10 tl bar elements: truss analysis

10TL Bar Elements:

Truss Analysis

10–1

Chapter 10: TL BAR ELEMENTS: TRUSS ANALYSIS

TABLE OF CONTENTS

Page§10.1 Introduction . . . . . . . . . . . . . . . . . . . . . 10–3

§10.2 The Two Nonlinear Analysis Levels . . . . . . . . . . . . 10–3

§10.3 Plane Truss Assemblers . . . . . . . . . . . . . . . . . 10–3§10.3.1 Plane Truss Master Internal Force . . . . . . . . . . 10–4§10.3.2 Plane Truss Master Tangent Stiffness . . . . . . . . . 10–5

§10.4 Space Truss Assemblers . . . . . . . . . . . . . . . . 10–5§10.4.1 Space Truss Master Internal Force . . . . . . . . . . 10–5§10.4.2 Space Truss Master Tangent Stiffness . . . . . . . . . 10–6

§10.5 Boundary Condition Applicators . . . . . . . . . . . . . . 10–7§10.5.1 Modified Force Vector . . . . . . . . . . . . . . 10–8§10.5.2 Reduced Force Vector . . . . . . . . . . . . . . 10–8§10.5.3 Reduced Stiffness Matrix . . . . . . . . . . . . . 10–8

§10.6 The Mises Truss . . . . . . . . . . . . . . . . . . . . 10–9§10.6.1 TL Plane Bar Model of Mises Truss . . . . . . . . . 10–9§10.6.2 How Will the Mises Truss Fail? . . . . . . . . . . . 10–10§10.6.3 Mises Truss Equilibrium Analysis . . . . . . . . . . 10–10§10.6.4 Mises Truss Equilibrium Solutions . . . . . . . . . . 10–10§10.6.5 Mises Truss Equilibrium Paths . . . . . . . . . . . 10–12§10.6.6 Mises Truss Equilibrium Paths in 3D . . . . . . . . . 10–14§10.6.7 Having Some Fun: Snapshots and Animation . . . . . . 10–14§10.6.8 Mises Truss Stiffness Analysis . . . . . . . . . . . . 10–14§10.6.9 Mises Truss Critical Points . . . . . . . . . . . . 10–16§10.6.10 Mises Truss Load Capacity . . . . . . . . . . . . . 10–19

§10.7 Pyramidal Trusses . . . . . . . . . . . . . . . . . . 10–20§10.7.1 Pyramidal Truss FEM Model . . . . . . . . . . . . 10–21§10.7.2 Pyramidal Truss With Three Members . . . . . . . . 10–21§10.7.3 Pyramidal Truss With Arbitrary Number of Members . . . . 10–23§10.7.4 Primary Path Response and Critical Points . . . . . . . 10–25§10.7.5 Which Pyramidal Truss is Best? . . . . . . . . . . . 10–26

§10. Exercises . . . . . . . . . . . . . . . . . . . . . . 10–27

10–2

§10.3 PLANE TRUSS ASSEMBLERS

§10.1. Introduction

This chapter presents a couple of tiny finite element programs that makes use of the TL plane bar andspace bar elements formulated and implemented in Chapter 9. The programs are applied to analyzesymbolically simple truss structures. The exact symbolic solutions are useful as benchmarks to testthe numerical solvers presented later. The implementation is done in Mathematica. The use of acomputer algebra system (CAS) as prototyping language has several advantages:

(i) The code is concise on account of compact array and list notation. This simplifies translationto other high-level, general-purpose scientific languages such as, e.g., Matlab or Maple.

(ii) Debugging is often quicker than with conventional programming languages because of theinterpretative nature of the language front end. This is of course platform dependent: theimplementation of Mathematica on a Mac or PC is friendler than on a Unix workstation.

(iii) Graphic output is part of the language. Mathematica plot cells can be inserted into TEXdocuments such as this one via PostScript export and Adobe Illustrator editing.

(iv) The language can manipulate both symbolic and numerical expressions. The latter may bedone exactly or in floating-point arithmetic. This flexibility is convenient for several scenarios,such as extracting symbolic output for inline coding or C/C++ conversion, or getting exactanalytical solutions that can be included in a document. The examples illustrate this point.

The main disadvantage of Mathematica is computational inefficiency. This is inherent in interpre-tative systems. The slowdown can reach several orders of magnitude with respect to a compiledlanguage such as Fortran or C. But for simple example problems, such as those presented here, thataspect is unimportant. Human efficiency, flexibility and portable graphics is what counts.

§10.2. The Two Nonlinear Analysis Levels

The computer implementation of linear FEM under the Direct Stiffness Method revolves aroundthe master stiffness matrix equation: K u = f. This is both a force equilibrium equation and a(trivial) incremental rate equation. (See Remark 3.3 in Chapter 3.) The programming work focuseson one level. In nonlinear problems, two levels1 are of primary interest:

Equilibrium level, also called residual level.

Rate level, also called stiffness level.

Which level should be implemented first? For simple programs, such as those presented here, bothcan be done simultaneously. An advantage of “task concurrency” is that much of the level-1 codecan be reused at level 2. For more complicated models it is recommended to start at level 1, anduse that for checking level 2 with finite difference techniques, as done for instance in §9.2.9.

1 The energy level is occasionally useful to derive and/or check the equilibrium level implementation using symbolicor numeric differentiation; see §9.2.9 for an example. But it is used primarily as an auxiliary tool, so it will not beemphasized here. Higher order levels are rarely used in practice since storage-hungry “3D matrices” appear.

10–3


TLPlaneTrussMasterIntForce[nodxyz_,elenod_,elemat_,elefab_,elestr_, noddis_,numer_]:=Module[{numele=Length[elenod],numnod=Length[nodxyz], neldof,e,eftab,enodXY,edisXY,ni,nj,i,ii,Em,A0,s0, pe,p}, p=Table[0,{2*numnod}]; For [e=1, e<=numele, e++, {ni,nj}=elenod[[e]]; eftab={2*ni-1,2*ni,2*nj-1,2*nj}; enodXY={nodxyz[[ni]],nodxyz[[nj]]}; edisXY={noddis[[ni]],noddis[[nj]]}; Em=elemat[[e]]; A0=elefab[[e]]; s0=elestr[[e]]; pe=TLPlaneBarIntForce[enodXY,Em,A0,s0,edisXY,numer]; neldof=Length[pe]; For [i=1, i<=neldof, i++, ii=eftab[[i]];

p[[ii]]+=pe[[i]] ]; ]; If [!numer, p=Simplify[p]]; Return[p]];

TLPlaneTrussMasterTanStiff[nodxyz_,elenod_,elemat_,elefab_,elestr_, noddis_,numer_]:=Module[{numele=Length[elenod],numnod=Length[nodxyz], neldof,e,eftab,enodXY,edisXY,ni,nj,i,j,ii,jj,Em,A0,s0, KeM,KeG,KM,KG}, KM=KG=Table[0,{2*numnod},{2*numnod}]; For [e=1, e<=numele, e++, {ni,nj}=elenod[[e]]; eftab={2*ni-1,2*ni,2*nj-1,2*nj}; enodXY={nodxyz[[ni]],nodxyz[[nj]]}; edisXY={noddis[[ni]],noddis[[nj]]}; Em=elemat[[e]]; A0=elefab[[e]]; s0=elestr[[e]]; {KeM,KeG}=TLPlaneBarTanStiff[enodXY,Em,A0,s0,edisXY,numer]; neldof=Length[KeM]; For [i=1, i<=neldof, i++, ii=eftab[[i]];

For [j=i, j<=neldof, j++, jj=eftab[[j]]; KM[[jj,ii]]=KM[[ii,jj]]+=KeM[[i,j]]; KG[[jj,ii]]=KG[[ii,jj]]+=KeG[[i,j]] ];

]; ]; If [!numer, {KM,KG}=Simplify[{KM,KG}]]; Return[{KM,KG}]];

Figure 10.1. Internal force and tangent stiffness assembler modules for a plane truss structure.

§10.3. Plane Truss Assemblers

This section presents the assembler modules for a plane truss structure modeled with the TL barelement described in §9.2.

Recall that the total residual force vector is the difference between internal and external forces:

r = p − f. (10.1)

As in Chapter 9, we assume that applied forces are conservative and proportional. That makesthe external force vector trivial: f = λ q, where q is user specified. So the programming workmust focus on the master internal force vector p. As regards the master stiffness matrix, theassembly process is similar to that of the linear case, except that the material and geometric stiffnessare returned separately for convenience in stability analysis. The two assembler modules, calledTLPlaneTrussMasterIntForce and TLPlaneTrussMasterTanStiff, are listed in Figure 10.1.

In the following usage description, the number of nodes and elements are denoted by N and Ne,respectively.

10–4

§10.4 SPACE TRUSS ASSEMBLERS

§10.3.1. Plane Truss Master Internal Force

The plane truss internal force assembler is invoked by the calling sequence

p = TLPlaneTrussMasterIntForce[nodxyz,elenod,elemat,elefab,

elestr,noddis,numer];(10.2)

The arguments in the calling sequence arenodxyz A list of node coordinates in the reference configuration, configured as

{ { X1,Y1 },{ X2,Y2 },...,{ X N , YN } }elenod Element nodes: a list configured as { { n(1)

1 ,n(1)2 },{ n(2)

1 ,n(2)2 },...,{ nNe

1 ,nNe2 } },

in which ne1 and ne

2 are the end nodes of the eth bar element, with ne1 < ne

2.elemat Material parameters: a list configured as { E (1), E (2),...,E Ne }, in which Ee is the

axial elastic modulus of the eth bar element. This is configuration independent.elefab Fabrication parameters: a list configured as { A(1)

0 , A(2)0 ,...,ANe

0 }, in which Ae0 is

the cross section area of the eth bar element in the reference configuration.elestr Initialization parameters: a flat vector configured as { s(1)

0 , s(2)0 ,...s Ne

0 }, in whichse

0 is the initial axial stress of the eth bar element in the reference configuration.noddis Node displacements from the reference to the current configuration, configured as

{ { u X1, uY 1 },{ u X2, uY 2 },...,{ u X N , uY N } }. Note that this is not a flat list.numer A logical flag: True to request numerical floating point work. Else set to False.

