Compucrs & Sfrucrures Vol. 62, No. 2, pp. 243-251, 1991 Copyright 0 1996 Elsevier .Sciena Ltd

PII: soo45-7949(%)00180-0 Printed in Gmt Britain. All rights reserved

00457949/97 $17.00 + 0.00

MOORE-PENROSE INVERSE METHOD OF TOPOLOGICAL VARIATION OF FINITE ELEMENT SYSTEMS

Ping Liang, Su Hum Chen and Cheng Huang Department of Mechanics, Jilin University of Technology, Chang Chun 130025,

People’s Republic of China

(Received 23 June 1995)

Abstract-A topological modification method for structural variations of finite element system is studied in this paper. The Moore-Penrose (M-P) inverse theory and a new factorization of a stiffness matrix are used. A set of explicit formulations of variations are obtained. Two numerical examples are given to illustrate the validity of the present method. Copyright 0 1996 Elsevier Science Ltd.

INTRODUCTION

The design process of an engineering structure is an iterative structural modification process. Structural modification falls into three types according to its modified pattern: parameter, shape, topology. Par- ameter modification involves modifying structural parameters such as cross-sectional area, mass and rigidity, etc. Much work on the subject has been carried out over the past 30 years. Some representa- tive books and papers [l-5] have been published. Shape modification involves modifying the bound- ary under fixed topology of the structure. Signifi- cantly less effort has been devoted to this field [6, 71. Topology modification concerns modifying the topology of a structure. There has been relatively little work on the subject [8,9]. In spite of this, it is recognized that topology and shape modification are important for improving the design of a structure.

With the development of computers, finite element models have been widely used to predict structural responses (displacement, stress, strain, etc.) in structure mechanics. A finite element model must be subjected to many modifications in order to satisfy the design requirement. In this situation the modification of a finite element model in order to match the design requirement is often a time-consum- ing and frustrating job, because one has to assemble and solve a set of simultaneous equations and to analyze sensitivity iteratively.

In order to present a solution to this problem, Refs [lo] and [ll] present a new method-a topological variation method of a structure. Its characteristic is that a set of simultaneous equations are not assembled and solved; instead all are realized through the following three types of elementary topological

variations: type I-add a new element to the system or remove an old one from it; type II-change the rigidity of an element; type III-add a new support to the system or remove an old one from it. It is apparent that through those three types of elementary variations one can change a system into any other one and restore it to the original system.

In this paper a new factorization of stiffness matrix is introduced, and using the Moore-Penrose inverse theory a set of explicit formulations of variations different from Ref. [lo] are obtained to predict the responses of a varied system. The following sections will discuss this in detail, taking the two-dimensional constant strain triangular element system in linear isotropic elasticity as their representative, and the final results are also valid for other finite element model in plate, shell and three-dimensional solids, etc.

BASIC CONCEF’TS AND LEMMAS

Consider a finite element system of n nodes and m

triangular elements. Use a, /?, . . . , to denote the element number and i, j, k its vertices as shown in Fig. 1. The formulations of finite element system are given as follows (see e.g. Ref. [12]):

e = [tx, 4, cry]’ = BQ, (1)

u = [cT,,cT~,c.~~]~ = DC, (2)

K” = AtB’DB, (3)

f” = K=Q, (4)

243

244 Ping Liang et al

K=iK=, ?=I

KQ=P,

B= %, 0 h, 0 hi 0 ’ 0 c, 0 L’, 0 c!.

c, b, c, h, ci h, 1

(5)

(6)

(7)

(8)

h, = .l’, - ,L’i , c, = - .Y, + _Yk . (i. i. k), (9)

where t and u are strain and stress vectors, respectively; D is the elastic matrix; E is the Young’s modulus; v is Poisson’s ratio; A is the area of element cc; K’ is the stiffness matrix of element a; x,, y, are the coordinates of the vertices of element 3. K, Q and P are the global stiffness matrix, nodal displacement vector and applied load vector, respectively.

