Truss Design Art

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Truss Design and Convex Optimization

RobertM.Freund

April 13, 2004

c2004MassachusettsInstituteof Technology.

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1 Outline

• PhysicalLawsof aTrussSystem• TheTrussDesignProblem• Second-OrderConeOptimization• TrussDesignandSecond-OrderConeOptimization• SomeComputationalResults• Semi-DefiniteOptimization

• TrussDesignandSemi-DefiniteOptimization• TrussDesignandLinearOptimization

• Extensionsof theTrussDesignProblem

2 Physical Laws of a Truss System

Atruss isastructureind= 2 or d=3dimensions,formedbynnodesandmbars joiningthesenodes. Figure1showsanexampleof atruss.

2

3

4

5

6

Figure 1: Example of a truss in d = 2 dimensions, with n = 6 nodes andm= 13 bars.

Examplesof trusses includebridges,cranes,andtheEiffelTower.

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Thedatausedtodescribeatruss is:

• asetof nodes(given inphysicalspace)• asetof bars joiningpairsof nodeswithassociateddataforeachbar:

– the lengthLk of bark

– theYoung’smodulusE k of bark

– thevolumetk of bark

• anexternalforcevectorF onthenodes

Nodes

can

be

static

(fixed

in

place)

or

free

(movable

when

the

truss

is stressed). The allowable movements of the nodes defines the degrees of freedom(dof)of thetruss. WesaythatthetrusshasN degreesof freedom.Of course,N ≤ nd.

Movements of nodes in the truss will be represented by a vector u of N displacements ,whereu∈ .

N Theexternalforceonthetruss isgivenbyavectorF ∈ .Displacementsof thenodesinthetrusscauseinternalforcesof compres-

sionand/orexpansiontoappearinthebarsinthetruss. Letf k denotetheinternalforceof bark. Thevectorf

∈ m isthevectorof forcesof thebars.

Figure

2

shows

an

example

of

a

truss

problem.

In

the

figure,

there

is

a

singleexternal forceappliedtonode3 inthedirection indicated. Thiswillresult in internal forcesalongthebars inthetrussandwillsimultaneouslycausesmalldisplacements inallof thenodes.

In Figure 2, nodes 1 and 5 are fixed, and the other nodes are free. Tosimplify notation we will label the 13 different bars by the nodes that heylink to: in generalthebark will joinnodes iand j. Thebarswillbe: 12,13, . . .,56.

The internalforcef k of barkcanbepositiveornegative:

•

If

f k ≥ 0,bark hasbeenexpandedand its internal forcecounteractstheexpansionwithcompression,seeFigure3.

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2

3

4

5

6

1

f

f

F13

12

3

Figure2: Atrussproblem.

∆ > 0 L k

Figure

3:

A

bar

under

expansion.

L k

∆ < 0

Figure4: Abarundercompression.

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• If f

k

≤0,barkhasbeencompressedanditsinternalforcecounteracts

the

compression

with

expansion,

see

Figure

4.

2.1 Physical Laws

2.1.1 Force Balance Equations

Inastatictrusstheinternalforceswillbalancetheexternalforcesineverydegreeof freedom. That is,theforcesaroundeach freenodemustbalance.This is a law of conservation of forces. (This is akin to material balanceequations innetworks,forexample.)

For

example,

consider

the

balance

of

forces

on

node

3:

xcoordinate: −f 13 −f 23 cos(π/4) +f 35 +f 36 cos(π/4) =−F 3xy coordinate: +f 23 sin(π/4) +f 34 +f 36 sin(π/4) =−F 3y

FortheentiretrusswewriteN linearequationsthatrepresentthebal-anceof forcesineachdegreeof freedom.

Inmatrixnotationthiscanbewrittenas

Af =−F

whereA isanN

×mmatrix,andwhereeachcolumnof A,denotedasak,

is

the

projection

of

the

bar

onto

the

degrees

of

freedom

of

the

nodes

that

barkmeets.

Inourexample,wehave:

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12 13 14 16 23 24 25 34 35 36 45 46 56

1√ 1 2√

0 0 0 0

0 0 0 0 0 0 2x52

−1 0 0 0 − 1√ 0 − 1√ 52

0 0 0 0 0 0

2y

0 −1 0 0 − 1√ 0 0 0 1 1√

22

0 0 0 3x 1√ 0 0 1 0 1√ 0 0 0 0 0 0 0 3yA=

22

1√ 0 0 −1 0 0 0 0 1√ 0 0 1 0 4x − 22

1√ 0 0 0 0 −1 0 0 − 1√ 0 0 0 0 4y− 22

2√ 0 0 0 0 0 − 1√ 0 0 0

6x0 −1 0−25

1√ 0

0

0

0 0

−1

√ 0 0

0

−

0 0

−1 6y25

2.1.2 Constitutive Relationship between Forces and Distortion

of a Bar

If bar k has length Lk and cross-sectional area Ack, Young’s modulus E k,

and its length ischangedby∆k,thenthe internalforcef k isgivenby:

∆kf k =E kA

c .kLk

However,

for

our

purposes

it

will

be

easier

to

work

instead

with

the

volume tk of bark,which is:

kLk .tk =Ac

Wethereforecanwritetheconstitutiverelationshipas:

E kf k =

L2tk∆k .

k

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2

2.1.3 Distortion and Displacements

Displacements in the nodes will cause the lengths of the bars to change.SupposethatL12 =L13 =L35 = 1.0,withallotherbarsmeasuredpropor-tionally. InFigure5,weshowadisplacementof:

u= (−,−,0,0,0,0,0,0) .

