Z,(x) are the moment of inertia along the beam in the two bending planes, v and w are the lateral displacements, P,(x) and P,(x) are the distributed lateral load along the member in the two directions, respectively, and R(x) = El(x) for the two directions. The solution for the general case of polynomial variation of A(x), J(x), L(x), Z,(X), Q(x), 7”(x), P,,(x) and P,(x) along the beam is not generally available. Using the finite element technique, it is possible to derive the terms in the stiffness matrix. We assume that the shape functions for the element are polynomials and we have to find the approp~ate coefficients. It is widely known that the exact terms will result, if one uses the solution of the differential equation as the shape functions, for the derivation of the terms in the stiffness matrix. In this work, ‘exact’ shape functions are used, to derive the exact stiffness coefficients. These shape functions are ‘exact’ up to the accuracy of the computer, or up to a preset value set by the analyst.

u(r) = f u,<‘. i-0 Calculating all the derivatives and substituting expressions back into eqn (8) we have

the

coi

- C C (i-k

+ l)(i -k

+2)~~~,_.~+~<~

i-Ok-0

(12) To satisfy this equation for every value of 5, we must have -kio(i-k+l)(i-k+2)=&aj-&+l

AXIAL STIFFNESS

-

We take the coefficients in eqn. (1) as the following ~lyno~al variation along the beam

(11)

i

(k +

l)(i -k

+ l)a&+iai_&+l =Qi

(13)

k=O

or

-k~l(i-kf~)(i-k+2)a,u,-,+2-k~o(k+~)(~-~+~k+,~,-k+,-q, u i+2=

(i -t

l)(i i2)uo

(14)

761

General variable cross-section members

so that in eqn (I 1) we have all the a, coefhcients, except for the first two, which should be found using the boundary conditions. The terms for ut+* converge to 0 as i + co. As we intend to use this formulation for the derivation of the element stiffness matrix, all the boundary conditions will be displacements. At c = 0 we have

so the 6rst term is known from the boundary conditions. The term u1 is found as follows: All the u,s are linearly dependent on the first two, and we can write u(1) = f tlj = c&l + c,u, + f C+qp i-o i-0

(16)

The coefficients Co, C, , and C, are functions of all the coetIicients in u(e), 6(e), and q(t). Co for example, is the value of u(l) when a,, = 1 and a, = q, = 0 calculated from eqn (11) using the recurrence formula in eqn (14). In general we can write all the C coefficients as follows:

k-0

(b) For the second shape function N,(l) = I] take (iu, = 0 and then

‘4/,& 1.2

[iv;(O) = 0;

(21)

and then all the 4Vi,2are evaluated using eqn (14). The subscripts in the terms for Q, in the above equations have the following meaning: i is the term number in the series for shape function number k. The shape functions that are found using this technique have the special propery that they are the ‘exact’ solution for the differential equation. The word exact in the previous sentence stands for ‘as exact as we can get on a digital computer’. This is so since the calculation of the C coefficients is stopped according to a preset criteria: it could be until the contribution of the next element is less than an arbitrary small L (in most of the cases E was chosen as 10-18) or until the C values converge completely (for the accuracy of the computer). This property of the shape functions enables us to find the terms in the stiffness matrix in a much faster alternative way [rather than using eqn (1911 as

k-2

with z&k [from eqn (14)] based on u, = 1, u~,~ = q. = 0; i,k=0,1,2 ,..., o,n=O,1,2 ,..., co,and

(23)

(18) with uk [from eqn (14)] based on ui = 0; i = 0, 1, and using the values qi for the particular loading. Then, knowing all the terms in eqns (17) and (18), the value of or,[eqn (1511,and the boundary conditions at x = L (t = 1) we can solve eqn (16) and fmd the unknown z+. Thus, for any given variable polynomial functions [eqns (S)-(6)] we can find all the coefficients ui in eqn (11). The terms in the stiffness matrix are found in the traditional finite element method using the following relation

(2.5) The terms that are calculated using these equations are exactly equal to the terms that are obtained using eqn (19). For the ~lculations, we need only find S(1, 1) and then use the equilibrium conditions S(1, 1) = S(7,7) = -S(l, 7) = -5(7, 1). Nodal loads are found similarly, using eqns (20) and (21), but with the current leadings and u(0) = u(l) = 0. TORSIONAL

where N’(t) are the derivatives of the basis functions. The basis functions (also called shape functions) are found using eqns (I l), (15), and (16) for an unloaded member [i.e. q(x) = 0] with boundary conditions as follows: (a) For the tirst shape function [N*(O)= 1; N,(l) = 0] take ao., = 1 and then

and then all the 4Y,‘,,are evaluated using eqn (14).

