14.3
743
Fast Reanalysis Techniques
Thus the equilibrium equations of the structure can be expressed as
[K]Y = P
(E 3)
where
Y=
•y5 _
• y 6 ••_
••y _
7
•y _
_
(a) At the base design, A
of Eqs. (E
3)
and
P=
=A
=
2
= 2 in. 2, A
3
=A
•
•
•
•
_
0_
•
_
_
•
•
_
8
1
• p5 _
• p6 ••
••p __
7
• p8 _
4
Š1000
=_1 in. 2, and
0 the exact solution
gives the displacements of nodes 3 and 4 as
Y base =
•y5 _
• y6 ••
••y __
7
• y _ base
_
(b) At the perturbed design, A 1 = A
=
•
•
•
•
•
Š_
0.07032
•
2, A
2 = 2.4 in. •
8
0_
0 .001165__
0 .002329•_
in.
0 .05147
3
=A
4
•
_
= 1.2 in. 2, and the exact
_
solution of Eq. (E 3) gives the displacements of nodes 3 and 4 as
Y perturb =
•y5 _
•y 6 ••
•y 7__
• y8 _ perturb
•
•
•
•
=
0
_
•
.0009705
0 .001941 • in.
_
0 .04289
_
• Š 0.05860 •
• = A = 1.2•
(c) The values of A 1 = A 2 = 2.4
and A
4
at the perturbed
_
_ 3
in.2 design are used to compute the new stiffness
in.2matrix as [K
= [K] + [_K],
perturb]
which is then used to compute _Y 1, _Y 2, . . . using the approximation procedure, 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 approximate 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 number 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]
Y =
m×m m×1
P
m×1
where [K] is the stiffness matrix, Y the vector of displacements, and P the load vector.
Let the structure have n design variables denoted by the design vector
X=
•x1_ _
• •_
•
•x _2
•.._
•• . _ •
x
• n•
_
_
(14.23)
744
Practical Aspects of Optimization
Table 14.2
Exact Y 0 =
Value of i
•
•
•
•
•
•
•
•
0 .116462E Š 02_ _
0 .232923E Š 02_ •
0 .514654E Š 01_
Š 0.703216E Š 01
•
•_
•
Exact (Y 0 + _Y) =
•Yi
Yi
•
•
•
•
•
_
•Š0.232922E Š 03_
• Š 0.465844E Š 03_ •
_
•
•
Š 01_
•Š0.102930E
0 .140642E
Š 01 •
•
_
_
_
1
_
2
3
4
•
•
•
•
•
0 .465842E Š 04_
0 .186335E Š 05_
•
= Y 0 +•
•
0 .970515E Š 03_ _
0 .194103E Š 02_ •
0 .428879E Š 01_
Š 0.586014E Š 01
_
i
k=1
•
•
•
•Yk
_ .
0 .931695E Š 03_ _
0 .186339E Š 02_ •
0 .411724E Š 01_
•
•_
•_
• Š 0.562573E Š 01
•
• 0 .978279E Š 03_
•_ 0 .195656E Š 02_ •
•
•
• 0 .432310E Š 01_
• Š 0.590702E Š 01 _•
•_
•
• 0 .968962E Š 03_
• 0 .193792E Š 02_ •
_
•
•
• 0 .428193E Š 01_
• Š 0.585076E Š 01 •_
•_
•
• 0 .970825E Š 03_
• 0 .194165E Š 02_ •
_
•
•
• 0 .429016E Š 01_
• Š 0.586201E Š 01 •
•
•
_
•
•
• 0 .931683E Š 04_ ••
•
• 0 .205859E Š 02_
• Š 0.281283E Š 02 •
•
•
_
•Š0.931678E Š 05_
• Š 0.186335E Š 04_ _•
_
•
•
_
Š 03_
•Š0.411716E
0 .562563E
Š 03 •
•
_
_
_
•
•
•
•
•
•
0 .372669E Š 05_ •
0 .823429E Š 04_
Š 0.112512E Š 03 •
•
_
•
If we find the exact solution
at r basic design vectors X 1, X_2, . . . , X r, the corresponding
•
• by solving the equations
solutions, Y i, are found
_
[K i]Y i = P,
i = 1, 2, . . . , r
(14.24)
where the stiffness matrix, [K i], is determined at the design vector X i. If we consider a
new design vector, X N, in the neighborhood of the basic design vectors, the equilibrium
can be expressed as
equations at XN
[K N]YN
=P
where [K N] is the stiffness matrix evaluated at X N. By approximating YN
(14.25)
as a linear
combination of the basic displacement vectors Y i, i = 1, 2, . . . , r, we have
YN _ c 1Y 1 + c 2Y 2 + · · · + c rY r = [Y ]c
where [Y ] = [Y 1, Y 2, · · · , Y r] is an n × r matrix and c = {c 1, c 2, · · · , c r}T
r-component column vector. Substitution of Eq. (14.26) into Eq. (14.25) gives
[K N][Y ]c = P
(14.26)
is an
(14.27)
14.4
745
Derivatives of Static Displacements and Stresses
By premultiplying Eq. (14.27) by [Y ]T
we obtain
[ _K] c = P_
r×r
r×1
(14.28)
r×1
where
[ _K] = [Y ]T [K N][Y ]
(14.29)
T
(14.30)
P = [Y ] P
_
can be obtained by solvIt can be seen that an approximate displacement vector YN
ing a smaller (r) system of equations, Eq. (14.28), instead of computing the exact
solution YN by solving a larger (n) system of equations, Eq. (14.25). The foregoing
method is equivalent to applying the Ritz-Galerkin principle in the subspace spanned
by the set of vectors Y 1, Y 2, . . . , Y r. The assumed modes Y i, i = 1, 2, . . . , r, can be
considered to be good basis vectors since they are the solutions of similar sets of
equations.
