Optimization design of structures subjected to transient loads using

Shock and Vibration 19 (2012) 1445–1461 1445DOI 10.3233/SAV-2012-0685IOS Press

Optimization design of structures subjected totransient loads using first and secondderivatives of dynamic displacement andstress

Qimao Liua,∗, Jing Zhangc and Liubin Yanb

aDepartment of Civil and Structural Engineering, Aalto University, Espoo, FinlandbCollege of Civil and Architecture Engineering, Guangxi University, Nanning, ChinacDepartment of Mechanical Engineering, Indiana University-Purdue University Indianapolis, Indianapolis, IN,USA

Received 16 September 2010

Revised 19 July 2011

Abstract. This paper developed an effective optimization method, i.e., gradient-Hessian matrix-based method or second ordermethod, of frame structures subjected to the transient loads. An algorithm of first and second derivatives of dynamic displacementand stress with respect to design variables is formulated based on the Newmark method. The inequality time-dependent constraintproblem is converted into a sequence of appropriately formed time-independent unconstrained problems using the integral interiorpoint penalty function method. The gradient and Hessian matrixes of the integral interior point penalty functions are alsocomputed. Then the Marquardt’s method is employed to solve unconstrained problems. The numerical results show that theoptimal design method proposed in this paper can obtain the local optimum design of frame structures and sometimes is moreefficient than the augmented Lagrange multiplier method.

Keywords: Dynamic response optimization, gradient, Hessian matrix, time-dependent constraint

1. Introduction

Dynamic response analysis of structures subjected to transient loads often requires much more demanding com-putational cost than static analysis. Therefore the efficiency of the optimization method becomes critical to dy-namic response optimization problems. Mathematically, there are three types of dynamic response optimizationmethods: zero order methods (nongradient-based algorithms), first order methods (gradient-based algorithms) andsecond order methods (gradient-Hessian matrix-based algorithms). Generally, the first order methods are moreefficient than the zero order methods, and the second order methods are more efficient than the first order methods.Zero order methods require only the information of dynamic responses to construct optimal algorithms. A lot ofnongradient-based algorithms [1–4] have been developed to solve the optimization problem of structures subjectedto transient loads. First order methods require the information of dynamic responses and their first derivatives withrespect to design variables to construct optimal algorithms. Therefore, it is very crucial to calculate efficiently the

∗Corresponding author: Qimao Liu, Department of Civil and Structural Engineering, Aalto University, Espoo, FI-00076 Aalto, Finland. Tel.:+358 9 470 23701; Fax: +358 9 470 23758; E-mail: [email protected].

ISSN 1070-9622/12/$27.50 2012 – IOS Press and the authors. All rights reserved

1446 Q. Liu et al. / Optimization design of structures subjected to transient loads using first and second derivatives

first derivatives of structural dynamic responses for gradient-based algorithms. However, many algorithms have beendeveloped to calculate dynamic response first derivatives with respect to design variables (also called sensitivity anal-ysis [5–12]). Today, the gradient-based algorithms are the mainstream pattern [13] to solve the optimization problemof structures subjected to transient loads. Much research [14–17] has been done to solve the structural dynamicresponse optimization problem with the gradient-based algorithms. Second order methods require the information ofdynamic responses, their first and second derivatives with respect to design variables to construct optimal algorithms.The Newton’s method and Quasi Newton’s method are the commonly used second order method. Although manyresearchers have calculated dynamic response first derivatives with respect to design variables, there is little literaturepublished on dynamic response second derivative analysis (also called Hessian matrix analysis). Second derivativeanalysis is more complicated than first derivative analysis. However, the efficiency of the optimization using secondderivative can be greatly improved when the gradient and Hessian matrix can be calculated efficiently. In this paper,a gradient-Hessian matrix-based algorithm for optimization problem of structures subjected to transient loads isdeveloped based on gradient and Hessian matrix calculations.

The purpose of this paper is to develop a gradient-Hessian matrix-based optimization method for structuressubjected to dynamic loads. The main work of this paper is as follows: First, we formulate an algorithm to calculatedynamic responses and their first and second derivatives with respect to design variables. The algorithm is achievedby direct differentiation and only a single dynamics analysis, based on Newmark-β method [18], is required. Second,we formulate the time-dependent structural optimization model. In this model, total mass of the structure is theobjective function. The dynamic responses including stresses of the beam element and nodal displacements areconstraints. Third, the time-dependent optimization model is converted into a sequence of appropriately formedunconstrained integral mathematic models using the interior penalty function method. The gradient and Hessianmatrixes of the interior penalty functions are also calculated using the first and second derivatives of dynamicresponses. Fourth, Marquardt’s method [19], a gradient-Hessian matrix-based algorithm, is employed to solve theunconstrained integralmathematicmodel. Finally, as the illustration of the developed approach, optimization designsof a plane frame subjected to the horizontal dynamic loads are demonstrated.

2. Calculation of first and second derivatives

This section presents two subjects. Section 2.1 introduces the calculations of first and second derivatives ofdynamic responses (i.e., nodal displacement and normal stress) with respect to design variables of structures.Section 2.2 discusses how to calculate first and second derivatives of structural mass with respect to structural designvariables.

2.1. Calculation of first and second derivatives of dynamic displacement and stress

The first and second derivatives of dynamic response of structures are the prerequisites of an efficient gradient-Hessian matrix-based algorithm. We will show here that, in a single structural dynamic analysis, the first and secondderivatives of a dynamic response can be derived simultaneously.

Plane beam element is used extensively in engineering. Therefore, the dynamic optimal design method is demon-strated with the plane beam element in this work. The element cross-section shown in Fig. 1 is H structural-steelshape. The design variables of element e are de

1, de2, de

3 and de4. Point c is the centroid of the element cross-section.

Axes y and z are the principal centroidal axes.The cross-sectional area is

Ae = 2de1d

e3 + de

2de4 (1)

The cross-sectional moment of inertia is

Iez =

de4d

e3

2

12+ 2

[de1d

e33

12+

(de3 + d2e

2

)2

de1d

e3

](2)

The farthest distance y from the neutral axis is

Q. Liu et al. / Optimization design of structures subjected to transient loads using first and second derivatives 1447

Fig. 1. Element cross-section and element design variables.

yem = ±

(de3 +

de2

2

)(3)

The first moment of the shaded area with respect to the neutral axis is

Se∗z max = de

1de3

de3 + de

2

2+

de2

2de4

de2

4(4)

The structure is divided with N elements. The design variable vector is defined as.

d =[d11, d

12, d

13, d

14, · · · , de

1, de2, d

e3, d

e4, · · · , dN

1 , dN2 , dN

3 , dN4

]T(5)

Consider the equations of motion for a linear system subjected to dynamic forces

Mx + Cx + Kx = f (t) (6)

with the following initial conditions:{x (0) = x0

x (0) = x0(7)

where K, M, and C are stiffness matrix, mass matrix , and damping matrix, respectively. x (t), x (t) and x (t)are unknown nodal displacement, velocity and acceleration vectors, and f(t) is the load vector. Suppose that thedynamic loads act on the nodes.

