Engineering Structures xxx (2012) xxx–xxx

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Engineering Structures

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

Beam model refinement and reduction

0141-0296/$ - see front matter � 2012 Elsevier Ltd. All rights reserved.http://dx.doi.org/10.1016/j.engstruct.2012.10.004

⇑ Corresponding author.E-mail address: senad_medic@yahoo.com (S. Medic).

Please cite this article in press as: Medic S et al. Beam model refinement and reduction. Eng Struct (2012), http://dx.doi.org/1j.engstruct.2012.10.004

Senad Medic a,⇑, Samir Dolarevic a, Adnan Ibrahimbegovic b

a Faculty of Civil Engineering, Patriotske lige 30, 71000 Sarajevo, Bosnia and Herzegovinab Ecole Normale Supérieure, Cachan, LMT, 61 avenue du president Wilson, 94235 Cachan, France

a r t i c l e i n f o

Article history:Available online xxxx

In memory of Prof. Ognjen Jokanovic

Keywords:Beam model reductionShear deformationHingesLength-invariance

a b s t r a c t

In this paper we present a method for systematic construction of the stiffness matrix of an arbitrary spatialframe element by performing a series of elementary transformations. The procedure of this kind is capableof including a number of element refinements (addition of shear deformation, variable cross-section, etc.)that are not easily accessible to standard displacement-based method. We also discuss the necessarymodifications of the element stiffness matrix in order to accommodate different constraints, such as pointconstraints in terms of joint releases (or hinges) for moments or shear forces. This is obtained by means ofmodel reduction providing a more effective approach than the alternative one in which the global numberof degrees of freedom has to be increased by one for each new release. Finally, we elaborate upon theglobal constraints imposing the length-invariant deformation of frame elements with an arbitrary positionin space. Several numerical examples are used to illustrate the performance of the proposed procedures.The computations are carried out by a modified version of computer code CAL.

� 2012 Elsevier Ltd. All rights reserved.

1. Introduction

The finite approximations, which is carried out by using thefinite element methods, translates the continuous system analysisinto discrete system analysis. In a mathematical sense, we switchfrom a set of differential equations to a set of algebraic equationswithin the framework of static analysis [1]. The basic advantageof such an approximation procedure is the possibility of creatinga complete discrete model as a set of its constituent (finite)elements no matter what the complexity of the structural systemis [2]. In this paper we briefly discuss the structural frame analysismethodology which pertains to stiffness assembly. The synthesis ofstiffness matrices is illustrated using the computer program CAL[3].

In this paper, the special attention is dedicated to 3D frameanalysis and enhancements of the basic beam element used formodelling. The construction of the stiffness matrix of an arbitrary(straight or curved) beam is derived by performing a series of ele-mentary transformations. Also, we present possible modificationsof the element stiffness matrix in the presence of joint releases(zero fields). The technique is described using simple 3D frameas an example. At the end, we discuss the implementation issuesrelevant for the technical displacement method which assumesaxially rigid rods.

2. Method of direct stiffness assembly

Static analysis of a discrete model which employs the assemblyprocedure inevitably reduces to solution of a system of linear alge-braic equations classically written as

Ku ¼ f ð1Þ

where K is the global stiffness matrix, u the displacement vectorand f the external force vector. In the method of stiffness assemblythe global stiffness matrix K is obtained as the assembly of the ele-ment stiffness matrices Ke as follows

K ¼X

e

LeT KeLe ð2Þ

Pe represents a symbolic summation of elementary stiffness matri-

ces Ke which are placed in the corresponding slots of the globalstiffness matrix K, as determined by the transformation of displace-ments between local and global coordinate systems specified by theeach element connectivity matrix Le (e.g. see [10,11]).

If local and global displacements are defined in the same coor-dinate system (usually the perpendicular Cartesian coordinate sys-tem), then the summation in (2) is carried out using the Booleanmatrices (with either unit or zero entries), i.e. only identifyingthe corresponding local and global displacements. In that casethe element stiffness matrix Ke must be defined in the global coor-dinate system, which is shown in the next section.

0.1016/

Fig. 2. Local degrees of freedom in the global coordinate system.

2 S. Medic et al. / Engineering Structures xxx (2012) xxx–xxx

3. Element stiffness matrix

The element stiffness matrix of a beam element can be derivedby employing the so-called Hermite polynomials for the real andvirtual displacement fields (e.g. see [10,11]), to obtain the weakform interpretation of equilibrium equations for each element thatcan formally be written as:

fp ¼ Kpp ð3Þ

where fp are the generalized forces, p1 to p12 are the generalized no-dal displacements shown in Fig. 1, and Kp is the corresponding stiff-ness matrix.

The results are recalled here for 3D case:

Kp ¼

EAl 0 0 0 0 0 �EA

l 0 0 0 0 012EIz

l30 0 0 6EIz

l20 �12EIz

l20 0 0 6EI

l2

12EIy

l30 �6EIy

l20 0 0 �12EIy

l30 �6EIy

l20

GJl 0 0 0 0 0 �GJ

l 0 04EIy

l 0 0 0 6EIy

l20 2EIy

l 04EIz

l 0 �6EIz

l20 0 0 2EIz

l

EAl 0 0 0 0 0

12EIz

l30 0 0 �6EIz

l2

sym: 12EIy

l30 6EIy

l20

GJl 0 0

4EIy

l 04EIz

l

0BBBBBBBBBBBBBBBBBBBBBBBBBBBB@

1CCCCCCCCCCCCCCCCCCCCCCCCCCCCA

ð4Þ

The polynomials of this kind are the exact solution to each elemen-tary load case where only one of the generalized displacements is ofthe unit value, whereas all the other ones remain equal to zero.Moreover, due to property of superconvergence (e.g. see [10,11])the generalized nodal displacements will also be exact for moregeneral load case, as long as the equivalent nodal loads are alsocomputed with these Hermite polynomials.

Furthermore, Eq. (3) can be transformed into global coordinatesystem with axes X, Y and Z (see Fig. 2). This is a usual elementtransformation from local to global coordinate system [7].

With that purpose, we first establish the relation between thelocal generalized displacements p1 to p12 defined in the local coor-dinate system x, y, z and the local generalized displacements q1 toq12 defined in the global coordinate system X, Y, Z as follows:

p ¼ Tpqq; qT ¼ hq1; . . . ; q12i ð5Þ

where Tpq is a 12 � 12 block diagonal matrix whose 3 � 3 blocksrepresent the inclinations of the local coordinate system with re-spect to the global one [7]. Thereafter, the principle of virtual dis-placements (shape-wise equivalent to the real displacement field)leads to

fq ¼ Kqq; Kq ¼ TTpqKpTpq ð6Þ

Since the element stiffness matrix Kq was derived for local degreesof freedom q in the global coordinate system X, Y, Z, it can directlybe assembled into the structural stiffness matrix K in Eq. (2),

Fig. 1. Set of degrees of freedom in local coordinate system of 2-node beamelement.

Please cite this article in press as: Medic S et al. Beam model refij.engstruct.2012.10.004

Kq � Ke ð7Þ

However, the result for element stiffness matrix in the previous sec-tion remains valid only for a beam with uniform geometrical/mechanical properties. Moreover, we have shown in [12] that evenaccounting for shear deformation would require non-conventionalinterpolations with the increase of computational cost due to addi-tional degrees of freedom.