Introduce a matrix S into eqn (3) such that it makes STDS diagonal:

where

H’ = ABTS-TW’ = ;

X

L

c, w; ll,w; c,w; tl,w; ckw; bkWf T b, wf - c,w; b, w; - c,tc; b, w; - QW; , h, w: c, w; b, rv: L.,w; bkw; c,, w; 1 (12)

w’ = diag(n;, M.;, Ml;) =

“’ = J Et 2A(1 + v)’

w; = J Et 2A(l + I’)’

where then eqn (3) can be rewritten as

‘+‘: = Et

2,4(1 _ v)’ (14)

Denote each column in HZ by a vector h:, s = 1, 2, 3. Thus

H” = [h;, h;, h;]. (15)

K” = H’(H”)‘,

KY= i K:, (16)

(11)

K: = h:(h:)*. s = 1, 2, 3. (17)

K: is called a subelement and is denoted by the symbol (:), J = I, 2, 3.

Let

k

x

L; 0

Fig. 1. Triangular element where H is called the factor of K. Its column h; is called a subfactor of K or factor of K:, s = 1, 2, 3.

H=[H’,H’,..., H’“], (18)

where HZ (c( = I, 2, 3) has been extended to a matrix of the same dimension as HZnxjm by inserting zero entries in appropriate location. When matrices (or vectors) of different dimensions appear together in an operation, this rule also is followed.

From eqn (18) we can obtain

K= f K”=HHT, (19) 1= I

Topological variation 245

Use G to denote the Moor+Penrose inverse H+ of factor H, i.e. G and H satisfy the following four conditions:

HGH = H,

GHG = G,

(HG)T = HG,

(GH)T = GH. (20)

Apparently G and HT have the same dimensions. Use g to denote the row in G, whose location is the same as (h,“)T in HT, s = 1,2, 3. G” = [(g;)’ (g;)’ (g”;)T]T, G is called the inverse factor of K, ps” is called the inverse subfactor. Introduce the following subscripts:

dflp = &h:, dy = Gsh:, a:: = Gh:. (21)

The results from eqns (22)-(26) can be proved easily:

K+ = (H+)TH+ = GTG, (22)

rank(K) = rank(K+) = rank(H+) = rank(H), (23)

where K+ denotes the M-P inverse of K. Let U be an elementary matrix that alternates two

columns of H, then

(HU)+ = UH’, (24)

(H’)TGTG = G’, (h:)TGTG = g;, s = 1,2,3,

(25)

HHT(CI)T=Hz, HHr(&Z)‘=h:, s= 1,2,3.

(26)

If there is a support in a degree of freedom, then the row and column (row) that respond the degree in K (or in H) is deleted. From now on it is assumed that the rows (and the columns) that respond the degrees which have supports are already deleted from H (or K). For this reason the stiffness matrix of the stable structure is nonsingular, and K+ = K-1.

With G” known from eqns (1), (2), (6), (12), (25), the displacement vector Q, the stress vector u and the strain vector L are calculated by

Q = K-‘P = GTGP,

L = (l/A)S(W”)-‘G”P,

u = (l/r)S-W”G”P.

(27)

(28)

(29)

Lemma 1 (Greoille)

Let fi = [H i h], d = H+h, c = h - Hd, then

fi+= H+-dg [ 1 g ’ (30)

where

Remark: c = 0 o he R(H) o rank[H : h] = rank H. Proof: see Ref. [13].

Lemma 2

Let

fi= Ii HI [ 1 0 H2 ’

where rank(H : HI) = rank(H), Hz is a matrix, then

-H+H H;’ H_: . * 1

nonsingular

(32)

Proof: it can be easily seen that fi and fi+ satisfy eqn (20) (here HI = HH+H, is used).

