force methods for trusses with elastic boundary conditions

International Journal of Mechanical Sciences 66 (2013) 202–213

Contents lists available at SciVerse ScienceDirect

International Journal of Mechanical Sciences

0020-74

http://d

n Corr

E-m1 Cu

journal homepage: www.elsevier.com/locate/ijmecsci

Force methods for trusses with elastic boundary conditions

Hoang Chi Tran 1, Jaehong Lee n

Department of Architectural Engineering, Sejong University 98 Kunja Dong, Kwangjin Ku, Seoul 143-747, Republic of Korea

a r t i c l e i n f o

Article history:

Received 17 July 2012

Received in revised form

29 October 2012

Accepted 15 November 2012Available online 29 November 2012

Keywords:

Complementary energy

Force method

Indeterminate trusses

Compatibility conditions

Elastic boundary conditions

03/$ - see front matter Crown Copyright & 2

x.doi.org/10.1016/j.ijmecsci.2012.11.009

esponding author. Tel.: þ82 2 3408 3287; fax

ail address: [email protected] (J. Lee).

rrently Postdoctoral research associate.

a b s t r a c t

A structural analysis using force method is presented for trusses accounting for the flexibilities of the

elastic boundary conditions. Novel zero-length dummy members to transform the truss with boundary

conditions into an equivalent free-standing model without boundary conditions are employed. Newly

general equilibrium equations and general kinematic relations in which the reaction forces and the

displacements at the elastic boundary conditions are, respectively, taken into account are clearly

derived. The compatibility conditions in terms of redundant forces in which the flexibilities of the

elastic boundary conditions are considered are obtained from the minimum of complementary energy.

The difference between with and without considering the effect of the elastic boundary conditions lies

in the formulation of redundant forces. Both member and reaction forces are simultaneously and

directly obtained. Finally, numerical examples are presented to demonstrate the efficiency and

accuracy of the proposed method.

Crown Copyright & 2012 Published by Elsevier Ltd. All rights reserved.

1. Introduction

The computational structural analysis is usually carried outusing two main methods known as displacement method (stiff-ness method) and force method (flexibility method) with force asthe primary unknown. The well-known displacement methodcurrently can be considered as dominated structural analyzerdue to its generality and simplicity in the computer implementa-tion. As a demonstration, a lot of commercial softwares have beendeveloped based on this method. In comparison with the dis-placement method, the force method has its own advantages:(a) easier basic concept; (b) generating accurate stress results [1,2]even for modest finite element models since both equilibrium andcompatibility conditions are simultaneously satisfied [3]; (c) simpleformulation for optimization problems including stress constraints[4–9]; (4) the properties of members of a structure most oftendepend on the member forces rather than joint displacements.These advantages have made the force method as an importantpart in the field of the structural mechanics.

In the force method, equilibrium equations themselves arenot sufficient to solve general structural analysis problems sincethey are indeterminate in nature. Accordingly, determinacy for acontinuum mechanics (a unique solution of analysis) is onlyachieved by employing the sufficient compatibility conditionsfor solving the structural analysis problems. In general, the force

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: þ82 2 3408 3331.

method can be classified into four different approaches: (i) topolo-gical force method; (ii) integrated force method; (iii) mixedalgebraic-combinatorial force method; and (iv) algebraic forcemethod. Topological methods were proposed by several investiga-tors [10–12] for rigid-jointed skeletal structures using manualselection of the cycle bases of their graph models. The integratedforce method was suggested by Patnaik [13], which was thensuccessfully extended to dynamic and nonlinear analysis of struc-tural systems [14,15]. Mixed methods were investigated by Gilbert[16]. Moreover, mixed algebraic-graph theoretical methods werealso suggested by Kaveh [17] for different finite element models.Force method which can be considered to has explicit merits overthe displacement approach in many situations, especially in opti-mization was developed. However, most of the available approachesdid not consider the effects of the elastic boundary conditions (BCs)in the structural analysis using the force method [18–20]. Accord-ingly, an efficient numerical method for structural analysis of trussesusing the force method in which the flexibilities of the elastic BCsare taken into account is the motivation of this research.

In this paper, a numerical procedure is proposed for structuralanalysis of trusses accounting for the flexibilities of the elastic BCsthrough the force method. Statically determinate and all types ofstatically indeterminate trusses are investigated. Novel zero-length dummy members are utilized to transform the truss withBCs into an equivalent free-standing model without BCs. Newlygeneral equilibrium equations and general kinematic relations inwhich the reaction forces and the displacements at the elastic BCsare, respectively, taken into account are formulated. The compat-ibility equations in terms of redundant forces in which theflexibilities of the elastic BCs are considered are derived from

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H. Chi Tran, J. Lee / International Journal of Mechanical Sciences 66 (2013) 202–213 203

the minimum of complementary energy. The vector of redundantforces is directly obtained without the need to select any consistentredundant members as the classical force method [21]. The differ-ence between with and without considering the effect of the elasticBCs lies in the formulation of the redundant forces. Both memberand reaction forces are simultaneously and directly obtained. Thenall nodal displacements including constrained nodes are backcomputed based on the obtained member and reaction forces.