4. Flexibility approach for accounting for variability of beamgeometric/mechanical properties along the beam and sheardeformation

We show here that the same element stiffness matrix can be ob-tained at neither extra cost nor additional degrees of freedom. Themain idea is to start with the reduced model pertaining to thedeformation space, which allows us to define the inverse of thecorresponding stiffness matrix accounting (exactly) for any varia-tion of mechanical/geometric beam properties and the shear defor-mation. The reduced deformation space matrix is then mappedinto the full 3D form by employing the corresponding transforma-tion between different sets of generalized displacements.

First, we define the natural local coordinate system for a given 3Dbeam (see Fig. 3) where x axis is directed along the beam and y and zaxes are perpendicular to it as well as mutually perpendicular.

The generalized displacements (see Fig. 3) are defined accordingto the following: v1 and v2 are rotations around y axis at the ele-ment start and end, v3 and v4 are the corresponding rotationsaround z axis, v5 is the rotation around x of the element end andfinally v6 represents the extension of the element end. The dis-placements v1 to v6 are six generalized (independent) beam dis-placements induced by deformation, which are separated fromthe six degrees of freedom corresponding to the rigid body modes.The rigid body displacements are restrained with the element sup-ports (see Fig. 3). The idea of distinguishing between generalizeddeformations and rigid body modes is similar to the one presentedin [4], which is given for large displacement and small deformationframe analysis, but the choice of deformation degrees of freedomin our case is different from the one in [4]. The relation between

Fig. 3. Local coordinate system of a beam with deformation degrees of freedom.

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S. Medic et al. / Engineering Structures xxx (2012) xxx–xxx 3

generalized displacements and the work-conjugate forces can bewritten as

fv ¼ Kvv ð8Þ

where

Kv ¼

4EIy

l2EIy

l 0 0 0 02EIy

l4EIy

l 0 0 0 0

0 0 4EIzl

2EIzl 0 0

0 0 2EIzl

4EIzl 0 0

0 0 0 0 GJl 0

0 0 0 0 0 EAl

0BBBBBBBBBB@

1CCCCCCCCCCAð9Þ

First four equations, for our choice of generalized displacements(deformations), directly stem from the equations of Takabeya (see[5]), whereas the 5th and 6th equation represent the well knowntorsional and axial beam stiffness.

The stiffness matrix Kv in Eq. (9) is given for a straight beamwith constant mechanical and geometrical properties. For a curvedbeam with variable properties Kv will generally be fully populatedmatrix. However, Eq. (8) can be derived for an arbitrarily curvedbeam in space as well, with the unknown displacements describedin Fig. 3. Geometrically exact relation of that sort (exact within theframework of infinitesimal displacements) can be constructed firstby finding the 6 � 6 flexibility matrix Fv and then inverting in orderto get

Kv ¼ F�1v ð10Þ

The computation of the entries of the flexibility matrix Fv for anarbitrary (straight or curved) spatial beam can be performed byemploying some established method, for example Mohr’s analogy(the method of conjugate structure, see [6]).

From Figs. 1 and 3 one can develop mapping between displace-ments v1 to v6 and displacements p1 to p12 in the following format

v ¼ Tvpp; vT ¼ hv1; . . . ;v6i; pT ¼ hp1; . . . ;p12i ð11Þ

where the transformation matrix Tvp is given with

Tvp ¼

0 0 � 1l 0 1 0 0 0 1

l 0 0 00 0 � 1

l 0 0 0 0 0 1l 0 1 0

0 1l 0 0 0 1 0 � 1

l 0 0 0 00 1

l 0 0 0 0 0 � 1l 0 0 0 1

0 0 0 �1 0 0 0 0 0 1 0 0�1 0 0 0 0 0 1 0 0 0 0 0

0BBBBBBBB@

1CCCCCCCCAð12Þ

If we apply the principle of virtual displacements (for example see[7]) on the previously shown beam, constructing virtual displace-ments (denoted with hat) in the same fashion as real displacements(Figs. 1 and 3), then we have

pT fp ¼ vT fv ð13Þ

where fp is a set of generalized forces which are work-conjugate togeneralized displacements described in Fig. 1. Consecutivelyexploiting Eqs. (11) and (8) and again (11) we wind up with

pT fp � TTvpKvTvpp

� �¼ 0 ð14Þ

Since virtual displacements are mutually independent (which isequivalent to the fundamental lemma of variational calculus in dis-crete formulation [8]), Eq. (14) results with

fp ¼ Kpp; Kp ¼ TTvpKvTvp ð15Þ

Please cite this article in press as: Medic S et al. Beam model refij.engstruct.2012.10.004

In order to include shear deformation for Euler–Bernoulli beam,one can exploit the possibilities of flexibility approach, namely Eq.(10). For a 3D beam element with deformation degrees of freedomshown in Fig. 3 we can evaluate the flexibility matrix coefficientsincluding displacements due to shear and obtain:

Fv ¼

l3EIyþ 1

kzGAl � l6EIyþ 1

kzGAl 0 0 0 0

� l6EIyþ 1

kzGAll

3EIyþ 1

kzGAl 0 0 0 0

0 0 l3EIzþ 1

kyGAl � l6EIzþ 1

kyGAl 0 0

0 0 � l6EIzþ 1

kyGAll

3EIzþ 1

kyGAl 0 0

0 0 0 0 lGJ 0

0 0 0 0 0 lEA

0BBBBBBBBBBB@

1CCCCCCCCCCCAð16Þ

After inverting we get Kv that includes shear deformability. Finally,exploiting Eqs. (12) and (15) we finally determine the element stiff-ness matrix Kp which is in case of horizontal beam equal to Ke:

Ke ¼

Kx 0 0 0 0 0 �Kx 0 0 0 0 0Ky1 0 0 0 Ky2 0 �Ky1 0 0 0 Ky2

Kz1 0 �Kz2 0 0 0 �Kz1 0 �Kz2 0T 0 0 0 0 0 �T 0 0

Kz3 0 0 0 Kz2 0 Kz4 0Ky3 0 �Ky2 0 0 0 Ky4

Kx 0 0 0 0 0Ky1 0 0 0 �Ky2

sym: Kz1 0 Kz2 0T 0 0

Kz3 0Ky3

0BBBBBBBBBBBBBBBBBBBBBBBB@

1CCCCCCCCCCCCCCCCCCCCCCCCAð17Þ

where

Kx ¼ EAl Ky1 ¼ 12EIz

ð1þ/yÞl3 Ky2 ¼ 6EIz

ð1þ/yÞl2

Ky3 ¼ ð1þ0:25/yÞ4EIz

ð1þ/yÞlKy4 ¼ ð1�0:5/yÞ4EIz

ð1þ/yÞl/y ¼ 12EIz

kyGAl2

T ¼ GJl Kz1 ¼ 12EIy

ð1þ/zÞl3� � �

ð18Þ

The remaining coefficients are easily obtained by subscript inter-change. We mention that kyA is the effective shear for transverseshear deformation in y direction (analogous for direction z).