Lemma 3

, heR(@, then

fi+ = B + dg, (33)

where

d = (1 - gh)-‘Bh. (34)

Proof: From lemma 1 h = fid, d = fi+h, B = 8+ - dg, and symmetry of H+H, we have gh = gild = (Bh)Td = (1 - gh)dTd, hence 1 - gh # 0, and d = (l/( 1 - gh))Bh. Then the conclusion is obtained.

Lemma 4

Let

rank[H i HI] = rank(H), dimensions of H and BT are the same and Hz is nonsingular, then

H+ =B. (35)

246 Ping Liang ut al.

1

d e

Fig. 2. (a) Connecting element; (b) branching element of having two supports in directions of .Y and J of new node k; (c) branching element of having no support in any direction of new node k; (d) branching element of having one support in the s-direction of new node k; (e) branching element of having one

support in the y-direction of new node k.

Pro& easily obtained from Lemma 2.

Lemma 5

Let HER” “, eER”, fcR’, R = H + ef’, and let

k, = H+e, k2 = fTH’, u = (I - HH*)e,

v = fT(I - H+H), a = 1 + fW-e,

(note that u = 0 o et-R(H), v = 0 + fgR(HT), a = 0 0 rank(H + efT) = rank H - I), then

8+ = H+ - k,k,+H+ - H+k+k 2 2

+ (k:H+k:)krk2. (u = 0, v = 0, a = 0), (36)

and

a+ = H+ + i vTk:H+ - ; ~1, (u = 0, a it: 01, (37)

where

p = k:k,vT + k,, qT = (vvT/a)k:H+ + kl,

h = k:k,vvT + a’.

Proof‘: see Ref. [14].

THEOREMS OF TOPOLOGICAL VARIATIONS

At first consider variations of type I. There are two kinds of elements: one is a connecting element, which connects three existing nodes i, j, k without any new node occurring as shown in Fig. 2a, the other one is a branching element, which branches out from two original nodes i, j and a new node k occurs at the same time (Fig. 2b-e). For convenience, the following four cases are discussed.

Case 1

This is the addition or removal of connecting or branching elements that have supports in the directions of x and y of k (Fig. 2a or b).

(1) Adding case. In this case, a new element TV can be added by adding subelements one after another. Notice that rank[Hi HE] = rank(H), from Lemma 1

Topological variation 247

after adding subfactor h,” the varied inverse factor becomes

$” = (1 + (d::)‘d::)-‘(d:f)‘G, (38)

@=gfl-dcg, f # ;. 0 0

(39)

(2) Removal case. In this case an element a can be removed by removing subelements one after another. The varied factor fi obtained from the original one H by removing h: , (s = 1, 2, 3) satisfies the conditions of Lemma 3, and then from eqns (24), (33) and (34) the varied inverse factor is

Case 2

The adding or removing of the branching element that has no support in any direction of k (Fig. 2~).

(1) Addition case. This variation can be completed by the following steps.

Step I: adding K; and K; to the original system, then by alternating rows and columns, the varied factor becomes

fi= I-I HI [ 1 0 H2'

where

biw; CjW; b,w; T -c,w; b,w; 1 -c,w; ’

From Lemma 3 we obtain

R+ = [“a -$“I,

H;l=; bk [ 1

2 Ir -ck w;(c; + b:)’

R=L c,’ + b:

bit, - bkci C,C~ + b,bk b,ck - b& cick+b,bk b,cj-b,Ck c,ck+ bjbt ’

From eqn (24) one gets

[I :; = (0 i H;'), (41)

O! = k! j -%Rh (B Z a). (42)

Step 2: add K; to former system. From Lemma 1 or eqns (38) and (39) the final varied inverse factor is obtained.

(2) Removal case. Similarly to case (l), we remove subelements by two steps.

Step 1: remove K; from the system. From Lemma 3 or eqn (40) the varied inverse factor can be obtained.

