14.3 Fast Reanalysis Techniques 743
Thus the equilibrium equations of the structure can be expressed as
[K]Y=P (E 3 )
where
Y=
y 5
y 6
y 7
y 8
and P=
p 5
p 6
p 7
p 8
=
0
0
0
− 1000
(a) At the base design,A 1 =A 2 = in. 2 2 , A 3 =A 4 = in. 1 2 , and the exact solution
ofEqs. (E 3 ) gives the displacements of nodes 3 and 4 as
Ybase=
y 5
y 6
y 7
y 8
base
=
0. 001165
0. 002329
0. 05147
− 0. 07032
in.
(b) At the perturbed design,A 1 =A 2 = 2. 4 in.^2 , A 3 =A 4 = 1. 2 in.^2 , and the exact
solution of Eq. (E 3 ) gives the displacements of nodes 3 and 4 as
Yperturb=
y 5
y 6
y 7
y 8
perturb
=
0. 0009705
0. 001941
0. 04289
− 0. 05860
in.
(c) The values ofA 1 =A 2 = 2. 4 in.^2 andA 3 =A 4 = 1. 2 in.^2 at the perturbed
design are used to compute the new stiffness matrix as [K]perturb= [K]+[K],
which is then used to computeY 1 , Y 2 ,... using the approximation proce-
dure, Eqs. (14.13) and (14.21). The results are shown in Table 14.2. It can be
seen that the solution given by Eq. (14.20) converged very fast.
14.3.2 Basis Vector Approach
In structural optimization involving static response, it is possible to conduct an approx-
imate analysis at modified designs based on a limited number of exact analysis results.
This results in a substantial saving in computer time since, in most problems, the num-
ber of design variables is far smaller than the number of degrees of freedom of the
system. Consider the equilibrium equations of the structure in the form
[K]
m×m
Y
m× 1
= P
m× 1 (14.23)
where [K] is the stiffness matrix,Ythe vector of displacements, andPthe load vector.
Let the structure havendesign variables denoted by the design vector
X=
x 1
x 2
..
.
xn