F3

2

3

4

5

6

1

Figure5: Exampleof asmalldisplacement inanode.

Table

1

shows

the

new

lengths

of

the

bars

under

the

displacement

u

=

(−,−,0,0,0,0,0,0). Thethirdcolumnof thetableshowsthelinearization(first-order Taylor approximation) of the lengths of the bars, and the lastcolumnshowsthecomputationof theinnerproduct:

(ak)T u .

Notethatif issmall,thenthedistortionof barkisnicelyapproximatedbythe linearexpression:

∆k =−(ak)T u .

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Bark OldLength NewLength LinearizedLength LinearizedChange∆k

(ak

)T u

12 1

1 + 2( −1) 1− − 13 1 1 1 0 014 1√ 1√ 1√ 0 016

23

5√ 2

5√ 2 + 22

5√ 2

0

0

0

0 24

25

1√ 5

1 + 2( + 1)

5 + 2( + 1)

1 + √ 5 + 1√

5

1√

5

−1−√ 5

34 1 1 1 0 035 1√ 1√ 1√ 0 036 2

√

2

√

2

√

0 0

45

2

2

2

0

0

46 1 1 1 0 056 1 1 1 0 0

Table1: Changes inthe lengthsof thebarsundernodaldisplacement

Wethereforewritetheinternalforceonbark duetoafeasibledisplace-mentu as:

E kf k =− 2 tk(ak)T uLk

LetthematrixB bethefollowingdiagonalmatrix:

L21

E 1t10

E 1t1

L21

0

B = ... , B−1 = ... .

0 L2m

E mtm0 E mtm

L2m

Thenwecanwritetheaboverelationshipas:

f =−B−1AT u ,

which

is:

Bf +AT u = 0 .

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2.1.4 Equilibrium Conditions and Compliance

Nowwecanwritethephysical lawsof thetrusssystemas:

Af = −F (conservationof forces)

Bf +AT u = 0 (forces,distortions,anddisplacements)

The compliance of the truss is the work (or energy) performed by thetruss,which isthesumof theforcestimesdisplacements:

1F T u .

2

“1

nience.

Notice that the larger the compliance, the more work or displacementthatthetrussundergoesunderthe loadF . Ideally,wewould likethecom-pliancetobeasmallnumber.

Combiningtheequilibriumconditions,wewrite:

We really do not need the fraction2

”, but we will keep it for conve-

F

=

−Af

= AB−1AT u

= Ku

wherethematrixK isdefinedtobe:

K =AB−1AT .

ThematrixK iscalledthestiffness matrix of thetruss. NotethatK isanSPSDmatrix.

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Usingthis,wecanwritethecomplianceas:

1 1F T u = uT Ku .

2 2

Notefromthisthatthecompliancewillalwaysbeanonnegativequantity,asintuitionsuggests. Moregenerally,of course,wewouldexpectthestiffnessmatrix K to be SPD, and so the compliance will generally be a positivequantity.

2.1.5 Solving the Equations and Computing the Compliance

Wesolvetheequationsystem:

Ku =F

toobtainthenodaldisplacementsu,andthenobtaintheforcesonthebarsbycomputing:

f =−B−1AT u . Foreachbark,this lastexpression issimply:

E kf k =

−

2 tk(ak)T

u .

Lk

Thecompliance isthencomputedas:

1 1F T u = uT Ku .

2 2

2.2

Viewing the Equilibrium Conditions as the Solution to

an Optimization Problem

We can also view the truss equilibrium conditions as the solution to anoptimizationproblem. Considerthefollowingoptimizationproblem:

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m1 1 L2

OP: minimizef ̃

k f ̃22 tk E k

k

k=1

m

s.t. akf ̃k =−F . k=1

ProblemOPseeksaforcevectorf ̃ thatsatisfiestheforcebalanceequa-tions:

m

−F = akf ̃k =Af̃ , k=1

andthatminimizestheweightedsumof squaresof theforces:

m 1 L21 k f ̃22 tk E k

k .k=1

ProblemOP isaconvexquadraticproblem. TheKKToptimalitycon-ditionsforthisproblemare:

m

akf k =−F k=1

L2 T k f k +aku= 0 for k= 1, . . . , m . tkE k

Noticethatwecanrewritetheseoptimalityconditionsas:

Af =−F

Bf +AT u= 0 ,

which

are

precisely

the

equilibrium

conditions

for

the

truss

system.

Thisshowsthatthetrusssystemisnature’ssolutiontoanoptimizationproblem.

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Theoptimalobjectivefunctionvalueof theoptimizationproblemOPis:

m1 1 L2 1 1 1 1k f 2 = f T Bf =− f T AT u=− uT Af = F T u ,2 tk E k

k 2 2 2 2k=1

which is the compliance of the truss. Therefore we see that the optimalobjectivevalueof OP isthecomplianceof thetrusssystem.

3 The Truss Design Problem

Let us now consider the volumes tk of the bars k to be design parametersthatwewishtodetermine. Wecanwritethesystemof equations:

F =Ku

as:

F = Ku

= AB−1AT u

m

B−1AT = (ak) uk

k=1

m

= −(ak)f kk=1

m E k

= tkL2

(ak)(ak)T u

k=1 k

= K (t)u

wherethestiffnessmatrix isnowwrittenas:

m E k

K (t) = tkL2

(ak)(ak)T .

k=1 k

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NoteinthisexpressionthatK =K (t)isaweightedsumof rank-onematri-ces

(weighted

by

the

volumes

tk). Putanotherway,K =K (t) isaweightedsumof outer-productmatrices.