Step 2: remove other two subelements. From Lemma 4 and eqn (24), the final varied inverse factor is obtained by removing two rows that respond to those subelements from the former inverse factor.

Case 3

This involves adding or removing branching element that has one support in the x-direction of k (Fig. 2d).

(1) Addition case. Two steps are taken: first we consider bk # 0.

Step 1: add KY to the system, the varied inverse factor can be calculated from Lemma 2 and eqn (24). In this case, HI = (w;/~)[c,, b,, c,, b,F, HI = (btw;/2), and from Lemma 2,

H;’ = [2/(&w?)], R = (l/bk)[C,, b,, cj, bjlT> (43)

hence

g; = [O, . , 0, 2/&W;)], (44)

@ = (g, -gfR), /? # a, r = 1,2, 3. (45)

Step 2: add K: (s = 2, 3) to the former system, from eqns (38) and (39) the final inverse factor is obtained.

If bt = 0, then ck # 0 and the following steps are taken.

Step 1: add K; to the system. Similar to eqns (42) and (43), the varied inverse factor becomes

t; = [O, . . .,o, 2/(ckw;)l, (46)

g!=(g!, -g!R), /?#a, r= 1,2,3, (47)

R = (l/Ck)[-bj, c;, -b,, c,]‘. (48)

Step 2: add K: (s = 1, 3) to the former system, and the final inverse factor can be. calculated from eqns (38) and (39).

248 Ping tiang rt al

(2) Remocal case. If bk # 0, then we remove subelements B;, K; at first, and the varied inverse factor G can be calculated from eqn (40). After this, the final inverse factor is the matrix obtained by removing the inverse subelement 8; from G:.

If bi = 0, then we remove subelements KY, K:, and the varied inverse factor G can be calculated from eqn (40). After this, the final inverse factor is the matrix obtained by removing the inverse subelenlent @ from c.

Now consider variation of type II, i.e. how the inverse factor varies after w: is changed into 1;: = 1~: + Air: = tt‘:( 1 + m:). where my = A~~~/w,~, (m: can be assumed to be nonzero). When rn: = - 1,

i.e. removing h: from H, it is discussed above. Consider the case of rn; # 0, - I only.

Let e = m:h:, and h,: is the 1 column of H, f is a vector of having unique nonzero component f(l) = 1, then H = H + ef’.

From Lemma 5, we have

Case 4 k,=m:Gh:, kz=g:, u=O for[eoR(H)], This involves adding or removing a branching

element that has one support in the y-direction of node k (Fig. 2e). Similarly to Case 3. formulae are obtained as follows.

v = fT - &‘H.

(I) A&Q cuse. If cA # 0, then the following steps are taken.

If m: + - I, we have rank(H) = rank H, then LI = 1 f rn,;e; # 0. Using eqns (25) and (26) one obtains

Step I: add KY to the system, and the varied inverse factor is

k,’ = (l/k:k,)k:, g; = [O, , 0, 2i(crM’;)], (49)

gf = (gi’, -_%R), ,8 # 3, Y = l,2, 3, (50)

R = (1 /cl )I(;. h,. is,, b,]‘. (51)

Step 2: add K: (s = 2,3) to the former system, and from eqns (38) and (39) the final inverse factor is obtained.

vvr = fff - 2g:h: + g:HHT(gs’)T = 1 - 4:.

Therefore from the second equation of Lemma 5, we have

However, if cA = 0 then br f 0 and the following steps are taken.

where

Step I: add K; to the system, and varied inverse factor becomes

$j; = [O, . . , 0, 2,‘(bkn;)]. (52)

f I I) = k$ v’ + m:Gh:, q’ = (

I-6” ---Lm:+ 1 g:,

a >

h = (m:)‘c( I - 4:) + u:.

$=(g!, -g;R), /?#a, r= 1,233, (53) The inverse subfactors are

R = (I:bk)[b,. -c,, b,. -c,]‘. (54)

Step 2: add K: (s = I, 3) to the former system, and from eqns (36) and (37) the final inverse factor is obtained.