2. Formulation

2.1. Equilibrium equations

For a d-dimensional (d¼2 or 3) truss structure with m members,n free nodes and nc constrained nodes (BCs or supports), its topologycan be expressed by a connectivity matrix CsðARm�ðnþnc ÞÞ asdiscussed in [22]. Suppose member k connects nodes i and j (io j),then the ith and jth elements of the kth row of Cs are set to 1 and�1, respectively, as follows:

Csðk,pÞ ¼

1 for p¼ i

�1 for p¼ j

0 otherwise

8><>: ð1Þ

If the free nodes are numbered first, then to the constrained nodes,Cs can be divided into two parts as

Cs ¼ ½C Cc� ð2Þ

where CðARm�nÞ and CcðARm�nc Þ describe the connectivities of the

members to the free and constrained nodes, respectively. For asimple two-dimensional truss as shown in Fig. 1(a), which consists

Fig. 1. (a) A 5-bar statically indeterminate plane truss; (b) a 5-bar statically indetermi

free body diagram of the two systems in (a) and (b); (d) the equivalent free-standing mo

of five members (m¼5) and four nodes including two free nodes(n¼2) and two constrained nodes (nc¼2), the connectivity matrixCsðAR5�4

Þ is given in Table 1.Let x, y, zðARn

Þ and xc , yc , zcðARnc Þ denote the nodal coordi-nate vectors of the free and constrained nodes, respectively, in x-,y- and z-directions. The coordinate difference vectors lxð ¼ lxkÞ,lyð ¼ lykÞ and lzð ¼ lzkÞARm

ðk¼ 1,2, . . . ,mÞ of the m members inx-, y- and z-directions, respectively, given by

lx ¼ CxþCcxc ð3aÞ

ly ¼ CyþCcyc ð3bÞ

lz ¼ CzþCczc ð3cÞ

Let l (¼ lk ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðlxkÞ

2þðlykÞ

2þðlzkÞ

2q

) ARm denote the length vec-tor; Lx ¼ diagðlxÞ, Ly ¼ diagðlyÞ, Lz ¼ diagðlzÞ and L¼ diagðlÞARm�m

are the diagonal square matrices of the coordinate differencevectors in x-, y-, z-directions and the member length vector,respectively. The equilibrium equations of the free nodes in eachdirection of a general pin-jointed structure given by [23,24] canbe stated as

CT LxL�1f ¼ px ð4aÞ

CT LyL�1f ¼ py ð4bÞ

CT LzL�1f ¼ pz ð4cÞ

where px, py and pz (ARn) are the vectors of external loadsapplied at the free nodes in x-, y- and z-directions, respectively;the symbol, ð�ÞT , denotes the transpose of a matrix or vector; and f(ARm) is the unknown vector of member forces. The equilibriumequations, Eq. (4), of the free nodes of the general pin-jointed

nate plane truss with the presence of elastic BCs (modeled via spring BCs); (c) the

del of the two systems in (a) and (b) with dummy members 6–9 to remove the BCs.

Table 1The incidence matrix of a 5-bar statically indeterminate plane truss.

Member/Node 1 2 3 4

1 1 0 �1 0

2 0 1 0 �1

3 1 �1 0 0

4 0 1 �1 0

5 1 0 0 �1

H. Chi Tran, J. Lee / International Journal of Mechanical Sciences 66 (2013) 202–213204

structure can also be reorganized as

Af ¼ p ð5Þ

where p¼ ðpTx , pT

y , pTz Þ

T ARdn; A (ARdn�m) is known as theequilibrium direction-cosine matrix [22] that transforms thevector of member forces f of the system to the vector of externalloads p of the free nodes, defined by

A¼

CT Lx

CT Ly

CT Lz

0BB@

1CCAL�1

ð6Þ

It should be noted that the equilibrium equations given in Eq. (5)do not include any BCs. There are three types of staticallyindeterminate trusses which are (i) externally, (ii) internally,and (iii) mixed (i.e., both externally and internally) staticallyindeterminate trusses. Let re, ri and r denote the numberof external, internal and total indeterminacies, respectively.The external indeterminacy is related to the BCs while the internalindeterminacy is associated with the number of members in thetruss. They are defined, respectively, as follows:

re ¼ c�rbZ0 ð7aÞ

ri ¼m�dnsþrb ð7bÞ

where c, rbð ¼ dðdþ1Þ=2Þ and nsð ¼ nþncÞ are the number of BCs,the number of independent rigid-body motions and total number ofnodes of the truss, respectively. Eq. (7a) indicates that all of rigid-body motions of the structure at least must be eliminated. Thenumber of total statically indeterminacy is given by

r¼ reþri ð8aÞ

or

r¼ms�dns ð8bÞ

where ms (¼mþc) is total number of members and BCs. Accordingto the values of r, there are the following two cases:

Case 1. (i) r¼0; and (ii). r40. If r¼0, the number of equationsand the number of the unknowns of member and reaction forcesare identical, i.e., the general equilibrium matrix A in Eq. (10)below is square and full rank. The equilibrium equations aresufficient to find the member and reaction forces; and the trussesare said to be determinate. If r40, the number of the unknownsof member and reaction forces exceeds the number of equations,i.e., the general equilibrium matrix A is rectangular. This meansthat the equilibrium equations themselves are not sufficientto find the member and reaction forces; and the trusses are saidto be indeterminate. In this context, additional equations ofcompatibility conditions are needed to determine the unknownmember and reaction forces.

Case 2. ro0, there is at least a mechanism that may cause thestructure unstable, which is not considered in this paper. In otherwords, from Eqs. (7) and (8), the trusses can be only staticallydeterminate or three types of statically indeterminate as men-tioned earlier, which are the concern of this study.