Indesigningthetruss,wemustchoosethevolumestk of thebarssubjectto linearconstraintsonthevolumesof thebars:

M t ≤ d

t ≥ 0 ,

wheretypicallytheseconstraintsincludeupperandlowerboundsonthe

volumes

of

certain

bars,

as

well

as

an

overall

cost

or

volume

constraint:

m

tk ≤ V . k=1

The criterion intruss design is to choose the volumes tk of the bars soastominimizethecomplianceof thetruss,namely:

1minimizet,u F

T u .2

Such

a

truss

will

be

the

most

resistant

to

the

external

force

F .

Thesingle-loadtrussdesignproblemcanbestatedas:

1(TDP

1): minimizet,u F

T u2

s.t. K (t)u=F

M t ≤ d

t≥ 0mu∈ N , t ∈ .

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whichisthesameas:

1(TDP

1): minimizet,u F

T u2

m

s.t. tkE k (ak)(ak)

T u=F L2

k=1 k

M t ≤ d

t≥ 0m

u∈

N

, t∈ .

Thedecisionvariableshereareuandt. However,oneshouldreallythinkof thedecisionvariablesastonly,sinceoncetischosen,uwillbedeterminedbythesolutiontothesystemof equations:

m

tkE k

(ak)(ak)T u=F .

L2k=1 k

Note that as written, the truss design problem TDP is not a convex

problem.

4 A Convex Version of the Truss Design Problem

Recallthatforgivenvolumestk onthebarsof thetruss,thatthecomplianceof thetruss istheoptimalobjectivevalueof thequadraticproblem:

m1 1 L2

OP: minimizef ̃

k f ̃22 tk E k

k

k=1

m

s.t.

akf ̃k =

−F .

k=1

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The truss design problem is then to choose the values of the volumest

= (t1, . . . , tm) on the bars so that the optimal solution of OP (which wehave shown is the compliance of the truss) is minimized, subject to theconstraintsont:

M t ≤ d

t ≥ 0 .

Wewritethisallas:

1 1 L2kf 2

(TDP2)

:

minimizef,t 2

tk E k ktk>0

s.t. akf k =−F tk>0

Mt ≤ d

t≥ 0mf ∈ m, t ∈ .

Notice here that we have made two modifications to the optimizationproblem. First,wehavemodifiedthesummation intheobjectivefunction,sothatbarswithzerovolume(tk =0)arenolongercounted,sincetheywillnotexist. Second,wechangedthenotation,usingf insteadof f ̃. Becausethe summations with the tk > 0 is rather awkward mathematically, wefurtherre-writeTDP2 asfollows:

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m1(TDP

2): minimizef,t,s

2sk

k=1

m

s.t. akf k =−F k=1

M t ≤ d

L2k f 2 ≤ tksk fork= 1, . . . mE k k

t≥ 0, s ≥ 0mf ,t ,s∈ .

Inthisproblem,wehavereplacedthequantity

1 L2 k f 2tk E k

k

withthenewvariablesk subjecttothecondition:

L2k f 2E k

k ≤ tksk .

Wecanfurtherre-writethisnowas:

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m1(TDP

2): minimizef,t,s

2sk

k=1

s.t. Af =−F

M t ≤ d

L2k f 2 ≤ tksk fork= 1, . . . mE k k

t

≥0, s

≥0

mf ,t ,s∈ .

Thislastproblemhasalinearobjectivefunction,andlinearconstraintsexceptfortheconstraints:

L2k f 2E k

k ≤ tksk .However, it isprettyeasytoshowthattheconstraints:

L2k f 2 ≤ tksk, tk ≥ 0, sk ≥ 0E k

k

describe

a

convex

region,

and

so

this

formulation

is

actually

a

convex

opti-mizationproblem.

5 Second-Order Cone Optimization

Asecond-orderconeoptimizationproblem(SOCP)isanoptimizationprob-lemof theform:

T SOCP: minx c x

s.t.

Ax

=

b

Qix +di ≤ giT x +hi , i = 1, . . . , k .

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Inthisproblem,thenormvisthestandardEuclideannorm:√

v := vT v .

The norm constraints in SOCP are called “second-order cone” con-straints.

NoticefirstthatSOCP isaconvexproblem,sincethefunction:

Qix +di − giT

x +

hi

isaconvexfunction.

Noticealsothat linearoptimization isaspecialcaseof SOCP, if weset

Qi = 0, di = 0, hi = 0, and gi tobetheith unitvectorfori= 1, . . . , n.

Alsonoticethatanyconvexquadraticconstraintcanbeconverted intoasecond-orderconstraint. Toseethis,supposewehaveaconstraint:

1xT Qx +q T x +r≤ 0 ,

2

whereQisSPSD.Wecanfactor

Q=M T M

forsomematrixM ,andthenwriteourconstraintas:

1√ 2

M x , q tx +r + 1

2

≤

−q tx − r+ 1 2

.

If yousquarebothsidesandcollectterms,youwillseethatthisisindeedequivalenttotheoriginalconvexquadraticconstraint.

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Finally,notethatwecanalways formulateaconvexquadraticproblemas

an

SOCP,

since

we

can

create

a

new

variable

xn+1 andwriteourquadraticproblemas:

minx,xn+1 xn+1

s.t....

T 1xT Qx +q x≤ xn+1 ,2

which

can

be

further

re-written

as

an

SOCP.