Rayleigh damping is used in this work, the structural damping matrix is

C = α1M + α2K (8)

where

α1 =2

(ς1ω1

− ς2ω2

)1

ω21− 1

ω22

(9)

α2 =2 (ς2ω2 − ς1ω1)

ω22 − ω2

1

(10)

where ω1 and ω2 are the first and second natural frequency of the structure, respectively. ς1 and ς2 are the dampingratio. In this work, ς1 = ς2 = 0.02. α1 and α2 are constants.

Equations (6) and (7) must be satisfied for all time period t ∈ [0, Γ]. Γ is the duration of the dynamic loads.In practice, the solution of this initial-boundary-value problem (IBVP) requires integration through time. This isachieved numerically by discretising in time the continuous temporal derivatives that appear in the equation. Anyone of the time integration procedures can be used for this purpose. The most widely used family of direct timeintegration methods for solving Eq. (6) is the Newmark family of methods. The Newmark method can be formulatedby considering equilibrium at any discrete time t + Δt, and is given by the following equation:

Mx (t + Δt) + Cx (t + Δt) + Kx (t + Δt) = f (t + Δt) (11)


The nodal displacements, nodal accelerations and nodal velocities can be achieved at any time t + Δt by solvingEq. (11).

Perhaps the most widely used direct method for the equation of motion (6) is the Newmark method, an im-plicit technique, which consists of the following finite difference assumptions with regard to the evolution of theapproximate solution:

x (t + Δt) = x (t) + Δtx (t) + Δt2[(

12− β

)x (t) + βx (t + Δt)

](12)

x (t + Δt) = x (t) + Δt [(1 − δ) x (t) + δx (t + Δt)] (13)

where any particular choice of the parameters β and δ determines the stability and accuracy characteristics of thesolution. In this work, parameters δ � 0.5 and β = 0.25 (0.5 + δ)2. We define the integral constants: a0 = 1

βΔt2 ,

a1 = δβΔt , a2 = 1

βΔt , a3 = 12β − 1, a4 = δ

β − 1,a5 = Δt2

(δβ − 2

), a6 = Δt (1 − δ), a7 = δΔt. The parameters β

and δ will be replaced by those constants in the following formulas.In addition to Eqs (12) and (13) the equilibrium Eq. (11) at time station t+Δt is considered. This way a system of

equations is formed for the determination of the three unknowns x (t + Δt), x (t + Δt) and x (t + Δt), assumingthat the displacement, velocity, and acceleration vectors at the previous time station t have already been computed.Thus, the solution for the displacement vector is

K∗x (t + Δt) = F∗ (t + Δt) (14)

where

K∗ = K + a0M + a1C (15)

and

F∗ (t + Δt) = f (t + Δt) + M [a0x (t) + a2x (t) + a3x (t)] + C [a1x (t) + a4x (t) + a5x (t)] (16)

The accelerations, x (t + Δt), which are required for the computations at the next time station, can be computedas

x (t + Δt) = a0 [x (t + Δt) − x (t)] − a2x (t) − a3x (t) (17)

while the velocities, x (t + Δt), can be obtained directly from equation (13), as follows.

x (t + Δt) = x (t) + a6x (t) + a7x (t + Δt) (18)

When the dynamic loads act on the nodes, we only need to determine the internal forces and internal couples atthe center and two ends of the element. The element nodal force vector, Fe (t + Δt), is

Fe (t + Δt) = KeTeδe (t + Δt) (19)

where Ke is the element stiffness matrix in a local coordinate system, Te is the element coordinate transformationmatrix, δe (t + Δt) is the element nodal displacement vector in a local coordinate system.

The internal force vector at the center of element, FeM (t + Δt), is

FeM (t + Δt) =

[−Fe1 (t + Δt) −Fe

2 (t + Δt) le2 Fe

2 (t + Δt) − Fe3 (t + Δt)

]T(20)

where le is the length of element. The three terms, i.e., −Fe1 (t + Δt), −Fe

2 (t + Δt) and le2 Fe

2 (t + Δt) −Fe

3 (t + Δt), are axial force, shear force and bending moment, respectively. Fe1 (t + Δt), Fe

2 (t + Δt) andFe

3 (t + Δt) are the first, second and third term of the element nodal force vector, Fe (t + Δt), respectively.The maximum stress at the center and two ends, i.e. the i end and the j end, of the element are calculated as

follows.The maximum normal stress at the i end of the element is

σeix (t + Δt) =

Fe1 (t + Δt)

Ae+

Fe3 (t + Δt) ye

m

EIez

(21)


The maximum normal stress at the j end of the element is

σejx (t + Δt) =

Fe4 (t + Δt)

Ae+

Fe6 (t + Δt) ye

m

EIez

(22)

where Fe4 (t + Δt) and Fe

6 (t + Δt) are the fourth and sixth term of the element nodal force vector, Fe (t + Δt),respectively.

The maximum normal stress at the center of the element is

σeMx (t + Δt) =

FeM1 (t + Δt)

Ae+

FeM3 (t + Δt) ye

m

EIez

(23)

where FeM1 (t + Δt) and Fe

M3 (t + Δt) are the first and third term of the internal force vector at the center ofelement, Fe

M (t + Δt), respectively.The maximum shear stress at the i end of the element is

τeixy (t + Δt) =

Fe2 (t + Δt)Se∗

z max

de4I

ez

(24)

The maximum shear stress at the j end of the element is

τejxy (t + Δt) =

Fe5 (t + Δt)Se∗

z max

de4I

ez

(25)

where Fe5 (t + Δt) is the fifth term of the element nodal force vector, Fe (t + Δt).

The maximum shear stress at the center of the element is

τeMxy (t + Δt) =

FeM2 (t + Δt)Se∗

z max

de4I

ez

(26)

where FeM2 (t + Δt) is the second term of the internal force vector at the center of element, Fe

M (t + Δt).

2.1.1. Formulas for first derivatives of dynamic displacement and stressNow we will derive the formulas for the first derivatives of dynamic response, i.e., dynamic displacement and

stress. Differentiating Eq. (14) with respect to the design variable di, we have

K∗ ∂x (t + Δt)∂di

=∂F∗ (t + Δt)

∂di− ∂K∗

∂dix (t + Δt) (27)

where∂K∗

∂di=

∂K∂di

+ a0∂M∂di

+ a1∂C∂di

(28)

and

∂F∗ (t + Δt)∂di

=∂M∂di

[a0x (t) + a2x (t) + a3x (t)] + M[a0

∂x (t)∂di

+ a2∂x (t)∂di

+ a3∂x (t)∂di

]

+∂C∂di

[a1x (t) + a4x (t) + a5x (t)] + C[a1

∂x (t)∂di

+ a4∂x (t)∂di

+ a5∂x (t)∂di

](29)

After the first derivatives of displacement vector at time t + Δt is obtained from Eq. (27), through differentiatingEq. (17) with respect to design variables di, we have

