15.2.1 Reduced Basis Technique

In the optimum design of certain practical systems involving a large number of (n) design variables, some feasible design vectors X ₁, X ₂, …, X _r may be available to start with. These design vectors may have been suggested by experienced designers or may be available from the design of similar systems in the past. We can reduce the size of the optimization problem by expressing the design vector X as a linear combination of the available feasible design vectors as

(15.1)

where c ₁, c ₂, …, c _r are the unknown constants. Then the optimization problem can be solved using c ₁, c ₂, …, c _r as design variables. This problem will have a much smaller number of unknowns since r ≪ n. In Eq. (15.1), the feasible design vectors X ₁, X ₂, …, X _r serve as the basis vectors. It can be seen that if c ₁ = c ₂ = ⋯ = c _r  = 1/r, then X denotes the average of the basis vectors.

15.2.2 Design Variable Linking Technique

When the number of elements or members in a structure is large, it is possible to reduce the number of design variables by using a technique known as design variable linking [15.17]. To see this procedure, consider the 12‐member truss structure shown in Figure 15.1. If the area of cross section of each member is varied independently, we will have 12 design variables. On the other hand, if symmetry of members about the vertical (Y) axis is required, the areas of cross section of members 4, 5, 6, 8, and 10 can be assumed to be the same as those of members 1, 2, 3, 7, and 9, respectively. This reduces the number of independent design variables from 12 to 7. In addition, if the cross‐sectional area of member 12 is required to be three times that of member 11, we will have six independent design variables only:

(15.2)

Once the vector X is known, the dependent variables can be determined as A ₄ = A ₁, A ₅ = A ₂, A ₆ = A ₃, A ₈ = A ₇, A ₁₀ = A ₉, and A ₁₂ = 3A ₁₁. This procedure of treating certain variables as dependent variables is known as design variable linking. By defining the vector of all variables as

the relationship between Z and X can be expressed as

(15.3)

where the matrix [T] is given by

(15.4)

The concept can be extended to many other situations. For example, if the geometry of the structure is to be varied during optimization (configuration optimization) while maintaining (i) symmetry about the Y axis and (ii) alignment of the three nodes 2, 3, and 4 (and 6, 7, and 4), we can define the following independent and dependent design variables:

Independent variables: X ₅, X ₆, Y ₆, Y ₇, Y ₄

Dependent variables:

Thus the design vector X is

(15.5)

The relationship between the dependent and independent variables can be defined more systematically, by defining a vector of all geometry variables, Z, as

which is related to X through the relations

(15.6)

where f _i denotes a function of X.

15.3.1 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

(15.7)

or

(15.8)

where [K ₀] is the stiffness matrix corresponding to the design vector, X ₀. When the design vector is changed to X ₀ + ΔX, let the stiffness matrix of the system change to [K ₀] + [ΔK], the displacement vector to Y ₀ + ΔY, and the load vector to P ₀ + ΔP. The equilibrium equations at the new design vector, X ₀ + ΔX, can be expressed as

(15.9)

or

(15.10)

Subtracting Eq. (15.7) from Eq. (15.10), we obtain

(15.11)

By neglecting the term [ΔK]ΔY, Eq. (15.11) can be reduced to

(15.12)

which yields the first approximation to the increment in displacement vector ΔY as

(15.13)

where [K ₀]⁻¹ is available from the solution in Eq. (15.8). We can find a better approximation of ΔY by subtracting Eq. (15.12) from Eq. (15.11):

(15.14)

or

(15.15)

By defining

(15.16)

Eq. (15.15) can be expressed as

(15.17)

Neglecting the term [ΔK] ΔY ₂, Eq. (15.17) can be used to obtain the second approximation to ΔY, ΔY ₂, as

(15.18)

From Eq. (15.16), ΔY can be written as

(15.19)

This process can be continued and ΔY can be expressed, in general, as

(15.20)

where ΔY _i is found by solving the equations

(15.21)

Note that the series given by Eq. (15.20) may not converge if the change in the design vector, ΔX, is not small. Hence it is important to establish the validity of the procedure for each problem, by determining the step sizes for which the series will converge, before using it. The iterative process is usually stopped either by specifying a maximum number of iterations and/or by prescribing a convergence criterion such as

(15.22)

where ||ΔY _i || is the Euclidean norm of the vector ΔY _i and ε is a small number on the order of 0.01.