With this we finalize the development of the element stiffnessmatrix Ke through a series of elementary transformations whichenhance the stiffness matrix of the basic element (the one withall rigid body modes restrained). This facilitates the constructionof Ke matrix as well as the understanding of the principles of thatconstruction. Additionally, this approach enables us to form a stiff-ness matrix for a special beam element that has one or more jointreleases (zero fields). Before we elaborate on such a possibility, wediscuss the inclusion of shear strain in different Timoshenko beamelements in the next section.

5. Timoshenko beam element for including shear strain

As opposed to adding flexibility on conventional Euler–Ber-noulli beam, one can introduce independent interpolation for no-dal rotations which produces Timoshenko beam formulation. Thebasic assumptions are that the sections still remain plane, but nolonger normal to the neutral axis after bending. In this way it ispossible to take into account the shear deformation (and stress)necessary for modelling thick beams. Following the kinematichypotheses u(x) = �yh(x) and h(x) = dv(x)/dx � c(x), where c(x) rep-resents constant transverse shear strain, we interpolate deflections

nement and reduction. Eng Struct (2012), http://dx.doi.org/10.1016/

4 S. Medic et al. / Engineering Structures xxx (2012) xxx–xxx

and rotations with linear shape functions (so called equal-orderinterpolation) expressed as:

vhðnÞ ¼ N1ðnÞv1 þ N2ðnÞv2

hhðnÞ ¼ N1ðnÞh1 þ N2ðnÞh2

ð19Þ

The element stiffness matrix Ke and the load vector fe correspond-ing to uniform loading obtained by standard displacement based fi-nite element procedure using two Gauss points (in this caseequivalent to analytically integrated results) have the followingform:

Ke ¼

kGAl

kGA2 � kGA

lkGA

2kGA

2EIl þ kGAl

3 � kGA2

kGAl6 � EI

l

� kGAl � kGA

2kGA

l � kGA2

kGA2

kGAl6 � EI

l � kGA2

EIl þ kGAl

3

0BBBB@1CCCCA

fe ¼ ql2 0 ql

2 0� �T

ð20Þ

It can be shown that this element is very stiff and that it yields poorresults (shear locking) when the beam becomes slender. Further-more, there are no forces on the nodal rotations which additionallydisturbs the solution quality. Thus, the formulation is not generallyapplicable. This problem can be overcome in several ways and thefirst remedy is selective reduced integration. If we integrate thebending stiffness terms using appropriate order Gaussian quadra-ture, whereas the shear stiffness term is under-integrated by evalu-ating the shear strain at the element midpoint, we wind up with thefollowing stiffness matrix which does not exhibit locking:

Ke ¼

kGAl

kGA2 � kGA

lkGA

2kGA

2EIl þ kGAl

4 � kGA2

kGAl4 � EI

l

� kGAl � kGA

2kGA

l � kGA2

kGA2

kGAl4 � EI

l � kGA2

EIl þ kGAl

4

0BBBB@1CCCCA ð21Þ

The other possibilities pertain to enhancing the displacement fieldwith additional terms for example:

vhðnÞ ¼ N1ðnÞv1 þ N2ðnÞv2 þ N3ðnÞDvhhðnÞ ¼ N1ðnÞh1 þ N2ðnÞh2 ð22Þ

The interpolation functions are Lagrangian polynomials superim-posed hierarchically, hence the quadratic N3(n) = 1 � n2 is addedto linear N1ðnÞ ¼ 1

2 ð1� nÞ and N2ðnÞ ¼ 12 ð1þ nÞ. N3 corresponds to

transverse displacement of the middle node, invisible to neighbour-ing elements which enables us to perform static condensation andagain obtain 4 � 4 element stiffness matrix Ke. First we determinethe coefficient of the additional term, which is chosen such that iteliminates the linear term in shear strain responsible for shear lock-ing. Thus, we have

vhðnÞ ¼ N1ðnÞv1 þ N2ðnÞv2 þ N3ðnÞl8ðh1 � h2Þ ð23Þ

Plugging the assumed solution in the weak form of equilibriumequations, we get the following element equations:

kGAl

kGA2 � kGA

lkGA

2kGA

2EIl þ kGAl

4 � kGA2

kGAl4 � EI

l

� kGAl � kGA

2kGA

l � kGA2

kGA2

kGAl4 � EI

l � kGA2

EIl þ kGAl

4

0BBBB@1CCCCA

v1

h1

v2

h2

0BBB@1CCCA ¼

ql2

ql2

12ql2

� ql2

12

0BBBBB@

1CCCCCA ð24Þ

The stiffness matrix is identical to the one obtained with reducedintegration, however here we have employed exact integration. Inaddition, the equivalent load vector gives forces on the nodal rota-tions, as was the case with conventional beam element.

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Finally, we present the element that gives exact solutions of thegoverning differential equations for uniform Timoshenko beamwith no distributed loads (so called linked interpolation [10]). Itis developed using the standard displacement based formulationby enhancing the displacements in the following manner:

vhðnÞ ¼ N1ðnÞv1 þ N2ðnÞv2 þ N3ðnÞl8ðh1 � h2Þ þ N4ðnÞalDh

hhðnÞ ¼ N1ðnÞh1 þ N2ðnÞh2 þ N3ðnÞDhð25Þ

where N4(n) = n(1 � n2) and Dh represents the rotation of the ele-ment middle node. The coefficient a ¼ 1

6 is determined from thecondition that the shear strain is constant within the element.Now we can easily write the relationship between strains and dis-placements connected through B matrix as

jc

� �¼

0 � 1l 0 1

l �4 nl

� 1l � 1

21l � 1

2 � 23

! v1

h1

v2

h2

Dh

0BBBBBB@

1CCCCCCA ð26Þ

Using the material elastic property array D

D ¼EI 00 kGA

� �ð27Þ

and the jacobian j ¼ l2 derived from x(n) = N1(n)x1 + N2(n)x2, we can

evaluate the element stiffness matrix Ke and the equivalent loadvector f e for constant q

Ke ¼Z þ1

�1BT DBjdn

fe ¼Z þ1

�1NT qjdn

ð28Þ

where

N ¼ 12 ð1� nÞ l

8 ð1� n2Þ 12 ð1þ nÞ � l

8 ð1� n2Þ l6 nð1� n2Þ

� �ð29Þ

and obtain the following results

Ke ¼

kGAl

kGA2 � kGA

lkGA

22kGA

3kGA

2kGAl

4 þ EIl � kGA

2 � kGAl4 � EI

lkGAl

3

� kGAl � kGA

2kGA

l � kGA2 � 2kGA

3kGA

2kGAl

4 � EIl � kGA

2kGAl

4 þ EIl

kGAl3

2kGA3

kGAl3 � 2kGA

3kGAl

34kGAl

9 þ 16EI3l

0BBBBBBB@

1CCCCCCCAfe ¼ ql

2ql2

12ql2 � ql2

12 0� �T

ð30Þ

The load vector is identical to the load vector of the Euler–Bernoullitheory, except for the last 0. Since Dh is inner degree of freedomassociated with each element individually, we can reduce the stiff-ness matrix to 4 � 4 by exploiting the static condensation algorithmpresented in the next chapter. In this case, rz is equal to Dh. With /previously defined, the element stiffness matrix can finally be sim-plified to the same result already shown in Eq. (17), assuming 2Dcase and deliberately omitting axial terms.