∂x (t + Δt)∂di

= a0

[∂x (t + Δt)

∂di− ∂x (t)

∂di

]− a2

∂x (t)∂di

− a3∂x (t)∂di

(30)

Then, the first derivatives of acceleration vector at time t + Δt is obtained from Eq. (30). Differentiating Eq. (18)with respect to design variable di, we have

∂x (t + Δt)∂di

=∂x (t)∂di

+ a6∂x (t)∂di

+ a7∂x (t + Δt)

∂di(31)


Differentiating Eq. (19) with respect to the design variable di, we have

∂Fe (t + Δt)∂di

=∂Ke

∂diTeδe (t + Δt) + KeTe ∂δe (t + Δt)

∂di(32)

where ∂δe(t+Δt)∂di

can be calculated from ∂x(t+Δt)∂di

.Differentiating Eq. (20) with respect to the design variable di, we have

∂FeM (t + Δt)

∂di=

[−∂Fe

1(t+Δt)∂di

−∂Fe2(t+Δt)∂di

le2

∂Fe2(t+Δt)∂di

− ∂Fe3(t+Δt)∂di

]T

(33)


∂σeix (t + Δt)

∂di=

∂Fe1 (t + Δt)

∂di

1Ae

− Fe1 (t + Δt)

Ae2

∂Ae

∂di

+∂Fe

3 (t + Δt)∂di

yem

EIez

+Fe

3 (t + Δt)EIe

z

∂yem

∂di− Fe

3 (t + Δt) yem

EIe2z

∂Iez

∂di(34)


∂σejx (t + Δt)

∂di=

∂Fe4 (t + Δt)

∂di

1Ae

− Fe4 (t + Δt)

Ae2

∂Ae

∂di

+∂Fe

6 (t + Δt)∂di

yem

EIez

+Fe

6 (t + Δt)EIe

z

∂yem

∂di− Fe

6 (t + Δt) yem

EIe2z

∂Iez

∂di(35)


∂σeMx (t + Δt)

∂di=

∂FeM1 (t + Δt)

∂di

1Ae

− FeM1 (t + Δt)

Ae2

∂Ae

∂di

+∂Fe

M3 (t + Δt)∂di

yem

EIez

+Fe

M3 (t + Δt)EIe

z

∂yem

∂di− Fe

M3 (t + Δt) yem

EIe2z

∂Iez

∂di(36)


∂τeixy (t + Δt)

∂di=

∂Fe2 (t + Δt)

∂di

Se∗z max

de4I

ez

+∂Se∗

z max

∂di

Fe2 (t + Δt)

de4I

ez

−∂de4

∂di

Fe2 (t + Δt) Se∗

z max

de24 Ie

z

− ∂Iez

∂di

Fe2 (t + Δt)Se∗

z max

de4I

e2z

(37)


∂τejxy (t + Δt)

∂di=

∂Fe5 (t + Δt)

∂di

Se∗z max

de4I

ez

+∂Se∗

z max

∂di

Fe5 (t + Δt)

de4I

ez

−∂de4

∂di

Fe5 (t + Δt) Se∗

z max

de24 Ie

z

− ∂Iez

∂di

Fe5 (t + Δt)Se∗

z max

de4I

e2z

(38)


∂τeMxy (t + Δt)

∂di=

∂FeM2 (t + Δt)

∂di

Se∗z max

de4I

ez

+∂Se∗

z max

∂di

FeM2 (t + Δt)

de4I

ez

−∂de4

∂di

FeM2 (t + Δt)Se∗

z max

de24 Ie

z

− ∂Iez

∂di

FeM2 (t + Δt) Se∗

z max

de4I

e2z

(39)


2.1.2. Formulas for second derivatives of dynamic displacement and stressNow we will derive the formulas for the second derivatives of dynamic response, i.e., dynamic displacement and

stress. Further differentiating Eq. (27) with respect to the design variable dj , we have

K∗ ∂2x (t + Δt)∂di∂dj

=∂2F∗ (t + Δt)

∂di∂dj− ∂2K∗

∂di∂djx (t + Δt) − ∂K∗

∂di

∂x (t + Δt)∂dj

− ∂K∗

∂dj

∂x (t + Δt)∂di

(40)

where∂2K∗

∂di∂dj=

∂2K∂di∂dj

+ a0∂2M

∂di∂dj+ a1

∂2C∂di∂dj

(41)

and∂2F∗ (t + Δt)

∂di∂dj=

∂2M∂di∂dj

[a0x (t) + a2x (t) + a3x (t)] +∂M∂di

[a0

∂x (t)∂dj

+ a2∂x (t)∂dj

+ a3∂x (t)∂dj

]

+∂M∂dj

[a0

∂x (t)∂di

+ a2∂x (t)∂di

+ a3∂x (t)∂di

]+ M

[a0

∂2x (t)∂di∂dj

+ a2∂2x (t)∂di∂dj


]

+∂2C

∂di∂dj[a1x (t) + a4x (t) + a5x (t)] +

∂C∂di

[a1

∂x (t)∂dj

+ a4∂x (t)∂dj

+ a5∂x (t)∂dj

]

+∂C∂dj

[a1

∂x (t)∂di

+ a4∂x (t)∂di

+ a5∂x (t)∂di

]+ C

[a1

∂2x (t)∂di∂dj



](42)

After ∂2x(t+Δt)∂di∂dj

is computed from Eq. (40), with further differentiating Eq. (30) with respect to the designvariables dj , we have

∂2x (t + Δt)∂di∂dj

= a0

[∂2x (t + Δt)

∂di∂dj− ∂2x (t)

∂di∂dj

]− a2

∂2x (t)∂di∂dj

− a3∂2x (t)∂di∂dj

(43)

Then ∂2x(t+Δt)∂di∂dj

is obtained from Eq. (43). Further differentiating Eq. (31) with respect to the design variablesdj , we obtain

∂2x (t + Δt)∂di∂dj

=∂2x (t)∂di∂dj


+ a7∂2x (t + Δt)

∂di∂dj(44)

Further differentiating Eq. (32) with respect to the design variable dj , we have

∂2Fe(t+Δt)∂di∂dj

= ∂2Ke

∂di∂djTeδe (t + Δt) + ∂Ke

∂diTe ∂δe(t+Δt)

∂dj

+ ∂Ke

∂djTe ∂δe(t+Δt)

∂di+ KeTe ∂2δe(t+Δt)

∂di∂dj

(45)

Further differentiating Eq. (33) with respect to the design variable dj , we obtain

∂2FeM (t + Δt)∂di∂dj

=[−∂2Fe

1(t+Δt)∂di∂dj

−∂2Fe2(t+Δt)

∂di∂dj

le2

∂2Fe2(t+Δt)

∂di∂dj− ∂2Fe

3(t+Δt)∂di∂dj

]T

(46)


∂2σeix (t + Δt)∂di∂dj

=∂2Fe

1 (t + Δt)∂di∂dj

1Ae

− 1Ae2

∂Fe1 (t + Δt)