STIFFNESS

The differential equation for the torsion of a member [eqn (211is very similar to the one for axial elongation [eqn (l)], If we replace eqns (5), (6), (9), (IO), (1 I), and (14) with the following equations J(x) = i

J,x’ 1-O

(26)

T(x) = g T,x’ 1-O

(27)

j(t)= i GJ,L’C’= i j,t’ I-0

1-O

(28)

(29)

+f

t: (i-k+l)(i-k+2)(i-k+3) i-0 k-0

(30)

-k~,(i-k-bl)(i-k-b2&6i_k+, e i+2=

x (i -k

(37)

+4)rky,-k+,5’=~o~,~‘.

-k;o'k+

I)(i-k+l~k+lei-k+l-fi

(31)

(i + l)(i + 2)jo

Then we can derive in a similar way the torsional stiffnesses as S(4,4) = - F and S(4,4) = S(i0,

a,*,

10) = -S(4,

To satisfy this equation for every value of C, we must have

(32)

10) = S(10,4).

x (i -k + 2)rk+2yl-k+2 BENDING

STIFFNIiSBBB

The bending in the two principal planes is governed by the same differential equation [eqns (3) and (411 with with different subscripts. In the following derivatin y will be used for Dor w, and the supscripts will be dropped and should be added only to the final formulas for the appropriate stiffness calculations. For bending, the differential equation is a fourth order equation and the expressions will be longer and more complex. Following the procedure as for the previous two cases, we use the same local variable and pol~omial variation of properties along the beam, so that eqns (3) or (4) can be written as

d2 @ [

wdy, 1=P(a d*y

+ i

2(k+l)(i-k+l)(i-k+2)

k-0

X

(j -k + 3)rk+lYi-&+3

+ i

(i-k+l)(i-k+2)(i-k+3)

k-0

(38)

or 1 “+‘=

(i + l)(i + 2)(i + 3)(i + 4)ro

(33) x

with

[

pt-k$o(k+I)(k+2)(i-k+i)

(W i-0

i-0

x (i--k

(35) Now we choose the solution y(() infinite power series

i-0

Calculating all the derivatives and substituting expressions back into eqn (33) we have

+%+ZYI-k+2ti

+f

2(k+l)(i-k+l)(i-k+2)

i

i-Ok-0

+ l)(i -k+I)(i-k+2)

as the following

Y(t) = f Yit'.

X(j -k

-*c,2(k I

+%'k+2Yi-k+2

(36)

the

x (j -k

+4)rk%-k+4

1

and as for the previous derivations, we have all the y1 coefficients except for the first four, that should be found using the boundary conditions. The terms for yj+4 converge to 0 as i+co. For this case we choose as degrees of freedom in the formulation the lateral deflection and rotation at the two ends of the beam element. At t: = 0 we have n-y(O)

WJ)

Geaeral

variablecross-sectionmembers

769

the @iBness matrixare definedas the holding actions at~~~of~~d~to~t~8tion~ (411 rotation, at each of the four degrees of fi-eedom, one Yl = Y’(O) at 8 time. Thus, there are four sets of boundary conditions as follows: (1) y(O)- 1; Y’(~~~Y(l) so the firsttwo terms8re readily known from the = y’(1) = 0; (2) y’(0) = 1; y(0) = y(1) = y’(1) - 0; (3) boundary conditions. y(l)= 1; y{0)=y’(0)=y’(1)=0; (4) y’(l)= 1; The terms y, and yp are found as follows: all the y(0) -y’(O) r-iy(1) = 0; Correslxulding to these four y,s are linearly dependent on the tirst two, and we sets there are four solutions r,, i = 1,2,3,4 for y(t) can write which are found using eqns (36), (39), and (40)-(43). Then, the holding actions will be and

Y (1)=

5Yt = COY,+ CIY,+ f;2Y2

I-0

+

Y’U) = f

c3Y3+

iyt =

i-1

+

The

(42)

i c*lFi i-0

c;y*+ c;y, + c;y, (49)

GY,

+

i I-0

c;d+.