(2) Remocing case. if ck # 0, then remove K: (s = 1, 3) from the system at first, and the varied inverse factor G is obtained from eqn (40). Then remove g; from G to obtain the final varied inverse factor. Otherwise remove K (s = 2,3) from the system at first; the varied factor G is obtained from eqn (40) and then remove g; from G to obtain the final varied inverse factor.

Now we consider variation of type III. First consider the case of adding support. The varied factor H is obtained by removing the row that responds to the support from the original factor H. Two cases will be discussed. Let h(l) be the I row removed from H, g(f) is I column of G which is the M-P inverse of H.

It is noticeable that when adding element ~1, if there are supports in some degrees of freedom, then the rows that respond those degrees should be removed from matrices H”, H,, R, etc. in the above formulae, and assume that the original system is stable.

Case 1. Rank(H) = rank H. Replace H with H in Lemma 3 and use eqn (24), and the columns of G are

&I’) = g(l’) + I ~~~~~ g(f), I’ # f. (57)

Topological variation 249

M-P INVERSE VARIATION METHOD FOR STRUCIWRAL ANALYSIS

Case 2. Rat&(@ = rank H - 1. Let e be a vector of having a unique nonzero component e(l) = - 1, fT = h(l), then H + efT is the matrix obtained by replacing entries of II(/) with zeros. From Lemma 5, one obtains

k, = -g(l), k2 = h(l)G, u = e + Hg(Z),

uTu = 1 - h(l)g(f),

v = 0 [for II(~)~ER(HT)], a= 1 - h(l)g(l)= uTu.

From rank(H + efT) = rank H - 1, one has a = 0 and u = 0. Hence from the first equation of Lemma 5 and eqns (25) and (26) one has

Based on the concepts and equations presented in the preceding sections, a new way can be created for structural analysis. We selected a stable (not necessarily a statically determinate) element arbitrar- ily from the system, and treat it as the initial structure from which will grow out other elements. The inverse factor of initial structure can be calculated from eqns (38) and (39), i.e. first calculate (h;)+, then calculate [h;, h;, &]+ by using eqns (38) and (39) iteratively. Through three types of topological variations, i.e. using eqns (38)-(60), one can calculate the inverse factor of a given system, and the displacement vector Q, the stress vector u and the strain vector E can be obtained from eqns (27x29).

1 (H + efT)+ = G - gT(l)g(l) g(OgTU)G.

i.e.

g(r) = g(l’) - $+$ g(f), I’ # I (58)

Now we discuss the case of support removal. In this case the varied factor fi is obtained by adding a row h(l) that responds to that support to the original in an appropriate location. By replacing H with H’ in Lemma 1. we can obtain

g(f) = c # 0,

;; + dTd)-‘H+d c = 0,

H

[ 1 k(l) = L-HI’ - g(l)dT, :, g(l)l,

where dT = h(l)H+, c = h(f) - d’H. Hence from eqn (24) we have

g(‘) = ;; + dT,,-Ia c # 0, c = 0,

Node 2x Node 3x Node 3y 0.000000 2.280351 0.000000

G’ = 1.140175 0.000000 -1.140175 ) 0.836660 0.000000 0.836660

1 where the zero components at the degrees of having

(59) supports are ignored as well as in the following steps.

Step 2. Add subelements K: and K$ to the initial

(60) structure; from eqns (41) and (42), one has

Example 1

The system (plane stress) shown in Fig. 3b has E = 1.0, v = 0.3 and t = 1.0; the task is to find the stress u and the displacements Q produced by the load P shown in Fig. 3b. From eqns (9), (13) and (14), one has the initial data for element 1: b, = - 1, b2 = 1, 61 = 0, cl = - 1, cZ = 0, c, = 1 and a = 0.5. For element 2: bJ= -1, bl=O, b.,=l, cl=O, cz = - 1, cd = 1 and a = 0.5. W’ = W2 = diag(,/lO/

13, J10/13, J10/7). The solution procedures is as follows.