Ke ¼

12EIð1þ/Þl3

6EIð1þ/Þl2

� 12EIð1þ/Þl3

6EIð1þ/Þl2

6EIð1þ/Þl2

ð1þ0:25/Þ4EIð1þ/Þl � 6EI

ð1þ/Þl2ð1�0:5/Þ2EIð1þ/Þl

� 12EIð1þ/Þl3

� 6EIð1þ/Þl2

12EIð1þ/Þl3

� 6EIð1þ/Þl2

6EIð1þ/Þl2

ð1�0:5/Þ2EIð1þ/Þl � 6EI

ð1þ/Þl2ð1þ0:25/Þ4EIð1þ/Þl

0BBBBBB@

1CCCCCCA ð31Þ

As the beam becomes ‘thin’ (the shear stiffness increases), the ratio/ ? 0 yielding the element stiffness matrix exactly the same as that

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S. Medic et al. / Engineering Structures xxx (2012) xxx–xxx 5

for conventional beam element. The element will not lock, thus it isthe most effective Timoshenko beam element among the formula-tions considered in this paper.

6. Beam element stiffness with joint releases

If we construct the beam element shown in Fig. 2 in a way thatit has a spatial hinge (which precludes the existence of all threebending moment components), then Eq. (6) can be written as

fq;n

0

� �¼

Kenn Ke

nz

Kezn Ke

zz

!qn

qz

� �ð32Þ

where the vector qTn ¼ hq1; . . . ; q9i represents the local independent

beam displacements (which also defines fq,n as conjugate forces)whereas qT

z ¼ hq10; q11; q12i is the vector of dependent generalizeddisplacements that are to be eliminated. The procedure for elimina-tion of qz is identical to static condensation [7,9].

fq;n ¼ bKennqn;

bKenn ¼ Ke

nn � KenzKe�1

zz Kezn ð33Þ

and can be implemented as a partial decomposition of Ke matrixemploying Gaussian elimination, which is described in [9] (see Sec-tion 8).

The proposed method to modify the element stiffness matrix inthe presence of joint releases can be applied to any of the three ver-sions of the stiffness matrix discussed in the previous section, i.e.Kv, Kp or Kq. For example, the joint release described by a spatialhinge has the completely identical effect on the modification ofKp matrix, as the one given in Eq. (19) for Kq (since the total rota-tion vector is decomposed into another component set). For a jointrelease of the local hinge type at the element end, one needs tomodify Kv matrix by eliminating degrees of freedom v2 (hingearound the local y axis) and v4 (hinge around the local z axis) fromEq. (8).

For a joint release pertaining to shear force at an element end,which modifies the matrix Kp, it is necessary to eliminate the gen-eralized displacements p8 and p9 from Eq. (15). Following the sameprinciple, in the matrix Kp we will obtain the joint release for axialforce by eliminating the displacement p7, whereas we will imposethe joint release for torsional moment by eliminating the displace-ment p10.

The structural analysis of frames with joint releases can againbe performed using the previously described assembly procedure,however we have to employ the element stiffness matrices modi-fied for the presence of joint releases. Symbolically it can bewritten

bK ¼Xe

bKe ð34Þ

With this procedure, we have eliminated the dependent displace-ments at the local level, i.e. at the element (beam) level, and weconsider this to be more effective analysis approach for larger struc-tural systems.

The alternative to the previous method for analysis of frameswith joint releases implies the elimination of dependent general-ized displacements at the global level, i.e. the modification of thestructural matrix K. Namely, the presence of release imposes dif-ferent generalized displacements on one and the other side ofthe considered node. It is understood that this pertains only tothe generalized displacements influenced by the release, for exam-ple two sets of rotations for the spatial hinge connection (seeexample). This approach considers every beam element as stan-dard (without the presence of joint release) and the assembly isconducted according to Eq. (2). If we denote the basic generalizeddisplacements of the joints with rn, and the additional set of

Please cite this article in press as: Medic S et al. Beam model refij.engstruct.2012.10.004

generalized displacements stemming from presence of joint re-leases with rz, then the system equilibrium equation can be written

fr

0

� �¼

Knn Knz

Kzn Kzz

� �rn

rz

� �ð35Þ

where fr represents the vector of generalized nodal forces. The pro-cedure of static condensation can now be applied on the entiresystem

fr ¼ bKnnrn; bKnn ¼ Knn � KnzK�1zz Kzn ð36Þ

The global stiffness matrix bKnn is identical to the global stiffnessmatrix in Eq. (34) obtained by the assembly procedure of the mod-ified stiffness matrices, hence the resulting displacements are thesame. The condensation of the stiffness matrix at the global levelis achieved in the triangular decomposition phase of the Gausselimination method. Since this is the most costly part, the gain isnegligible compared to solving system of algebraic equations with-out any reduction (see Section 8 for illustration of these ideas).

7. Reduction method for invariant length constraint

The displacement method (DM) for frame analysis has pushedout numerous iterative methods from engineering practise. Withthe increase of computational power the solution of the algebraicsystem (with the typical size for frame structures) was not a majorproblem, which led to the displacement based frame modelsimplemented in a number of computer programs. In planar DM

analysis the degrees of freedom (DOF) for each node are two trans-lations and one rotation.

However, this is no longer the case in technical displacementmethod (TDM) which assumes axially rigid rods, and thus reducesthe number of independent displacements. The TDM remains pop-ular in the structural mechanics courses, because of its simplicityfor analysis by hand. Moreover, this kind of method can also im-prove the system conditioning if the axial deformation is signifi-cantly smaller from the flexural one, where one should ratherexclude it completely. This is the main motivation for further elab-orating upon the appropriate implementation of such reducedmodel within the standard framework of the finite elementsmethod.

The assembly of the global stiffness matrix K in TDM for frameswith orthogonal beams can easily be implemented reducing thetransformation matrix T (6 � 6) to identity (4 � 4) since the un-known translations are either vertical or horizontal (see Eq. (24)).

h1

h2

v1

v2

0BBB@1CCCA ¼

1 0 0 0 0 00 1 0 0 0 00 0 � sina cos a 0 00 0 0 0 � sin a cos a

0BBB@1CCCA

h1

h2

u1

v1

u2

v2

0BBBBBBBB@

1CCCCCCCCAð37Þ

However, the solution of frames with inclined members is morecomplicated. We want to generalize the stiffness matrix computa-tion based on the assembly procedure as follows. Let us considera system with n nodes, e elements, r supports and c unknown rota-tions. The starting point is the well-known equation for assembly ofthe global stiffness matrix:

K ¼X

e

LeT TeT KeTeLe ð38Þ

The system to be solved in displacement method Ku = f has there-fore the following unknowns:

u ¼ h �uð ÞT ð39Þ

nement and reduction. Eng Struct (2012), http://dx.doi.org/10.1016/

Fig. 4. Timoshenko beam – geometrical and mechanical properties with loading.

0 1 2 3 4 5 6 7 8 9 100

0.2

0.4

0.6

0.8

1

1.2

Slenderness ratio l/h

Rel

ativ

e de

flect

ion

v/v ex

act

Shear locking for different interpolations

Equal−orderEnhancedLinkedEuler−BernoulliExact

Fig. 5. Locking problem for different Timoshenko beam formulations.