2.2. Boundary condition treatment

There are two kinds of BCs investigated in this study. The firstone is a rigid BC, and the other is an elastic BC. The elastic BCs canbe modeled by the spring BCs. If the stiffness of a spring BCincreases to infinite, this spring will become a rigid BC. Letrc (ARc) denotes the unknown vector of reaction forces employedto remove all the BCs. By doing this, the structure becomes a freebody diagram. Accordingly, the constrained nodes can now betreated as the free nodes. That is, the second terms in the right-hand side of Eqs. (2) and (3) which are related to the constrainednodes vanish. In the meantime, the reaction forces at theconstrained nodes can be considered as external nodal loads. Inthis context, the equilibrium equations for all the nodes (includ-ing the constrained nodes) of the discrete truss after moving theunknown reaction forces to the left-hand side can be written asfollows:

ð9Þ

or

A f ¼ p ð10Þ

where f ð ¼ ffT ,rTc g

T ÞARms is the unknown member and reactionforce vector; p (ARdns ) is the external load vector of all nodes.A (¼ ½A9Ac�) ARdns�ms is called general equilibrium matrix of allnodes in this paper, which transforms the vector of member andreaction forces f of the system to the vector of external loads p;Ac (ARdns�c) is the coefficient matrix of c unknown reactionforces in which its c columns correspond to the last c unknowncomponents of f , defined as follows:

Acðdns ,cÞ ¼�1 at the constrained degrees of freedom ðdof’sÞ

0 otherwise

�ð11Þ

In fact, the c columns of the matrix Ac are vectors with all oftheir entries being zero except the entries in the rows correspond-ing to the constrained dof’s have negative unit values (i.e., �1).As an example, consider the 5-bar plane truss supported by twohinges, and the one supported by one hinge and two elastic BCs(modeled via two spring BCs) with the stiffness k3x and k3y asshown in Fig. 1(a,b), respectively. These two trusses have the samefree body diagram (Fig. 1c) in which four unknown reaction forces(i.e., r3x, r4x, r3y and r4y in x- and y-directions) are employed toremove all the BCs. The connectivity matrix of this free bodydiagram is given in Table 1. Based on the free body diagram, theequilibrium equations (Eq. (9)) of all the nodes for these two trussesafter moving the unknown reaction forces to the left-hand side canbe written as

ð12Þ


The main challenge in the force method for analysis of thetrusses with the elastic BCs is that the unit elastic BC does nothave the full information concerning its static and kinematicproperties in the available force formulation. For this reason, anovel dummy member that has zero length and same stiffnesswith the elastic BC is introduced to replace each elastic BC, whichgenerally enables the analysis of the trusses accounting for theflexibilities of the elastic BCs. In fact, from Eq. (12), Fig. 1(c,d), itcan be seen that each translational BC is modeled as a noveldummy member having zero length and the same stiffness ki asthat translational BC. Hence, Fig. 1(d) where the four noveldummy members 6–9 (which are represented by thick lines)having zero lengths (i.e., li¼0, i¼ 629) and the stiffness k62k9 areutilized to substitute the four unknown reaction forces r3x, r4x, r3y

and r4y, respectively, can be considered as the equivalent free-standing model of the two structures in Fig. 1(a,b). In other words,by using the concept of the novel zero-length dummy members,the truss with the elastic BCs can be converted into the equivalentfree-standing structure without BCs. After implementation ofstructural analysis with the elastic BCs using the proposedmethod, the novel zero-length dummy members 6–9 will beremoved to transform the nodes 3 and 4 back to the BCs.

It is known that the flexibility of a rigid BC is zero, i.e., itsstiffness is infinite (k¼1). Accordingly, its substituted dummymember should have the flexibility of zero. For instance, theflexibilities of all four novel dummy members 6–9 in the equiva-lent free-standing model (Fig. 1(d)) of the truss without springBCs (Fig. 1(a)) must be assigned zero values. While for the trusswith spring BCs at node 3 (Fig. 1(b)), only the flexibilities of twonovel dummy members 7 and 9 in its equivalent free-standingmodel (Fig. 1(d)) are assigned zero values. For a two-dimensionaltruss, a spring BC (or a roller BC) and hinged BCs are modeledusing one and two novel dummy members, respectively. Thesenovel dummy members should be numbered and used in theformulation of the general equilibrium matrix A at the correctdof’s.

2.3. Kinematic relations

Let eðARms Þ and dðARdns Þ denote elongation (extension)vector of all members (m members and c dummy ones) anddisplacement vector of all nodes, respectively, due to the externalloads p. The vectors e and d can be divided into two parts as

e¼ feT1 , eT

2gT ð13aÞ

d¼ fdT1, dT

2gT ð13bÞ

where e1ðARmÞ and e2ðARc

Þ are the elongation vectors of the m

members and the c novel dummy members in which themembers should be numbered first, then to the dummy ones,respectively; similarly, d1ðARdn

Þ and d2ðARdnc Þ are the displace-ment vectors of the n free nodes and the nc constrained nodes,respectively. Once the flexibility of a spring BC decreases to zero(i.e., its stiffness increases to infinite), the spring BC will become arigid one. In this situation, the elongation of the correspondingnovel dummy member and the translational displacement of thesupported node reduce to zero.

The elongations of all members e are related to the displace-ments of all nodes d by the following kinematic relations:

e¼ Bd ð14Þ

where B (ARms�dns ) is called a general kinematic matrix in thisstudy, which transforms the vector of all nodal displacements d ofthe system to the vector of elongations of all members e.

The work done by external loads (W) and the complementaryenergy (Uc) stored in the structures are, respectively, given by

W ¼1

2pT d ð15aÞ

Uc ¼1

2f

Te ð15bÞ

From the conservation of energy, the work done by theexternal loads should be equal to the increase of complementaryenergy stored in the structure for the elastic conservative sys-tems, i.e.,

12 pT d¼ 1

2fTe ð16Þ

By substituting Eqs. (10) and (14) into Eq. (16), the followingexpression is obtained:

fTA

Td¼ f

TBd ð17Þ

Based on the principle of virtual work [25,26], the above equationcan be ensured to be true only if the following relation holds:

B¼AT

ð18Þ

Hence, the kinematic relations, Eq. (14), can be rewritten as

e¼ATd ð19Þ

The expression given by Eq. (19) is the general form of thedeformation–displacement relationship of the discrete system. Inthis equation, ms elongations of all members e are expressed in dns

displacements of all nodes d. Thus, there are r (¼ms�dns) compat-ibility equations which can be achieved through the elimination ofthe dns nodal displacements from the ms elongations.

