1

MS5019 – FEM 1

MS5019 – FEM 2

2

MS5019 – FEM 3

MS5019 – FEM 4

4.1. Beam Stiffness MatrixWe will derive the stiffness matrix for a simple beam element. A beamis a long, slender structural member generally subjected to transverse loading that produces significant bending effects as opposed to twisting or axial effects.The bending deformation is measured as a transverse displacement and a rotation. Hence, the d.o.f considered per node are a transverse displacement and a rotation (as opposed to only an axial displacement for a bar element in Chapter 3).

shown. as s'ˆby moments bending theand s'ˆby given are forces nodal

local The .'ˆby rotation theand s'ˆby given are ntsdisplaceme transverselocal The .ˆ coordinate local e transversand ˆcoordinat local axialwith

Llength of is beam The 1.-4 Figurein shown element beam heConsider t

iiy

iiy

mf

sdyx

φ

3

MS5019 – FEM 5

Figure 4-1 Beam element with positive nodal displacements and nodal forces(We initially neglect all axis effects)

direction. ˆ positivein positive arent displaceme and Forces 2.direction.CCW in positive are rotations and Moments 1.

:used are sconventionsign following thenodes, allAt

y

MS5019 – FEM 6

.ˆ moments bending and ˆ forcesshear positivefor theory beam simplein used sconventionsign theindicates 2-4 Figure

mV

Figure 4-2 Beam theory sign conventions for shear forces and bending moments

have webeam, theofelement aldifferenti a of mequilibriumoment and force From gth).(force/len )ˆ( loading ddistribute a tosubjected

3-4 Figurein shown beam heConsider t follows. as derived isbehavior beam elastic-linear elementary governingequation aldifferenti The

xw

4

MS5019 – FEM 7

Figure 4-3 Beam under distributed load

)b1.1.4(ˆ

or0ˆ

)a1.1.4(ˆ

or0ˆ)ˆ(

xddMVdMxdV

xddVwdVxdxw

==+

−==−−

MS5019 – FEM 8

axes). ˆ and ˆ thelar toperpendicu is axis ˆ the(where axis ˆabout the inertia ofmoment principal theis and ,elasticity of modulus the

is 4b),-4 Figure (seedirection ˆ in thefunction nt displaceme transverse theis ˆ 4b,-4 Figurein shown curve deflected theof radius theis where

)c1.1.4(1bymoment thetorelatedisbeam theof curvature theAlso,

yxzzI

Eyv

EIM

ρρ

κ

κ

==

Figure 4-4 Deflected curve beam

5

MS5019 – FEM 9

matrix. stiffness beam thedevelop to1Chapter in steps thefollow will weNow,

)1.1.4(0ˆˆ

becomes (4.1.1f) Eq. moments, and forces nodalonly with and EIconstant For

)f1.1.4()ˆ(ˆˆ

ˆ

obtain we(4.1.1a), Eq.and (4.1.1b) Eq. intoresult thissubstuting and Mfor (4.1.1e) Eq. Solving

)e1.1.4(ˆˆ

obtain we(4.1.1c), Eq.in (4.1.1c) Eq. Using

)d1.1.4(ˆˆ

bygiven is ˆˆ slope small afor curvature The

4

4

2

2

2

2

2

2

2

2

gxdvdEI

xwxdvdEI

xdd

EIM

xdvd

xdvd

xdvd

=

−=⎟⎟⎠

⎞⎜⎜⎝

⎛

=

=

=

κ

θ

MS5019 – FEM 10

Step 1 Select Element Type

Represent the beam by labeling nodes at each end and in general by labeling the element number (see Figure 4-1).

Step 2 Select a Displacement Functions

Assume the transverse displacement variations through the element length to be

)2.1.4(ˆˆˆ)ˆ(ˆ 432

23

1 axaxaxaxv +++=

The complete cubic displacement approximation function Eq. (4.1.2) is appropriate because there are four d.o.f (a transverse displacement and a small rotation at each node). The cubic function also satisfies the basic beam differential equation – further justifying its selection. In addition, the cubic function also satisfies the condition of displacement and slope continuity at nodes shared by two elements.