6 S. Medic et al. / Engineering Structures xxx (2012) xxx–xxx

where u(1 �m) comprises unknown rotations h and translations offree nodes �uð1� 2n� rÞ. For technical displacement method Ke are(4 � 4) matrices, Te are (4 � 6) transformation matrices and Le are(6 �m) location matrices, where

m ¼ c þ 2n� r ð40Þ

Due to the assumption of axial rigidity, the total number of un-known translations ntransl is reduced to:

ntransl ¼ 2n� e� r ð41Þ

Now we can write the global equilibrium condition and unknowndisplacements:

KTDMuTDM ¼ fTDM uTDM ¼ h Dð ÞT ð42Þ

where D comprises unknown translations. Exploiting the basic the-orem of rigid body kinematics that the projections of nodal dis-placements on the bar axis have to be the same, we obtain ehomogeneous linear equations with 2n � r unknown projectionson the global axes:

W�u ¼ 0 ð43Þ

The previous equation can be written as:

W1 W2ð Þ u1 Dð ÞT ¼ 0 ð44Þ

where u1 is the vector of displacements to be eliminated through D.Elements of D are chosen such that the matrix W1 is regular. Thefirst element of vector D will be set to 1 and all others to zero.The solution of these equations represents the first column of ma-trix G which relates �u and D:

GD ¼ �u ð45Þ

Other columns of G will be obtained by successively setting nextelements of D to 1. Assembly procedure for the global stiffness ma-trix for the technical displacement method is now defined by:

KTDM ¼ GTX

e

LeT TeT KeTeLe

!G ¼ GT KG ð46Þ

where

G ¼I 00 G

� �uDM ¼ GuTDM ð47Þ

Nodal force vector is defined by:

fTDM ¼ GT f ð48Þ

The internal nodal forces for each element can be determined afterthe corresponding reduction is applied:

fe ¼ KeTeLeGuTDM ð49Þ

8. Numerical examples

8.1. Shear locking in Timoshenko beam element

As an example to illustrate shear locking phenomenon perti-nent to Timoshenko beam element, we consider a 2D beam of var-iable length clamped at one end and free at the other (see Fig. 4).We will compare the calculated vertical displacements for differentelements with the exact deflection of the cantilever tip equal to

vexact ¼ vEB þ v shear ¼Fl3

3EIþ Fl

kGAð50Þ

where vEB is the displacement portion corresponding to the bendingof Euler–Bernoulli beam. The analyzed beam is discretized usingone element for each case.

Please cite this article in press as: Medic S et al. Beam model refij.engstruct.2012.10.004

The evaluated results are displayed in Fig. 5. From the diagramone can see that the element with linear shape functions (equal-or-der interpolation) yields acceptable results only for very deepstructural elements. Otherwise, the element exhibits shear locking,being unable to adequately represent the deformation pattern inbending dominated problems. The displacement field enhancedwith additional terms only partially solves the problem, resultingwith cca. 20% stiffer response for slender elements. For compari-son, Euler–Bernoulli beam element captures the relevant displace-ment in problems dominated by bending strain, however predictspoor results in shear dominated tasks. Finally, the beam elementwith linked interpolation successfully combines Euler–Bernoullimodel for slender elements with the advantages of Timoshenkomodel for deep elements yielding accurate displacements regard-less of l/h ratio.

8.2. Joint releases

In this section we present the results of several numerical sim-ulations, which were carried out by the elements described hereinand implemented in the computer program CAL [3]. In order tobetter understand the procedure, the list of used CAL commandsis briefly explained in Appendix A. We analyze a simple 3D frame(see Fig. 6 for its properties), first for the case of moment release(hinge) placed at the elements’ connection, second for the case ofshear force release at the same location. Then we consider a 3Dframe with larger number of degrees of freedom. Before advancingwith the examples, we recall the number of operations needed forsolving a system of algebraic equations by Gauss elimination (see[11]). Triangular decomposition is computationally the mostexpensive phase, dictating the total cost of the method to beapproximately 2

3 n3 for large n (exactly 23 ðn3 � nÞ þ 5

2 nðn� 1Þ þ n).Also, the condensation of the stiffness matrix of order n to order(n � k) is carried out by k steps of LU decomposition. If one element

nement and reduction. Eng Struct (2012), http://dx.doi.org/10.1016/

Fig. 6. Spatial frame: geometrical and mechanical properties with loading.

(a) (b)Fig. 7. Discrete frame models with moment releases: global degrees of freedom.

S. Medic et al. / Engineering Structures xxx (2012) xxx–xxx 7

has s releases, we have taken it into account as s elements having 1release. We remark that forward reduction and back substitutionphases are quadratic functions of number of unknowns n.

8.2.1. Joint release for momentsIn the case of joint release for moments, two analyses were con-

ducted: first in which the components of the rotation vector leftand right from the hinge in the given node were treated as inde-pendent degrees of freedom, and second in which we have modi-fied the element stiffness matrix for the presence of joint releasein one of the beams. In the first analysis, the global displacements7–9 were eliminated at the structural level. For both models, theglobal displacements are marked in Fig. 7. In accordance with theprevious explanations, the total number of operations required toperform the first analysis (global reduction) equals 540. On theother hand, the element reduction costs 980 operations which ismuch larger than the global reduction as well as the solution ofthe system without any reduction (equal to 669 exactly). We canconclude that both the element and the global reduction do notplay significant role for small structural systems.

Next, CAL commands for solving the problem in Fig. 7a aregiven.

Pj.

C FIGURE 7A-FRAME WITH

MOMENT RELEASES

lease cite this article in pressengstruct.2012.10.004

ADDK KK K1 ID N = 1

ADDK KK K2 ID N = 2

C NODAL COORDINATES

C CONDENSATION OF GLOBAL

STIFFNESS MATRIX

LOAD XYZ R = 3 C = 3

LOADI IDKT R = 1 C = 9

10 0 0

4 5 6 7 8 9 1 2 3

10 10 0

TRAN IDKT IDK

0 10 0

ZERO KKP R = 9 C = 9

C FORM ELEMENT

STIFFNESS MATRICES

ADDK KKP KK IDK N = 1

C (Ki(12 � 12))

ZERO A R = 9 C = 1

FRAME3 K1 T1 E = 30000

A = 0.16 J = 0.001

SOLVE KKP A EQ = 3

as: Medic S et al. Beam model refine

I = 0.003,0.003 G = 12000

N = 1,2 P = 1,0

ment and reduction. Eng Struc

DUPSM KKP KKC R = 6 C =6 L =

4,4

FRAME3 K2 T2 E = 30000 A

= 0.16 J = 0.001

C LOAD VECTOR

I = 0.003, 0.003 G = 12000

N = 2,3 P = 1,0

LOAD RT R = 1 C = 6

C LOCAL - GLOBAL

DISPLACEMENTS

30 20 �10 0 0 0

LOADI IDT R = 2 C = 12

TRAN RT R

0 0 0 0 0 0 1 2 3 4 5 6

C SOLVE SYSTEM OF

EQUILIBRIUM EQUATIONS

1 2 3 7 8 9 0 0 0 0 0 0

SOLVE KKC R

TRAN IDT ID

C DISPLACEMENTS

C ASSEMBLE GLOBAL

STIFFNESS MATRIX

P R

ZERO KK R = 9 C = 9

RETURN

Now we display CAL code for solving the problem in Fig. 7b.