Step 1. Take element 1 as the initial structure (Fig. 3a). The inverse factor G’ can be calculated from eqns (30) and (31) or other methods quite easily.

M Fig. 3. (a) Initial structure; (b) plane system in example I.

250 Ping Liang el al.

Nodes 2.x 3x 3y

Node 2x Node 3x Node 3~ Node 4x Node 4y

0.~000 2.280351 0.000#0 1.140175 - l.I401175

I.140175 0.~00~ - 1.140175 0.000~ 0.~~0

G: = 0.836660 0.000000 0.836660 0.836660 0.836660

0.000000 0.000000 0.000000 1.140175 1.1401175

0.000000 0.000000 0.000000 1.140175 - I.1401175

Step 3. Add subelement K: to the former system; from eqns (38) and (39) one has

d: = (- I .362770, 0.000000, I .OOOOOO, I .326770, O.OOOOOO)T.

Node 2x Node 3s Node 3~ Node 4x Node 4y

0.199531 I .539237 0.199531 1.339707 -0.199531

1.140175 0.000000 - I.140175 0.000000 0.000000

Gfi,, 0.690245 0.543829 0.690245 0.690244 0.146416 =

-0.199531 0.741114 - 1.199531 0.940645 0.199531

0.~~00 0.0~~0 0.00~~ 1.140175 - 1.140175

0.146416 -0.543829 1.146416 0.146416 0.690244

Step 4. Load the system with P = [OS, 0.51T applied at degrees of y-directions of nodes 3 and 4 to obtain CT and Q. At first calculate GP:

GP = (0.~0000, -0.570088, 0.418330, 0.~0000, - 0.57~88, 0.418330)T.

From eqns (29) and (27), one has

6’ = (0.000000. 1 .ooooooo, 0.000000)‘,

(I2 = (0.0~000, 1 .~OO~, O.O~~O)T.

Node 2x Node 3x Node 3y Node 4x Node 4,~

Q = GTGP = (- 0.300000, 0.000000, 1 .OOOOOO, -0.300000, I .OOOOOO)T.

The results calculated above are the same as results in Ref. [lo].

Example 2

The system shown in Fig. 3b has added a support in the x-direction of node 4 (Fig. 4). The data of this system is the same as example 1. The inverse factor of the system is obtained as follows.

Step 1. Take element I as the initial structure (Fig. 3a). The inverse factor of initial structure is already obtained in step 1 of example 1.

Node 2x Node 3x Node 33’ Node 4y

0.000~0 2.280351 0.000000 0.~0000

1.140175 0.~~ - 1.140175 0.~~

0.836660 0.000000 0.836660 1.673320

0.000000 0.000000 0.000000 2.28035 1 I

Step 2. Add subelement K: to the initial structure, Step 3. Add subelement K: to the former system, for that b, # 0 from eqns (43)--(45), one has from eqn (38) and (39) one has

nodes 2x 3x 3y

R = (-1.0, 0.0, -l.O)T,

G: =

Topological variation 251

Fig. 4. Plane system in example 2.

(1.000000, 0.000000, -0.733799, - 1.000000)r,

(1) This paper presents a new method of M-P inverse topological variation using a new factoriz- ation of stiffness matrix and its inverse in a finite element system. Using this factorization a new point treatment of support is presented, i.e. a support in a degree of freedom responds to a row of factor H of stiffness matrix K.

G; =

Node 2x Node 3x Node 3y Node 4y

-0.173505 1.635904 -0.173505 -0.991457

1.140175 0.000000 -1.140175 0.000000

0.709342 -0.472895 0.709342 0.945790 .