The module output is

p Master internal force vector: a flat list configured as { pX1,pY 1,pX2,...pY N }.This includes all 2N state freedoms, before application of zero-displacement BC.

§10.3.2. Plane Truss Master Tangent Stiffness

The plane truss tangent stiffness assembler is invoked by the calling sequence

{ KM,KG } = TLPlaneTrussMasterTanStiff[nodxyz,elenod,elemat,elefab,


The arguments are exactly the same as those for TLPlaneTrussMasterIntForce, which aredescribed in §10.3.1. The module output is

{ KM,KG } Master material and geometric stiffnesses, both stored as 2N × 2N matrices. Themaster tangent stiffness is the sum of these two.

§10.4. Space Truss Assemblers

This section presents the assembler modules for a space truss structure modeled with the TL barelement described in §9.3. The two assembler modules, called TLSpaceTrussMasterIntForceand TLSpaceTrussMasterTanStiff, are listed in Figure 10.2. The modules are very similar tothe plane truss assemblers covered in §10.3. Thus only usage deviations are pointed out.

10–5


TLSpaceTrussMasterIntForce[nodxyz_,elenod_,elemat_,elefab_,elestr_, noddis_,numer_]:=Module[{numele=Length[elenod],numnod=Length[nodxyz], neldof,e,eftab,enodXYZ,edisXYZ,ni,nj,i,ii,Em,A0,s0, pe,p}, p=Table[0,{3*numnod}]; For [e=1, e<=numele, e++, {ni,nj}=elenod[[e]]; eftab={3*ni-2,3*ni-1,3*ni,3*nj-2,3*nj-1,3*nj}; enodXYZ={nodxyz[[ni]],nodxyz[[nj]]}; edisXYZ={noddis[[ni]],noddis[[nj]]}; Em=elemat[[e]]; A0=elefab[[e]]; s0=elestr[[e]]; pe=TLSpaceBarIntForce[enodXYZ,Em,A0,s0,edisXYZ,numer]; neldof=Length[pe]; For [i=1, i<=neldof, i++, ii=eftab[[i]];

p[[ii]]+=pe[[i]] ]; ]; If [!numer, p=Simplify[p]]; Return[p]];

TLSpaceTrussMasterTanStiff[nodxyz_,elenod_,elemat_,elefab_,elestr_, noddis_,numer_]:= Module[{numele=Length[elenod],numnod=Length[nodxyz], neldof,e,eftab,enodXYZ,edisXYZ,ni,nj,i,j,ii,jj,Em,A0,s0, KeM,KeG,KM,KG}, KM=KG=Table[0,{3*numnod},{3*numnod}]; For [e=1, e<=numele, e++, {ni,nj}=elenod[[e]]; eftab={3*ni-2,3*ni-1,3*ni,3*nj-2,3*nj-1,3*nj}; enodXYZ={nodxyz[[ni]],nodxyz[[nj]]}; edisXYZ={noddis[[ni]],noddis[[nj]]}; Em=elemat[[e]]; A0=elefab[[e]]; s0=elestr[[e]]; {KeM,KeG}=TLSpaceBarTanStiff[enodXYZ,Em,A0,s0,edisXYZ,numer]; neldof=Length[KeM]; For [i=1, i<=neldof, i++, ii=eftab[[i]];

For [j=i, j<=neldof, j++, jj=eftab[[j]]; KM[[jj,ii]]=KM[[ii,jj]]+=KeM[[i,j]]; KG[[jj,ii]]=KG[[ii,jj]]+=KeG[[i,j]] ];

]; ]; If [!numer, {KM,KG}=Simplify[{KM,KG}]]; Return[{KM,KG}]];

Figure 10.2. Internal force and tangent stiffness assembler modules for a space truss structure.

§10.4.1. Space Truss Master Internal Force

The space truss internal force assembler is invoked by the calling sequence

p = TLSpaceTrussMasterIntForce[nodxyz,elenod,elemat,elefab,


The arguments are identical to those for the plane truss assembler described in §10.3, with only twoformatting changes due to the additional space dimension:nodxyz A list of node coordinates in the reference configuration, configured as

{ { X1,Y1, Z1 },{ X2,Y2, Z2 },...,{ X N , YN , Z N } }noddis Node displacements from the reference to the current configuration, configured as

{ { u X1, uY 1, uZ1 },{ u X2, uY 2, uZ2 },...,{ u X N , uY N , uZ N } }. Note that this is nota flat list.

The module output isp Master internal force vector: a flat list configured as { pX1, pY 1, pZ1, pX2,...,pZ N }.

This includes all 3N state freedoms, before application of zero-displacement BC.

10–6

§10.5 BOUNDARY CONDITION APPLICATORS

ModifiedForceVector[nodtag_,nodval_,K_,f_]:=Module[{numdof,tags,vals, fixdof,i,j,k,d,fmod=f}, tags=Flatten[nodtag]; vals=Flatten[nodval]; numdof=Length[tags]; fixdof=Flatten[Position[tags,_?(#>0 &)]]; For [k=1,k<=Length[fixdof],k++, i=fixdof[[k]]; d=vals[[i]]; fmod[[i]]=d; If [d==0, Continue[]]; For [j=1,j<=numdof,j++,

If [tags[[j]]==0,fmod[[j]]-=K[[i,j]]*d]]; ]; ClearAll[tags,vals,fixdof]; Return[fmod]];

ReducedForceVector[nodtag_,f_]:=Module[{numdof,tags,i,ii, nred,retdof,fred}, tags=Flatten[nodtag]; numdof=Length[tags]; retdof=Flatten[Position[tags,_?(#==0 &)]]; nred=Length[retdof]; If [nred<=0, Return[Null]]; fred=Table[0,{nred}]; For [i=1,i<=nred,i++, ii=retdof[[i]]; fred[[i]]=f[[ii]]]; ClearAll[tags,retdof]; Return[fred]];

ReducedStiffMatrix[nodtag_,K_]:=Module[{numdof,tags,i,ii,j,jj, nred,retdof,Kred}, tags=Flatten[nodtag]; numdof=Length[tags]; retdof=Flatten[Position[tags,_?(#==0 &)]]; nred=Length[retdof]; If [nred<=0, Return[Null]]; Kred=Table[0,{nred},{nred}]; For [i=1,i<=nred,i++, ii=retdof[[i]];

For [j=1,j<=nred,j++, jj=retdof[[j]]; Kred[[i,j]]=K[[ii,jj]] ]];

ClearAll[tags,retdof]; Return[Kred]];

Figure 10.3. Modules that apply zero-displacement boundary conditions by explicit reduction.

§10.4.2. Space Truss Master Tangent Stiffness

The space truss tangent stiffness assembler is invoked by the calling sequence

{ KM,KG } = TLSpaceTrussMasterTanStiff[nodxyz,elenod,elemat,elefab,


The arguments are exactly the same as those for TLSpaceTrussMasterIntForce, which aredescribed in §10.4.1. The module output is

{ KM,KG } Master material and geometric stiffnesses, both stored as 3N × 3N matrices. Themaster tangent stiffness is the sum of these two.

§10.5. Boundary Condition Applicators

This section describes modules that apply prescribed displacement boundary conditions. This isdone by explicitly removing the associated equations from both the assembled internal force vectorand stiffness matrix.2 If there are nonzero prescribed displacements, the reduction is preceding bya force modification operation.The three modules that apply displacement BC areModifiedForceVector, ReducedForceVector,and ReducedStiffMatrix. They are listed in Figure 10.3. All three operate at the assembledequation level. Consequently they are independent of element information. Thus there is no need

2 This reduction method is in contrast with linear FEM, in which the force vector and stiffness matrix are modified butthe number of DOF remains the same. See Chapter 3 of [273]. In nonlinear FEM the modification method becamesless attractive, particularly in symbolic computations, since inlining of symbolic output is simplified. It also helps inpreparing for stability eigenvalue computations.

10–7


to have separate modules for plane and space trusses. Although some arguments are configuredas node-by-node tables, they are internally converted to flat lists through the Flatten function,whence any dependence on DOF-per-node is erased.

§10.5.1. Modified Force Vector

This module is only called if nonzero displacements are prescribed, to produce a modified (alsocalled effective) force vector. This vector may be subsequently reduced by ReducedForceVector.The calling sequence is

fmod = ModifiedForceVector[nodtag,nodval,K,f]; (10.6)

The arguments in the calling sequence arenodtag BC tags arranged node-by-node. For a plane truss the configuration is

{ { tagX1,tagY 1 },...,{ tagX N,tagY N } }. For a space truss it is{ { tagX1,tagY 1,tagZ1 },...,{ tagX N,tagY N,tagZ N } }. If the tag is 0 the forceon that DOF is prescribed, whereas if 1 the displacement is prescribed.

nodval BC values specified node by node. Configured exactly as nodtag. For a plane truss:{ { valX1,valY 1 },...,{ valX N,valY N } }. For a space truss:{ { valX1,valY 1,valZ1 },...,{ valX N,valY N,valZ N } }. An entry correspond-ing to a 0-tag is the force value whereas that for a 1-tag it is the displacement value.