C FIGURE 7B-FRAME WITH

MOMENT RELEASES

t

ADDK K2P K2 IDK N = 1

SOLVE K2P A EQ = 3

C NODAL COORDINATES

C LOCAL–GLOBAL

DISPLACEMENTS

LOAD XYZ R = 3 C = 3

LOADI IDT R = 2 C = 12

10 0 0

0 0 0 0 0 0 1 2 3 4 5 6

10 10 0

0 0 0 1 2 3 0 0 0 0 0 0

0 10 0

TRAN IDT ID

C FORM ELEMENT

STIFFNESS MATRICES

C ASSEMBLE GLOBAL

STIFFNESS MATRIX

C (Ki(12 � 12))

ZERO KK R = 6 C = 6

FRAME3 K1 T1 E = 30000 A

= 0.16 J = 0.001

ADDK KK K1 ID N = 1

I = 0.003,0.003 G = 12000

N = 1,2 P = 1,0

ADDK KK K2P ID N = 2

FRAME3 K2 T2 E = 30000 A =

0.16 J = 0.001

C LOAD VECTOR

I = 0.003,0.003 G = 12000

N = 2,3 P = 1,0

LOAD RT R = 1 C = 6

C STATIC CONDENSATION

FOR

30 20 �10 0 0 0

C BEAM ELEMENT 2

TRAN RT R

LOADI IDKT R = 1 C = 12

C SOLVE SYSTEM OF

EQUILIBRIUM EQUATIONS

4 5 6 1 2 3 7 8 9 10 11 12

SOLVE KK R

TRAN IDKT IDK

C DISPLACEMENTS

ZERO A R = 12 C = 1

P R

ZERO K2P R = 12 C = 12

RETURN

(2012), http://dx.doi.org/10.1016/

8 S. Medic et al. / Engineering Structures xxx (2012) xxx–xxx

8.2.2. Joint release for shear force

Very similar to the previous procedure is the one that we imple-

ment in the presence of joint release (zero field) for shear force atthe elements’ connection (see Fig. 8).

In this case, one model is created such that the transverse dis-placement components are independent considering sections leftand right from the shear force joint release. Displacement compo-nents 7 and 8 are eliminated using static condensation algorithmat the global stiffness matrix level. The second model, however,employs the special beam element which is enhanced for the pres-ence of the shear force release. The listing of CAL commands imple-mented in the solution process of the problem described in Fig. 8ais given next:

Pj.

C FIGURE 8A - FRAME

WITH SHEAR FORCE

lease cite this article inengstruct.2012.10.004

ADDK KK K1 ID N = 1

C RELEASE

ADDK KK K2 ID N = 2

C NODAL

COORDINATES

C CONDENSATION OF

GLOBAL STIFFNESS

MATRIX

LOAD XYZ R = 3 C = 3

LOADI IDKT R = 1 C = 8

10 0 0

3 4 5 6 7 8 1 2

10 10 0

TRAN IDKT IDK

0 10 0

ZERO KKP R = 8 C = 8

C FORM ELEMENT

STIFFNESS

MATRICES

ADDK KKP KK IDK N = 1

C (Ki(12 � 12))

ZERO A R = 8 C = 1

FRAME3 K1 T1 E =

30000 A = 0.16 J =

0.001

SOLVE KKP A EQ = 2

I = 0.003,0.003

G = 12000 N = 1,2

P = 1,0

DUPSM KKP KKC R = 6 C =

6 L = 3,3

FRAME3 K2 T2

E = 30000 A = 0.16

J = 0.001

C LOAD VECTOR

I = 0.003,0.003

G = 12000 N = 2,3

P = 1,0

press as: Medic S et al.

LOAD RT

R = 1

C = 6

(a) (b)

C LOCAL - GLOBAL

DISPLACEMENTS

30 20 -10 0 0 0

Fig. 8. Frame model with shear force releases: global degrees of freedom.

LOADI IDT R = 2 C = 12 TRAN RT R

0 0 0 0 0 0 1 2 3 4 5

6

C SOLVE SYSTEM OF

EQUILIBRIUM

EQUATIONS

7 1 8 4 5 6 0 0 0 0 0

0

SOLVE KKC R

TRAN IDT ID

C DISPLACEMENTS

C ASSEMBLE GLOBAL

STIFFNESS

MATRIX

P R

ZERO KK R = 8 C = 8

RETURN

Fig. 9. 3 D frame with 288 DOFs without reduction.

As opposed to the previous, the listing of CAL commands usedto solve the problem in Fig. 8b is

Beam model refine

C FIGURE 8B - FRAME WITH

SHEAR FORCE

ment and reduction. Eng Struct

I = 0.003,0.003 G = 12000

N = 1,2 P = 1,0

C RELEASE

FRAME3 K2 T2 E = 30000

A = 0.16 J = 0.001

C NODAL COORDINATES

I = 0.003,0.003 G = 12000

N = 2,3 P = 1,0

LOAD XYZ R = 3 C = 3

C STATIC CONDENSATION

10 0 0

LOADI IDKT R = 1 C = 12

10 10 0

2 3 1 4 5 6 7 8 9 10 11 12

0 10 0

TRAN IDKT IDK

C FORM ELEMENT

STIFFNESS MATRICES

ZERO A R = 12 C = 1

C (Ki(12 � 12))

ZERO K2P R = 12 C = 12

FRAME3 K1 T1 E = 30000

A = 0.16 J = 0.001

ADDK K2P K2 IDK N = 1

SOLVE K2P A EQ = 2

LOAD RT R = 1 C = 6

C LOCAL - GLOBAL

DISPLACEMENTS

C LOAD VECTOR

LOADI IDT R = 2 C = 12

30 20 �10 0 0 0

0 0 1 4 5 6 0 0 0 0 0 0

TRAN RT R

0 0 0 0 0 0 1 2 3 4 5 6

C SOLVE SYSTEM OF

EQUILIBRIUM EQUATIONS

TRAN IDT ID

SOLVE KK R

C ASSEMBLE GLOBAL

STIFFNESS MATRIX

C DISPLACEMENTS

ZERO KK R = 6 C = 6

P R

ADDK KK K1 ID N = 1

RETURN

ADDK KK K2P ID N = 2

The analysis output is given in Appendix B. One can see that for

both cases of joint releases (moment and shear force) the corre-sponding displacement components calculated from model a andmodel b are identical.

8.2.3. 3D frameIn the next example we analyze a 3D frame from the aspect of

number of necessary operations to solve the system, hence thegeometrical and mechanical properties are of no interest. Theframe consists of continuous beams interconnected with beams

(2012), http://dx.doi.org/10.1016/

S. Medic et al. / Engineering Structures xxx (2012) xxx–xxx 9

that have spatial hinges at both ends (see Fig. 9) and has a muchhigher number of degrees of freedom compared to the previousexample. Here we have the total number of unknown displace-ments equal to 288, which, however, can be reduced to 144. Thecost of solving the system without any reduction equals approxi-mately 16 � 106, with the cost of global reduction slightly below.However, the element reduction costs only �2 � 106 operationsand the advantage is obvious.