6

MS5019 – FEM 11

( ) ( )

( ) ( ) )4.1.4(ˆˆˆˆˆˆ21ˆˆ3

ˆˆˆ1ˆˆ2)ˆ(ˆ

have we(4.1.2), Eq. into stituting-sub and d.o.f nodal theof in terms through for (4.1.3) Eqs. Solving

)3.1.4(

23ˆˆ

)(ˆˆ)(ˆ

ˆˆ

)0(ˆˆ)0(ˆ

:follows as ˆ and ,ˆ ,ˆ ,ˆ d.o.f nodal theoffunction a as ˆ express we2.2,Section in described as procedure same the Using

112

21213

3212213

41

322

12

432

23

12

31

41

2121

yyy

yy

y

y

yy

dxxL

ddL

xL

ddL

xv

aa

aLaLaxdLvd

aLaLaLadLv

axd

vdadv

ddv

++⎥⎦⎤

⎢⎣⎡ +−−−+

⎥⎦⎤

⎢⎣⎡ ++−=

++==

+++==

==

==

φφφ

φφ

φ

φ

φφ

MS5019 – FEM 12

[ ]{ }

{ }

[ ] [ ]

( ) ( )( ) ( )

)7.1.4(ˆˆ1)ˆ(ˆ3ˆ21)ˆ(

ˆˆ2ˆ1)ˆ(ˆ3ˆ21)ˆ(

with)6.1.4()ˆ()ˆ()ˆ()ˆ()ˆ( and

)6.1.4(

ˆˆˆˆ

ˆwhere

)5.1.4(ˆ)ˆ()ˆ(ˆas(4.1.4)Eq.express weform,matrix In

22334

2333

322332

32331

4321

2

2

1

1

LxLxL

xNLxxL

xN

LxLxLxL

xNLLxxL

xN

bxNxNxNxNxN

ad

d

d

dxNxv

y

y

−=+−=

+−=+−=

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

=

=

φ

φ

7

MS5019 – FEM 13

2. nodefor results analogous have and functionsShape 1. nodeat evaluated when 1)( (4.1.7), Eqs. of second

thefrom have, we,ˆ wuth associated is Because 2. nodeat evaluated when0 and 1 nodeat evaluated when 1 element, beam For the

element. beam afor thecalled are, and,,,

43

2

12

11

4321

NNdxdNN

NNNNNN

=

==

φ

function shape

Step 3 Define the Strain/Displacement and Stress/Strain RelationshipsAssume the following axial strain/displacement relationship to be valid:

bynt displaceme e transvers thetoplacement -dis axail therelate we5,-4 Figurein shown beam theof figuration-con deformed theFrom function.nt displaceme axial theis ˆ where

)8.1.4(ˆˆ

)ˆ,ˆ(

uxdudyxx =ε

MS5019 – FEM 14

Figure 4-5 Beam segment (a) before deformation and (b) after deformation;(c) angle of rotation of cross-section ABCD

8

MS5019 – FEM 15

as empresent th now wematrix,stiffnesselement beam theof derivation in the ipsrelationsh these

use will weSince function.nt displaceme e transvers the torelated areforceshear andmoment bending the theory,beam elementary From

)10.1.4(ˆˆˆ)ˆ,ˆ(

obtain we(4.1.8), Eq,in (4.1.9) Eq. Using).ˆˆ( anglean through rotate general,in and,n deformatio

afterplanar remain n deformatio bending beforeplanar that ABCD)section cross as(such beam theof sections cross that assumptionbasic e theory thbeam elementary from recall should wewhere

)9.1.4(ˆˆˆˆ

2

2

xdvdyyx

xdvd

xdvdyu

x −=

−=

ε

MS5019 – FEM 16

)11.1.4(ˆˆˆ

ˆˆ

)ˆ(ˆ3

3

2

2

xdvdEIV

xdvdEIxm ==

Step 4 Derive the Element Matrix and Equations

First, derive the element stiffness matrix and equations using a direct equilibrium approach. We now use the nodal and beam theory sign conventions for shear forces and bending moments, along with Eqs. (4.1.4) and (4.1.11), to obtain