6 Truss Design and Second-Order Cone Optimiza

tion

Wenowpresentasecond-orderconeoptimizationproblemthatisequivalentto the truss design problem TDP2. Recall this version of the truss designproblem:

m1(TDP

2): minimizef,t,s

2sk

k=1

s.t. Af =−F

M t ≤ d

L2k f 2 ≤ tksk fork= 1, . . . mE k

k

t≥ 0, s ≥ 0mf ,t ,s∈ .

Letusnowmakethesimplechangeof variables:

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1wk = 2 tk +1sk2

yk = −1 tk +1sk2 2

Thensk =wk +yk andtk =wk− yk, and so:

2 2tksk = (wk− yk)(wk +yk) = wk− yk,

andsotheconstraintL2k

f 2

k ≤ tkskE k

becomes:

L2kf 2 2E k

k ≤ wk− y2 .kWecanrearrangethisandtakesquarerootstoobtain:

L2k f 2 +y2 ≤ wk ,E k

k k

whichwecan inturnwriteas:

Lk

yk,√ f k ≤ wk .

E k

Thereforewecanre-writeTDP2 as:

1(CTDP): minimizef,w,y e

T (w +y)2

s.t. Af =−F

yk,√ Lk f k

≤ wk , k = 1, . . . , mE k

M (w

− y)

≤ d

mw ,y ,f ∈ .

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NoticethatCTDP isasecond-orderconeproblem. Theobjectivefunc-tion and the first and third set of constraints are linear equations and in-equalities. Thesecondsetof constraintsaresecond-order-coneconstraints,sincetheLHS isthenormof a2-dimensionalvector:

(v1, v2):=

yk ,

Lk√ E k

f k

,

andtheRHSisthe linearexpression:

wk.

Proposition 6.1 Suppose that (f ,w ,y) is a feasible or optimal solution of CTDP.

Let:

t=w −y and s=w+y.Then

(f ,t ,s) is the corresponding feasible or optimal solution of TDP 2.

6.1

Other Formulations and Formats for the Truss Design

Problem

We

have

just

seen

how

to

formulate

the

truss

design

problem

as

the

convex

optimization problem TDP2 and more specially as the second-order coneoptimization problem CTDP. In addition to these two formulations of theproblem,thereisawaytoformulatethetrussdesignproblemasaninstanceof amoregeneraltypeof convexproblemcalleda“semi-definiteoptimizationproblem” (SDO for short). The topic of SDO and its application to trussdesign isdiscussedhereininSections8and9.

Finally,whentheonlyconstraintsonthevolumevariablest1, . . . , tm arenonnegativityconditionsandatotal-volumeconstraintof theform:

m

tk ≤V , k=1

then the truss design problem can actually be converted to a linear opti-mizationproblem. This isshownherein inSection10.

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7 Some Computational Results

Inthissectionwereportsomecomputationalresultsonsolvingtrussdesignproblems. Wesolvedthreedifferenttypesof trussdesignproblems. Thefirsttype is a basic bridge design, and is illustrated in Figure 6. In the figure,thearrowscollectivelycomprisetheforcevectorF appliedtothestructure,andthecircles(intheleftlowerandrightlowercorners)arethefixednodesof the structure. The stars in the figures are the other nodes in the trussstructure. Figure7showsthesetof barsthatcanbeusedtoconstructthebridge. These bars were generated by considering bars between any twonodeswhose“distance”fromoneanotherwasthreenodesor less.

7

6

5

4

3

2

1

0

0 2 4 6 8 10 12 14 16

Figure6: Theforcesandnodesforthebasicbridgedesignmodel.

Thesecondtypeof problemthatwesolvedistheproblemof designingahangingsupportforaloadsuchasatrafficsign,andisillustratedinFigure

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0 2 4 6 8 10 12 14 16

0

1

2

3

4

5

6

7

Figure 7: The set of possible bars for the basic bridge design problem.

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8. The circles on the left part of the figure correspond to the fixed nodes, andthe arrow on the right side is a force that will be applied (the gravitationalforce of the traffic sign).

0 5 10 15 20 25 30

0

2

4

6

8

10

12

14

16

18

20

Figure 8: The forces and nodes for the hanging sign design model.

The third type of problem that we solved is the problem of designinga crane to handle a gravitational load, and is illustrated in Figure 9. Thecircles on the bottom of the figure correspond to the fixed nodes, and thearrow on the right side is a force that will be applied (the gravitational forceof the traffic sign).

For the problems shown in Figures 6, 8, and 9, the truss design modelsolutions are shown in Figures 10, 11, and 12. In these figures, the thickness

of the image of the bars is drawn proportional to the volumes of the bars inthe optimal solutions.

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40

35

30

25

20

15

10

5

0

0 2 4 6 8 10 12 14 16 18 20

Figure9: Theforcesandnodesforthecranedesignmodel.

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7

6

5

4

3

2

1

0

0 2 4 6 8 10 12 14 16

Figure10: Optimalsolutiontothebasicbridgedesignproblem.

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20

18

16

14

12

10

8

6

4

2

0

0 5 10 15 20 25 30

Figure11: Optimalsolutiontothehangingsigndesignproblem.

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40

35

30

25

20

15

10

5

0

0 2 4 6 8 10 12 14 16 18 20

Figure12: Optimalsolutiontothecranedesignproblem.

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Notice inFigure10thatthebasicbridgedesigndoesnotautomaticallyinclude

horizontal

bars

for

a

“road

surface”

on

the

base

of

the

structure.