Fig. 10. Geometry and loads (left) an

Fig. 11. Displacement scheme 1

Fig. 12. Diagrams of bending momen

Please cite this article in press as: Medic S et al. Beam model refij.engstruct.2012.10.004

8.3. Global reduction to length-invariant beam model

As an example we look at the frame shown in Fig. 10 with thegeometry and loads on the left and restrained independent trans-lations on the right. Young’s modulus is E = 3 � 107 kN/m2. Conse-quently, the displacement schemes are displayed in Fig. 11.

In contrast to the unknown displacements vector for displace-ment method (DM)

d restrained translations (right).

(left) and scheme 2 (right).

t (left) and normal force (right).

nement and reduction. Eng Struct (2012), http://dx.doi.org/10.1016/

10 S. Medic et al. / Engineering Structures xxx (2012) xxx–xxx

uDM ¼ h3 h4 h5 h6 u3 v3 u4 v4 u5 v5 u6 v6ð ÞT

ð51Þ

exploiting the axial rigidity constraint in technical displacementmethod (TDM) we wind up with the algebraic system with 6 un-knowns that correspond to vector

uTDM ¼ h3 h4 h5 h6 u4 v6ð ÞT ð52Þ

Following the previously described procedure, some characteristicarrays for the analyzed example are given below:

u1 ¼ u3 v3 v4 u5 v5 v6ð ÞT

W ¼

1 0 �1 0 0 0 0 00 1 0 0 0 0 0 00 0 0 1 0 0 0 00 0 0 1 0 0 0 �10 0 0 0 1 0 �1 0

0:8 0:6 0 0 �0:8 �0:6 0 0

0BBBBBBBB@

1CCCCCCCCAD1 ¼ 1 0ð ÞT D2 ¼ 0 1ð ÞT

GT ¼1 0 1 0 0 4=3 0 00 0 0 0 1 �4=3 1 0

� �

ð53Þ

In Fig. 12 we show the calculated bending moment and normalforce diagrams.

9. Concluding remarks

In this work we have addressed several issues pertinent to con-structing model refinements in materially and geometrically linearstructural analysis of spatial frames. First we derived the elementstiffness matrix starting from the reduced model pertaining tothe deformation space, which allows to define the correspondingflexibility matrix, the inverse of the stiffness matrix. The proposedapproach enables us to easily account for shear deformation in amore direct manner than the standard alternative based uponthe displacement-based Timoshenko beam element, which re-quires non-conventional interpolations and the increase of compu-tational cost (with additional degrees of freedom) to deliver thesame quality of results. The merit of the proposed flexibility ap-proach can further become clear when deriving the stiffness matri-ces of beam elements with varying cross section or Young’smodulus, where standard cubic interpolation functions would pro-duce rather poor results. However, the flexibility approach is basedon the principle of virtual forces which assumes that there existsequilibrium between external and internal set of infinitesimalforces before the application of real loads/displacements. In thecase of nonlinear analysis this principle cannot be used becausethe linear relationship between external and internal forces doesnot hold after the application of real loads. On the other hand, Tim-oshenko elements can be extended for material nonlinearity, butthe additive decomposition of total rotations is not in accord withthe multiplicative nature of large three-dimensional rotations (forpossible extensions to geometrical nonlinearity see [13,14]). Wehave also addressed the corresponding model reduction issueswith the element stiffness matrix in the presence of joint releases(zero fields) for moments and shear force. This is an alternative ap-proach to doubling the corresponding nodal degrees of freedom forevery release and it is applicable for nonlinear problems. The num-ber of independent displacements is reduced compared to thestandard beam element, thus yielding a system with lesser numberof equations. In the case of structural system with many members,the element reduction is much cheaper than solving the systemwithout condensation or with condensation at the global level.Hence we consider it to be computationally more effective. Addi-

Please cite this article in press as: Medic S et al. Beam model refij.engstruct.2012.10.004

tional model reduction pertaining strictly to linear analysis is dis-cussed to provide the systematic computer implementation oftechnical displacement method with no change of element length.A detailed illustration is provided for implementation of the tech-nical displacement method for frames with inclined members byexploiting the fundamental theorem from rigid body kinematics.Global reduction of the beam model which introduces length-invariance results in a system with fewer unknown displacements,and it could also be interesting for improving the system condi-tioning when the axial deformation remains very small.

Appendix A

A.1. Partial list of program CAL commands

LOAD M1 R = ? C = ?load matrix M1 of real numbers with ‘R=’ number of rows and

‘C=’ number of columnsLOADI M1 R = ? C = ?load array M1 of integer numbersZERO M1 R = ? C = ?load array M1 with all entries equal to zeroP M1list array named M1TRAN M1 M2transpose matrix M1 to form matrix M2DUPSM M1 M2 R = ? C = ? L = L1,L2duplicates submatrix M2 from location M1(L1,L2) M2 is NR x

NCSOLVE M1 M2 EQ=?solve symmetric system of equations M1 x = M2; EQ = number

of equations to be decomposedFRAME3 Ki Ti E = ? A = ? I = I3,I2 J = ? G = ? N = Ni,Nj P = P1,P2forms a (12 � 12) spatial beam element stiffness matrix; E is

Young’s modulus, A is cross-sectional area, I = I2,I3 are second mo-ments of inertia around axes 2 and 3, J is torsional moment of iner-tia, N=Ni,Nj are node numbers of element start and end used toextract coordinates from the previously defined coordinate matrixXYZ, P = P1,P2 are also defined by XYZ and are used to specify thedirection of z axis in the local coordinate system; the degrees offreedom are 3 translations and 3 rotations at the element startand 3 translations and 3 rotations at the element end

ADDK K Ki ID N = ?assembly od element stiffness matrices Ki into the global stiff-

ness matrix K; ID is the connectivity matrix, N specifies the columnnumber in ID corresponding to the considered element