( )( )( )

( )

)12.1.4(

ˆ4ˆ6ˆ2ˆ6ˆ

)(ˆˆˆ

ˆ6ˆ12ˆ6ˆ12ˆ

)(ˆˆˆ

ˆ2ˆ6ˆ4ˆ6ˆ

)0(ˆˆˆ

ˆ6ˆ12ˆ6ˆ12ˆ

)0(ˆˆˆ

22

212

132

2

2

221133

3

2

22

212

132

2

1

221133

3

1

φφ

φφ

φφ

φφ

LdLLdLLEI

xdLvdEImm

LdLdLEI

xdLvdEIVf

LdLLdLLEI

xdvdEImm

LdLdLEI

xdvdEIVf

yy

yyy

yy

yyy

+−+===

−+−−=−=−=

+−+=−=−=

+−+===

9

MS5019 – FEM 17

where the minus signs in the second and third of Eqs. (4.1.12) are the result of opposite nodal and beam theory positive bending momentconventions at node 2 as seen by comparing Figure 4-1 and 4-2. Equations (4.1.12) relate the nodal forces to the nodal displacement. In matrix form, Eqs. (4.1.12) become

)14.1.4(

12626612612

2646612612

ˆ

theiselement beam theofmatrix stiffness thewhere

)13.1.4(

ˆˆˆˆ

12626612612

2646612612

ˆ

ˆˆ

ˆ

2

22

3

2

2

1

1

2

22

3

2

2

1

1

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−−−−

−−

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−−−−

−−

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

LLLLL

LLLLLL

LEI

d

d

LLLLL

LLLLLL

LEI

mfmf

y

y

y

y

k

φ

φ

MS5019 – FEM 18

4.2. Example of Assemblage of Beam Stiffness Matrices

Consider the beam in Figure 4-6 as an example to illustrate the procedure for assemblage of beam element stiffness matrices. Assume EI to be constant throughout the beam. A force of 1000 lb and a moment of 1000 lb-ft are applied to the beam at midlength. The left end is a fixed support and the right end is a pin support.

2k

Figure 4-6 Fixed-hinged beam subjected to a force and a moment

4.2. Example of Assemblage of Beam Stiffness Matrices

Consider the beam in Figure 4-6 as an example to illustrate the procedure for assemblage of beam element stiffness matrices. Assume EI to be constant throughout the beam. A force of 1000 lb and a moment of 1000 lb-ft are applied to the beam at midlength. The left end is a fixed support and the right end is a pin support.

10

MS5019 – FEM 19

First, we discrretize the beam into two elements with nodes 1, 2, and 3 as shown. We include a node at midlength because applied force and moment exist at midlength and, at this time, loads are assumed to be applied only at nodes. (Another procedure for handling loads applied on elements will be discussed in Section 4.4).

)1.2.4(

4626612612

2646612612

ˆ

bygiven are elements twofor the matrices stiffness global the(4.1.14), Eq. Using

22

22

3)1()1(

2211

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−−−

−−

=

LLLLLL

LLLLLL

LEI

dd yy

kk

φφ

MS5019 – FEM 20

)2.2.4(

4626612612

2646612612

ˆ

and

22

22

3)2()2(

3322

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−−−

−−

==

LLLLLL

LLLLLL

LEI

dd yy

kk

φφ

where the d.o.f. associated with each element are indicateed by the usual labels above the columns in each element stiffness matrix. Here the local coordinate axes for each element coincide with global x and y axes of the whole beam. Consequently, the local and global element stiffnessmatrices are identical.