K The master tangent stiffness matrix containing all DOF, Must not be a reducedstiffness matrix, because entries associated with prescribed nonzero displacements(which disappear upon reduction) are used in the modification process.

f The force vector to be modified. Configured as a flat list that includes all DOF, thusmust not be a reduced force vector. May be the total residual r, the internal force por the external force vector f; but be careful with signs since r = p − f.

The function return isfmod Modified (a.k.a. effective) force vector as flat list containing all DOF. The prescribed-

displacement entries are set to the specified values to support linear FEM. In nonlinearFEM those entries are subsequently removed by ReducedForceVector.

§10.5.2. Reduced Force Vector

The module that reduces a force vector by removing all prescribed DOF is invoked as

fred = ReducedForceVector[nodtag,f]; (10.7)

The arguments in the calling sequence arenodtag Same argument as for ModifiedForceVector. See §10.5.1.f The force vector to be reduced, configured as a flat list, including all DOF. May be

the total residual, the internal force, or the external force vector, depending on theimplementation of higher levels. If nonzero prescribed displacements are specified,this must be the modified force vector produced by ModifiedForceVector.

The function return isfred Reduced force vector, with all entries pertaining to prescribed displacements re-

moved.

10–8

§10.6 THE MISES TRUSS

��

α α

X, x

Y, y(2)(1)

1

2

3

uX

uY

fY = λ

0E, A 0E, A

S

H

0L

Here λ > 0 means crownload is applied upward

Figure 10.4. The Mises truss modeled with two TL plane bar elements. It has two degreesof freedom: u X and uY , which are the displacements of node 2 (truss crown).

§10.5.3. Reduced Stiffness Matrix

The module that reduces a stiffness matrix by removing all prescribed DOF is invoked as

Kred = ReducedStiffMatrix[nodtag,K]; (10.8)

The arguments in the calling sequence arenodtag Same as for ModifiedForceVector or ReducedForceVector. See §10.5.1K The stiffness matrix to be reduced, which includes all DOF. May be the tangent

stiffness, the material stiffness or the geometric stiffness matrix.3

The function return isKred Reduced stiffness matrix, with all rows and columns pertaining to prescribed dis-

placements removed.

§10.6. The Mises Truss

The two-member truss truss configuration pictured in Figure 10.4 is known as the Mises truss[853,854] in the structural mechanics literature.4 It is the plane truss version of the two-spring trussconsidered in Exercise 6.3.The truss is widely used as an example in the literature since it is one of the simplest structuresthat can exhibit both limit-point (snap-through) and bifurcation behavior. The order in which thesecritical states occur can be controlled by varying the rise angle α. It is used here to illustrate theuse of the plane truss modules described in previous sections in the determination of failure modes.The entire analysis is carried out symbolically.

§10.6.1. TL Plane Bar Model of Mises Truss

The FEM idealization of the Mises truss is shown in Figure 10.4. The two truss members aremodeled with TL bar elements labeled (1) and (2) that join nodes 1-2 and 2-3, respectively. The

3 The last two cases occur when KM and KG are kept separated, as in stability analysis.4 R. von Mises (1883-1953), was a German mechanician well known for his contributions to structural analysis as well as

plasticity theory.

10–9


modulus of elasticity E and the bar areas A0 are the same for both elements. The truss geometry isdefined by the span S and initial height H . The reference configuration, which is unloaded, is takento be stress free. Both bars are assumed to be sufficiently stiff so that Euler buckling is precluded.Because nodes 1 and 3 are pinned to the ground, the only active degrees of freedom (DOF) are thehorizontal and vertical displacements of node 2. To avoid clutter these will be denoted as u X anduY rather than u X2 and uY 2, respectively. The vertical motion uY is positive upward.The components of the applied and internal force vector at node 2 are denoted simply by ( fX , fY )and (pX ,pY ), respectively. Displacement and force components for nodes 1 and 3 are explicitlyremoved from the governing equations.

Remark 10.1. Several variants of this problem can be found in the literature. A common variant is to attach avertical spring or a third TL bar at the crown. One modification of this nature is used in Exercise ?.

.§10.6.2. How Will the Mises Truss Fail?

The external force is parametrized as

f =[

fX

fY

]= λ q = λ

[01

], (10.9)

whence pX = 0 and pY = λ. Note that λ has the dimension of force. To get pY acting downward,λ must be negative. Recall that in Exercise 6.3 only symmetric motions of the truss under avertical load were considered. That constraint effectively reduces the model to a single DOF: uY ,which allows only limit point behavior. When two DOF are considered, however, a richer set ofpossibilities emerges; notably lateral bifurcation under the same load system.The main conclusion from the ensuing study will be that a sufficiently shallow truss fails by snap-through (i.e., on reaching a limit point) whereas a deep truss fails by bifurcation.

§10.6.3. Mises Truss Equilibrium Analysis

Equilibrium analysis is based on solving the total force residual equation r = p − f = 0 or,equivalently, the internal-equal-external force balance: p = f. In FEM work we always work withthe latter. While the external force f is given by (10.9), the internal force is formed by callingmodules TLPlaneTrussMasterIntForce and TLReducedForceVectors, which are describedin §10.3.1 and §10.5.2, respectively.The script shown at the top of Figure 10.5 forms the master internal force 6-vector p for all sixDOF, and then reduces it to a 2-vector pred for the crown DOF. The results are shown at the bottomof Figure 10.5. Translating to ordinary notation after some simplifications:

pred =[

pX

pY

]= E A0

L30

[u X

(S2/2 + u2

X + 2H uY + u2Y

)(H + uY )

(u2

X + uY (2 H + uY )) ]

, (10.10)

in which L0 =√

H 2 + S2/4 = H/ cos α = 12 S/ sin α is the original length of both truss members;

cf. Figure 10.4. Expression (10.10) is valid for arbitrary inputs except for the assumption thats0 = 0.

10–10


ClearAll[S,H,Em,A0,s0,uX,uY]; nodxyz={{-S/2,0},{0,H},{S/2,0}}; elenod={{1,2},{2,3}}; elemat={Em,Em}; elefab={A0,A0}; elestr={0,0}; noddis={{0,0},{uX,uY},{0,0}}; nodtag={{1,1},{0,0},{1,1}}; p= TLPlaneTrussMasterIntForce[nodxyz,elenod,elemat,elefab,elestr, noddis,False]; pred=ReducedForceVector[nodtag,p]; Print["p=",p//MatrixForm," pred=",pred//MatrixForm];

p=

- 2 A0 Em S+2 uX S uX+uX2 +uY 2 H+uY

(

(( (

((

(

(

(

(

(

((

((

(

(

(

(

(

(

(

( (

4 H2 +S2 )

))

))

)

))

)

))

- 4 A0 Em H+uY

H

H S uX+uX2 +uY H

H

2 H+uY

LL

L

(4 H2 +S2 )

4 A0 Em uX S2 +2 uX2 +uY 2 H+uY

4 H2 +S2 )

8 A0 Em H uX2 +2 H2 uY+uX2 uY+3 H uY2 +uY3

4 H2 +S2 )

2 A0 Em S-2 uX -S uX+uX2 +uY 2 H+uY

4 H2 +S2 )

)

)

)

)

)))

)

- 4 A0 Em H+uY -S uX+uX2 +uY 2 H+uY

4 H2 +S2

, pred=

4 A0 Em uX S2 +2 uX2 +uY 2 H+uY

4 H2 +S2

8 A0 Em H uX2 +2 H2 uY+uX2 uY+3 H uY2 +uY3

4 H2 +S2

3/2

3/2

3/2

3/2

3/2

3/2

3/2

3/2

Figure 10.5. Top of figure shows script for symbolically computing the internal force of the Mises truss.The output results are shown at the bottom. shown at bottom in blue text.

pX=0 roots: uX 0 , uX − − S2

2− 2 H uY − uY2 , uX − S2

2− 2 H uY − uY2

λ= A0 Em uY (2 H2 + 3 H uY + uY2)L03

, ,

⇒ ⇒ ⇒

2

− A0 Em S (H + uY) 2 L03

− A0 Em S (H + uY) 2 L03

ClearAll[Em,A0,L0,S,H,uX,uY,pX,pY,λ];pX=(Em*A0/L0^3)*uX*(S^2/2+uX^2+2*H*uY+uY^2);pY=(Em*A0/L0^3)*(H*uX^2+2*H^2*uY+3*H*uY^2+uX^2*uY+uY^3);roots=Simplify[Solve[pX==0, uX]]; Print["pX=0 roots: ",roots];λ=Simplify[pY/.roots]; Print["λ=",λ];

Figure 10.6. Top cell is a script that illustrates the use of Solve to obtain the 3 roots of pX = 0 in termsof u X , where pX is given in (10.10). These roots are then replaced into pY = λ with pY from the sameequation, to get the expression of the three equilibrium paths in the form λ = λ(uY ). Results shown in bottom

cell.