Appendix B. analysis results

Fig. 5a – frame with moment release

ne

LOAD XYZ R = 3 C = 3

ment and reduction. Eng Struct (2

NUMBER OF COLUMNS = 9

ARRAY NAME = XYZ NUMBER OF

ROWS = 3

TRAN IDKT IDK

NUMBER OF COLUMNS = 3

ZERO KKP R = 9 C = 9

FRAME3 K1 T1 E = 30000

A = 0.16 J = 0.001

ADDK KKP KK IDK N = 1

I = 0.003,0.003 G = 12000

N = 1,2 P = 1,0

ZERO A R = 9 C = 1

FRAME3 K2 T2 E = 30000

A = 0.16 J = 0.001

SOLVE KKP A EQ = 3

I = 0.003,0.003 G = 12000

N = 2,3 P = 1,0

TOTAL SOLUTION OF Ax = B

LOADI IDT R = 2 C = 12

DUPSM KKP KKC R = 6 C = 6

012), http://dx.doi.org/10.1016/

S. Medic et al. / Engineering Structures xxx (2012) xxx–xxx 11

Please cite this article in press as:j.engstruct.2012.10.004

L = 4,4

ARRAY NAME = IDT NUMBER OF

ROWS = 2

LOAD RT R = 1 C = 6

NUMBER OF COLUMNS = 12

ARRAY NAME = RT NUMBER

OF ROWS = 1

TRAN IDT ID

NUMBER OF COLUMNS = 6

ZERO KK R = 9 C = 9

TRAN RT R

ADDK KK K1 ID N = 1

SOLVE KKC R

ADDK KK K2 ID N = 2

TOTAL SOLUTION OF Ax = B

LOADI IDKT R = 1 C = 9

P R

ARRAY NAME = IDKT NUMBER OF

ROWS = 1

COL# = 1

ROW 4 �2.7778 ROW 1 .62465E�01 ROW 5 .00000

ROW 2 .41643E�01

ROW 6 �.93697E�02 ROW 3 �18.519 RETURN

Fig. 5b – frame with moment release

LOAD XYZ R = 3 C = 3

ZERO KK R = 6 C = 6

ARRAY NAME = XYZ NUMBER OF

ROWS = 3

ADDK KK K1 ID N = 1

NUMBER OF COLUMNS = 3

ADDK KK K2P ID N = 2

FRAME3 K1 T1 E = 30000

A = 0.16 J = 0.001

LOAD RT R = 1 C = 6

I = 0.003,0.003 G = 12000

N = 1,2 P = 1,0

ARRAY NAME = RT NUMBER

OF ROWS = 1

FRAME3 K2 T2 E = 30000

A = 0.16 J = 0.001

NUMBER OF COLUMNS = 6

I = 0.003,0.003 G = 12000

N = 2,3 P = 1,0

TRAN RT R

LOADI IDKT R = 1 C = 12

SOLVE KK R

ARRAY NAME = IDKT NUMBER OF

ROWS = 1

TOTAL SOLUTION OF Ax = B

NUMBER OF COLUMNS = 12

P R

TRAN IDKT IDK

ZERO A R = 12 C = 1

COL# = 1

ZERO K2P R = 12 C = 12

ROW 1 .62465E�01 ADDK K2P K2 IDK N = 1 ROW 2 .41643E�01 SOLVE K2P A EQ = 3 ROW 3 �18.519 TOTAL SOLUTION OF Ax = B ROW 4 �2.7778 LOADI IDT R = 2 C = 12 ROW 5 .00000

ARRAY NAME = IDT NUMBER OF ROWS = 2

ROW 6 �.93697E�02 NUMBER OF COLUMNS = 12 RETURN

TRAN IDT ID

Fig. 6a – frame with shear force release

LOAD XYZ R = 3 C = 3

ZERO A R = 8 C = 1

ARRAY NAME = XYZ NUMBER OF

ROWS = 3

SOLVE KKP A EQ = 2

NUMBER OF COLUMNS = 3

TOTAL SOLUTION OF Ax = B

FRAME3 K1 T1 E = 30000 A

= 0.16 J = 0.001

DUPSM KKP KKC R = 6 C = 6

L = 3,3

I = 0.003,0.003 G = 12000 N =

1,2 P = 1,0

LOAD RT R = 1 C = 6

FRAME3 K2 T2 E = 30000 A

ARRAY NAME = RT NUMBER

Medic S et al. Beam model refine

= 0.16 J = 0.001

ment and reduction. Eng Struct (2

OF ROWS = 1

I = 0.003,0.003 G = 12000 N =

2,3 P = 1,0

NUMBER OF COLUMNS = 6

LOADI IDT R = 2 C = 12

TRAN RT R

ARRAY NAME = IDT NUMBER OF

ROWS = 2

SOLVE KKC R

NUMBER OF COLUMNS = 12

TOTAL SOLUTION OF Ax = B

TRAN IDT ID

P R

ZERO KK R = 8 C = 8

ADDK KK K1 ID N = 1

COL# = 1

ADDK KK K2 ID N = 2

ROW 1 13.889

LOADI IDKT R = 1 C = 8

ROW 2 .41667E�01

ARRAY NAME = IDKT NUMBER OF

ROWS = 1

ROW 3 �33.769

NUMBER OF COLUMNS = 8

ROW 4 �4.9020

TRAN IDKT IDK

ROW 5 .00000

ZERO KKP R = 8 C = 8

ROW 6 .34266E�15

ADDK KKP KK IDK N = 1

RETURN

Fig. 6b – frame with shear force release

LOAD XYZ R = 3 C = 3

FRAME3 K2 T2 E = 30000 A

= 0.16 J = 0.001

ARRAY NAME = XYZ NUMBER

OF ROWS = 3

I = 0.003,0.003 G = 12000 N

= 2,3 P = 1,0

NUMBER OF COLUMNS = 3

LOADI IDKT R = 1 C = 12

FRAME3 K1 T1 E = 30000 A =

0.16 J = 0.001

ARRAY NAME = IDKT NUMBER

OF ROWS = 1

I = 0.003,0.003 G = 12000

N = 1,2 P = 1,0

NUMBER OF COLUMNS = 12

TRAN IDKT IDK

NUMBER OF COLUMNS = 6

ZERO A R = 12 C = 1

TRAN RT R

ZERO K2P R = 12 C = 12

SOLVE KK R

ADDK K2P K2 IDK N = 1

TOTAL SOLUTION OF Ax = B

SOLVE K2P A EQ = 2

P R

TOTAL SOLUTION OF Ax = B

LOADI IDT R = 2 C = 12

COL# = 1

ARRAY NAME = IDT NUMBER

OF ROWS = 2

ROW 1 13.889

NUMBER OF COLUMNS = 12

ROW 2 .41667E�01 TRAN IDT ID ROW 3 �33.769 ZERO KK R = 6 C = 6 ROW 4 �4.9020 ADDK KK K1 ID N = 1 ROW 5 .00000

(continued on next page)

012), http://dx.doi.org/10.1016/

12 S. Medic et al. / Engineering Structures xxx (2012) xxx–xxx

Pj.

ADDK KK K2P ID N = 2

lease cite this article in pressengstruct.2012.10.004

ROW 6 .34266E�15

LOAD RT R = 1 C = 6 RETURN

ARRAY NAME = RT NUMBER

OF ROWS = 1

References

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[2] Ibrahimbegovic A. Unified approach to structural system modeling: finiteelements with rotational degrees of freedom. Yugoslav J Eng Model 1990;1/2:21–7 [in Bosnian].

[3] Wilson EL. CAL – a computer analysis language for teaching structural analysis.J Comput Struct 1979;10:127–32.

[4] Argyris JH et al. Finite element method – the natural approach. Comput MethAppl Mech Eng 1979;17/18:1–106.

[5] Jokanovic O. Deformation method. Svjetlost, Sarajevo, 1987 [in Bosnian].

as: Medic S et al. Beam model refi

[6] Ibrahimbegovic A, Jokanovic O. Method of conjugate structure: generalizationof Mohr’s analogy. Almanac of research papers in the field of materials andstructures, 19, 1990 [in Bosnian].

[7] Bathe KJ, Wilson EL. Numerical methods in finite element analysis. EnglewoodCliffs (New Jersey): Prentice-Hall Inc.; 1976.

[8] Hildebrand FB. Methods of applied mathematics. Englewood Cliffs (NewJersey): Prentice-Hall Inc.; 1965.

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