11

MS5019 – FEM 21

)3.2.4(

00

bygiven thusare 6-4 Figurein beam for the equations governing the(4.2.2), and (4.2.1) Eqs. andion superposit Usingnts.displaceme nodal

global the torelated are forces nodal global external theassembled,been hasmatrix stiffness toal When themethod. stiffnessdirect the

usingby beamfor theassembledbenowcan matrix stiffness totalThe

3

1

2

1

1

1

3

3

2

2

1

1

22

2222

22

3

462606126120

26)44()66(26612)66()1212(61200264600612612

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−

−=

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

−++−

−+−+−−−−

φ

φ

φ

y

y

y

y

y

y

d

d

d

LLLLLL

LLLLLLLLLLLL

LLLLLL

MFMFMF

LEI

MS5019 – FEM 22

solution. final thedetermine toencourage are students The . and rotations nodal

unknown theand nt displaceme nodalunknown for thetaneously -simul (4.2.5) Eq. solve now could Weequations. ofset reduced theinto

dsubstitutebeen have 0ft,-lb 1000lb, 1000 where

)5.2.4(

obtain we(4.2.4), Eqs. using and d.o.funknown with rows the togrrespondin

-co (4.2.3) Eq. of equationssixth and fourth, third, thegconsiderinOn )4.2.4(000

have we3, nodeat support hinge theand1 nodeat support fixed theof s,constraintor BC, thegconsiderin Now

21

2

322

22

23

311

3

2

2

4262806024

010001000

φφ

φ

φφ

y

y

yy

d

MMF

LEI

dd

yd

LLLLLL

==−=

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡=

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

===

−

12

MS5019 – FEM 23

4.3. Example of Beam Analysis Using theDirect Stiffness Method

We will now perform complete solutions for beams with various boundary support and loads to further illustrate the use of the equations developed in Section 4.1.

EXAMPLE 5.1, 5.2, and 5.3

MS5019 – FEM 24

4.4. Distributed Loading

Figure 4-7 Fixed-fixed beam subjected to a uniformly distributed load

Figure 4-8 Fixed-end reactions for the beam in Figure 4-7

4.4. Distributed LoadingBeam members can support distributed loading as well as concentrated nodal loading. Therefore, we must be able to account for distributed loading. Consider the fixed-fixed beam subjected to a uniformly distributed loading w shown in Figure 4-7. The reactions, determined from structural analysis theory, are shown in Figure 4-8. These reactions called fixed-end reactions.

13

MS5019 – FEM 25

Guided by the results from structural analysis for the case of a uniformly distributed load, we replace the load by concentrated nodal forces and moments tending to have the same effect on the beam as the actual ditributed load. Figure 4-9 illustrates this idea for a beam. We have replaced the uniformly distributed load by an equivalent force system consisting of a concentrated nodal force and moment at each end of the member. This equivalent forces are always of opposite sign from the fixed-end forces. To determine the maximum deflection and maximum moment in the beam span, a node is needed at midspan of the beam 2-3.

Figure 4-9 (a) Beam with a distributed load and (b) the equivalent nodal force system.

MS5019 – FEM 26

Work Equivalent MethodThis method is based on the concept the work of the distributed load is equal to that of the discrete load replacement for arbitrary nodal displacements. To illustrate the method, we consider the sample shown in Figure 4-10. The work due to the distributed load is given by

Figure 4-10 (a) Beam element subjected to a general and (b) the equivalent nodal force system.

(4.1.4). Eq.by given nt displaceme e transvers theis )ˆ(ˆ where

)1.4.4(ˆ)ˆ(ˆ)ˆ(0

ddistribute

xv

xdxvxwWL

∫=

14

MS5019 – FEM 27

The work due to the discrete nodal forces is given by

.ˆ and ,ˆ,ˆ,ˆnt displacemearbitrary for setting,by is, thateequivalenc work ofconcept theusingby load ddistribute thereplace toused ˆ and ,ˆ,ˆ,ˆ forces and moments nodal thedeterminecan then We

)2.4.4(ˆˆˆˆˆˆˆˆ

2121discreteddistribute

2121

22112211discrete

yy

yy

yyyy

ddWW

ffmm

dfdfmmW

φφ

φφ

=

−

+++=

Example of Load ReplacementTo illustrate more clearly the concept of work equivalence, we consider a beam subjected to a specified (uniformly) distributed as shown in Figure 4-11. The support conditions are not shown because they are not relevant to the replacement scheme.