∂di

∂Ae

∂dj− 1

Ae2

∂Fe1 (t + Δt)∂dj

∂Ae

∂di

+2Fe

1 (t + Δt)Ae3

∂Ae

∂dj

∂Ae

∂di− Fe

1 (t + Δt)Ae2

∂2Ae

∂di∂dj+

∂2Fe3 (t + Δt)∂di∂dj

yem

EIez

+∂Fe

3 (t + Δt)∂di

∂yem

∂dj

1EIe

z (47)

−∂Fe3 (t + Δt)

∂di

yem

EIe2z

∂Iez

∂dj+


1EIe

z

∂yem

∂di− ∂Ie

z

∂dj

Fe3 (t + Δt)EIe2

z

∂yem

∂di+

Fe3 (t + Δt)

EIez

∂2yem

∂di∂dj

−∂Fe3 (t + Δt)∂dj

yem

EIe2z

∂Iez

∂di− ∂ye

m

∂dj

Fe3 (t + Δt)EIe2

z

∂Iez

∂di+

∂Iez

∂dj

2Fe3 (t + Δt) ye

m

EIe3z

∂Iez

∂di− Fe

3 (t + Δt) yem

EIe2z

∂2Iez

∂di∂dj



∂2σejx (t + Δt)∂di∂dj

=∂2Fe


1Ae

− 1Ae2

∂Fe4 (t + Δt)

∂di

∂Ae

∂dj− 1

Ae2


∂Ae

∂di

+2Fe

4 (t + Δt)Ae3

∂Ae

∂dj

∂Ae

∂di− Fe

4 (t + Δt)Ae2

∂2Ae

∂di∂dj+

∂2Fe6 (t + Δt)∂di∂dj

yem

EIez

+∂Fe

6 (t + Δt)∂di

∂yem

∂dj

1EIe

z (48)

−∂Fe6 (t + Δt)

∂di

yem

EIe2z

∂Iez

∂dj+


1EIe

z

∂yem

∂di− ∂Ie

z

∂dj

Fe6 (t + Δt)EIe2

z

∂yem

∂di+

Fe6 (t + Δt)

EIez

∂2yem

∂di∂dj

−∂Fe6 (t + Δt)∂dj

yem

EIe2z

∂Iez

∂di− ∂ye

m

∂dj

Fe6 (t + Δt)EIe2

z

∂Iez

∂di+

∂Iez

∂dj

2Fe6 (t + Δt) ye

m

EIe3z

∂Iez

∂di− Fe

6 (t + Δt) yem

EIe2z

∂2Iez

∂di∂dj


∂2σeMx (t + Δt)∂di∂dj

=∂2Fe

M1 (t + Δt)∂di∂dj

1Ae

− 1Ae2

∂FeM1 (t + Δt)

∂di

∂Ae

∂dj− 1

Ae2

∂FeM1 (t + Δt)

∂dj

∂Ae

∂di

+2Fe

M1 (t + Δt)Ae3

∂Ae

∂dj

∂Ae

∂di− Fe

M1 (t + Δt)Ae2

∂2Ae

∂di∂dj+

∂2FeM3 (t + Δt)∂di∂dj

yem

EIez

+∂Fe

M3 (t + Δt)∂di

∂yem

∂dj

1EIe

z

−∂FeM3 (t+Δt)

∂di

yem

EIe2z

∂Iez

∂dj+

∂FeM3 (t+Δt)

∂dj

1EIe

z

∂yem

∂di− ∂Ie

z

∂dj

FeM3 (t+Δt)

EIe2z

∂yem

∂di+

FeM3 (t+Δt)

EIez

∂2yem

∂di∂dj

−∂FeM3 (t + Δt)

∂dj

yem

EIe2z

∂Iez

∂di− ∂ye

m

∂dj

FeM3 (t + Δt)

EIe2z

∂Iez

∂di

+∂Ie

z

∂dj

2FeM3 (t + Δt) ye

m

EIe3z

∂Iez

∂di− Fe

M3 (t + Δt) yem

EIe2z

∂2Iez

∂di∂dj(49)


∂2τeixy (t + Δt)∂di∂dj

=∂2Fe


Se∗z max

de4I

ez

+∂Fe

2 (t + Δt)∂di

∂Se∗z max

∂dj

1de4I

ez

− ∂Fe2 (t + Δt)

∂di

∂de4

∂dj

Se∗z max

de24 Ie

z

−∂Fe2 (t + Δt)

∂di

∂Iez

∂dj

Se∗z max

de4I

e2z

+∂2Se∗

z max

∂di∂dj

Fe2 (t + Δt)

de4I

ez

+∂Se∗

z max

∂di


1de4I

ez

−∂Se∗z max

∂di

∂de4

∂dj

Fe2 (t + Δt)de24 Ie

z

− ∂Se∗z max

∂di

∂Iez

∂dj

Fe2 (t + Δt)de4I

e2z

− ∂2de4

∂di∂dj

Fe2 (t + Δt)Se∗

z max

de24 Ie

z

−∂de4

∂di


Se∗z max

de24 Ie

z

− ∂de4

∂di

∂Se∗z max

∂dj


z

+∂de

4

∂di

∂de4

∂dj

2Fe2 (t + Δt)Se∗

z max

de34 Ie

z

+∂de

4

∂di

∂Iez

∂dj

Fe2 (t + Δt)Se∗

z max

de24 Ie2

z

− ∂2Iez

∂di∂dj

Fe2 (t + Δt)Se∗

z max

de4I

e2z

− ∂Iez

∂di

∂Fe2 (t + Δt)

∂dj

Se∗z max

de4I

e2z

−∂Iez

∂di

∂Se∗z max

∂dj

Fe2 (t + Δt)de4I

e2z

+∂Ie

z

∂di

∂de4

∂dj

Fe2 (t + Δt)Se∗

z max

de24 Ie2

z

+∂Ie

z

∂di

∂Iez

∂dj

2Fe2 (t + Δt)Se∗

z max

de4I

e3z

(50)


∂2τejxy (t + Δt)∂di∂dj

=∂2Fe


Se∗z max

de4I

ez

+∂Fe

5 (t + Δt)∂di

∂Se∗z max

∂dj

1de4I

ez

− ∂Fe5 (t + Δt)

∂di

∂de4

∂dj

Se∗z max

de24 Ie

z

−∂Fe5 (t + Δt)

∂di

∂Iez

∂dj

Se∗z max

de4I

e2z

+∂2Se∗

z max

∂di∂dj

Fe5 (t + Δt)

de4I

ez

+∂Se∗

z max

∂di


1de4I

ez


−∂Se∗z max

∂di

∂de4

∂dj


z

− ∂Se∗z max

∂di

∂Iez

∂dj

Fe5 (t + Δt)de4I

e2z

− ∂2de4

∂di∂dj

Fe5 (t + Δt)Se∗

z max

de24 Ie

z

−∂de4

∂di


Se∗z max

de24 Ie

z

− ∂de4

∂di

∂Se∗z max

∂dj


z

+∂de

4

∂di

∂de4

∂dj

2Fe5 (t + Δt)Se∗

z max

de34 Ie

z

+∂de

4

∂di

∂Iez

∂dj

Fe5 (t + Δt)Se∗

z max

de24 Ie2

z

− ∂2Iez

∂di∂dj

Fe5 (t + Δt)Se∗

z max

de4I

e2z

− ∂Iez

∂di

∂Fe5 (t + Δt)