3. Structural analysis

3.1. Member and reaction forces

The member and reaction forces f based on the standardforce method, which is the general solution of Eq. (10), can beexpressed by

f ¼ B0pþB1q ð20Þ

or in a matrix form

f ¼ ½B0 B1�p

q

( )ð21Þ

where qðARrÞ is a vector of r redundant forces which can be

obtained by minimizing the complementary energy; B0p andB1qARms ðB0ARms�dns ,B1ARms�r

Þ are the particular and homoge-neous solutions, respectively. Furthermore, B0 and B1 are general-ized (or Moore–Penrose pseudo) inverse and null (or the transposeof self-stress matrix) matrices of A, respectively, such that

AB0 ¼ I ð22aÞ

AB1 ¼ 0 ð22bÞ

Regarding a statically determinate structure which is the caseof r¼0, the general equilibrium matrix A in Eq. (10) is square andfull rank. In this situation, Eq. (22) yields

B0 ¼ ðAÞ�1

ð23aÞ

B1 ¼ 0 ð23bÞ

Eq. (20) then becomes

f ¼ ðAÞ�1p ð24Þ


Regarding a statically indeterminate structure (r40), there arevarious numerical algorithms for the generation of matrices B0

and B1 in using the force method such as Gauss Jordan procedure[27,28], LU [29], REDUC [20] and singular value decomposition(SVD) [19]. In this study, the SVD is adopted as an effective andnumerically stable algorithm. In order to obtain the matrices B0

and B1, the SVD [30] is performed on the general equilibriummatrix A

A ¼UVWTð25Þ

where UðARdns�dns Þ ¼ ½u1 u2 � � � udns� and WðARms�ms Þ ¼

½w1 w2 � � � wms � are the orthogonal matrices. VðARdns�ms Þ is adiagonal matrix with nonnegative singular values of A in des-cending order in which the last r singular values are zero as

v1Zv2Z � � �Zvms�r 4s1 ¼ � � � ¼ sr ¼ 0 ð26Þ

In this case, the matrix W from Eq. (25) can be expressed as

W¼ ½w1 w2 � � � wdns9s1 s2 � � � sr � ð27Þ

where the transpose of the last r vectors si (ARms ) in Wcorresponding to the last r zero singular values si ði¼ 1,2, . . . ,rÞin V form a null basis B1 of the matrix A. Once the SVD of thegeneral equilibrium matrix A is calculated, the matrices B0 and B1

are obtained as follows:

B0 ¼ Aþ¼WVþUT

ð28aÞ

B1 ¼ ½s1 s2 � � � sr�T ð28bÞ

where a diagonal matrix Vþ (ARms�dns ) is the pseudo-inverse ofV by taking the reciprocal of each nonzero singular value on thediagonal of V, leaving the zeros in place, and transposing theresulting matrix. Assuming linear elastic material, elongation-force relation is given as

e¼ Gf ð29Þ

where G (ARms�ms ) is the diagonal flexibility matrix. By substitutingEq. (29) into Eq. (15b), the following expression is obtained:

Uc ¼1

2f

TGf ð30Þ

The complementary energy can also be rewritten as follows bysubstituting f from Eq. (21) into Eq. (30)

Uc ¼1

2

p

q

( )T

Mp

q

( )ð31Þ

where the matrix MðARms�ms Þ is defined by

M¼ ½B0 B1�T G½B0 B1� ð32aÞ

or in an explicit form

M¼BT

0GB0 BT0GB1

BT1GB0 BT

1GB1

" #ð32bÞ

Decomposing the matrix M into four submatricesM p p ðARdns�dns Þ, MpqðARdns�r

Þ, Mqp ðARr�dns Þ and MqqðARr�rÞ

which are given, respectively, as

Mp p ¼ BT0GB0 ð33aÞ

Mpq ¼ BT0GB1 ð33bÞ

Mqp ¼ BT1GB0 ð33cÞ

Mqq ¼ BT1GB1 ð33dÞ

The complementary energy in Eq. (31) can now be expressed asfollows:

Uc ¼12ðp

T Mpp pþpT MpqqþqT Mqp pþqT MqqqÞ ð34Þ

According to Castigliano’s second theorem, the complemen-tary energy for a system of internal forces which satisfies thecompatibility conditions is minimum. Thus, the derivative of Uc

with respect to the unknown redundant forces q should be zero

@Uc

@qi

¼ 0 ði¼ 1,2, . . . ,rÞ ð35Þ

Since the matrix M is symmetric, i.e., Mqp ¼Mpq , Eq. (35) yields

Mqp pþMqqq¼ 0 ð36Þ

Hence, if the matrices B0, B1 are found, the redundant forces qcan be directly obtained as follows without the need to select anyconsistent redundant members as the classical force method [21]:

q¼�ðMqqÞ�1Mqp p ð37Þ

or in an explicit form by inserting Eqs. (33c) and (33d) intoEq. (37) as follows:

q¼�ðBT1GB1Þ

�1ðBT

1GB0Þp ð38Þ

The diagonal flexibility matrix G and the matrices B0, B1 canalso be, respectively, decomposed into