§10.6.4. Mises Truss Equilibrium Solutions

From now on pred will be called simply p. The total residual force equilibrium equations arer = p − f = 0 or p = f. From (10.9) and (10.10) it follows that the equilibrium equations are

pX = 0, pY = λ. (10.11)

Expression (10.10) tells us that pX = 0 is a cubic equation that has the trivial root u X = 0 inaddition to the two roots of the quadratic S2/2 + u2

X + 2 H uY + u2Y = 0. Rather than solving

these by hand, let us exemplify the use of Mathematica to solve a polynomial equation using thebuilt-in Solve function. These roots are then substituted, using the /. operator, into the otherequation λ = pY to get the equations of the 3 equilibrium paths. The script and results are shownin Figure 10.6. The output shows that the three roots in terms of u X are

u pX = 0, us1,s2

X = ±√

− 12 S2 − 2H uY − u2

Y . (10.12)

10–11


ClearAll[Em,A0,S,H,uX,uY,pX,pY,λ]; L0=Sqrt[H^2+S^2/4];λP= Em*A0*uY*(2*H^2+3*H*uY+uY^2)/L0^3;λS=-Em*A0*S^2*(H+uY)/(2*L0^3);case1={H->Sqrt[3]/3,S->2,A0->1,Em->1};case2={H->Sqrt[2], S->2,A0->1,Em->1};case3={H->Sqrt[3], S->2,A0->1,Em->1};case4={H->3, S->2,A0->1,Em->1};style={{AbsoluteThickness[2],RGBColor[0,0,0]},

{AbsoluteThickness[2],Dashing[{0.02,0.02}],RGBColor[1,0,0]}};Plot[{λP/.case1,λS/.case1},{uY,0,-2*H/.case1}, Frame->True, PlotStyle->style, ImageSize->300, AxesLabel->{"uY","λ"}, PlotLabel->"Case 1: H=Sqrt[3]/3", DisplayFunction->DisplayChannel];Plot[{λP/.case2,λS/.case2},{uY,0,-2*H/.case2}, Frame->True, PlotStyle->style, ImageSize->300, AxesLabel->{"uY","λ"}, PlotLabel->"Case 2: H=Sqrt[2]", DisplayFunction->DisplayChannel];Plot[{λP/.case3,λS/.case3},{uY,0,-2*H/.case3}, Frame->True, PlotStyle->style, ImageSize->300, AxesLabel->{"uY","λ"}, PlotLabel->"Case 3: H=Sqrt[3]", DisplayFunction->DisplayChannel];Plot[{λP/.case4,λS/.case4},{uY,0,-2*H/.case4}, Frame->True, PlotStyle->style, ImageSize->300, AxesLabel->{"uY","λ"}, PlotLabel->"Case 4: H=3", DisplayFunction->DisplayChannel];

Figure 10.7. Mathematica script to generate equilibrium path plots for four rise angle cases.

The solution u pX , in which superscript p stands for “primary,” corresponds to a symmetric defor-

mation of the truss in its primary equilibrium path. This represents the kind of response studiedin Exercise 6.3. A nonzero u X is possible if the roots us1

X and us2X , in which supercripts stand for

“secondary,” given by (10.12) are real. This can occur only if the discriminant verifies

= − 12 S2 − 2H uY − u2

Y > 0. (10.13)

Because both S and H are positive, condition (10.13) can only happen for negative uY , that is,downward displacement of the truss. Furthermore, given S and H , condition (10.13) requires (theanalysis using Mathematica is not shown) that

−H +√

H 2 − 12 S2 ≥ uY ≥ −H −

√H 2 − 1

2 S2. (10.14)

This obviously can only happen if H 2 ≥ S2/2, or

H

S/2= tan α ≥

√2. (10.15)

If the foregoing equations are verified, the equilibrium equation pX = 0 has three real roots:u p

X = 0, us1,s2X = ±√

. We shall see that the last two pertain to a secondary equilibrium path thatintersects the primary equilibrium path u X = 0 at two bifurcation points.

§10.6.5. Mises Truss Equilibrium Paths

The three equilibrium paths given in the bottom cell of Figure 10.6 are

λp = E A0

L30

uY (2H 2 + 3H uY + u2Y ), λs1 = λs2 = E A0

2L30

S2(H + uY ). (10.16)

The primary path is cubic in uY . The two secondary equilibrium paths coalesce and are linear inuY . The Mathematica script in Figure 10.7 produces the four response plots of λ versus uY shown

10–12


−1 −0.8 −0.6 −0.4 −0.2 0−0.2

−0.1

0

0.1

0.2

−0.6

0.6

−2.5 −2 −1.5 −1 −0.5 0

−0.4

−0.2

0

0.2

0.4

−3.5 −3 −2.5 −2 −1.5 −1 −0.5 0−0.4

−0.2

0

0.2

0.4

−6 −5 −4 −3 −2 −1 0−0.3−0.2−0.1

00.10.20.3

Case 1: H = Sqrt[3]/3 Case 2: H = Sqrt[2]

Case 3: H = Sqrt[3] Case 4: H = 3

uY

uY

uYuY

uY

λ λ

λ λ

Figure 10.8. Mises truss equilibrium paths plotted on the λ versus uY planegenerated by the script of Figure 10.7.

in Figure 10.8. For all plots E = 1, A0 = 1 and S = 2. The height H is set to√

3/3 (rise angleα = 30◦, as 6.3),

√2/2,

√3 and 3. Those are labeled cases 1 through 4 respectively.

All plots show the symmetric-response (u X = 0) primary path with three zero load equilibriumpoints at (λ = 0, uY = u X = 0), (λ = 0, uY = −H, u X = 0), and (λ = 0, uY = −2H, u X = 0).The straight dashed line is the projection on (λ, uY ) the plane on which the secondary equilibriumpath lies.5 This path has a closed elliptical shape quite similar to that sketched in Figure 5.3 inChapter 5; see also Figure 10.12 below. A study of the location of the secondary path with respectto the primary path leads to the following conclusions.(a) If H <

√2, or equivalently α < 54.74◦, the secondary path does not interesect the primary

path. The structure has only two critical points, both of which are limit points. This isexemplified by Case 1 of Figure 10.8, for which α = 30◦.

(b) If√

2 ≤ H ≤ √3, the secondary path intersects the primary path at two bifurcation points,

which occur between the limit points. When the truss is loaded downward from λ = 0, the firstlimit point is encountered first, then the first bifurcation point. The truss will then follow thesecondary path as it bifurcates unsymmetrically, until reaching the second bifurcation point,and deflects symmetrically from then on. The structure has 4 critical points: two limit andtwo bifurcation, but the load capacity is determined by the limit points. Cases 2 and 3 inFigure 10.8 illustrates the minimum and maximum heights for which this kind of behavioroccur.

(c) If H >√

3, which corresponds to α = 60◦, the structure has again 4 critical points: two limitand two bifurcation, but the limit points are now “inside” the bifurcation ones. Physically the

5 The secondary path lies on a plane in control-state space (λ, u X , uY ) because λs is linear in uY . Its projection on the λ

versus uY plane is a straight line, as illustrated in Figure 10.8.

10–13


truss will bifurcate first, then deflect unsymmetrically until it reaches the second bifurcationpoint and continues to deform symmetrically from then on. Thus no limit point can bephysically reached. Case 4 in Figure 10.8, in which H = 3, exemplifies this kind of behavior.

The transition between (b) and (c) occurs at rise angle α = 60◦, which is interesting because thelimit and bifurcation points coalesce, as shown in Case 3 of Figure 10.8. See also Figure 10.10.

§10.6.6. Mises Truss Equilibrium Paths in 3D

Plotting the equilibrium paths in 3D control-space space (λ, u X , uY ) require more programmmingeffort because the graphic capabilities of Mathematica begin to be strained. Figure 10.9 illustratesthe kind of script needed to get decent results. The logic will not be explained here because it wouldnot be intelligible unless the reader has expertise in advanced graphics. The results of runningthis procedure for the interesting case H = √

3, which gives a rise angle α = 60◦ are shown inFigure 10.10. This diagram clearly displays the coalescence of limit and bifurcation points for thatparticular rise angle.

§10.6.7. Having Some Fun: Snapshots and Animation

To illustrate what happens to the truss as it traverses its equilibrium path, configuration “snapshots”and animations are invaluable and fun to do. The program shown in Figure 10.11 does the firstkind of display, showing snapshots as the truss traverses the equilibrium path by incrementing uY .The program also depicts (very roughly) the magnitude of the applied vertical force. The result ofrunning this program on an truss with S = 2, H = 3, E = A0 = 1 is shown in Figure 10.12.A plot such as that of Figure 10.12 is a bit confusing because all configurations are in the same cell,which appears on the screen as a flash. Animations are more instructive. Mathematica animationsare like movies or videos: a sequence of pictures that, when displayed in rapid succession, appearto move. In Mathematica versions endowed with a Notebook interface, a sequence of graphic cellscan be animated by simply doubly clicking the mouse on one of the pictures (any one will do). Thecell group containing the pictures will be selected and the animation started automatically.The script shown in Figure 10.13 is a variant of that listed in Figure 10.11. The logic has beenrearranged to produce a sequence of graphic cells, one per configuration, to make the animationpossible. Unfortunately the result cannot be shown on paper, only demonstrated from a computer.