∂dj

Se∗z max

de4I

e2z

−∂Iez

∂di

∂Se∗z max

∂dj

Fe5 (t + Δt)de4I

e2z

+∂Ie

z

∂di

∂de4

∂dj

Fe5 (t + Δt)Se∗

z max

de24 Ie2

z

+∂Ie

z

∂di

∂Iez

∂dj

2Fe5 (t + Δt)Se∗

z max

de4I

e3z

(51)


∂2τeMxy (t+Δt)∂di∂dj

=∂2Fe

M2 (t+Δt)∂di∂dj

Se∗z max

de4I

ez

+∂Fe

M2 (t+Δt)∂di

∂Se∗z max

∂dj

1de4I

ez

− ∂FeM2 (t+Δt)

∂di

∂de4

∂dj

Se∗z max

de24 Ie

z

−∂FeM2 (t + Δt)

∂di

∂Iez

∂dj

Se∗z max

de4I

e2z

+∂2Se∗

z max

∂di∂dj

FeM2 (t + Δt)

de4I

ez

+∂Se∗

z max

∂di

∂FeM2 (t + Δt)

∂dj

1de4I

ez

−∂Se∗z max

∂di

∂de4

∂dj

FeM2 (t + Δt)

de24 Ie

z

− ∂Se∗z max

∂di

∂Iez

∂dj

FeM2 (t + Δt)

de4I

e2z

− ∂2de4

∂di∂dj


z max

de24 Ie

z

−∂de4

∂di

∂FeM2 (t + Δt)

∂dj

Se∗z max

de24 Ie

z

− ∂de4

∂di

∂Se∗z max

∂dj

FeM2 (t + Δt)

de24 Ie

z

+∂de

4

∂di

∂de4

∂dj

2FeM2 (t + Δt)Se∗

z max

de34 Ie

z

+∂de

4

∂di

∂Iez

∂dj

FeM2 (t + Δt)Se∗

z max

de24 Ie2

z

− ∂2Iez

∂di∂dj

FeM2 (t + Δt)Se∗

z max

de4I

e2z

− ∂Iez

∂di

∂FeM2 (t + Δt)

∂dj

Se∗z max

de4I

e2z

−∂Iez

∂di

∂Se∗z max

∂dj

FeM2 (t + Δt)

de4I

e2z

+∂Ie

z

∂di

∂de4

∂dj


z max

de24 Ie2

z

+∂Ie

z

∂di

∂Iez

∂dj

2FeM2 (t + Δt)Se∗

z max

de4I

e3z

(52)

2.1.3. Computation procedure of the first and second derivatives of dynamic displacement and stressIn this work, we suppose that the initial conditions are x (0) = 0, x (0) = 0, x (0) = 0, and the dynamic responses

(nodal displacements and stresses), their first and second derivatives with respect to design variables are equal tozero.

Section 2.1.2 gives the formulas for calculating the first and second derivatives of the dynamic response. Thissection provides the detailed computation procedure as follows.

Procedure of calculating dynamic response first and second derivatives:

Step 1 Initial calculations:Step 1.1 x (0) = 0,x (0) = 0,x (0) = 0,σe

ix (0) = 0,σejx (0) = 0,σe

Mx (0) = 0,τeixy (0) = 0,τe

jxy (0) = 0,

τeMxy (0) = 0.

Step 1.2 ∂x(0)∂di

= 0, ∂x(0)∂di

= 0, ∂x(0)∂di

= 0, ∂σeix(0)∂di

= 0,∂σe

jx(0)

∂di= 0,∂σe

Mx(0)∂di

= 0,∂τe

ixy(0)

∂di= 0,

∂τejxy (0)∂di

= 0,∂τe

Mxy (0)∂di

= 0.

Step 1.3 ∂2x(0)∂di∂dj

= 0, ∂2x(0)∂di∂dj

= 0, ∂2x(0)∂di∂dj

= 0,∂2σeix(0)

∂di∂dj= 0,

∂2σejx(0)

∂di∂dj= 0,∂2σe

Mx(0)∂di∂dj

= 0,

∂2τeixy (0)

∂di∂dj= 0,

∂2τejxy (0)

∂di∂dj= 0,

∂2τeMxy (0)

∂di∂dj= 0.


Step 1.4 Set Δt.Step 1.5 Compute integral constants:

a0 =1

βΔt2, a1 =

δ

βΔt, a2 =

1βΔt

, a3 =12β

− 1, a4 =δ

β− 1, a5 =

Δt

2

(δ

β− 2

), a6 = Δt (1 − δ) ,

a7 = δΔt.

where integral parameters δ � 0.5 and β = 0.25 (0.5 + δ)2.Step 1.6 Solve Eq. (15)⇒ K∗.

Step 2 Calculations for each time step t + Δt:Step 2.1 Solve Eq. (14)⇒ x (t + Δt).Step 2.2 Solve Eq. (17)⇒ x (t + Δt), solve Eq. (21)⇒ σe

ix (t + Δt), solve Eq. (22)⇒ σejx (t + Δt),

solve Eq. (23)⇒ σeMx (t + Δt), solve Eq. (24)⇒ τe

ixy (t + Δt), solve Eq. (25)⇒ τejxy (t + Δt),

and solve Eq. (26)⇒ τeMxy (t + Δt).

Step 2.3 Solve Eq. (18)⇒ x (t + Δt).Step 2.4. Solve Eq. (27)⇒ ∂x(t+Δt)

∂di.

Step 2.5 Solve Eq. (30)⇒ ∂x(t+Δt)∂di

, solve Eq. (34)⇒ ∂σeix(t+Δt)

∂di,

solve Eq. (35)⇒ ∂σejx(t+Δt)

∂di, solve Eq. (36)⇒ ∂σe

Mx(t+Δt)∂di

,

solve Eq. (37)⇒ ∂τeixy(t+Δt)

∂di, solve Eq. (38)⇒ ∂τe

jxy(t+Δt)

∂di,

and solve Eq. (39)⇒ ∂τeMxy(t+Δt)

∂di.

Step 2.6 Solve Eq. (31)⇒ ∂x(t+Δt)∂di

.

Step 2.7 Solve Eq. (40)⇒ ∂2x(t+Δt)∂di∂dj

.


, solve Eq. (47)⇒ ∂2σeix(t+Δt)

∂di∂dj,

solve Eq. (48)⇒ ∂2σejx(t+Δt)

∂di∂dj, solve Eq. (49)⇒ ∂2σe

Mx(t+Δt)∂di∂dj

,

solve Eq. (50)⇒ ∂2τeixy(t+Δt)

∂di∂dj, solve Eq. (51)⇒ ∂2τe

jxy(t+Δt)

∂di∂dj,

and solve Eq. (52)⇒ ∂2τeMxy(t+Δt)

∂di∂dj.