(43)

r(l) d’bcr 1 dr(1) d$, Y(l)= - I;S.@-sq-dt

ten C coefficients (C,, C,, C,, C,, CA, C;, C;,

C;, Cpr,and C;) are functions of all the coe&ients in t(C) and p(5). C, for example, is the value of y(1) wheny,=l andy,=y~=y3=pr=Ocal~la~from eqn (36) using the recurrence formula in eqn (39). In general we can write all the C coe%cients as follows: c,=y(l)=~y,=l+~y* k-0

C;=y’(l)=

k-4

5 !cyk==i+ i ky, t-4 k-l

(44)

(45)

(50)

M(l) where V is the shear force and A4 is the moment. The stiffness for tire u-o plane are

both with yk [from eqn (39)] based on yi = 1, yk+* i,k-0,1,2 ,... *Co, n=0,1,2,...,Co, =&I---0; and (53) S(8, i) = - Ff3&k (47) &l=Y’w=ki4ky, _ I -F*QR both with yt [from eqn (3911 based on yI= 0; i = 1,2,3, and using the values pi for the particular loading. Then, knowing all the terms in eqns (44)~(47), the values of ya and yt [eqns (40)-(41)], and the boundary tuition at n = L (t = 1) we c8n solve eqns (42) and (43) and find the unknowns yr and y, . Thus, for any given variable polynomial ftions ~n~~~9~~~3S~]we can find all the coefhcients yI in As was shown for the axial stiffnesses there is no need to use the traditional &rite element formulation [eqn (1911,and terms in the stiffness matrix can be found directly from tbe shape functions. The terms in

S(l2, i) i=i~~~2k(k =

- l)(k - 2)1y,&

_

- l)Y,&

W

- l)~,+u,

(54)

(55)

where pls are calctdated using the r, coe&ients. And the stifbresses for the u-w plane are S(3,i)=6ypL3+2%$br,

(56)

R,(O) S(5, i) = -2 7 &Q

m

770

M. ~~ S(9, i) = - J?$Q(k

-

r-J..-.-

ykf2 f k-2

and r in rows 4 and 11 is

- l)(k -2)&Q

k(k - l)&&

w

-

(59)

1)Yi.k

where Y,~ are calculated using the rw coefficients. Nodal loads are found similarly, using eqns (43) and (44), but with the current loadings and y(O) = y’(O) = y(1) = y’(1) = 0. The actual calculations can be shorter by evaluating only the terms S(i, k) with i, k < 6, and utilizing the equilibrium equations to find all the other terms in the matrix.

(62)

2(1 + v)

(58)

and IC is the correction factor for twist for square cross section, and v is Poisson’s ratio for the material. From equilibrium considerations one should have: s,=s,+q. sr=s3+ss; s3 =s,+s,; The displacements due to unit loads in the six directions at the tip of the cantilever are

Ed,=O.4058063lOO3506 Ed,, = Ed,= 0.33153209878944 Erd,,= 0.59647171959271

APPLICA~ON OF THE THEORY

Ed,, =EdE2=0.63588683131572.(63)

The stiffness matrix and the equivalent nodal loads that were derived in the previous sections have been tested on numerous examples. The values of the stiffnesses were computed to the exact values given by Karabalis and Beskos [3] and by Banerjee and Williams [2], for the types of variation that they solved. For other variations along the member, where exact solutions were not available, a comparison was made to the solutions obtained from the three approximate methods given by Eisenberger and Reich 151.In all these cases ail the results were exactly the same. As an example, a cantilever with variable square cross section was solved. The cantilever was fixed at t = 0 (x = 0). The beam of unit length, and the variation of the width and depth is given by b(x) = h(x) = 2 -x2.