ð39Þ

ð40Þ

ð41Þ

where G1ðARm�mÞ and G2ðARc�c

Þ are the diagonal flexibilitymatrices of the m members and the c dummy ones of theequivalent system, respectively; B01ðARm�dns Þ, B02ðARc�dns Þ andB11ðARm�r

Þ, B12ðARc�rÞ are submatrices of B0 and B1, respec-

tively. Inserting Eq. (39) into (38) yields the following expression:

q¼�ðBT11G1B11þBT

12G2B12Þ�1ðBT

11G1B01þBT12G2B02Þp ð40Þ

It should be noted that if all constrained nodes are fully rigid (i.e.,there exists no spring BC in the structure), all the diagonalcomponents of G2 must be zero (i.e., G2 ¼ 0). Eq. (40) thenbecomes

q¼�ðBT11G1B11Þ

�1ðBT

11G1B01Þp ð41Þ

Eq. (40) shows the relationship between the redundant forces qand the flexibility of the elastic BCs G2. Whereas, Eq. (41)indicates that these redundant forces are independent of theeffects of the fully rigid BCs. Accordingly, the difference betweenwith and without considering the effect of the elastic BCs lies inthe formulation of the redundant forces.

As an example, consider again the 5-bar plane truss supported bytwo hinges (Fig. 1(a)), and the one supported by one hinge and twoelastic BCs (modeled via two spring BCs) (Fig. 1(b)) with the stiffnessk3x ¼ k3y ¼ 4448:2 kN. The number of total statically indeterminacyof these two trusses is one (r¼1), which implies only one redundantforce obtained from the compatibility condition is required todetermine the unknown member and reaction force vector f . Theyshare the same free body diagram (Fig. 1(c)), which leads theirobtained general equilibrium matrices A and generated matricesB0, B1 to be identical. They also have the same member propertiesand all members have the same axial stiffness of eiai ¼ 44482:221 kN(i¼ 528); i.e., they have the same matrix G1.


The only difference between these two trusses is the BCs atnode 3. This indicates that the flexibility matrices G2 of all thenovel dummy members (6–9) of their equivalent model (Fig. 1(d))are different from each other. Matrix G2 of all the dummymembers in the equivalent model of the truss without springBCs (Fig. 1(a)) is zero, whereas that of the ones in the equivalentmodel of the truss with the presence of spring BCs (Fig. 1(b)) isdifferent from zero. In particular, two nonzero diagonal compo-nents of G2 corresponding to the two dummy members 6 and 8(Fig. 1(d)) representing for the two spring BCs, which areexpressed with respect to the global diagonal flexibility matrixG, are Gð6,6Þ ¼ Gð8,8Þ ¼ 4448:2�1 kN�1. Consequently, the redundantforces q of the truss without spring BCs (Fig. 1(a)) differ fromthose of the one with the presence of spring BCs (Fig. 1(b)). In fact,the redundant forces q of the truss without spring BCs and theone with the presence of spring BCs can be, respectively, calcu-lated by Eqs. (41) and (40) and obtained as: 2.9387 and 5.6105.

Once the general equilibrium matrix A is established, thematrices B0 and B1 are then determined. They are obviouslyindependent of member properties and the types of elastic BCs.Hence, in practical design using the force method, in order tore-evaluate the member and reaction forces f with either newcross-sections or different types of elastic BCs to reach a rationaldesign, only Eq. (40) with r unknown redundant forces q (which isoften less than dns unknown displacements) is needed to besolved again. This is considered as an advantage of the forcemethod in comparison with the displacement approach.

3.2. Nodal displacements

Once the matrix B0, and member and reaction forces f aredetermined, the displacements can be then obtained as

d¼ BT0Gf ð42Þ

The member and reaction forces f and all nodal displacements dfor the trusses accounting for the flexibilities of the elastic BCs canbe defined by the proposed method through the followingprocedure.

Procedure

�
Step 1: Convert the truss with BCs into the equivalent free-standing model. � Step 2: Determine C, A and Ac by Eqs. (1), (6) and (11),
respectively. Then determine A by combining A and Ac .
� Step 3: Calculate r by Eq. (8). If r40, generate the matrices B0
and B1 by Eq. (28). Otherwise, generate B0 and B1 by Eq. (23)and go to Step 5.
� Step 4: Determine q by Eq. (40) or Eq. (41) for the case with or
without spring BCs, respectively.
� Step 5: If r40, compute f by Eq. (20). Otherwise, compute f by
Eq. (24).
� Step 6: Calculate d by Eq. (42).
4. Numerical examples

Numerical examples are presented for statically determinateand all types of indeterminate trusses considering the flexibilitiesof the elastic BCs using the above developed procedure. In orderto perform a detailed structural analysis, all members areassumed to be steel pipes with the same axial stiffness ofeiai ¼ 44482:221 kN through all numerical examples. The elasticBCs which are modeled by the spring BCs are also assumed toexist at specified BCs. The results obtained are verified foraccuracy and efficiency by comparing with those generated bythe commercial software [31] based on the displacement method.

4.1. Plane truss

4.1.1. 4-bar statically determinate plane truss

Example 1. Consider a 4-bar truss as shown in Fig. 2(a) and itsfree body diagram is depicted in Fig. 2(c). The truss is subjected toa vertical load of p1y ¼�444:82 kN at node 1. In order to obtainthe equivalent free-standing model, the constrained nodes areconverted into the free nodes by using the novel zero-lengthdummy members as mentioned in Section 2.2. After implementa-tion of structural analysis with the elastic BCs using the proposedmethod, the novel zero-length dummy members will be removedto transform the free nodes back to the BCs. By using the novelzero-length dummy members 5–8 , both constrained nodes 3 and4 supported by hinges can now be treated as the free nodes. Theequivalent free-standing model of the 4-bar truss (Fig. 2(a)) isdescribed in Fig. 2(d) where thick lines represent the novel zero-length dummy members. Since all the constrained nodes are fullyrigid (i.e., ki ¼1, i¼ 528), the flexibilities of all the novel dummymembers must be zero (i.e., G2 ¼ 0Þ.