.

Step 3 Repetition for the next time step. Replace t by t + Δt and implement steps 2.1 to 2.9 for the next timestep.

2.2. First and second derivatives of structural mass

We use structural mass as the objective function in optimization. The objective function or the structural mass canbe expressed as

w (d) =N∑

e=1

ρAele (53)

The first derivatives of the structural mass are obtained by differentiating Eq. (53) with respect to the designvariables,

∂w (d)∂di

=N∑

e=1

ρle∂Ae

∂di(54)

The second derivatives of the structural mass can be obtained by further differentiating Eq. (54) with respect tothe design variables,


∂2w (d)∂di∂dj

=N∑

e=1

ρle∂2Ae

∂di∂dj(55)

3. Optimization mathematical model

In general, we aim to minimize the total mass w (d) of the structure. At the same time, the normal stresses, shearstresses, and nodal displacements should satisfy the constraints in the duration of transient loads. We formulate theoptimization problem of the structures subjected to transient loads as follows.

Find d

min w (d)

s.t. − [σ] � σe+ix (d, t) � [σ] (e = 1, 2, · · · , N, t ∈ [0, Γ])

− [σ] � σe−ix (d, t) � [σ] (e = 1, 2, · · · , N, t ∈ [0, Γ])

− [σ] � σe+jx (d, t) � [σ] (e = 1, 2, · · · , N, t ∈ [0, Γ])

− [σ] � σe−jx (d, t) � [σ] (e = 1, 2, · · · , N, t ∈ [0, Γ])

− [σ] � σe+Mx (d, t) � [σ] (e = 1, 2, · · · , N, t ∈ [0, Γ])

− [σ] � σe−Mx (d, t) � [σ] (e = 1, 2, · · · , N, t ∈ [0, Γ])

− [τ ] � τeixy (d, t) � [τ ] (e = 1, 2, · · · , N, t ∈ [0, Γ])

− [τ ] � τejxy (d, t) � [τ ] (e = 1, 2, · · · , N, t ∈ [0, Γ])

− [τ ] � τeMxy (d, t) � [τ ] (e = 1, 2, · · · , N, t ∈ [0, Γ])

− [xk] � xk (d, t) � [xk] (k = 1, 2, · · · , Nf)

dJ � dJ � dJ (J = 1, 2, · · · , 4N) (56)

where ‘+’ and ‘−’ are tension and compression, respectively. [σ] is allowable normal stress. [τ ] is allowable shearstress. Γ is the duration of transient loads. Nf is the number of degree of freedom. [xk] is allowable displacementon the kth degree of freedom. dJ is the lower limit of the J th design variable. dJ is the upper limit of the J th designvariable.

Normalizing the constraints of the optimal model Eq. (56), we obtain the new equivalent mathematic model asfollows.

Find d

min w (d)

s.t. ge (d, t) =σe+

ix (d,t)[σ] − 1 � 0 (e = 1, 2, · · · , N ; t ∈ [0, Γ])

ge+N (d, t) = −σe+ix (d,t)

[σ] − 1 � 0 (e = 1, 2, · · · , N ; t ∈ [0, Γ])

ge+2N (d, t) =σe−

ix (d,t)[σ] − 1 � 0 (e = 1, 2, · · · , N ; t ∈ [0, Γ])

ge+3N (d, t) = −σe−ix (d,t)

[σ] − 1 � 0 (e = 1, 2, · · · , N ; t ∈ [0, Γ])

ge+4N (d, t) =σe+

jx (d,t)[σ] − 1 � 0 (e = 1, 2, · · · , N ; t ∈ [0, Γ])

ge+5N (d, t) = −σe+jx (d,t)

[σ] − 1 � 0 (e = 1, 2, · · · , N ; t ∈ [0, Γ])


ge+6N (d, t) =σe−

jx (d,t)[σ] − 1 � 0 (e = 1, 2, · · · , N ; t ∈ [0, Γ])

ge+7N (d, t) = −σe−jx (d,t)

[σ] − 1 � 0 (e = 1, 2, · · · , N ; t ∈ [0, Γ])

ge+8N (d, t) =σe+

Mx(d,t)[σ] − 1 � 0 (e = 1, 2, · · · , N ; t ∈ [0, Γ])

ge+9N (d, t) = −σe+Mx(d,t)

[σ] − 1 � 0 (e = 1, 2, · · · , N ; t ∈ [0, Γ])

ge+10N (d, t) =σe−

Mx(d,t)[σ] − 1 � 0 (e = 1, 2, · · · , N ; t ∈ [0, Γ])

ge+11N (d, t) = −σe−Mx(d,t)

[σ] − 1 � 0 (e = 1, 2, · · · , N ; t ∈ [0, Γ])

ge+12N (d, t) =τe

ixy(d,t)[τ ] − 1 � 0 (e = 1, 2, · · · , N ; t ∈ [0, Γ])

ge+13N (d, t) = − τeixy(d,t)

[τ ] − 1 � 0 (e = 1, 2, · · · , N ; t ∈ [0, Γ])

ge+14N (d, t) =τe

jxy(d,t)[τ ] − 1 � 0 (e = 1, 2, · · · , N ; t ∈ [0, Γ])

ge+15N (d, t) = − τejxy(d,t)

[τ ] − 1 � 0 (e = 1, 2, · · · , N ; t ∈ [0, Γ])

ge+16N (d, t) =τe

Mxy(d,t)[τ ] − 1 � 0 (e = 1, 2, · · · , N ; t ∈ [0, Γ])

ge+17N (d, t) = − τeMxy(d,t)

[τ ] − 1 � 0 (e = 1, 2, · · · , N ; t ∈ [0, Γ])

gk+18N (d, t) =xk(d,t)

[xk] − 1 ≤ 0 (k = 1, 2, · · · , Nf )

gk+Nf+18N (d, t) = −xk(d,t)[xk] − 1 � 0 (k = 1, 2, · · · , Nf )

gJ (d) = dJ

dJ− 1 � 0 (J = 1, 2, · · · , 4N)

gJ+4N (d) = 1 − dJ

dJ� 0 (J = 1, 2, · · · , 4N) (57)

4. Transformation of mathematical model

The key feature of the transformationmethod is to transform a constrained problem into an unconstrained problem.Thus, we minimize only one function in the transformation method. This is an attractive aspect in that manytime-dependent constraints and an objective function can be merged into a single function. The representatives ofthe transformation method are the augmented Lagrange multiplier method [16] and the exterior penalty functionmethod [20] in structure optimal design under dynamic loads. However, the augmented Lagrange multiplier functionand the exterior penalty function are discontinuous functions, therefore, the gradient and Hessian matrix calculationsof these functions are difficult when the direct differentiation method is employed to obtain the first and secondderivatives. Compared to the augmented Lagrange multiplier function and the exterior penalty function, the interiorpenalty function is a continuous function, so the gradient and Hessian matrix calculations of the interior penaltyfunction are relatively easy when the direct differentiation method is used to obtain the first and second derivatives.The interior penalty function method requires a feasible initial design point. Typically, it may be difficult to obtaina feasible initial design in a complex problem. However, in structural optimization problems, a feasible designpoint can be found in the structures with the large cross-sectional areas. Therefore, in this paper the interior penaltyfunction is employed to transform the inequality constraint optimization problem.