15.3.2 Basis Vector Approach

In structural optimization involving a 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

(15.23)

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

If we find the exact solution at r basic design vectors X ₁, X ₂, …, X _r , the corresponding solutions, Y _i , are found by solving the equations

(15.24)

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 equations at X _N can be expressed as

(15.25)

where [K _N ] is the stiffness matrix evaluated at X _N . By approximating Y _N as a linear combination of the basic displacement vectors Y _i , i = 1, 2, …, r, we have

(15.26)

where [Y] = [Y ₁, Y ₂, ⋯, Y _r ] is an n × r matrix and c = {c ₁, c ₂, ⋯, c _r }^T is an r‐component column vector. Substitution of Eq. (15.26) into Eq. (15.25) gives

(15.27)

By premultiplying Eq. (15.27) by [Y]^T we obtain

(15.28)

where

(15.29)

(15.30)

It can be seen that an approximate displacement vector Y _N can be obtained by solving a smaller (r) system of equations, Eq. (15.28), instead of computing the exact solution Y _N by solving a larger (n) system of equations, Eq. (15.25). 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 [3] applied this method for the analysis of a 124‐member, 96‐degree‐of‐freedom space truss (shown in Figure 15.3). By using a 5‐degree‐of‐freedom approximation, they observed that the solution of Eq. (15.28) required 0.653 second while the solution of Eq. (15.25) required 5.454 seconds without exceeding 1% error in the maximum displacement components of the structure.

15.5.1 Derivatives of λ _i

Premultiplication of Eq. (15.39) by gives

(15.41)

Differentiation of Eq. (15.41) with respect to the design variable x _j gives

(15.42)

where Y _i,j  = ∂ Y _i /∂x _j . In view of Eq. (15.39), Eq. (15.42) reduces to

(15.43)

Differentiation of Eq. (15.40) gives

(15.44)

where ∂[K]/∂x _j and ∂[M]/∂x _j denote the matrices formed by differentiating the elements of [K] and [M] matrices, respectively, with respect to x _j . If the eigenvalues are normalized with respect to the mass matrix, we have [10]

(15.45)

Substituting Eq. (15.44) into Eq. (15.43) and using Eq. (15.45) gives the derivative of λ _i with respect to x _j as

(15.46)

It can be noted that Eq. (15.46) involves only the eigenvalue and eigenvector under consideration and hence the complete solution of the eigenvalue problem is not required to find the value of ∂λ _i /∂x _j .

15.5.2 Derivatives of Y _i

The differentiation of Eqs. (15.39) and (15.45) with respect to x _j results in

(15.47)

(15.48)

where ∂[P _i ]/∂x _j is given by Eq. (15.44). Equations (15.47) and (15.48) can be shown to be linearly independent and can be written together as

(15.49)

By premultiplying Eq. (15.49) by

we obtain

(15.50)

The solution of Eq. (15.50) gives the desired expression for the derivative of the eigenvector, ∂ Y _i /∂x _j , as

(15.51)

Again it can be seen that only the eigenvalue and eigenvector under consideration are involved in the evaluation of the derivatives of eigenvectors.

For illustration, a cylindrical cantilever beam is considered [4]. The beam is modeled with three finite elements with six degrees of freedom as indicated in Figure 15.4. The diameters of the beam are considered as the design variables, x _i , i = 1, 2, 3. The first three eigenvalues and their derivatives are shown in Table 15.3 [4].

i	Eigenvalue, λ i
1	24.66	0.3209	−0.1582	1.478	−2.298
2	974.7	3.86	−0.4144	0.057	−3.046
3	7782.0	23.5	21.67	0.335	−5.307

15.7.1 Sensitivity Equations Using Kuhn–Tucker Conditions

The Kuhn–Tucker conditions satisfied at the constrained optimum design X ^* are given by [see Eqs. (2.73) and (2.74)]

(15.64)

(15.65)

(15.66)

where J ₁ is the set of active constraints and Eqs. (15.64,15.65,15.66) are valid with X = X ^* and λ _j  =  . When a problem parameter changes by a small amount, we assume that Eqs. (15.64,15.65,15.66) remain valid. Treating f, g _j , X, and λ _j as functions of a typical problem parameter p, differentiation of Eqs. (15.64) and (15.65) with respect to p leads to

(15.67)

(15.68)

Equations (15.67) and (15.68) can be expressed in matrix form as

(15.69)

where q denotes the number of active constraints and the elements of the matrices and vectors in Eq. (15.69) are given by

(15.70)

(15.71)

(15.72)

(15.73)

(15.74)

The following can be noted in Eq. (15.69):

1. Equation (15.69) denotes (n + q) simultaneous equations in terms of the required sensitivity derivatives, ∂x _i /∂p (i = 1, 2, …, n) and ∂λ _j /∂p (j = 1, 2, …, q). Both X ^* and λ^* are assumed to be known in Eq. (15.69). If λ^* are not computed during the optimization process, they can be computed using Eq. (7.263).
2. Equation (15.69) can be solved only if the system is nonsingular. One of the requirements for this is that the active constraints be independent.
3. Second derivatives of f and g _j are required in computing the elements of [P] and a.
4. If sensitivity derivatives are required with respect to several problem parameters p ₁, p ₂, …, only the vectors a and b need to be computed for each case and the system of Eq. (15.69) can be solved efficiently using the techniques of solving simultaneous equations with different right‐hand‐side vectors.