§10.6.8. Mises Truss Stiffness Analysis

Having largely exhausted the force equilibrium analysis, we proceed to the next level, which dealswith the first-order rate equations. The tangent stiffness matrix appears. We will not go into how tonumerically integrate the rate ODE, as that is covered later. Instead, our main use of this level is forprecise determination of critical points using the singular stiffness criterion introduced in §5.2.3.The top cell of Figure 10.14 is a script to compute the 2 × 2 reduced tangent stiffness of the Misestruss. Results for two kinematic assumptions are shown in the bottom cell.In what follows Kred will be simply denoted as K. If u X = 0 so the truss crown displaces verticallyand the deflected configuration stays symmetric about Y , the script gives after some simplifications:

Ks = E A0

L30

[12 S2 + 2H uY + u2

Y 00 2H 2 + 6H uY + 3u2

Y

], (10.17)

in which L0 =√

H 2 + S2/4 as before. Here subscript s stands for “symmetric.” For a general

10–14


ClearAll[S,H,Em,A0,uX,uY]; H=Sqrt[3.];λmax[S_,H_,Em_,A0_]:=3.0792*A0*Em*H^3/(4*H^2+S^2)^(3/2);style1={AbsoluteThickness[2],RGBColor[0,0,0]};style2={AbsoluteThickness[2],RGBColor[1,0,0]};style3={AbsoluteThickness[2],RGBColor[1,0,0]};pripath[S_,H_,Em_,A0_,uY_]:=Module[{}, Return[{0,uY,8*A0*Em*uY*(H+uY)*(2*H+uY)/(4*H^2+S^2)^(3/2),style1}]];secpath1[S_,H_,Em_,A0_,uY_]:=Module[{c,uYB1,uYB2}, uYB1=Re[N[-H+Sqrt[H^2-S^2/2]]]; uYB2=Re[N[-H-Sqrt[H^2-S^2/2]]]; c=4*A0*Em*S^2/(4*H^2+S^2)^(3/2); If [N[uY]>=uYB1, Return[{0,uYB1,-c*(H+uYB1)}]]; If [N[uY]<=uYB2, Return[{0,uYB2,-c*(H+uYB2)}]]; Return[{Sqrt[-S^2/2-2*H*uY-uY^2],uY,-c*(H+uY),style2}]];secpath2[S_,H_,Em_,A0_,uY_]:=Module[{c,uYB1,uYB2}, uYB1=Re[N[-H+Sqrt[H^2-S^2/2]]]; uYB2=Re[N[-H-Sqrt[H^2-S^2/2]]]; c=4*A0*Em*S^2/(4*H^2+S^2)^(3/2); If [N[uY]>=uYB1, Return[{0,uYB1,-c*(H+uYB1)}]]; If [N[uY]<=uYB2, Return[{0,uYB2,-c*(H+uYB2)}]]; Return[{-Sqrt[-S^2/2-2*H*uY-uY^2],uY,-c*(H+uY),style3}]];λrange=1.1*{-λmax[2,H,1,1],λmax[2,H,1,1]};pp=ParametricPlot3D[pripath[2,H,1,1,uY], {uY,0,-2*H}, PlotPoints->201, PlotRange->{{-H,H},{0,-2*H},λrange}, BoxRatios->{1,2,1},AxesLabel->{"uX","uY","λ"}, Compiled->False,DisplayFunction->Identity];ps1=ParametricPlot3D[secpath1[2,H,1,1,uY], {uY,0,-2*H}, PlotPoints->201, PlotRange->{{-H,H},{0,-2*H},λrange}, BoxRatios->{1,2,1},AxesLabel->{"uX","uY","λ"}, Compiled->False,DisplayFunction->Identity];ps2=ParametricPlot3D[secpath2[2,H,1,1,uY], {uY,0,-2*H}, PlotPoints->201, PlotRange->{{-H,H},{0,-2*H},λrange}, BoxRatios->{1,2,1},AxesLabel->{"uX","uY","λ"}, Compiled->False,DisplayFunction->Identity];Show[pp,ps1,ps2, ViewPoint->{3,1,2},DisplayFunction->DisplayChannel];

Figure 10.9. Mathematica script used to generate response plots in the three-dimensionalcontrol-space space (λ, u X , uY ), such as the one shown in Figure 10.10.

−10

1−3

−2−1

0

−0.2

0

0.2

λ

01

−0.2

uY

uX

Figure 10.10. The Mises truss equilibrium paths for rise angle α = 60◦, plotted inthe three-dimensional control-state space (λ, u X , uY ).

10–15


ClearAll[S,H,Em,A0,uX,uY,a,c,d]; S=2.; H=3.; A0=1; Em=1;nmax =20; c=N[4*A0*Em/(4*H^2+S^2)^(3/2)];pelem1=Table[0,{nmax+1}];pelem2=Table[0,{nmax+1}];pload =Table[0,{nmax+1}];parrow=Table[0,{nmax+1}];Do [uY=-2.5*n*H/nmax; d= N[-S^2/2-2*H*uY-uY^2]; If[d <=0., uX=0; λ= N[2*c*uY*(H+uY)*(2*H+uY)], uX=Sqrt[d]; λ= N[-c*S^2*(H+uY)]]; pelem1[[n+1]]=Graphics[Line[{{-S/2,0},{uX,H+uY}}]]; pelem2[[n+1]]=Graphics[Line[{{ uX,H+uY},{S/2,0}}]]; pload[[n+1]] =Graphics[Line[{{ uX,H+uY},{uX,H+uY-λ}}]]; a = Min[.5*Abs[λ],0.2*H]; If [λ>0, parrow[[n+1]]=Graphics[{ Line[{{ uX,H+uY},{uX-a/2,H+uY-a}}], Line[{{ uX,H+uY},{uX+a/2,H+uY-a}}]} ], parrow[[n+1]]=Graphics[{ Line[{{ uX,H+uY},{uX-a/2,H+uY+a}}], Line[{{ uX,H+uY},{uX+a/2,H+uY+a}}]} ] ], {n,0,nmax}];Show[Graphics[Thickness[.004]],pelem1,pelem2, Graphics[Thickness[.005]],Graphics[RGBColor[1,0,1]], pload,parrow,ImageSize->300,DisplayFunction->DisplayChannel, PlotRange->{{-2*H,2*H},{-2*H,2*H}},AspectRatio->1];

Figure 10.11. A Mathematica script to generate and plot configuration snapshotsof the truss structure as it traverses equilibrium paths.

Figure 10.12. Result from running the script of Figure 10.11 for rise 2H/S = tan α = 3.

motion in which {u X , uY } are arbitrary, we get

K = E A0

L30

[12 S2 + 3 u2

X + uY (2 H + uY ) 2 u X (H + uY )

2 u X (H + uY ) 2 H 2 + u2X + 6 H uY + 3 u2

Y

]. (10.18)

In both cases the initial stress in both members is assumed zero.

§10.6.9. Mises Truss Critical Points

A detailed symbolic analysis, not reported here, for arbitrary S, H , E , A0, u X and uY shows thatunder a vertical load the determinant of K can only vanish if u X = 0. Thus any critical pointmust be on the fundamental path. Thus it is enough to restrict consideration to symmetric motions.The tangent stiffness is (10.17). Since Ks is diagonal, its determinant is simply the product of the

10–16


ClearAll[S,H,Em,A0,uX,uY,a,c,d]; S=2; H=.8; A0=1; Em=1;nmax =20; c= N[4*A0*Em/(4*H^2+S^2)^(3/2)];Do [uY=-2.5*n*H/nmax; d= N[-S^2/2-2*H*uY-uY^2]; If[d <=0., uX=0; λ= N[2*c*uY*(H+uY)*(2*H+uY)], uX=Sqrt[d]; λ= N[-c*S^2*(H+uY)]]; pelem1=Graphics[Line[{{-S/2,0},{uX,H+uY}}]]; pelem2=Graphics[Line[{{ uX,H+uY},{S/2,0}}]]; pload =Graphics[Line[{{ uX,H+uY},{uX,H+uY-λ}}]]; a = Min[.5*Abs[λ],0.2*H]; If [λ>0, parrow=Graphics[{ Line[{{ uX,H+uY},{uX-a/2,H+uY-a}}], Line[{{ uX,H+uY},{uX+a/2,H+uY-a}}]} ], parrow=Graphics[{ Line[{{ uX,H+uY},{uX-a/2,H+uY+a}}], Line[{{ uX,H+uY},{uX+a/2,H+uY+a}}]} ] ]; Show[Graphics[Thickness[.012]],pelem1,pelem2, Graphics[Thickness[.005]],Graphics[RGBColor[1,0,1]], pload,parrow,ImageSize->250, PlotRange->{{-2*H,2*H},{-2*H,2*H}},AspectRatio->1], {n,0,nmax},DisplayFunction->DisplayChannel];

Figure 10.13. A variant of the script of Figure 10.11. This one generates a sequenceof configuration plots (frames) that can be animated.