4.1. Interior penalty function method

In this paper£‹the inequality constraint optimal mathematic model Eq. (57) is converted to a sequence of appro-priately formed unconstrained integral mathematic model using the interior penalty function method. The interiorpenalty function method is adopted as follows.

P (d, rk) = w (d) − rk

⎛⎝2Nf +18N∑

J=1

∫ Γ

0

1gJ (d, t)

dt +8N∑I=1

1gI (d)

⎞⎠ (58)

In Eq. (58), penalty parameter rk is a sequence of numbers which are degressive. When rk → 0, the minimumof penalty function P (d, rk) approaches the minimum of the constraint problem. So the solution of the inequalityconstraint optimization model, Eq. (57), is transformed into a sequence of unconstraint problems:

Find d

min P (d, rk) = w (d) − rk

⎛⎝2Nf+18N∑

J=1

∫ Γ

0

1gJ (d, t)

dt +8N∑I=1

1gI (d)

⎞⎠ (59)

Initial penalty parameter r1 can be calculated by the following equation,

r1

⎛⎝2Nf+18N∑

J=1

∫ Γ

0

1gJ (d0, t)

dt +8N∑I=1

1gI (d0)

⎞⎠ =

p0

100w (d0) (60)

where d0 is the initial design point and p0 = 1 ∼ 50. In this work, we choose p0 = 50.The penalty parameter rk decreases according to the following rule:

rk+1 =rk

c(61)

where c = 10 ∼ 50 and c = 10 in this work. k is the number of penalty parameter which will be used in the processof search.

4.2. Calculation of gradient and Hessian matrix of interior penalty function

Now we calculate the first and second derivatives of the penalty function with respect to the structural designvariables. The time step and duration of dynamic loads are Δt and Γ, respectively. Let a = Γ

Δt .The first derivatives of penalty function can be obtained by differentiating Eq. (58) with respect to the design

variable di,

∂P (d, rk)∂di

=∂w (d)

∂di+ rk

⎛⎝2Nf+18N∑

J=1

∫ Γ

0

1g2

J (d, t)∂gJ (d, t)

∂didt +

8N∑I=1

1g2

I (d)∂gI (d)

∂di

⎞⎠ (62)

The second derivatives of penalty function is calculated by further differentiating Eq. (62) with respect to thedesign variable dj ,

∂2P (d, rk)∂di∂dj

=∂2w (d)∂di∂dj

+ rk

2Nf +18N∑J=1

∫ Γ

0

[ −2g3

J (d, t)∂gJ (d, t)

∂dj

∂gJ (d, t)∂di

+1

g2J (d, t)

∂2gJ (d, t)∂di∂dj

]dt

+ rk

8N∑I=1

−2g3

I (d)∂gI (d)

∂dj

∂gI (d)∂di

(63)

The integral terms in Eqs (58), (59), (60), (62), (63) are computed by using the trapezoidal form integral formula:∫ Γ

0

1gJ (d, t)

dt =a−1∑z=0

Δt

2

[1

gJ (d, zΔt)+

1gJ (d, (z + 1)Δt)

](64)


∫ Γ

0

1g2

J (d, t)∂gJ (d, t)

∂didt=

a−1∑z=0

Δt

2

[1

g2J (d, zΔt)

∂gJ (d, zΔt)∂di

+1

g2J (d, (z+1) Δt)

∂gJ (d, (z+1) Δt)∂di

](65)

∫ Γ

0

[ −2g3

J (d, t)∂gJ (d, t)

∂dj

∂gJ (d, t)∂di

+1

g2J (d, t)

∂2gJ (d, t)∂di∂dj

]dt =

a−1∑z=0

Δt

2

{[ −2g3

J (d, zΔt)∂gJ (d, zΔt)

∂dj

∂gJ (d, zΔt)∂di

+1

g2c (d, zΔt)

∂2gJ (d, zΔt)∂di∂dj

]+ (66)

[ −2g3

J (d, (z + 1)Δt)∂gJ (d, (z + 1)Δt)

∂dj

∂gJ (d, (z + 1)Δt)∂di

+1

g2J (d, (z + 1)Δt)

∂2gJ (d, (z + 1)Δt)∂di∂dj

]}

The first and second derivatives of the penalty function with respect to the structural variables are calculated. Thenthe gradient and Hessian matrix can be achieved.

5. Solving optimization problems

The inequality constraint optimization model, Eq. (57), is converted into a sequence of the appropriately formedunconstrained integral model, Eq. (59). Marquardt’s method, a gradient-Hessian matrix-based algorithm, is adoptedto solve the unconstrained problem, taking advantages that the gradient and Hessian matrixes are fully used inthis optimal method. Marquardt’s method combines Cauchy’s and Newton’s methods in a convenient manner thatexploits the strengths of both but does require second-order information. The major merit of Marquardt’s method isits simplicity, descent property, excellent convergence rate near the optimum, and absence of a line search. Basedon Marquardt’s method, the computation procedure of solving the mathematic model Eq. (56) is as follows.

The computer procedure of solving the mathematic model Eq. (56):

Step 1. Chose the initial feasible design point d0, calculate r1 by solving Eq. (60), define convergence criterionε1, let k = 1.

Step 2. Start from design point dk−1, solve the mathematic model Eq. (59) with Marquardt’s method to obtainthe optimum design dk.The steps of solving the mathematic model Eq. (59) with Marquardt’s method isfrom Step 2.1. to Step 2.11.:Step 2.1. Let d(0)

k−1 = dk−1. DefineMI = maximum number of iterations allowedε2 =convergence criterionI =identity matrixStep 2.2. Set i = 0. λ(0) = 105.

Step 2.3. Calculate ∇P(d(i)

k−1, rk

).

Step 2.4. Is∥∥∥∇P

(d(i)

k−1, rk

)∥∥∥ � ε2?Yes: Go to step 2.11.No: Continue.Step 2.5. Is i � MI?Yes: Go to step 2.11.No: Continue.

Step 2.6. Calculate S(d(i)

k−1

)= −

[∇2P

(d(i)

k−1, rk

)+ λ(i)I

]−1

∇P(d(i)

k−1, rk

).

Step 2.7. Set d(i+1)k−1 = d(i)

k−1 + S(d(i)

k−1

).

Step 2.8. Is P(d(i+1)

k−1 , rk

)< P

(d(i)

k−1, rk

)?

Yes: Go to step 2.9.


No: Go to step 2.10.Step 2.9. Set λ(i+1) = 1

2λ(i) and i = i + 1. Go to step 2.3.Step 2.10. Set λ(i) = 2λ(i). Go to step 2.6.Step 2.11. Set dk = d(i)

k−1. Go to step 3.Step 3. Is P (dk−1, rk) − P (dk, rk) � ε1?