The number of external indeterminacy (Eq. (7a)) is re ¼ 4�3¼ 1,

while the number of internal indeterminacy (Eq. (7b)) ri ¼ 4�2�

4þ3¼�1. Thus, the number of redundancy (Eq. (8)) is

r¼ 1�1¼ 0. The structure is said to be statically determinate.

Hence, both the member forces and reaction forces can be

simultaneously and directly calculated based on only the equili-

brium equations, Eq. (24). That is, there is no need any informa-

tion about the compatibility conditions in terms of the redundant

forces q for structural analysis of this 4-bar statically determinate

truss. The obtained member and reaction forces f , and all nodal

displacements d, as presented in Tables 2 and 3, respectively,

agree well with those of SAP2000 program. As expected, the

obtained displacements of both hinged nodes 3 and 4 as listed in

Table 3 are all zero (i.e., d2 ¼ 0), which confirms the accuracy of

proposed method.

Example 2. The hinged BCs at node 3 of the truss in Fig. 2(a) arenow replaced by two elastic BCs (modeled via two spring BCs)having the same flexibility of 4448:2�1 kN�1 as shown in Fig. 2(b).Its equivalent free-standing model by applying the novel dummymembers is depicted in Fig. 2(d) which is the same with the oneof the truss without spring BCs (Fig. 2a). However, the flexibilitymatrix of the novel dummy members in this example is no longerzero (i.e., G2a0). Particularly, two nonzero diagonal componentsof G2 corresponding to the two dummy members 5 and 7(Fig. 2d) representing for the two spring BCs, which are expressedwith respect to the global diagonal flexibility matrix G, areGð5,5Þ ¼ Gð7,7Þ ¼ 4448:2�1 kN�1. In SAP2000, the spring stiffness atnode 3 in both x- and y-directions are set as k3x ¼ k3y ¼ 4448:2 kN.

Tables 2 and 3 show the comparisons between the member and

reaction forces f , and all the nodal displacements d obtained from

the proposed procedure and those of SAP2000 program, respec-

tively, which verify the accuracy and efficiency of the present

method. From Table 2, it is obvious that, for the statically

determinate truss, the distribution of member and reaction forces

is independent of the flexibilities of the members and elastic BCs

since their flexibilities are not included in Eq. (24) which is

sufficient to determine member and reaction forces. In other

words, any change of flexibilities of either some specified mem-

bers or elastic BCs cannot change the distribution of member and

reaction forces in the statically determinate truss except the

nodal displacements as indicated in Eq. (42) and Table 3.

Table 3Nodal displacements of (cm) the 4-bar statically determinate plane truss.

d1x d2x d3x d4x d1y d2y d3y d4y

Example 1

SAP2000 0 �2.5400 0 0 �12.2641 �9.7242 0 0

Present 0 �2.5400 0 0 �12.2641 �9.7242 0 0

Example 2

SAP2000 0.1000 �2.5400 0.1000 0 �12.4641 �9.9242 �0.1000 0

Present 0.1000 �2.5400 0.1000 0 �12.4641 �9.9242 �0.1000 0

Note: Example 2 with the presence of elastic BCs (modeled via spring BCs).

Table 2Member and reaction forces (kN) of the 4-bar statically determinate plane truss.

f1 f2 f3 f4 f5 f6 f7 f8

ð � r3xÞ ð � r4xÞ ð � r3y) ð � r4yÞ

Examples 1 and 2

SAP2000 0 �444.820 �444.820 629.070 �444.820 444.820 444.820 0

Present 0 �444.820 �444.820 629.070 �444.820 444.820 444.820 0

Note: f 52f 8: Novel dummy member forces; Example 2 with the presence of elastic BCs (modeled via spring BCs).

Fig. 2. (a) A 4-bar statically determinate plane truss; (b) a 4-bar statically determinate plane truss with the presence of elastic BCs (modeled via spring BCs); (c) the free

body diagram of the two systems in (a) and (b); (d) the equivalent free-standing model of the two systems in (a) and (b) with dummy members 5–8 to remove the BCs.


4.1.2. 5-bar statically indeterminate plane truss

Example 3. As a next example, consider a 5-bar truss subjectedto a vertical load of p1y ¼�444:82 kN as shown in Fig. 1(a). Bysubstituting the elastic BCs with the novel zero-length dummymembers 6–9 as mentioned in Section 2.2, the configurationof the equivalent free-standing system is described in Fig. 1(d).

Since all the elastic BCs are fully rigid (i.e., ki ¼1, i¼ 629), theflexibility matrix of the novel dummy members must be zero (i.e.,G2 ¼ 0Þ.

The number of external indeterminacy (Eq. (7a)) is re ¼ 4�3¼ 1,

while the number of internal indeterminacy (Eq. (7b)) ri ¼ 5�2�

4þ3¼ 0. Hence, the number of redundancy (Eq. (8)) is r¼

1�0¼ 1. This structure is said to be one degree of externally

Table 4Member and reaction forces (kN) of the 5-bar statically indeterminate plane truss.

f1 f2 f3 f4 f5 f6 f7 f8 f9

ð � r3xÞ ð � r4xÞ ð � r3yÞ ð � r4yÞ

Example 3

SAP2000 248.102 �196.718 �196.718 278.202 �350.869 �444.820 444.820 196.718 248.102

Present 248.102 �196.718 �196.718 278.202 �350.869 �444.820 444.820 196.718 248.102

Example 4

SAP2000 248.992 �195.828 �195.828 276.942 �352.128 �444.820 444.820 195.828 248.992

Present 248.992 �195.828 �195.828 276.942 �352.128 �444.820 444.820 195.828 248.992

Note: f 62f 9: Novel dummy member forces; Example 4 with the presence of elastic BCs (modeled via spring BCs).