ClearAll[S,H,Em,A0,s0,uX,uY];nodxyz={{-S/2,0},{0,H},{S/2,0}}; elenod={{1,2},{2,3}}; elemat={Em,Em};elefab={A0,A0}; elestr={0,0}; noddis={{0,0},{0,uY},{0,0}};nodtag={{1,1},{0,0},{1,1}};{KM,KG}=TLPlaneTrussMasterTanStiff[nodxyz,elenod,elemat,elefab,elestr, noddis,False];K=Simplify[KM+KG]; Kred=ReducedStiffMatrix[nodtag,K];Print["Kred for uX=0: ",Kred//MatrixForm];noddis={{0,0},{uX,uY},{0,0}};{KM,KG}=TLPlaneTrussMasterTanStiff[nodxyz,elenod,elemat,elefab,elestr, noddis,False];K=Simplify[KM+KG]; Kred=ReducedStiffMatrix[nodtag,K];Print["Kred for NZ uX: ",Kred//MatrixForm];

4 A0 Em (S +6 uX + 4H uY + 2 uY ) 16 A0 Em uX (H+uY) (4 H +S ) (4 H +S )

16 A0 Em uX (H+uY) 8 A0 Em (2 H + uX + 6 H uY + 3 uY ) (4 H +S ) (4 H +S )

0

0

2 2 2

2 2

2 2

3/2

3/2

3/2

3/2

2 2

2 2

2 2 2Kred for NZ uX:

4 A0 Em (S + 2 uY (2 H+uY)) (4 H +S )

8 A0 Em (2 H + 6 H uY + 3 uY ) (4 H +S )

2

2 2 3/2

3/22 2

2 2Kred for uX=0:

Figure 10.14. Top cell is a script to form the tangent stiffness matrix of the Mises truss, first withu X = 0, and then with arbitrary {u X , uY }. Bottom cell shows the printed results.

diagonal entries:

ds = det(Ks) = E2 A20

L60

(12 S2 + 2HuY + u2

Y

)(2H 2 + 6HuY + 3u2

Y ). (10.19)

The factor E2 A20/L6

0 is always nonzero and can be dropped. The resulting ds is the product of two

10–17


ClearAll[µ,c,s,α];detK=-((2+6*µ+3*µ^2)*c^2*(-2-2*µ-µ^2+(-2+2*µ+µ^2)*(c^2-s^2))*s^2)/2;detK=Simplify[detK/.c^2->1-s^2]; Print["detK=",detK];sol=Simplify[Solve[detK==0,µ],α>0]; {µ1,µ2,µ3,µ4}=µ/.sol;{µ1,µ2,µ3,µ4}=Simplify[{µ1,µ2,µ3,µ4}/.s->Sin[α]]; Print["roots µ1,µ2,µ3,µ4= ",TrigExpand[{µ1,µ2,µ3,µ4}]];Print["roots µ1,µ2 (numeric)= ",N[{µ1,µ2}]];{µ1,µ2,µ3,µ4}=Simplify[{µ1,µ2,µ3,µ4}/.α->angdeg*(Pi/180)];pstyleL={AbsoluteThickness[2],RGBColor[0,0,0]};pstyleB={AbsoluteThickness[2],RGBColor[1,0,0]};pstyle= {pstyleL,pstyleL,pstyleB,pstyleB}; Off[Plot::plnr]; Plot[{µ1,µ2,µ3,µ4},{angdeg,0,90},PlotStyle->pstyle, Frame->True,ImageSize->300,DisplayFunction->DisplayChannel];

detK=-s (-1+s )(2+6µ+3µ )(2+s (-2+2µ+µ ))

roots µ1,µ2,µ3,µ4= {−1− , −1+ ,

−1−Csc[α] −2+3 Sin[α] , −1+Csc[α] −2+3 Sin[α] }roots µ1,µ2 (numeric)= {−1.57735, −0.42265}

2 2

2 2

2 2 2

1

3

1

3

Figure 10.15. Top cell is a script to get the critical points of the symmetrically loaded Mises truss asroots of the tangent stiffness determinant evaluated on the primary equilibrium path u X = 0. Bottom

cell shows printed results. Edited plot output is displayed in Figure 10.16.

0 20 40 60 80−2

−1.5

−1

−0.5

0

Rise angle α in degrees

Dim

ensi

onle

ss v

erti

cal

defl

ecti

on µ

= u

/H

Y

Limit point 1

Limit point 2

Bifurcation point 1

Bifurcation point 2

Figure 10.16. Location of critical points (roots of stiffness determinant(10.20) for u X = 0) shown as µ = uY /H versus rise angle α. Plotproduced by script of Figure 10.15; labels added by Adobe Illustrator.

quadratic polynomials in uY , whence the equation ds = 0 has four roots in the crown deflection uY .Consequently there are four possible critical points, but these are not necessarily real. To study theroots in more detail it is convenient to reduce ds to the following dimensionless form:

s = −c2 s2 (2 + 6 µ + 3 µ2) (−2 − 2 µ − µ2 + (c2 − s2) (−2 + 2 µ + µ2)), (10.20)

Here µ = uY /H , c = cos α and s = sin α, and innocuous factors have been dropped. Remainingmanipulations can be followed in the script shown at the top of Figure 10.15. Note that the cosinecan be readily eliminated in favor of the sine using c2 = 1 − s2. The four roots of det(K ) = 0 in

10–18


1 2 3 4 5

0.05

0.1

0.15

0.2

0.25

0.3

0.35

Bifurcation ptfirst

Limit ptfirst

λL1

λL1

λB1

λB1

√2

H

√3H =|λ| at critical

point

Height H

Result for S = 2, A = 1, E = 10

0.2566

0.25

coalescence @

maximum

Figure 10.17. Plot of maximum load capacity as defined by λ (negative upwards)at the critical points for trusses with different heights H (S = 2, E = 1 and A0 = 1).

Limit point failure in red, bifurcation failure in blue.

terms of µ and α can be expressed, after some reordering, as

µL1 = −1 + 1√3

= −0.42265, µL2 = −1 − 1√3

= −1.57735,

µB1 = −1 +√

3s2 − 2

s, µB2 = −1 −

√3s2 − 2

s.

(10.21)

Subscripts L and B stand here for limit and bifurcation point, respectively. That classificationis readily done given the diagonal nature of Ks , and is relegated to Exercise 10.4. The printedoutput of this script is shown in the bottom cell of Figure 10.15. The plot output, following somepostprocessing by Adobe Illustrator, can be seen in Figure 10.16.

§10.6.10. Mises Truss Load Capacity

Since all critical points are on the fundamental path, from (10.21) the corresponding values ofthe load parameter λ are easily obtained by recovering uY and replacing in the fundamental pathequilibrium equation. The result is

λL1 = −λL2 = − 16E A0 H 3

3√

3√

(4H 2 + S2)3, λB1 = −λB2 = − 2

√2 E A0√

(4H 2 + S2)3S2

√2H 2 − S2.

(10.22)

These are used to produce the maximum load capacity plot shown in Figure 10.17 (the script todo it is not shown). This Figure displays λL1 and λB1 versus H upon setting E = A0 = 1 andS = 2, whence H = tan α. Negative λ (downward load) is plotted upward for convenience. Thecoalescence of bifurcation and limit points occurs for

H =√

3 (S/2) = 1.732 (S/2) ⇒ α = 60◦, |λL1| = |λB1| = E A0

4. (10.23)

The plot shows, however, that at α = 60◦, both |λL1| and |λB1| are still rising with H , and that|λB1| becomes the smaller of the two for α > 60◦. So the maximum load capacity actually happens

10–19


X,x

Z,z

Y,y

R

H

1

5

3

4

SupportCircle

Groundφ = 90

9090

90

Apex

(a) (b)

2

L0

(1)(2)(3) (4)

Figure 10.18. Pyramidal truss with 4 members: (a) physical structure; (b) FEM model inreference configuration, showing node and element numbering.

when |λB1| attains a maximum. Solving ∂λL1/∂ H = 0 yields

H =√

7/2 (S/2) = 1.871 (S/2), α = 61.87◦, |λmax | = 4

9√

3E A0 = 0.2566E A0.

(10.24)

This |λmax | is only about 2% higher than at coalescence: |λL1| = |λB1| = 0.25 E A0/4. But notethat in going from H = √

3 S/2 to H = √7/2 (S/2) the truss volume 2A0 L0 increases by about

9.3%. If the objective is to maximize load capacity with respect to a fixed structural volume, it isshown in Exercise 10.2 that the optimal angle is actually 60◦.This kind of optimality criterion (coalescence of critical points gives the strongest stable structure)is typical of those encountered in optimal design under stability and structural weight constraints.

§10.7. Pyramidal Trusses

Pyramidal trusses represent the simplest generalization of the Mises truss to 3D space. Pick an(X, Y, Z) RCC global system with (X, Y ) as ground level plane and Z as vertical axis. The trusscrown, henceforth called the apex, is placed on the Z axis at height H . On the (X, Y ) plane draw acircle of radius R with center at X = Y = 0. Place m ≥ 2 identical truss members going from theapex to the circle, spaced at equal circumferential angle φ = 2π/m = 360◦/m, and starting at theX axis. The reference (initial) length of each member is L0 = √

H 2 + R2. Members are fixed inall directions at the ground whereas the apex joint is allowed to move freely in any direction. Theelastic modulus and cross section area of all members are E and A0, respectively.The resulting structure is denoted by PT-m. Obviously PT-2 is the Mises truss studied in §10.6,with the radius R as the halfspan S/2. Being planar, PT-2 has no stiffness normal to its plane andrequires that motion to be precluded at the apex. On passing to 3D, stability in the reference staterequires m ≥ 3. Figure 10.18 depicts the case m = 4, in which φ is 90◦.The main attraction of this structure as benchmark for nonlinear solution methods is the appearanceof multiple bifurcation points, as well as a multiplicity-3 critical point for α = 60◦. See §10.7.4.

Remark 10.2. Pyramidal truss lattices are attracting attention as core components of sandwich beams andpanels. See sketch in Figure 10.19. Such configurations have shown high energy absorption capacity underextreme loading conditions such as high velocity impact and explosive shock waves. Fabrication considerationsfavor PT-4 as lattice unit cells, as in the figure.

10–20

§10.7 PYRAMIDAL TRUSSES

��

Figure 10.19. Pyramidal truss lattice structure for sandwich panel core.