(60)

The stiffness matrix for this member is given as

S=E

Sl 0 0 0 0 0 - $1 0 0 0

0 0

These values are exactly the converged values for the respective solutions using the approximation methods in [5]. The values of yi, the coefficients in the series for the final solution for unit load in the direction of d.o.f. number 8, are given in Table 1. These values are found as EYi = Edsz/i,l\+ Edirgi.4

with d2= & = 0.Here we make use of the reciprocal theorem of Maxwell, in that the rotation, at the tip of the cantilever, when unit load is applied in the direction of d.o.f. number 8, is equal to the deflection d,2 when a unit moment is applied in the direction of d.o.f. number 12. These values were identical in the two solutions. The sum of all the yi, is equal to the tip deflection, as is shown in the respective columns in the table. It can be seen that the convergence is very fast, demonstrating the convergence of terms

0

0

0

0

0

s2

0

0

0

s,

0

-s2

0 0 0 “d’

0 0 0 0

0 0 0 -sj 0

s3

0

0

0 % 0

0 0 $6

0 0 0

0 0 0

“0’ - s3

0 SJ 0

-Y3

s3

0

0

“0’

0 -sr 0

0 0

0 0 0

0 o-s,;

0 0 s5

-OS2 -s,r - s5 0 0 0

-s,

0

:

with s, = 2.46422979453821;

sr = 6.17072~897968~

sj = 4.52610390627345;

s, = 3.73893753839913;

sS = 1.64462278352342;

s, = 0.78716636787432;

s, = 0.85745641564910;

ss = 1.67652541965079

(64)

0 0 -s,

0

0

0

0

0

0

0

ss

-6’ sj 0 0 0

0 -s,r 0 0 0 0

0 "0" s,r s5 0 0 0

-Y” “d 0 0

"0'

"d

0 0 0 s, 0 -sg

(61)

0 0 0 s,

in eqn (39). From statics we have V(0) = 1 and M(O) = 1, and these check with the eqns (48) and (49), when we substitute the values of y, (from Table 1) as follows: V(O)=6$0.125=1

(65)

771

General variable cross-section members

Table 1. Coefficients in the 6nal solution for unit load in direction 8 i

Yi

c;yj

i

Yi

CbYj

0

o.ooooooooooooooe+oo

o.ooooooooooooooe+OO

55

0.13749707828866e-07

-.33153209827423e+OO

1

o.ooooooooooooooe+OO

O.oooooooooooooOe+OO

56

-.73659149083206e-O8

-.33153210564014e+OO

2

-.375oooooooooooc+OO

-.375oooooooooooe+OO

57

0.7107461753643le-08

-.33 153209853268e+OO

3 4

o.125oooooooooooe+oo -.125oooooooooooe+OO

-.25oooooooooooOe+OO -.375oooooooooooe+OO

58 59

-.37988157648779e-O8 0.36700423491195e-08

-.3315321023315Oc+00 -.33153209866145e+OO

5

0.74999999999999e-01

-.3ooooooooooooOe+OO

60

-.19573559195302e-08

-.33153210061881e+OO

6 7 8

-.62499999999999e-O1 OA4642857142857e-01 -.33482142857142e-O1

-.362SoooooooooOe+oo -.31785714285714e+OO -.35133928571428e+OO

61 62 63

0.18931803156113e-08 -.10076604905672e-08 0.97567126864449e-09

-.33153209872563e+OO -.33153209973329e+OO -.33 153209875762e+OO

9 10

0.26041666666666e-01 -.18229166666666e-O1

-.32529761904761e+O0 -.34352678571428e+CQ

64 65

-.51832536146734e-O9 0.50237688880682e-09

-.33 153209927595e+OO -.33153209877357e+OO

11 12

O.l4914772727273e-01 -.99431818181817e-O2

-.32861201298701e+OO -.33855519480519e+OO

66 67

-.26641198648845e-O9 0.25845938987686e-09

-.33153209903998e+M -.33153209878152e+OO

13

0.84134615384614e-02

-.33014173326673e+OQ

68

-.13683144169950e-09

-.33 153209891835e+OO

14

-.54086538461538e-02

-.33555038711288e+OO

69

0.13286531295459e-09

-.33153209878549e+OO

15

0.46874999999999e-02

-.33086288711288e+OO

70

-.70228808275994e-10

-.33153209885572e+OO

16

71

0.68250531986527e-10

-.33 153209878747e+OO

17 18

-.29296875OQOOOOe-O2 -.33379257461288e+OO -.33120755623053e+OO 0.25850183823529e-02 -.33278728968641e+OO -.15797334558823e-02