Table 5Nodal displacements of (cm) the 5-bar statically indeterminate plane truss.

d1x d2x d3x d4x d1y d2y d3y d4y

Example 3

SAP2000 1.4167 �1.1233 0 0 �5.4237 �4.3004 0 0

Present 1.4167 �1.1233 0 0 �5.4237 �4.3004 0 0

Example 4

SAP2000 1.5218 �1.1182 0.1000 0 �5.5432 �4.4250 �0.0440 0

Present 1.5218 �1.1182 0.1000 0 �5.5432 �4.4250 �0.0440 0



statically indeterminacy, which implies only one redundant force

(Eq. (41)) obtained from the compatibility condition is required to

determine the unknown member and reaction force vector f , as

presented in Section 3.1. The member and reaction forces f , and

all nodal displacements d obtained are given in Tables 4 and 5,

respectively. It can be seen that the results of the proposed

method are well compared with those of SAP2000 program.

Example 4. In order to show the capability of the proposedprocedure in capturing the flexibility effects of the elastic BCs,the hinged BCs at node 3 of the truss in Fig. 1(a) are nowsubstituted by two elastic BCs (modeled via two spring BCs)possessing the same flexibility of 4448.2 �1 kN�1 in the x- and y-directions as shown in Fig. 1(b). Its equivalent free-standingmodel by applying the novel dummy members, which is thesame with the one of the truss without spring BCs (Fig. 1(a)), isdepicted in Fig. 1(d).

However, the flexibility matrix of the novel dummy members

G2 in this case is no longer zero as mentioned in Section 3.1.

Accordingly, the compatibility equation in terms of redundant

force needs to be solved for q again but by Eq. (40) (not by

Eq. (41) as the case in Example 3). The obtained member and

reaction forces f , and all the nodal displacements d, as listed in

Tables 4 and 5, respectively, are in excellent agreements with

those of SAP2000 program, which confirm the accuracy and

efficiency of the proposed method. It can be observed from

Tables 4 and 5 that, as compared to the Example 3 of fully rigid

BCs, the flexibilities of the elastic BCs have important effects on

the distribution of member and reaction forces as well as the

nodal displacements of the structures.

4.2. Space truss

4.2.1. 25-bar statically indeterminate space truss

Example 5. As the last example, consider a 25-bar truss (Fig. 3(a))subjected to horizontal and vertical loads as listed in Table 6. Bysubstituting the elastic BCs with the novel zero-length dummy

members 26–37 , the configuration of the equivalent free-standing system is described in Fig. 4. The number of externalindeterminacy (Eq. (7a)) is re ¼ 12�6¼ 6, whereas the number ofinternal indeterminacy (Eq. (7b)) ri ¼ 25�3� 10þ6¼ 1. Thismeans the truss has mixed statically indeterminacy of seven inwhich one is internally statically indeterminacy, and the remain-ing are externally generated. Hence, seven redundant forces qdefined from seven equations of compatibility conditions (Eq.(41)) are needed to determine the unknown member and reactionforce vector f . The obtained member and reaction forces f , and allnodal displacements d, as given in Tables 7 and 8, respectively,coincide exactly with those of SAP2000 program.

Example 6. Similarly, to investigate the flexibility effects of theelastic BCs on the system, the hinged BCs at node 7 of the truss inFig. 3(a) are now replaced by three elastic BCs (modeled via threespring BCs) having the same flexibility of 4:4482�1 kN�1 in the x-,y- and z-directions as shown in Fig. 3(b). Its equivalent free-standing model by applying the novel zero-length dummy mem-bers, which is the same with the one of the truss without springBCs (Fig. 3(a)), is depicted in Fig. 4.

The obtained general equilibrium matrix A and its generated

matrices B0, B1 of the truss with the presence of spring BCs in this

example are the same as those of the one without spring BCs in

Example 5. However, due to the presence of spring BCs the

compatibility conditions in terms of redundant forces in this

example are different from the ones in Example 5. In fact, the

redundant forces q of the truss without spring BCs in Example 5

and the one with the presence of spring BCs in this example are

calculated by Eqs. (41) and (40) and obtained, respectively, as

follows:

qex:5 ¼ f�2:6296, �34:2363, �20:0093, 3:2944,

8:9115, �11:0752, 16:7479gT ð43aÞ

qex:6 ¼ f�66:4920, 36:3577, 18:2829, 0:5853,

23:0793, �61:8201, �3:5831gT ð43bÞ

Tables 7 and 8 show the excellent agreements between the

member and reaction forces f , and all the nodal displacements d

Fig. 3. (a) The 25-bar statically indeterminate space truss; (b) the 25-bar statically

indeterminate space truss with the presence of elastic BCs (modeled via

spring BCs).

Table 6Nodal load components (kN) for the 25-bar statically indeterminate space truss.

Node 1 2 3 6

x-dir 80 60 30 30

y-dir 120 100 0 0

z-dir 30 30 0 0

Fig. 4. The equivalent free-standing model of the two systems in (a) and (b) with

dummy members 26–37 to remove the BCs.


obtained from the proposed method and those of SAP2000

program, respectively, which demonstrate the accuracy and

efficiency of the present method.