Once Eq. (15.69) are solved, the sensitivity of optimum objective value with respect to p can be computed as

(15.75)

The changes in the optimum values of x _i and f necessary to satisfy the Kuhn–Tucker conditions due to a change Δp in the problem parameter can be estimated as

(15.76)

The changes in the values of Lagrange multiplier λ _j due to Δp can be estimated as

(15.77)

Equation (15.77) can be used to determine whether an originally active constraint becomes inactive due to the change, Δp. Since the value of λ _j is zero for an inactive constraint, we have

(15.78)

from which the value of Δp necessary to make the jth constraint inactive can be found as

(15.79)

Similarly, a currently inactive constraint will become critical due to Δp if the new value of g _j becomes zero:

(15.80)

Thus, the change Δp necessary to make an inactive constraint active can be found as

(15.81)

15.7.2 Sensitivity Equations Using the Concept of Feasible Direction

Here we treat the problem parameter p as a design variable so that the new design vector becomes

(15.82)

As in the case of the method of feasible directions (see Section 7.7), we formulate the direction finding problem as

subject to

(15.83)

where the gradients of f and g _j (j ∈ J ₁) can be evaluated in the usual manner. The set J ₁ can also include nearly active constraints (along with the active constraints), so that we do not violate any constraint due to the change, Δp. The solution of the problem stated in Eq. (15.83) gives a usable feasible search direction, S. A new design vector along S can be expressed as

(15.84)

where λ is the step length and the components of S can be considered as

(15.85)

so that

(15.86)

If the vector S is normalized by dividing its components by s _n+1, Eq. (15.86) gives λ = Δp and hence Eq. (15.85) gives the desired sensitivity derivatives as

(15.87)

Thus the sensitivity of the objective function with respect to p can be computed as

(15.88)

Note that unlike the previous method, this method does not require the values of λ^* and the second derivatives of f and g _j to find the sensitivity derivatives. Also, if sensitivities with respect to several problem parameters p ₁, p ₂, … are required, all we need to do is to add them to the design vector X in Eq. (15.82).

1. 1 Schmidt, L.A. Jr. and Farshi, B. (1974). Some approximation concepts for structural synthesis. AIAA Journal 12 (5): 692–699.
2. 2 Haug, E.J., Choi, K.K., and Komkov, V. (1986). Design Sensitivity Analysis of Structural Systems. New York: Academic Press.
3. 3 Fox, R.L. and Miura, H. (1971). An approximate analysis technique for design calculations. AIAA Journal 9 (1): 177–179.
4. 4 Fox, R.L. and Kapoor, M.P. (1968). Rates of change of eigenvalues and eigenvectors. AIAA Journal 6 (12): 2426–2429.
5. 5 Murthy, D.V. and Haftka, R.T. (1988). Derivatives of eigenvalues and eigenvectors of general complex matrix. International Journal for Numerical Methods in Engineering 26: 293–311.
6. 6 Nelson, R.B. (1976). Simplified calculation of eigenvector derivatives. AIAA Journal 14: 1201–1205.
7. 7 Rao, S.S. (1972). Rates of change of flutter Mach number and flutter frequency. AIAA Journal 10: 1526–1528.
8. 8 Sutter, T.R., Camarda, C.J., Walsh, J.L., and Adelman, H.M. (1988). Comparison of several methods for the calculation of vibration mode shape derivatives. AIAA Journal 26 (12): 1506–1511.
9. 9 Rao, S.S. (2018). The Finite Element Method in Engineering, 6e. Burlington, MA: Elsevier Butterworth Heinemann.
10. 10 Rao, S.S. (2017). Mechanical Vibrations, 6e. Upper Saddle River, NJ: Pearson Prentice Hall.
11. 11 Grandhi, R.V., Haftka, R.T., and Watson, L.T. (1986). Efficient identification of critical stresses in structures subjected to dynamic loads. Computers and Structures 22: 373–386.
12. 12 Greene, W.H. and Haftka, R.T. (1989). Computational aspects of sensitivity calculations in transient structural analysis. Computers and Structures 32 (2): 433–443.
13. 13 Kirsch, U., Reiss, M., and Shamir, U. (1972). Optimum design by partitioning into substructures. ASCE Journal of the Structural Division 98 (ST1): 249–267.
14. 14 Schmit, L.A., Jr., and Miura, H. (1976). Approximation Concepts for Efficient Structural Synthesis, NASA CR‐2552.
15. 15 Schmit, L.A. and Fleury, C. (1980). Structural synthesis by combining approximation concepts and dual methods. AIAA Journal 18: 1252–1260.
16. 16 Pan, T.S., Rao, S.S., and Venkayya, V.B. (1990). Rates of change of closed‐loop eigenvalues and eigenvectors of actively controlled structures. International Journal for Numerical Methods in Engineering 30 (5): 1013–1028.
17. 17 Vanderplaats, G.N. (1984). Numerical Optimization Techniques for Engineering Design with Applications. New York: McGraw‐Hill.
18. 18 Sobieszczanski‐Sobieski, J., Barthelemy, J.F., and Riley, K.M. (1982). Sensitivity of optimum solutions to problem parameters. AIAA Journal 20: 1291–1299.
19. 19 Kirsch, U. (1981). Optimum Structural Design. Concepts, Methods, and Applications. New York: McGraw‐Hill.
20. 20 Haftka, R.T. and Gürdal, Z. (1992). Elements of Structural Optimization, 3e. Dordrecht, The Netherlands: Kluwer Academic.