72 73

-.36021114103999e-10 0.3503423426553le-10

-.33153209882349e+OO -.33 153209878845e+OO

19 20

0.141344572.36842e-02 -.84806743421051e-03

-.331373843%273e+OO -.33222191139694e+OO

74 75

-.18463988329131e-10 0.179716153WO19e-10

-.33 153209880692e+OO -.33153209878894e+OO

21

0.76729910714284e-03

-~3314546122898Oe+Oil

76

-.945874489843 15e-11

-.3315320987984Oe+OO

22 23

-.45340401785713e-03 0.41397758152173e-03

-.33190801630765e+OO -.33149403872613e+W

77 78

0.92130632127567e-11 -.48427639964493e-11

-.33153209878919e+OO -.33153209879403e+OO

24

-.24148692255434e-O3

-.33 173552564868e+OO

79

0.47201623762852+11

-.33153209878931e+OO

25 26

0.222167%875OOOe-03 -.128173828125OOe-03

-.33 151335767993e+OO -.33164153150806e+OO

80 81

-.247808524755OOe-11 0.24168979574865e-11

-.33153209879179e+OQ -.33153209878937e+OO

27 28 29

0.1186794704861 le-03 -.6781684@277776e-04 0.63139816810343e-04

-.33152285203757e+00 -.33159066887785e+OO -.33152752906104e+OO

82 83 84

-.12673977094138e-11 O.l2368580055722e.-11 -.64787800291886e-12

-.33 153209879064e+OO -.3315320987894Oe+oO -.33153209879005e+OO

30 31 32

-.35779229525861e-O4 0.33470892137096e-04 -.18827376827116e-O4

-.33156330829057e+CiI -.33 152983739843e+OO -.33154866477526e+OO

85 86 87

0.63263381461469e-12 -.3310293216OU77e-12 0.32341945213857e-12

-.33 153209878942e+OO -.33 153209878975e+OO -.33 153209878943e+OO

33 34

0.17686323686079e-04 -.98835338245734e-O5

-.33153097845157e+Cil -.3315408619854&+00

88 89

-.16906016816338e-12 O.l6526106326077e-12

-.33 15320987896Oe+OO -.33153209878943e+OO

35 36

0.93187604631693e-05 -.51770891462051e-O5

-.33153154322493e+OO -.33153672031408e+OO

90 91

-.86302999702869e-13 0.84406230478593e-13

-.33 153209878952e+OO -.33153209878943e+OO

37

0.48972464896535e-05

-.33153182306759e+OO

92

-.44038033293192e-13

-.33153209878948e+OO

38

-.27063730600716e-O5

-.33 153452944065e+CO

93

0.4309097881374&-13

-.33153209878943e+OO

39

025675846980167e-05

-.33153196185595e+OO

94

-.22462318743344e-13

-.33153209878946e+OO

40 41 42

-.14121715839091e-05 0.13432851651819e-05 -.73560854283768e-O6

-.33153337402753e+00 -.33 153203074237e+OO -.33153276635091e+OO

95 96 97

0.2198942782242&-13 -.11452826990848e-13 O.l1216686228143e-13

-.33153209878943e+OO -.33153209878945e+OO -.33153209878943e+OO

43

0.70139419200803e-06

-.33153206495672e+OO

98

-.58372550779138e-14

-.33153209878944e+OO

44 45

-.38257865018619e-06 0.36557515462236c-06

-.33153244753537e+OO -.331532081%022e+OO

99 100

0.571933t-Y732901&-14 -.29740519811103e-14

46

-.198682149251Ule-0

-.33153228064237e+OO

101

0.29151598626704.e-14

-.33153209878943e+OO -.33153209878944e+OO -.33153209878944e+oo

47 48

0.19022758970867e-06 -.10303994442553e-&

-.33153209041478e+OO -.33153219345472e+OO

102 103

-.15147399286432e-14 O.l4853275028432e-14

-.33 153209878944e+OO

49 50

0.98834232408162e-W -.53370485500405eU7

-.33153209462049e+C0 -.33153214799097e+OO

104 105

-.77122774186133e-15 0.75653768%3479e-15

51 52 53 54

0.51277525284704e-07 -.27610975153301e-O7 0.2656905156261le-07 -.14268564728068eJJl