-0.173505 -0.644447 -0.173505 1.288894

_ -0.173505 -0.644447 -0.173505 -0.991457

Step 4. Add final subelement K: to the former system, from eqns (38) and (39) the final inverse factor Genal is obtained.

(2) This method has the advantage of eliminating the need of assembling stiffness matrix and solving simultaneous equations.

(3) This method is suitable for numerical compu- tation on a computer.

(4) The method is suitable for interface design be- cause any local modification of a structure can be easily done using this method, consuming much less time compared to other common finite element methods.

1.

d: = (- 1.570148, 0.000000, 0.847826, 2.

1.155392, -0.207378)T, GM =

3.

Node 2x Node 3x Node 3y Node 4~ - 0.026738 0.625155 0.026738 0.152787 -

1.140175 0.000000 -1.140175 o.OOOOOO

0.601218 0.072875 0.601218 0.327937

-0.320853 -0.099312 -0.320853 0.446903 ’

-0.147058 -0.777942 -0.147058 -0.840330

_ 0.127531 -0.643728 0.127531 0.728749

Step 5. Load the system with P = [0.5, 0.51T applied at the degrees of y-directions of nodes 3 and 4 to obtain u and Q. At first calculate GP:

4.

5.

6.

7.

8.

GP = (0.0897626, - 0.570088, 0.464578,

0.063025, -0.493694, 0.428140)T.

From eqns (27) and (29) one has

Q = Node 2x Node 3x Node 3y Node 4y

GTGP = (0.386293, 0.148575, 0.386293, 0.907391)T,

u’ = (0.055276, 1.055276, 0.078727)T,

a2 = (0.078727, 0.944724, 0.055276)T.

9. U. Kirsch, Optimal topologies of structures. Appl. Mech. Rev. 42(8), 223-238 (1989).

IO. Ting-Yu Rong and An-Qi Lii, Theory and method of structural variations of finite element systems. AIAA .I. 32(9), 1911-1919 (1994).

I I. T. Y. Rong, General theorems of topological variations of elastic structures and the method of tooolonical variation. Acta Mech. Solida Sinica 1, 2943’(198?).

12. K. J. Bathe and E. J. Wilson, Numerical Methods in Finite Element Analysis. Prentice-Hall, Englewood Cliffs, NJ (1976).

13. A. Ben-Israel and T. N. E. Greville, Generalized Inverse: Theory and Applications. Wiley, New York (1974).

14. R. E. Hartwig, Singular value decomposition and the MoorePenrose inverse of bordered matrices, SIAM J. appl. Math. 31, 3141 (1976).

In Ref. [lo], the variation of the basic displacement matrix of example 2 has to be calculated in three steps: add a simply-supported element, add a constraint subelement and then add a support subelement. However, using the present method in this paper, steps 24 were realized by adding subelements and no variation of adding support was used.

CONCLUDING REMARKS

REFERENCES

S. Chen, Matrix Perturbation Theory in Structural Dynamics. International Academic Publishers, Beijing (1993). E. J. Huag, V. Komkov and K. K. Choi, Design Sensitivity Analysis of Structural Systems. Academic Press, Orlando, FL (1985). H. M. Adelmen and R. T. Haftka, Sensitivity analysis of discrete structural systems. AIAA J. 24, 823-832 (1986). Z. S. Liu and S. H. Chen, Reanalysis of static response and its design sensitivity of locally modified structures. Commun. hkmer. Meth. Engng 8(11), 797-800 (1992). T. Y. Chen, Structural modification with freauencv response constraints for undamped MDOF systems. Comput. Struct. 36, 1013-1018 (1990). M. P. Bendsoe, Generating optimal topologies in structural design using a homogenization method. Comput. Meth. appl. Mech. Engng 71, 197-224 (1988). B. H. V. Topping, Shape optimization of skeletal structures: a review. J. Struct. Engng ASCE 109, 1933-1951 (1983). H. L. Cox, The Design of Structures of Least Weight. Pergamon, London (1956).