§10.7.1. Pyramidal Truss FEM Model

The FEM model of a PT-m structure uses the TL spatial bar element developed in ? for eachmember. The assembled model has m elements and m +1 nodes. The node and element numberingconvention is illustrated in Figure 10.18(b). The number of DOF is 3m +3 before application of thekinematic BC at the ground supports. Upon removing those constrained DOF, the reduced modelhas only three DOF: the apex displacements in the X , Y and Z directions. These are denoted byu X , uY and uZ , respectively. For convenience those DOF are recast in dimensionless form as

µX = u X/H, µY = uY /H, µZ = uZ/H, (10.25)

whence the scaled state vector is

u = [ µX µY µZ ]T . (10.26)

The applied force at the apex is vertical and proportional:

f = λ q = λ

[ 001

], (10.27)

in which λ has dimension for force.6 In the ensuing scripts, H and R are eliminated in favor of L0

and α by

H = L0 sin α, R = L0 cos α. (10.28)

Thus u X = µX H = L0 µX sin α, etc.

§10.7.2. Pyramidal Truss With Three Members

For the case m = 3 (the minimum number of bars that gives initial stability in 3D) the Mathematicascript listed in Figure 10.20 retrieves the reduced internal force vector and tangent stiffness matrixassociated with the state vector (10.26). The script uses the space truss assembly modules of §10.4and the BC applicators of §10.5. The raw output (not shown) was postprocesed and beautified with

6 The control parameter λ can be rendered dimensionless by redefining the incremental load vector as q = [ 0 0 E A0 ]T ,but that was not found necessary in the derivations.

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ClearAll[m,φ,α,R,H,Em,A0,L0,µX,µY,µZ,c,s]; m=3; numnod=m+1; numele=m; R=Cos[α]*L0; H=Sin[α]*L0;uX=µX*H; uY=µY*H; uZ=µZ*H; nodxyz=Table[0,{numnod}]; elenod=Table[0,{numele}];φ=2*Pi/m; ang=0; nodxyz[[1]]={0,0,H};For [i=1,i<=m,i++, nodxyz[[i+1]]=R*{Cos[ang],Sin[ang],0}; elenod[[i]]={1,i+1}; ang+=φ]; elemat=Table[Em,{numele}]; elefab=Table[A0,{numele}]; elestr=Table[0,{numele}]; rep1={Cos[α]->c,Sin[α]->s,Cos[2*α]->c^2-s^2,Sin[2*α]->2*s*c};rep2={c^2->1-s^2,Em->1,A0->1,L0->1};noddis=Table[{0,0,0},{numnod}]; noddis[[1]]={uX,uY,uZ};nodtag=Table[{1,1,1},{numnod}]; nodtag[[1]]={0,0,0};p=TLSpaceTrussMasterIntForce[nodxyz,elenod,elemat,elefab, elestr,noddis,False];pred=ReducedForceVector[nodtag,p]; pred=TrigExpand[pred];pred=Simplify[pred/.rep1]; pred=Simplify[pred/.rep2]; Print["pred=",pred//MatrixForm]; {KM,KG}=TLSpaceTrussMasterTanStiff[nodxyz,elenod,elemat,elefab, elestr,noddis,False]; K=Simplify[KM+KG]; Kred=ReducedStiffMatrix[nodtag,K]; Kred=TrigExpand[Kred];Kred=Simplify[Kred/.rep1]; Kred=Simplify[Kred/.rep2]; Print["Kred=",Kred//MatrixForm];

Figure 10.20. Script that forms the internal force p and tangent stiffnessmatrix K of pyramidal truss with 3 members (PT-3).

ClearAll[m,φ,θ,α,R,H,Em,A0,L0,uX,uY,uZ,µX,µY,µZ,c,s,cθ,sθ]; m=1; numnod=m+1; numele=m; R=Cos[α]*L0; H=Sin[α]*L0;uX=µX*H; uY=µY*H; uZ=µZ*H; nodxyz=Table[0,{numnod}]; elenod=Table[0,{numele}];φ=2*Pi/m; ang=θ; nodxyz[[1]]={0,0,H};For [i=1,i<=m,i++, nodxyz[[i+1]]=R*{Cos[ang],Sin[ang],0}; elenod[[i]]={1,i+1}; ang+=φ]; elemat=Table[Em,{numele}]; elefab=Table[A0,{numele}]; elestr=Table[0,{numele}]; rep1={Cos[α]->c,Sin[α]->s,Cos[2*α]->c^2-s^2,Sin[2*α]->2*s*c};rep2={c^2->1-s^2,Em->1,A0->1,L0->1};rep3={Cos[θ]->cθ,Sin[θ]->sθ,Cos[2*θ]->cθ^2-sθ^2,Sin[2*θ]->2*cθ*sθ};noddis=Table[{0,0,0},{numnod}]; noddis[[1]]={uX,uY,uZ};nodtag=Table[{1,1,1},{numnod}]; nodtag[[1]]={0,0,0};p=TLSpaceTrussMasterIntForce[nodxyz,elenod,elemat,elefab, elestr,noddis,False];pred=ReducedForceVector[nodtag,p]; pred=TrigExpand[pred];pred=Simplify[pred/.rep1]; pred=Simplify[(pred/.rep2)/.rep3]; Print["pred=",pred//MatrixForm]; {KM,KG}=TLSpaceTrussMasterTanStiff[nodxyz,elenod,elemat,elefab, elestr,noddis,False]; K=Simplify[KM+KG]; Kred=ReducedStiffMatrix[nodtag,K]; Kred=TrigExpand[Kred];Kred=Simplify[Kred/.rep1]; Kred=Simplify[(Kred/.rep2)/.rep3]; Print["Kred=",Kred//MatrixForm];

Figure 10.21. Script that constructs the internal force p and tangent stiffness matrix K ofpyramidal truss with one member that forms an angle θ with the X Z plane.

10–22


some face makeup. Denoting pred and Kred by p and K for brevity, the result is

p = 3

2E A0

s µX (1 + s2 C1)

s µY (1 + s2 C2)

s3 (1 + µZ ) C3

,

K = 3 E A0

2 L0

1 + s2 C4 2 s2 µX µY 2 s2 µX (1 + µZ )

2 s2 µX µY 1 + s2 C5 2 s2 µY (1 + µZ )

2 s2 µX (1 + µZ ) 2 s2 µY (1 + µZ ) s2 C6

.

(10.29)

Here c = cos α, s = sin α, and the C coefficients are given by

C1 = C2 = µ2X + µ2

Y + 2 µZ + µ2Z − 1, C3 = µ2

X + µ2Y + µ2

Z + 2µZ ,

C4 = 3 µ2X + µ2

Y + 2 µZ + µ2Z − 1, C5 = µ2

X + 3 µ2Y + 2 µZ + µ2

Z − 1,

C6 = µ2X + µ2

Y + 6 µZ + 3 µ2Z + 2.

(10.30)

For a vertical motion in which µX = µY = 0, the foregoing expressions reduce to

p = 3

2E A0

0

0

s3 µZ (1 + µZ ) (2 + µZ )

,

K = 3 E A0

2 L0

1 + s2 (2 µZ+µ2

Z−1) 0 0

0 1 + s2 (2 µZ+µ2Z−1) 0

0 0 s2 (µZ+3 µ2Z+2)

.

(10.31)

From these results the remainder of the deflection and stability analysis is easy (but not trivial). Butit is more instructive to consider the case of an arbitrary number of members. This is carried out inthe next subsection.

§10.7.3. Pyramidal Truss With Arbitrary Number of Members

Next we consider the case of arbitrary m ≥ 3. At first sight this goal appears to be impossible withFEM, since a truss model must have a unique m. But it can be attained in three steps.First, run a single member (i.e., m = 1) that goes from the apex to a circle support and formsan angle θ with X , with (µX , µY , µZ ) as scaled apex displacements. This is done with the scriptshown in Figure 10.21. The indigestible raw output is not shown,7 but upon some grooming thereduced internal force and tangent stiffness matrix of the member can be compactly expressed as

pθ = E A0

2

[ s F1 F3

s F2 F3

−s2 (1 + µZ ) F3

],

Kθ = E A0

2 L0

[ 1 + F4 + s2 F5 2 F1 F2 −2 s F1 (1 + µZ )

2 F1 F2 1 − F6 − s2 F7 −2 s F2 (1 + µZ )

−2 s F1 (1 + µZ ) −2 s F2 (1 + µZ ) s (−2 c F8 + s F9)

],

(10.32)

7 The raw output, printed in InputForm, is actually inlined on top of the script of Figure 10.22.

10–23


ClearAll[m,Em,A0,L0,µX,µY,µZ,c,s,cθ,sθ];pθ=(Em*A0/2)*{ s*(-c*cθ+s*µX)*(-2*c*(cθ*µX+sθ*µY)+s*(µX^2+µY^2+µZ*(2+µZ))), s*(-c*sθ+s*µY)*(-2*c*(cθ*µX+sθ*µY)+s*(µX^2+µY^2+µZ*(2+µZ))), s^2*(1+µZ)*(-2*c*(cθ*µX+sθ*µY)+s*(µX^2+µY^2+µZ*(2+µZ)))};Kθ=(Em*A0/(2*L0))*{{1-cθ^2*(s^2-1)-sθ^2-6*c*cθ*s*µX-2*c*s*sθ*µY+s^2*(-1+sθ^2+3*µX^2+ µY^2+2*µZ+µZ^2),2*(c*cθ-s*µX)*(c*sθ-s*µY),2*s*(-c*cθ+s*µX)*(1+µZ)},{2*(c*cθ-s*µX)*(c*sθ-s*µY),1+cθ^2*(s^2-1)+sθ^2-2*c*cθ*s*µX-6*c*s*sθ*µY+ s^2*(-1-sθ^2+µX^2+3*µY^2+2*µZ+µZ^2),2*s*(-c*sθ+s*µY)*(1+µZ)},{2*s*(-c*cθ+s*µX)*(1+µZ),2*s*(-c*sθ+s*µY)*(1+µZ), s*(-2*c*(cθ*µX+sθ*µY)+s*(2+µX^2+µY^2+6*µZ+3*µZ^2))}};Cp00=pθ/.{cθ->0,sθ->0}; Cp20=Coefficient[pθ,cθ^2];Cp02=Coefficient[pθ,sθ^2]; CK00=Kθ/.{cθ->0,sθ->0};CK20=Coefficient[Kθ,cθ^2]; CK02=Coefficient[Kθ,sθ^2];p=m*Simplify[Cp00+(Cp20+Cp02)/2]; Print["p=",p//MatrixForm];K=m*Simplify[CK00+(CK20+CK02)/2]; Print["K=",K//MatrixForm];