Yes: dk is the best design. Print result and stop.No: Continue.

Step 4. Calculate rk+1 by solving Eq. (61), k = k + 1. Go to step 2.

6. Numerical example

In this section, optimization design of the plane frame shown in Fig. 2 is performed using the method proposedin this paper. The plane frame is divided with three plane beam elements. The cross-section of the element is Hstructural-steel shape, shown in Fig. 1. We use typical steel materials properties, i.e., elastic modulus E = 210 GPaand material density ρ = 7850 kg/m3, for all elements. The structural damping ratios are ς1 = 0.02, ς2 = 0.02.The horizontal dynamic load, Fh (t) = 1500 sin 3π

4 t (kN), acts on the node 2. A duration of 3 s and an incrementaltime step of 0.01 s are considered in optimization procedures. The convergence criterion: ε1 = 10−3, ε2 = 10−3

and the maximum number of iterations allowed: MI = 5. Numerical tests show that these criteria are sufficient forachieving convergence in a reasonable time.

Fig. 2. Plane frame.

In the optimal mathematic model Eq. (57), the allowable normal stress [σ] = 200 MPa and allowable shearstress [τ ] = 100 MPa; the nodal allowable displacements are: [x1] = 0.001 m, [x2] = 0.001 m, [x3] = 0.1 rad,[x4] = 0.001 m, [x5] = 0.001 m, [x6] = 0.1 rad; the design variable vector is defined as: d = [d1

1, d12,

d13, d

14, d

21, d

22, d

23, d

24, d

31, d

32, d

33, d

34]

T . The design space is shown in Table 1.

Table 1Design space of frame (unit: mm)

Design variables d11 d1

2 d13 d1

4 d21 d2

2 d23 d2

4 d31 d3

2 d33 d3

4

Lower limit 100 100 4 4 100 100 4 4 100 100 4 4Upper limit 500 1000 70 70 500 1000 70 70 500 1000 70 70

The optimum designs of the plane frame are searched from two different initial feasible design points. Sequencesof optimum designs are shown in Tables 2 and 3.

From the optimum results shown in Tables 2 and 3, the masses of the optimum designs approach to the finaltargets, 26 kg and 16 kg, from the initial feasible design 1 (i.e., 153 kg) and 2 (i.e., 45 kg), respectively. It indicatesthat the optimization method presented in this paper is effective. However, the optimum designs are different if theinitial designs are not same. Therefore, the optimum designs obtained with the optimization method in this paperare local solutions, but not global solutions. We should find the different local solutions from as many initial designsas possible. Then we can choose the best design from the local designs as the effective design.


Table 2Initial feasible design 1 and optimum design (unit: mm)


2 d13 d1

4 d21 d2

2 d23 d2

4 d31 d3

2 d33 d3

4 Mass/kg Time/s

Initial design 1 450 900 60 60 450 900 60 60 450 900 60 60 153 0Method in this paper r1 = 251.14 342 801 45 46 263 785 42 39 294 794 45 44 63 2392

r2 = 25.114 341 800 33 34 261 785 23 11 292 793 34 29 26 4770r3 = 2.5114 341 800 33 34 261 785 23 11 292 793 34 29 26 7149

Augmented Lagrange multiplier 363 820 29 32 276 779 18 10 300 810 38 27 24 8426

Table 3Initial feasible design 2 and optimum design (unit: mm)


2 d13 d1

4 d21 d2

2 d23 d2

4 d31 d3

2 d33 d3

4 Mass/kg Time/s

Initial design 2 300 600 40 40 300 600 40 40 300 600 40 40 45 0Method in this paper r1 = 96.33 303 600 30 37 298 599 18 18 298 599 27 27 24 2118

r2 = 9.633 302 600 27 35 298 599 18 18 298 599 27 27 22 4235r3 = 0.9633 302 600 25 33 298 599 8 8 298 599 22 22 16 6558r4 = 0.09633 302 600 25 33 298 599 8 8 298 599 22 22 16 8896

Augmented Lagrange multiplier 300 596 26 31 298 598 8 7 294 593 25 27 16 11042

All simulations are performed on a personal computer with Window XP operating system. The computer hasa Pentium(R) 4 CPU, and frequency of the CPU is 2.8 GHz. The computer also has 512MB RAM. The iterationcourses shown in Fig. 3 indicate that the algorithm is convergent. From initial feasible design 1, the computationaltime of achieving the local optimum design is about 7149 s with the method in this paper, and about 8426 s withthe Augmented Lagrange multiplier method. From initial feasible design 2, the computational time of achievingthe local optimum design is about 8896 s with the method in this paper, and about 11042 s with the AugmentedLagrange multiplier method. The computational time shows that the method proposed in this paper is sometimesmore efficient than the Augmented Lagrange multiplier method.

0

20

40

60

80

100

120

140

160

180

0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000

Time/s

Mas

s/kg

Initial design 1Initial design 2

Fig. 3. Iteration courses.

It should be noted that the gradient and Hessian matrix calculation is very difficult for the dynamic optimizationproblem because it is difficult to calculate the dynamic response first and second derivatives. Generally, thegradient and Hessian matrix calculation requires much computational time that can not be accepted in the structuraloptimization. However, in this work we develop an algorithm, only a single dynamics analysis is required, toobtain the gradient and Hessian matrix. In addition, we use the interior penalty function method to transform aconstrained problem into an unconstrained problem. So many time-dependent constraints and an objective functioncan be merged into a sequence of appropriately formed unconstrained integral single time-independent functions.Those unconstrained integral single time-independent functions are continuous, so the gradient and Hessian matrixcalculations are easier than other discontinuous transform functions, the augmented Lagrange multiplier functionand exterior penalty function.


7. Conclusions

In this paper, we developed an optimization method of structures subjected to transient loads. The conclusionsare as follows.

(1) An algorithm is formulated to calculate structural dynamic responses, i.e., nodal displacements and stresses,and their first and second derivatives with respect to design variables. The algorithm is achieved by directdifferentiation and only a single dynamics analysis based on Newmark-β method is required.

(2) The interior penalty function method is used to transform a constrained problem into an unconstrainedproblem. So many time-dependent constraints and an objective function can be merged into a sequence ofappropriately formed unconstrained integral single time-independent functions. Those unconstrained integralsingle time-independent functions are continuous, so the gradient and Hessian matrix calculations are easierthan other discontinuous transform functions, the augmentedLagrangemultiplier function and exterior penaltyfunction.

(3) The gradient-Hessian matrix-based optimization method presented in this paper has the characteristics of itssimplicity, descent property, excellent convergence rate near the optimum, and absence of a line search.

(4) The numerical results show that the optimum designs obtained with the optimization method presented in thispaper are local solutions, but not global solutions and that the optimization method is effective.

(5) The numerical results also show that the optimal design method proposed in this paper is sometimes moreefficient than the augmented Lagrange multiplier method. Sometimes, it depends on the chosen initial design.

References

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