In

order to force there to be such a road surface, we must add lower boundsonthevolumesof thosebarsonthebaseof structure. If weaddsuchlowerbounds,theoptimalbridgedesign isasshown inFigure13.

7

6

5

4

3

2

1

0

0 2 4 6 8 10 12 14 16

Figure13: Optimalsolutionof thebridgedesignproblem,withlowerboundsonthe“roadsurface”barvolumes.

7.1 Details of Problems Solved

Forthebridgedesignproblem,we included lower-boundconstraintsonthevolumesof theroad-surfacebarsaswellasaconstraintonthetotalvolume

of

the

structure.

With

the

lower-bound

constraints

in

the

model,

the

model

mustbesolvedeitherasasecond-orderconeproblem(SOCP)orasasemi-definiteoptimizationproblem(SDO).Forthehangingsigndesignproblem

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Name Sizeof grid Nodes Arcs Degreesof freedom MaximumVolumen

m

N

Bridge-16×7 16×7 136 1,547 268 1,600Bridge-20×10 20×10 231 2,878 458 2,000Bridge-30×10 30×10 341 4,368 678 4,000Sign-10×20 10×20 231 2,878 444 1,000Sign-20×30 20×30 651 9,058 1,280 2,000Sign-30×40 30×40 1,271 18,438 2,512 3,000Crane-10×20 10×20 96 818 188 2,000Crane-20×40 20×40 291 3,188 578 5,000Crane-30×40 30×40 401 4,678 798 10,000

Table

2:

Dimensions

of

the

truss

design

problems.

aswellasforthecranedesignproblem,theonlyconstraintonthevolumeswas a constraint on the total volume of the truss structure. This made itpossibletosolvetheseproblemsusing linearoptimization.

We solved three different instances of each of the bridge, hanging sign,and cranedesignproblems, with differentnumbers of nodes correspondingtodifferentmeshes. Table2showsdetailsof themodelsthatwesolved. Theright-mostcolumninthetableshowsthevalueof thevolumeconstraintonthevolumeof thetrussstructure.

Table3showsthenumberof variablesandthenumberof inequalitiesintheseproblems,andTable4showsthenumberof variablesandthenumberof equations/constraints/matrixdimensionof theseproblems.

Table 5 shows the compliance of the optimal truss design for the nineproblemsthatweresolved.

Table 6 shows the iterations of the interior-point algorithms that wereused to solve the truss design problems. For the linear problems and thesecond-order cone problems, we used LOQO to solve the model. For the

semi-definite

optimization

models,

we

used

an

SDO

algorithm

called

SDPa.Therunningtimesof thesemethods isshown inTable7.

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ProblemLP

Model

SOCP

Model

Variables Inequalities Variables Inequalities2m 2m 3m m+ 1 + LBs

Bridge-16×7 † † 4,641 1,564Bridge-20×10 † † 8,634 2,899Bridge-30×10 † † 13,104 4,399Sign-10×20 5,756 5,756 8,634 2,879Sign-20×30 18,116 18,116 27,174 9,059Sign-30×40 36,876 36,876 55,314 18,439Crane-10×20 1,636 1,636 2,454 819Crane-20

×40 6,376 6,376 9,564 3,189

Crane-30×40

9,356

9,356

14,034

4,679

Table3: Numberof variablesand inequalitiesinthetrussdesignproblems.†The linearoptimizationmodelcannotbeused forthemodifiedbridgede-signmodel.

ProblemLPModel SOCPModel SDOModel

Variables Equations Variables Constraints MatrixDimension2m N 3m N +m+1+LBs N +m+ 2

Bridge-16×7

†

†

4,641

1,832

1,685Bridge-20×10 † † 8,634 3,357 3,111

Bridge-30×10 † † 13,104 5,077 4,711Sign-10×20 5,756 444 8,634 3,323 3,111Sign-20×30 18,116 1,280 27,174 10,339 9,711Sign-30×40 36,876 2,512 55,314 20,951 19,711Crane-10×20 1,636 188 2,454 1,007 916Crane-20×40 6,376 578 9,564 3,767 3,481Crane-30×40 9,356 798 14,034 5,477 5,081

Table 4: Number of variables and equations in the truss design problems.

†The linearoptimizationmodelcannotbeused forthemodifiedbridgede-

sign

model.

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Problem LP SOCP SDP

Bridge-16×7 † 27.43365 27.15510Bridge-20×10 † 52.58314 52.10850Bridge-30×10 † 132.46038 ‡Sign-10×20 0.77279 0.77293 0.77279Sign-20×30 2.52003 2.52056 ‡Sign-30×40 4.08573 4.08681 ‡Crane-10×20 28.67827 28.67840 28.67753Crane-20×40 286.24854 286.24954 286.19340Crane-30×40 596.73177 596.73324 596.63740

Table

5:

Compliance

of

the

optimized

truss

design

for

the

truss

design

prob-lems. †Thelinearoptimizationmodelcannotbeusedforthemodifiedbridgedesignmodel. ‡ThedatatoruntheSDOmodelcouldnotbeprepared forthisinstanceduetomemoryrestrictions.

Problem LP SOCP SDO

Bridge-16×7 † 38 36Bridge-20

×10

† 55 43

Bridge-30×10

†

47

‡

Sign-10×20 18 61 37Sign-20×30 21 47 ‡Sign-30×40 24 53 ‡Crane-10×20 17 52 34Crane-20×40 31 158 58Crane-30×40 31 144 60

Table 6: Number of iterations of the interior-point algorithm to solve thetruss design problems. †The linear optimization model cannot be used forthemodifiedbridgedesignmodel. ‡ThedatatoruntheSDOmodelcouldnot

be

prepared

for

this

instance

due

to

memory

restrictions.