Fox and Miura 14.3 applied this method for the analysis of a 124-member,
96-degree-of-freedom space truss (shown in Fig. 14.3). By using a 5-degree-of-freedom
approximation, they observed that the solution of Eq. (14.28) required 0.653 s while
the solution of Eq. (14.25) required 5.454 s without exceeding 1% error in the
maximum displacement components of the structure.
14.4
DERIVATIVES OF STATIC DISPLACEMENTS AND STRESSES
The gradient-based optimization methods require the gradients of the objective and
constraint functions. Thus the partial derivatives of the response quantities with respect
to the design variables are required. Many practical applications require a finite-element
31
z
y
25
32
34
19
26
20
13
33
28
35
22
16
23
17
30
15
2
11
24
9
6
40 in.
36
21
8
10
4
27
14
1
30 in.
29
7
40 in.
18
3
12
5
40 in.
6
x
Figure 14.3
Space truss [13.3].
746
Practical Aspects of Optimization
analysis for computing the values of the objective function and/or constraint functions
at any design vector. Since the objective and/or constraint functions are to be evaluated
at a large number of trial design vectors during optimization, the computation of the
derivatives of the response quantities requires substantial computational effort. It is
possible to derive approximate expressions for the response quantities. The derivatives
of static displacements, stresses, eigenvalues, eigenvectors, and transient response of
structural and mechanical systems are presented in this and the following two sections.
The equilibrium equations of a machine or structure can be expressed as
(14.31)
[K]Y = P
where [K] is the stiffness matrix, Y the displacement vector, and P the load vector.
By differentiating Eq. (14.31) with respect to the design variable x i, we obtain
•[K]
•xi
where _[K]/_x
i
Y + [K]
•Y
•xi
=
•P
•xi
(14.32)
denotes the matrix formed by differentiating the elements of [K] with
respect to x i. Usually, the matrix is computed using a finite-difference scheme as
•[K]
•[K]
•
•xi
•xi
=
Š [K]
[K
new]
•xi
(14.33)
where [K new] is the stiffness matrix evaluated at the perturbed design vector X + _X i,
where the vector _X i contains _x i in the ith location and zero everywhere else:
•X i = {0 0
. . . 0 _x
i
0 ... 0
(14.34)
T}
In most cases the load vector P is either independent of the design variables or a
known function of the design variables, and hence the derivatives, _P/_x i, can be
evaluated with no difficulty. Equations (14.32) can be solved to find the derivatives of
the displacements as
•Y
•xi
Since [K
= [K
Š] 1
/
•P
•xi
Š
•[K]
•xi
Y
0
(14.35)
Š1]
or its equivalent is available from the solution of Eqs. (14.31), Eqs. (14.35)
can readily be solved to find the derivatives of static displacements with respect to the
design variables.
The stresses in a machine or structure (in a particular finite element) can be determined using the relation
•
= [R]Y
(14.36)
where [R] denotes the matrix that relates stresses to nodal displacements. The derivatives of stresses can then be computed as
•
•xi
= [R]
•Y
•xi
(14.37)
where the matrix [R] is usually independent of the design variables and the vector
•Y/_x i is given by Eq. (14.35).