-.33153209671345e+OO -.33153212432442e+OO -.33 153209775537e+OO -.33153211202394e+OO

106 107 108

0.385205885879Ole-15 -.19973638527073e-15

-.33 153209878944e+OO -.33 1532098789#+00 -.33 153209878944e+OO -.33 153209878944e+OO -.33 153209878944e+OO

109

O.l%07149746741e-15

-.33 153209878944e+OO

-.39254314084847e-15

-.33 153209878944wOO

772

M. ~~~

M(0) = -2.;.(-0.375)=1.

(66)

The second example involves the calculations of the shortening and bending displacement of the cantilever due to its own weight. The results are given in terms of p the specific weight of the beam material. For the axial deformation (when the beam is viewed as a column) the equivalent end forces were calculated as 2.016165236379p and 0.85050141430287p, for the strong and weak ends of the beam, respectively, and the displacement at the tip was -0.34513884061785p/E. For bending of the member under its own weight, the values for the four equivalent forces were 1.99167795~87p, 0.875048871658Op

0.39112511014349p, and

-0.099507315133612p.

The tip deflection was -0.226831344076p/E. The results for the booing and deflection were exactly the same as found using the approximate methods, Here again we can check the shear and moment at the left end of the cantilever as follows: from statics we have V(O)=p

t(4-4X~+X3dx I0

‘(4-4x2fX4)Xd.X I0

(67) =fp

= l.l6666666*..p

V(0) = 6 *; * 0.35833333 - +* p = 2.86666666. . . p

(69)

;. ( - 0.4375)p

= 1.16666666 * ’ * p

In this work, exact terms (up to the accuracy of the computer) for the stiffness matrix for any polynomial variation of the properties along the beam, are derived. The work includes developing formally the terms of the axial, torsional, and bending stiffnesses for a tapered member. The advantages of the new method are as follows: 1. The shape functions are derived automatically. 2. The shape fuctions are the exact solutions for the differential equation, and thus, the solution is the exact solution. 3. For continuous variation of member properties and loads, only one element is needed for the exact solution of the unknown function and its derivatives. 4. The exact solution is guaranteed. 5. There is no need for mesh refinements. The main idea in this new exact element method is that higher order elements, in the sense that high-order shape functions are derived, can be used without increasing the number of degrees-of-freedom. The solutions is exact upto any desired accuracy (depending only on the accuracy of the computations on the digital computer). Acknowledgements-This study was done during the author’s sabbatical leave from Technion. The author wishes to thank the Department of Civil Engineering and the Engineering Design Research Center at Carnegie Mellon University for the computing facilities that were provided for the completion of this research. Their warm and kind hospi~lity is gratefully appreciated.

(68)

and using from the solution yz = -0.4375~ and y3 = 0.35833333 * * *p in eqns (48) and (49) we get

M(0) = -2.

SUMMARY

=;p

= 2.86666666 . . . p M(O)=p

space frame programs, and into optimization programs that present the shape of the member as power series.

(70)

which shows the correctness of the solution. The procedure for calculating the exact stiffness matrix can be readily incorporated into plans and

REFERENCES 1. J. R. Banerjee and F. W. Williams, Exact BemoulliEuler dynamic stiffness matrix for a range of tapered beams. ht. J. Numer. Meth. Engng 21, 2289-2302 (1985). 2. J, R. Banerjee and F. W. Williams, Exact BemoulliEuler static stiffness matrix for a range of tapered barns. Int. J. Numer. Meth. Ennna 23. 16151628 (1986). 3. D. L. Karabalis and D. E. Beskos, Static, dynamic, and stability analysis of structures composed of tapered beams. CornDut. Struct. 16, 731-748 (1983). 4. M. Eisenbe&er, Explicit itiffnesss &at&es for nonorismatic members. Commrt. Struct. 20.71%7200985). 5. ‘M. Eisenberger and Y.-Reich, Static, vibration, anh stability analysis of non-uniform beams. Comput. Struct. 31, 567-573 (1989). 6. M. Eisenberger, An exact element method. Znt. J. Numer. Meth. Engng 30, 363-370 (1990). 1-