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Problem LP SOCP SDO

Bridge-16×7 † 93.90 257.01Bridge-20×10 † 460.96 1,088.09Bridge-30×10 † 904.86 ‡Sign-10×20 3.33 513.37 1,081.54Sign-20×30 32.08 4,254.08 ‡Sign-30×40 111.03 26,552.74 ‡Crane-10×20 0.52 41.12 446.98Crane-20×40 6.46 1,299.12 33,926.20Crane-30×40 10.67 1,599.85 95,131.49

Table7: Runningtime(inseconds)of the interior-pointalgorithmtosolve

the

truss

design

problems.

†The

linear

optimization

model

cannot

be

used

for the modified bridge design model. ‡The data to run the SDO modelcouldnotbepreparedforthis instanceduetomemoryrestrictions.

8 Semi-Definite Optimization

If S isak × kmatrix,thenS isasymmetricpositivesemi-definite(SPSD)matrix if S issymmetric(S ij =S ji foranyi, j= 1, . . . , k) and

v

T

Sv

≥ 0

for

any

v

∈ k

.

If S isak × kmatrix,thenS isasymmetricpositivedefinite(SPD)matrixif S issymmetricand

vT Sv >0 forany v∈ k, v = 0.

LetS

k

denote

the

set

of

symmetric

k× k matrices, and letS

k

denote

the

+setof symmetricpositivesemi-definite(SPSD)k × kmatrices. SimilarlyletS k denotethesetof symmetricpositivedefinite(SPD)k× kmatrices.++

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LetS and X beanysymmetricmatrices. Wewrite “S

0”todenotethat

S

is

symmetric

and

positive

semi-definite,

and

we

write

“S

X ” to

denote that S − X 0. We write “S 0” to denote that S is symmetricandpositivedefinite,etc.

Remark 1 S k ={S ∈ S k | S 0} is a convex set in k2.+Proof: Suppose that S, X ∈ S k+. Pick any scalars α, β ≥ 0 for whichα +β =1. Foranyv∈ k, we have:

vT (αS +βX )v=αvT Sv +βv T Xv ≥ 0,

whereby

αS

+

βX

∈ S k

+.

This

shows

that

S k

is

a

convex

set.

+

q.e.d.

Asemi-definiteoptimizationproblem isanoptimizationproblemof thefollowingtype:

SDO: minimizey bT y

m

s.t. C + yiAi 0i=1

My

≥ g .

wherethematricesC, A1, . . . , Am aresymmetricmatrices. Oneconvenientway of thinking about this problem is as follows. Given values of the mscalarvariablesy1, . . . , ym,theobjective istominimizethelinearfunction:

m

biyi .i=1

Theconstraintsof SDOstatethatthevariablesy= (y1, . . . , ym)mustsatisfy

the

linear

inequalities:

M y ≥ g

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aswellastheconditionthatthematrixS ,definedby:

m

S :=C + yiAi ,i=1

mustbepositivesemi-definite. That is,m

S :=C + yiAi0.i=1

We illustratethisconstructionwiththefollowingexample:

SDO

:

minimizey1,y2 11y1 + 19y2

1 2 3 1 0 1 0 2 8

s.t. 2 9 0 +y10 3 7 +y22 6 0 =S 03 0 7 1 7 5 8 0 4

3y1 + 7y2 ≤12

2y1 +y2 ≤6 .

whichwecanrewrite inthefollowingform:

SDO: minimize 11y1 + 19y2

s.t.

1 + 1y1 + 0y2 2 + 0y1 + 2y2 3 + 1y1 + 8y2

2 + 0y1 + 2y2 9 + 3y1 + 6y2 0 + 7y1 + 0y2 03 + 1y1 + 8y2 0 + 7y1 + 0y2 7 + 5y1 + 4y2

3y1 + 7y2 ≤

12

2y1 +y2 ≤6.

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Remark 2 SDO is a convex minimization problem.

Proof: Theobjectivefunctionof SDOisalinearfunction,whichisconvex.yand˜Supposethat¯ yaretwofeasiblesolutionsof SDO,andlety=αȳ+βỹ ,

whereα,β ≥0 and α+β =1. Then:m

S :=C + ȳiAi 0i=1

and

m

X :=C + ỹiAi 0.i=1

Therefore,

m m

C +

yiAi = C +

(αȳ+β ̃y)Aii=1 i=1

m m

= α C + ȳiAi +β C + ỹiAii=1 i=1

=

αS

+

βX

0

.

Thisshowsthatthefeasibleregionof SDO isaconvexset.q.e.d.

Semi-definite optimization is a unifying model in optimization. Linearoptimization, quadratic optimization, second-order cone optimization, aswell as certain other convex optimization problems, can all be shown tobe special cases of semi-definite optimization. Furthermore, semi-definiteoptimizationhasapplicationsthatspanconvexoptimization,discreteopti-

mization,

and

control

theory.

In

fact,

semi-definite

optimization

may

very

wellbecomethecanonicalwaythatoptimizerswillthinkaboutoptimizationinthenextdecade.

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9 Truss Design and Semi-Definite Optimization

We now present an SDO problem that is equivalent to the truss designproblemTDP.Thisproblem is:

(STDP): minimizet,θ θ

s.t.θ

F T

m

F tkE k (ak)(ak)

T 0L2

k=1 k

M t

≤d

t≥ 0mθ∈ , t ∈ .