Figure 10.22. Script that forms the internal force p and tangent stiffness matrix K of pyramidal truss withm members from the one-member produced by the script of Figure 10.21 and the trigonometric sums (10.35).

in which c = cos α, s = sin α, cθ = cos θ , sθ = sin θ , and the F coefficients are given by

F1 = c cθ − s µX , F2 = c sθ − s µY ,

F3 = 2 c (cθ µX + sθ µY ) − s (µ2X + µ2

Y + µ2Z + 2 µZ ),

F4 = c2θ c2 − s2

θ − 6 c s cθ µX − 2 c s sθ µY ,

F5 = s2θ + 3 µ2

X + µ2Y + 2 µZ + µ2

Z − 1,

F6 = c2θ c2 − s2

θ + 2 c s cθ µX + 6 c s sθ µY ,

F7 = s2θ − µ2

X − 3 µ2Y − 2 µZ − µ2

Z + 1,

F8 = cθ µX + sθ µY , F9 = 2 + µ2X + µ2

Y + 6 µZ + 3 µ2Z .

(10.33)

Second, the master internal force and tangent stiffness of the PT with m members is formed byassembling the pθ and Kθ of (10.32) while letting θ run over (2π/m)(i − 1) for i = 1, 2, . . . m.For instance, if m = 4:

p = p(0◦)+p(90◦)+p(180◦)+p(270◦), K = K(0◦)+K(90◦)+K(180◦)+K(270◦). (10.34)

On expanding the assembled p and K, trigonometric sums with m terms that involve cθ , sθ , c2θ ,

cθ sθ , and s2θ are factored out. Those sums can be evaluated in closed form by using the identities

m∑i=1

cos θi =m∑

i=1

sin θi =m∑

i=1

sin θi cos θi = 0,

m∑i=1

cos2 θi =m∑

i=1

sin2 θi = 12 m, (10.35)

in which θi = 2(i−1)π/m and m ≥ 3. (They are only partly valid for m = 1, 2.) Expressions(10.35) are easily verified using geometric arguments. Futhermore the portion of p and K that isindependent of θ repeats, and for that one obviously

∑mi=1 1 = m.

10–24


The job is finished with the script listed in Figure 10.22. Therein the Coefficient function isused to extract trigonometric sum coefficients, and (10.35) applied. The final result is

p(m) = m

2E A0

s µX (1 + s2 C1)

s µY (1 + s2 C2)

s3 (1 + µZ ) C3

,

K(m) = m E A0

2 L0

1 + s2 C4 2 s2 µX µY 2 s2 µX (1 + µZ )

2 s2 µX µY 1 + s2 C5 2 s2 µY (1 + µZ )

2 s2 µX (1 + µZ ) 2 s2 µY (1 + µZ ) s2 C6

.

(10.36)

in which the C coefficients are given by (10.30), and the number of members is written as argument.For the symmetric motion µX = µY = 0 we get

ps(m) = m

2E A0

0

0

s3 µZ (1 + µZ ) (2 + µZ )

,

Ks(m) = m E A0

2 L0

1 + s2 (2 µZ+µ2

Z−1) 0 0

0 1 + s2 (2 µZ+µ2Z−1) 0

0 0 s2 (µZ + 3 µ2Z + 2)

.

(10.37)

As a quick check, if m = 3 we recover (10.29) and (10.31).

§10.7.4. Primary Path Response and Critical Points

From the results in §10.7.3, the primary (fundamental) equilibrium path under vertical load λ is

µX = µY = 0, λ = m

2E A0 s3 µZ (2 + µZ ) (1 + µZ ). (10.38)

This response is identical to that of the Mises truss, as studied in §10.6.4, except for a scaling factorm/2. In particular λ vanishes at µz = 0, 1, 2, which are the roots of the cubic in µZ . Equating thedeterminant of the diagonal Km in (10.37) provides the critical points at the µZ roots of K11 = 0,K22 = 0 and K33 = 0. These are

µZ L1 = −1 + 1√3

= −0.42265, µZ L2 = −1 − 1√3

= −1.57735,

µZ B1 = −1 +√

3s2 − 2

s(double), µZ B2 = −1 −

√3s2 − 2

s(double).

(10.39)

These scaled deflections coincide with the critical points (10.21) of the Mises truss, and are in-dependent of m. Recovering the force λ from the last of (10.38) shows that the critical loads aresimply scaled by the factor m/2.But there is a significant difference as regards the bifurcation points: these now have multiplicity 2because K11 = K22 in (10.21). Hence µZ B1 and µZ B2 are double roots. Physically the bucklingmode (the null eigenvector of Km for those roots) may be any combination of µX and µY while µZ

stays fixed. This is a consequence of the radial symmetry of the vertically loaded perfect structure.

10–25


The actual bifurcation direction will depend on geometric or loading imperfections. The union ofall secondary paths forms a surface of revolution about the Z axis.The rise angle at which limit and bifurcation points coalesce is again 60◦. In this instance one hasa critical point of multiplicity 3. This configuration can provide a stern test of numerical solutionprocedures for traversing critical points.

§10.7.5. Which Pyramidal Truss is Best?

An interesting design question can be posed here. The total structural material volume of a PT-mis obviously V = m L0 A0 = L0 Am , where Am = m A0 is the member cross-section area sum.Suppose that the modulus E , height H and structural volume V are given. The cross section of eachbar is A0 = A/m = V/(mL0). Find the rise angle α and member count m that give the highestload capacity for fixed H and V .The answer is: rise angle α = 60◦ and any m ≥ 3. This is a straightforward consequence of theprevious results. So the remaining decision is, which m to pick? If the cost of supports is significantm = 3 would be the best, but m = 4 may be preferred on account of fabrication considerations.(For example, in PT-lattice assemblies for sandwich panel cores; see Figure 10.19.)

10–26

Exercises

Homework Exercises for Chapter 10TL Bar Elements: Truss Analysis

EXERCISE 10.1 [D:10] The α = 30◦ (case 1) plot in Figure 10.8 shows a limit-point λ of about 0.048.Explain why that value could be different from that of the first limit point λL1 = 0.0638560 of the exactsolution obtained in Homework Exercise 6.3 for the same two-bar structure.

EXERCISE 10.2 [A/C:25] A problem in optimal design. For the Mises truss, the plot in Figure 10.17 makesplain that, given the span S, modulus E and cross section area A0, the largest vertical-downward-load capacityis obtained if the rise angle is α = 60◦, or H = (S/2) tan α = S

√3/2. But that might not be the best use

of material since as α increases, the bars get longer. Suppose that the design objective is to get the largestpossible λmax = |λ| per unit volume of material, with respect to the total volume. Mathematically: maximizeλmax/V0 with respect to V0, where λmax is the largest downward load sustained before failure by snap-throughor bifurcation. What would be the best α?Hints:(1) There are several ways to formulate the optimization problem, and they may lead to different answers.

Since this is an advanced graduate course, think about possible ways of setting up this problem.(2) The analytical expression for critical point values is given in equation (10.22). Of the 4 points, only the

first limit point λL1 and the first bifurcation point λB1 are of interest. Of these, λL1 controls the structuralload capacity for α ≤ 60◦, whereas λB1 does it for α ≥ 60◦.

EXERCISE 10.3 [A/C:30] (Tough) Recall that the internal force expression for the Mises truss obtained inequation (10.10), which is valid for arbitrary inputs except that s0 = 0, is

p =[

pX

pY

]= E A0

L30

[u X

(S2/2 + u2

X + 2H uY + u2Y

)(H + uY )

(u2

X + uY (2 H + uY )) ]

, (E10.1)

in which L0 =√

H 2 + S2/4. Suppose that the truss is loaded at node 2 (crown) by a horizontal force fX ,positive to the right, while the vertical force is zero. The external force vector is

f =[

fX

fY

]= λ q = λ

[10

]=

[λ

0

]. (E10.2)

Since p = f along any equilibrium path, the internal force pY must vanish if the truss is to be in staticequilibrium.

��

��

α α

X, x

Y, y (2)(1)

1

2

3

uX

uY

0

0

Here λ > 0 means crownload is applied to the right(the +X direction)

E, A 0E, A

S

H Numerical values forExercise: E = A = 1, S = H = 2

fX = λ fref

Figure E10.1. Mises truss under horizontal load for Exercise 10..3.

The configuration to be analyzed is that shown in Figure E10.1. Using the numerical data there, show thatthere are four critical points: two isolated limit points, and two multiple critical points at which a limit and abifurcation point coincide. Locate these points.

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EXERCISE 10.4 [A:15] Justify the L (limit point) and B (bifurcation point) labels in (10.21) using thecriterion of §5.3. Hint: the null eigenvectors at critical states are very simple due to the diagonal nature of Ks ;combine those with the q of (10.9).

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10 tl bar elements: truss analysis

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