1. 15.1 Consider the minimum‐volume design of the four‐bar truss shown in Figure 15.2 subject to a constraint on the vertical displacement of node 4. Let X ₁ = {1, 1, 0.5, 0.5}^T and X ₂ = {0.5, 0.5, 1, 1}^T be two design vectors, with x _i denoting the area of cross section of bar i(i = 1, 2, 3, 4). By expressing the optimum design vectors as X = c ₁ X ₁ + c ₂ X ₂, determine the values of c ₁ and c ₂ through graphical optimization when the maximum permissible vertical deflection of node 4 is restricted to a magnitude of 0.1 in.
2. 15.2 Consider the configuration (shape) optimization of the 10‐bar truss shown in Figure 15.6. The (X, Y) coordinates of the nodes are to be varied while maintaining (a) symmetry of the structure about the X axis, and (b) alignment of nodes 1, 2, and 3 (4, 5, and 6). Identify the independent and dependent design variables and derive the relevant design variable linking relationships.

   

3. 3 For the four‐bar truss considered in Example 15.1 (shown in Figure 15.2), a base design vector is given by X ₀ = {A ₁, A ₂, A ₃, A ₄}^T = {2.0, 1.0, 2.0, 1.0}^T in.². If ΔX is given by ΔX = {0.4, 0.4, −0.4, −0.4}^T in.², determine
   1. The exact displacement vector Y ₀ = {y ₅, y ₆, y ₇, y ₈}^T at X ₀
   2. The exact displacement vector (Y ₀ + ΔY) at (X ₀ + ΔX)
   3. The displacement vector (Y ₀ + ΔY) where ΔY is given by Eq. (14.20) with five terms
4. 4 Consider the 11‐member truss shown in Figure 5.1 with loads Q = −1000 lb, R = 1000 lb, and S = 2000 lb. If A _i  = x _i denotes the area of cross section of member i, and u ₁, u ₂, …, u ₁₀ indicate the displacement components of the nodes, the equilibrium equations can be expressed as shown in Eqs. (E1)–(E10) of Example 5.1. Assuming that E = 30 × 10⁶ psi, l = 50 in., x _i  = 1 in.²(i = 1, 2, …, 11), Δx _i  = 0.1 in.²(i = 1, 2, …, 5), and Δx _i  = −0.1 in.²(i = 6, 7, …, 11), determine
   1. Exact displacement solution U ₀ at X ₀
   2. Exact displacement solution (U ₀ + ΔU) at the perturbed design, (X ₀ + ΔX)
   3. Approximate displacement solution, (U ₀ + ΔU), at(X ₀ + ΔX) using Eq. (15.20) with four terms for ΔU
5. 15.5 Consider the four‐bar truss shown in Figure 15.2 whose stiffness matrix is given by Eq. (E2) of Example 15.1. Determine the values of the derivatives of y _i with respect to the area A ₁, ∂y _i /∂x ₁ (i = 5, 6, 7, 8) at the reference design X ₀ = {A ₁ A ₂ A ₃ A ₄}^T = {2.0, 2.0, 1.0, 1.0}^T in.².
6. 15.6 Find the values of ∂y _i  / ∂x ₂ (i = 5, 6, 7, 8) in Problem 5.
7. 15.7 Find the values of ∂y _i  / ∂x ₃ (i = 5, 6, 7, 8) in Problem 5.
8. 15.8 Find the values of ∂y _i  / ∂x ₄ (i = 5, 6, 7, 8) in Problem 5.
9. 9 The equilibrium equations of the stepped bar shown in Figure 15.7 are given by
   (1)1