NoticethatSTDP isasemi-definiteoptimizationproblem. Theequiva-lenceof TDPandSTDPisaconsequenceof thefollowingtwopropositions:

Proposition 9.1 Suppose that (t, u) is a feasible solution of TDP. Let:

θ

=

F T u .

Then (t, θ) is a feasible solution of STDP, and θ=F T u.

Proposition 9.2 Suppose that (t, θ) is a feasible solution of STDP. Then there exists a vector u for which (t, u) is feasible for TDP, and in fact:

F T u≤ θ .

Together, Propositions 9.1 and 9.2 demonstrate that any solution of STDPtranslates intoasolutiontoTDP.Therefore, inordertosolveTDP,

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∗ ∗wecansolveSTDP for(t , θ )andthensolvethe following linearequation

∗system

for

u

:

mt∗kE k T ∗

L2aka u =F .k

k=1 k

AsameanstoprovePropositions9.1and9.2,wefirstshowthefollowing:

Remark 3 Given a vector v, a square matrix M , and a scalar θ, then:

θ vT

Q

:=

0

v M

if and only if:

T M 0 , there exists u satisfying Mu =v , and θ≥v u .

Proof of Remark 3: Toseewhythisistrue,letusfirstsupposethatQ0.ThenclearlyM 0,sinceM isformedbyasubsetof thecomponentsthatform Q. Now, let us suppose that there is no u such that Mu = v. ThisimpliesthatthereexistsλsuchthatλT M =0andλT v > 0. Then:

−(−, λT )

θ vT

=

θ2 −

2v

T λ +

λT M λ

=

θ2 −

2v

T λ

<

0v M λ

for small enough, which is a contradiction. Therefore there exists usuchthatM u =v. Furthermore,forsuchau, we have:

−1 T 0≤(−1, u T ) θ vT

=θ−2vT u +uT M u =θ−v uv M u

T whichshowsthatθ≥v u.Now

let

us

prove

the

reverse

of

these

implications.

Suppose

that

M

0,

thereexistsusatisfyingMu =v, and θ≥vT u. Since M 0 we only have toprovethat:

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−1 T δ (x) := (−1, x T ) θ vT

=θ−2v x +xT Mx ≥0 ∀x . v M x

Butnoticethatδ (x)isaconvexquadraticfunctionof x(sinceinpartic-ularM 0),andtherefore

∇δ (x) = −2v+ 2M x = 0

is

a

necessary

and

sufficient

condition

for

the

unconstrained

minimization

of δ (x). Sinceusatisfiesthiscondition, it istruethat

T T δ (x)≥δ (u) = θ−2v u +uT M u =θ−v u≥0 ∀x .

q.e.d.

Nowletusprovethepropositions.

Proof of Proposition 9.1: Suppose that (t, u) is a feasible solution of TDP.Let:

θ=F T u ,

and let

m E kM = tk

L2(ak)(ak)

T ,k=1 k

and letv=F . Then

m E k

Mu = tkL2

(ak)(ak)T u=F ,

k=1 k

andnoticeaswellthat

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=

M

0

because M is the nonnegative sum of rank-one SPSD matrices. FromRemark3,

θ F T

θ vT

m

F tkE k (ak)(ak)

T v M 0 .

L2 k=1 k

Therefore (t, θ) is feasible for STDP with objective function value θ =F T u.

q.e.d.

Proof of Proposition 9.2: Suppose that (t, θ) is a feasible solution of STDP.ThenfromRemark3,thereexistsavectoruforwhich:

m

tkE k

(ak)(ak)T u=F ,

L2k=1 k

and

θ≥F

T

u.

Therefore

(t, u)

is

feasible

for

TDP

and

θ≥F

T

u.

q.e.d.

10 Truss Design and Linear Optimization

Considerthefollowingspecial,butnotunusual,instanceof thetrussdesignproblem:

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(TDP): minimizet,u F T u

m

s.t. tkE k (ak)(ak)

T u=F L2

k=1 k

m

tk ≤ V k=1

t≥ 0mu

∈ N , t

∈ .

In this problem, the only constraint on the volumes of the bars tk is avolume constraint limiting the total volume of the bars to not exceed thegivenvalueV . Substitutingthenotation:

m E kK (t) := tk

L2(ak)(ak)

T ,k=1 k

wecanalsoconvenientlywriteourproblemas:

(TDP):

minimizet,u F T u

s.t. K (t)u=F

m

tk ≤ V k=1

t≥ 0mu∈ N , t ∈ .

In

this

section,

we

show

that

when

TDP

has

this

particularly

simple

form,thenitcanbesolvedvialinearoptimization. Beforedoingso,wefirstwritedownadualproblemassociatedwithTDP inthisform:

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(DTDP): maximizev,z −2F T v− V z

s.t. 2 L2T kakv ≤ E k z , k = 1, . . . , m y∈ N .

InorderforDTDPtobeanhonestdualof TDP,wenowpresentaweakdualityresult:

Proposition 10.1 Suppose that (t, u)is feasible for TDP and that (v, z)is feasible for DTDP.Then:

F T u≥ −2F T v− V z .

Proof: For k= 1, . . . , m, we have

E k T tk ≥ 0 and L2

(akv)2 ≤ z.

k

Multiplyingandsummingterms,weobtain:

m

T T vT K (t)v=

tkE k(v ak) (akv)≤

m

tk z≤ V z. L2

k=1 k k=1

Also,

0≤ (u +v)T K (t)(u +v) = uT K (t)u + 2uT K (t)v+vT K (t)v= F T u + 2vT F +vT K (t)v

≤ F T u + 2vT F +V z. ThereforeF T u≥ −2F T v− V z.q.e.d.