4.2.2. 392-bar statically indeterminate double layer grid

Example 7. As the last example, consider a 392-bar double layergrid (Fig. 5) in which all nodes at the top layer are subjected tovertical loads Pð ¼�6:786 kN) calculated according to ASCE-2006[32]. The structure has mixed statically indeterminacy of 65 inwhich 59 (ri ¼ 392�3� 113þ6¼ 59) are internally staticallyindeterminacy, and the remaining are externally generated. Theelastic BCs (modeled via three spring BCs) at nodes 110–113 havethe same flexibility of 444:8222�1 kN�1 in the x-, y- and z-directions as shown in Fig. 5. Similarly, by substituting these 12spring BCs with corresponding 12 novel zero-length dummymembers 393–404 , its equivalent free-standing model can beeasily obtained.

Excellent agreements between the member and reaction forces

f , and all the nodal displacements d obtained from the proposed

method and those of SAP2000 program can be observed in

Tables 9 and 10, respectively, which again demonstrate the

accuracy and efficiency of the present method. In general, it can

be seen from Tables 2, 4, 7 and 9 that in contrast with the

statically determinate trusses, the distribution of member and

reaction forces in the statically indeterminate trusses is consider-

ably effected by the flexibilities of the elastic BCs. This is because

of the fact that the flexibilities of the elastic BCs G2 play an

important role in determination of the redundant forces q as

shown in Eq. (40) for indeterminate trusses through the compat-

ibility conditions.

5. Concluding remarks

A structural analysis procedure using the force method is pro-posed for trusses accounting for the flexibilities of the elasticboundary conditions. Novel zero-length dummy members to trans-form the truss with boundary conditions into the equivalent free-standing model without boundary conditions are employed. Staticallydeterminate and all types of statically indeterminate trusses are

30

Fig. 5. The 392-bar statically indeterminate double layer grid (the hidden lines represent the struts at the bottom layer) with the presence of elastic BCs (modeled via

spring BCs): (a) perspective view and (b) top view.


considered. The general equilibrium equations and general kinematicrelations in which the reaction forces and the displacements at theelastic boundary conditions are, respectively, taken into account are

formulated. The compatibility equations in terms of the redundantforces in which the flexibilities of the elastic boundary conditions areconsidered are clearly derived from the minimum of complementary

Table 8Typical nodal displacements of (cm) the 25-bar statically indeterminate space truss.

d1x d3x d7x d1y d3y d7y d1z d3z d7z

Example 5

SAP2000 3.3518 0.4188 0 5.0988 0.2551 0 1.1627 �0.3375 0

Present 3.3518 0.4188 0 5.0988 0.2551 0 1.1627 �0.3375 0

Example 6

SAP2000 3.5963 �0.1951 �2.7321 5.5390 0.1010 �3.0771 1.2462 �0.5185 �0.7678

Present 3.5963 �0.1951 �2.7321 5.5390 0.1010 �3.0771 1.2462 �0.5185 �0.7677


Table 7Typical member and reaction forces (kN) of the 25-bar statically indeterminate space truss.

f1 f5 f13 f22 f26 f29 f30 f33 f34 f37

ð � r7xÞ ð � r10xÞ ð � r7yÞ ð � r10yÞ ð � r7zÞ ð � r10zÞ

Ex.5

SAP2000 �14.6540 123.8730 �28.9590 223.8840 29.6440 �151.7010 30.1410 �134.9830 6.2480 �206.2480

Present �14.6536 123.8735 �28.9587 223.8836 29.6441 �151.7010 30.1412 �134.9832 6.2480 �206.2480

Ex.6

SAP2000 �12.0740 119.7850 �35.2080 245.3770 12.1530 �152.8900 13.6880 �116.0510 3.4150 �203.4150

Present �12.0743 119.7850 �35.2082 245.3768 12.1529 �152.8896 13.6875 �116.0507 3.4151 �203.4151

Note: f26, f29, f30, f33, f34 and f37: Novel dummy member forces; Example 6 with the presence of elastic BCs (modeled via spring BCs).

Table 9Typical member and reaction forces (kN) of the 392-bar statically indeterminate double layer grid.

f4 f85 f113 f210 f311 f368 f393 f399 f404

ð � r110xÞ ð � r112yÞ ð � r113zÞ

Ex.7

SAP2000 66.8250 12.7180 �143.9830 2.7140 �138.4500 �35.4810 144.2660 �144.2660 83.1280

Present 66.8252 12.7175 �143.9828 2.7144 �138.4495 �35.4809 144.2659 �144.2659 83.1285

Note: f393, f399 and f404: Novel dummy member forces.

Table 10Typical nodal displacements of (cm) the 392-bar statically indeterminate double layer grid.

d31x d85x d110x d31y d85y d110y d31z d85z d110z

Example 7

SAP2000 �0.1372 0 �0.3243 0.2253 0 �0.3243 �17.1388 �20.7903 �0.1869

Present �0.1372 0 �0.3243 0.2253 0 �0.3243 �17.1388 �20.7903 �0.1869


energy. The difference between with and without considering theeffect of the elastic boundary conditions lies in the formulation of theredundant forces. Both member and reaction forces are simulta-neously and directly obtained. Then all nodal displacements are backcalculated from the member and reaction forces. The flexibilities ofthe elastic boundary conditions have considerable effects on thedistribution of member and reaction forces as well as the nodaldisplacements in the statically indeterminate trusses. The numericalexamples are shown to illustrate the efficiency and accuracy of theproposed method.

Acknowledgments

The support of the research reported here by Basic ScienceResearch Program through the National Research Foundation of Korea

(NRF) funded by the Ministry of Education, Science and Technology(2012R1A2A1A01007405) is gratefully acknowledged.

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force methods for trusses with elastic boundary conditions

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