   with

   (2)2
   (3)3

   If A ₁ = 2 in.², A ₂ = 1 in.², E ₁ = E ₂ = 30 × 10⁶ psi, 2 l ₁ = l ₂ = 50 in., P ₁ = 100 lb, and P ₂ = 200 lb, determine

   1. Displacements, Y
   2. Values of ∂ Y /∂A ₁ and ∂ Y /∂A ₂ using the method of Section 15.4
   3. Values of ∂σ/∂A ₁ and ∂σ/∂A ₂, where σ  = {σ ₁, σ ₂}^T denotes the vector of stresses in the bars and σ ₁ = E ₁ Y ₁ /l ₁ and σ ₂ = E ₂(Y ₂ − Y ₁)/l ₂

      

10. 10 The eigenvalue problem for the stepped bar shown in Figure 15.7 can be expressed as [K]Y = λ[M]Y with the mass matrix, [M], given by
    

    where ρ _i , A _i , and l _i denote the mass density, area of cross section, and length of the segment i, and the stiffness matrix, [K], is given by Eq. (2) of Problem 9. If A ₁ = 2 in.², A ₂ = 1 in.², E ₁ = E ₂ = 30 × 10⁶ psi, 2 l ₁ = l ₂ = 50 in., and ρ ₁ g = ρ ₂ g = 0.283 lb/in.³, determine

    1. Eigenvalues λ _i and the eigenvectors Y _i , i = 1, 2
    2. Values of ∂λ _i /∂A ₁, i = 1, 2, using the method of Section 15.5
    3. Values of ∂ Y _i /∂ Y ₁, i = 1, 2, using the method of Section 15.5
11. 11 For the stepped bar considered in Problem 10, determine the following using the method of Section 15.5.
    1. Values of ∂λ _i /∂A ₂, i = 1, 2
    2. Values of ∂ Y _i /∂A ₂, i = 1, 2
12. 15.12 A cantilever beam with a hollow circular section with outside diameter d and wall thickness t (Figure 15.8) is modeled with one beam finite element. The resulting static equilibrium equations can be expressed as
    

    where I is the area moment of inertia of the cross section, E is Young's modulus, and l the length. Determine the displacements, Y _i , and the sensitivities of the deflections, ∂Y _i /∂d and ∂Y _i /∂t(i = 1, 2), for the following data: E = 30 × 10⁶ psi, l = 20 in., d = 2 in., t = 0.1 in., P ₁ = 100 lb, and P ₂ = 0.

    

13. 13 The eigenvalues of the cantilever beam shown in Figure 15.8 are governed by the equation
    

    where E is Young's modulus, I the area moment of inertia, l the length, ρ the mass density, A the cross‐sectional area, λ the eigenvalue, and Y = {Y ₁, Y ₂}^T = eigenvector. If E = 30 × 10⁶ psi, d = 2 in., t = 0.1 in., l = 20 in., and ρg = 0.283 lb/in.³, determine

    1. Eigenvalues λ _i and eigenvectors Y _i  (i = 1, 2)
    2. Values of ∂λ _i /∂d and ∂λ _i /∂t (i = 1, 2)
14. 15.14 In Problem 13, determine the derivatives of the eigenvectors ∂ Y _i /∂d and ∂ Y _i /∂t(i = 1, 2).
15. 15.15 The natural frequencies of the spring–mass system shown in Figure 15.9 are given by (for k _i  = k, i = 1, 2, 3 and m _i  = m, i = 1, 2)
    

    where ω ₁ and ω ₂ are the natural frequencies of vibration of the system and c ₁ and c ₂ are constants. The stiffness of each helical spring is given by

    

    where d is the wire diameter, D the coil diameter, G the shear modulus, and n the number of turns of the spring. Determine the values of ∂ω _i /∂D and ∂ Y _i /∂D for the following data: d = 0.04 in., G = 11.5 × 10⁶ psi, D = 0.4 in., n = 10, and m = 32.2 lb ‐s ²/in. The stiffness and mass matrices of the system are given by

    

    

16. 15.16 Find the sensitivities of , , and f ^* with respect to Young's modulus of the tubular column considered in Example 1.1.