Engineering Optimization: Theory and Practice, Fourth Edition

740 Practical Aspects of Optimization

14.3 Fast Reanalysis Techniques

14.3.1 Incremental Response Approach

Let the displacement vector of the structure or machine,Y 0 , corresponding to the load

vector,P 0 , be given by the solution of the equilibrium equations

[K 0 ]Y 0 =P 0 (14.7)

or

Y 0 =[K 0 ]−^1 P 0 (14.8)

where [K 0 ] is the stiffness matrix corresponding to the design vector,X 0. When the

design vector is changed toX 0 + X,let the stiffness matrix of the system change to

[K 0 ] +[K], the displacement vector toY 0 + Y,and the load vector toP 0 +P.

Theequilibrium equations at the new design vector,X 0 + X,can be expressed as

([K 0 ]+[K])(Y 0 +Y)=P 0 +P (14.9)

or

[K 0 ]Y 0 + [K]Y 0 +[K 0 ] Y+[K]Y=P 0 +P (14.10)

Subtracting Eq. (14.7) from Eq. (14.10), we obtain

([K 0 ] +[K])Y=P−[K]Y 0 (14.11)

By neglecting the term [K]Y, Eq. (14.11) can be reduced to

[K 0 ]Y≈P−[K]Y 0 (14.12)

which yields the first approximation to the increment in displacement vectorYas

Y 1 =[K 0 ]−^1 ( P−[K]Y 0 ) (14.13)

where [K 0 ]−^1 is available from the solution in Eq. (14.8). We can find a better approx-

imation ofYby subtracting Eq. (14.12) from Eq. (14.11):

([K 0 ]+[K])Y−[K 0 ]Y 1 = P−[K]Y 0 − (P−[K]Y 0 ) (14.14)

or

([K 0 ]+[K])(Y−Y 1 ) =−[K]Y 1 (14.15)

By defining

Y 2 =Y−Y 1 (14.16)

Eq. (14.15) can be expressed as

([K 0 ] +[K])Y 2 = − [K]Y 1 (14.17)