Considerthefollowingpairof primalandduallinearoptimizationmod-els:

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m √ Lk(LP): minimizef +,f − E k

f k+ +f k

−k=1

s.t. A(f + − f −) = −F

f + ≥ 0 , f − ≥ 0mf +, f − ∈ .

(LD): maximizey−

F T y

s.t.T −√ Lk ≤ aky≤ √ Lk , k= 1, . . . , mE k E k

y∈ N .

We will prove the following important result that shows how to usesolutionsof LPandLDtoconstructsolutionstoTDP:

¯Proposition 10.2 Suppose that (f ̄+, f −) is an optimal solution of LP and that ȳ is an optimal solution of LD, and consider the following assignment

of

variables:

m

¯R =

√ Lk f ̄+ +f k−E k kk=1t̄ k =

V √

Lk ¯f ̄+ +f k−

k= 1, . . . , m , R E k

k

Rū = − ȳ

V R

v̄ = ȳV R2

z̄ = V 2

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V

Then (¯u) solves TDP and (¯ z) solves DTDP.t, ¯ v, ¯

Proof: Notethatbythecomplementary slackness propertyof linear opti-mization,wehave:

T + Lk +akyf = √ f k, k = 1, . . . , m , k E kT − −Lk −akyf = √ f k, k = 1, . . . , m .k E k

m

We first show that (t, u) is feasible forTDP. Note that t≥0 and tk =k=1

R=V . Also

R

RK (t)u = − K (t)y

V tkE kR m T = −V L2

akakyk=1 k

Lk − T = −R V m

√ E k (f + +f k)akakykV R E k L2k=1 k√ m E k + T − T = − ak (f kaky+f kaky)Lkk=1 m −=

−

ak (f

+

−

f

−

k) =

−A(f

+

−

f

) =

F.kk=1

Wenextshowthat(v, z)isfeasibleforDTDP.Toseethis,observethatfork= 1, . . . , m, we have:

R2 R2 L2T T (akv)2 =

V 2(aky)

2 ≤ √ Lk2

=z k,V 2 E k E k

andso(v, z)isfeasible.Finally,weshowthatthesesolutionsexhibitstrongduality:

−2R 2R2 R2 R2 R−2F T v−V z= F T y−V R2

= − = = R R= (−F T y) = F T u.V V 2 V V V V V

q.e.d.

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11 Extensions of the Truss Design Problem

• Multiple Loads. We consider a finite set of external loads on thetruss:

F ={F 1, F 2, . . . , F J } ⊂ IRN .

In this case we might consider solving two types of design problems.Thefirstistodesignthetrussstructureconservatively,soastomini-mizethemaximumcompliance:

minimizet max j=1,...,J F T j u j

Mt ≤ d,t≥ 0 s.t. K (t)u j =F j.

Alternatively, we might consider the “average-case” design problem:letλ j denotetherelativefrequencyorimportanceassociatedwiththe

J

truss structure undergoing the external force F j, where λ j = 1.0. j=1

The following optimization model minimizes the average complianceof thetrussstructure:

J

minimizet λ jF jT u j

j=1

Mt ≤ d,t≥ 0 s.t. K (t)u j =F j.

• Self-weights. Thetrussdesignproblemthatwehaveformulatedpre-sumesthatthetrussstructureitself isnotaffectedbyitsownweight.Tocorrectforthis,wecansimplyaddanexternalforcecorresponding

to

the

gravitational

force

associated

with

bar

k,

and

linearly

propor-tionaltotk fork= 1, . . . , m. Letusdenotebygk ∈ IRN thevectorthatprojects the gravitational force of bar k onto the appropriate nodes.

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Then we can write the TDP of a truss under a single external loadvector

F

as:

T m(TDP): minimizet,u F + tkgk u

k=1

m

s.t. K (t)u= F + tkgkk=1

M t ≤ d

t≥

0

mu∈ N , t ∈ .

It is straightforward to convert this problem into a convex problem, justasforthebasecaseconsideredearlier.

• Reinforcement. Inthisproblem,wearegivenanexistingtrussstruc-ture,andwemustdeterminehowtostrengthenit. Tosolvethisprob-lem,wesimplyadd lowerboundsonthevolumesof theexistingbarsequaltothecurrentvolumesof thebars:

(TDP): minimizet,u F T u

s.t. K (t)u=F

M t ≤ d

t≥ 0

tk ≥ tk , k = 1, . . . , m mu

∈ N , t

∈ .

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• Robustness of Solutions. Theoptimalsolutionof theTDPmight

yield

a

truss

design

that

is

not

rigid

for

a

different

external

force

than

theoneused,andcouldchangeconsiderablyevenunderasmallchangein the external force. To obtain a robust solution, we must considermultiple external loads that will contain the possible loads that thetrusswillbesubjectto. Thiscanbemodeledbyconvexoptimizationaswell.

• Buckling Constraints. Thetrussdesignproblemthatwehavepre-sentedignoresthefactthatif abarisundergreatcompressionitmightactuallycollapse insteadof counteractingthatexternal forcewith itsinternalforce. Forthiswehavetoadd lowerboundsontheallowable

internal

forces

in

terms

of

the

design

variables

tk and

the

geometryof thebars. Theseconstraintscausetheresultingproblemto lose its

convexstructure.

• Other Considerations . . . .

47

Documents

Truss Design Art