12.4 Optimality Criteria Methods 687
Figure 12.6 Three-bar truss.
whereρis the weight density,Eis Young’s modulus,Umaxthe maximum permissible
displacement,x 1 the area of cross section of bars 1 and 3,x 2 the area of cross section
of bar 2, and the vertical displacement of nodeSis given by
U 1 =
P 1 l
E
1
x 1 +
√
2 x 2
(E 3 )
Find the solution of the problem using the optimality criteria method.
SOLUTION The partial derivatives offandgrequired by Eqs. (12.68) and (12.71)
can be computed as
∂f
∂x 1
= 08. 0445 ,
∂f
∂x 2
= 82. 3
∂g
∂zi
=
∂g
∂xi
∂xi
∂zi
=
∂g
∂xi
(−xi^2 ), i= 1 , 2
∂g
∂x 1
=
− 1
(x 1 +
√
2 x 2 )^2
,
∂g
∂x 2
=
−
√
2
(x 1 +
√
2 x 2 )^2
At any designXi, Eq. (12.70) gives
g 0 = g(Xi)+
∂g
∂x 1
∣
∣
∣
∣
Xi
xi 1 +
∂g
∂x 2
∣
∣
∣
∣
Xi
xi 2
=
1
xi 1 +
√
2 xi 2
− 1 −
xi 1
(xi 1 +
√
2 xi 2 )^2
−
√
2 xi 2
(xi 1 +
√
2 xi 2 )^2
Thus the values ofλand (x 1 , x 2 ) an be determined iteratively using Eqs. (12.71) andc
(12.68). Starting from the initial design (x 1 , x 2 ) =(2.0, 2.0) in^2 , the results obtained
are shown in Table 12.1.