MS5019 – FEM 28

Figure 4-11 (a) Beam subjected to an uniformly distributed loading and (b) the equivalent nodal forces to be determined.

( ) ( ) ( )

( ) ( ) )4.4.4(ˆ2

ˆˆˆ23

ˆˆˆˆ4

ˆˆ2

ˆ)ˆ(ˆ)ˆ(

as loadon distributi the todue work obtain the we(4.1.4), Eq. from)ˆ(ˆ and )ˆ( ngsubstitutiby (4.4.3) Eq. of side hand-left theEvaluating

)3.4.4(ˆˆˆˆˆˆˆˆˆ)ˆ(ˆ)ˆ(

have we,for (4.4.2) and (4.4.1) Eqs. g Usin

1

2

121

2

1221

2

210

221122110

discreteddistribute

LwdwLwL

ddLwwLddLwxdxvxw

xvwxw

dfdfmmxdxvxw

WW

y

yyyy

L

yyyy

L

−⎟⎟⎠

⎞⎜⎜⎝

⎛−++

−−+−−−=

−=

+++=

=

∫

∫

φφφ

φφ

φφ

15

MS5019 – FEM 29

)7.4.4(

22)1(ˆ

22)1(ˆ

obtain we, ˆ theand ˆfirst except zero toequal ntsdisplaceme nodal all letting Finally,

)6.4.4(1234

)1(ˆ

yields 0ˆ and ,0ˆ,1ˆ,0ˆ letting Similarly,

)5.4.4(1223

24

)1(ˆ

obtain then weand 0ˆ and ,0ˆ,0ˆ,1ˆlet wents,displacemenodalarbitrary for (4.4.4) and (4.4.3) Eqs. using Now

2

1

21

222

21

2121

2222

1

2121

LwLwLwf

LwLwLwLwf

dd

wLwLwLm

dd

wLwLwLwLm

dd

y

y

yy

yy

yy

−=−=

−=−+−=

=⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

====

−=⎟⎟⎠

⎞⎜⎜⎝

⎛+−−=

====

=

φφ

φφ

MS5019 – FEM 30

load. ddistribute thereplace toused moments)(and/or forces nodal edconcentratobtain the to(4.4.3) Eq. toaccording integrate then and )ˆ(ˆby multiply

can we,)ˆ(function loadgiven any for general,in that,concludecan Wexv

xw

Moreover, we can obtain the load replacement by using the concept of fixed-end reactions from structural analysis theory. Table of equivalent nodal forces has been generated for this text in the Table 4-1.

Hence, if a concentrated load is applied other than at the natural intersection of two elements, we can use the concept of equivalent nodal forces to replace the concentarted load by nodal concentrated values acting at the beam ends, instead of creating a node on the beam at the location where the load is applied.

16

MS5019 – FEM 31

Tabel 4-1 Equivalent joint forces fo for different type of loads.

MS5019 – FEM 32

Tabel 4-1 Equivalent joint forces fo for different type of loads (cont’d)

17

MS5019 – FEM 33

General FormulationIn general, we can account for distributed loads or concentrated loads acting on beam elements by starting with the following formulation application for a general structure

as (4.4.8) Eq. rewritecan we),(present not initially are forces nodal edconcentrat assume now weSince reactions. the

including forces, edconcentrat nodal global therepresents that Recall .components coordinate-global of in terms expresscan we,components

coordinate-local of in terms expressed ˆ forces nodal equivalent of 1-4Table Usingload. ddistribute the wouldas ntsdisplaceme same theyield

y that themagnitudesuch of are which ,components coordinate-globalof in terms expressed now forces, nodal equivalent thecalled are where

)8.4.4(

o

0F

FF

f

FFKdF

=

−=

o

o

o

MS5019 – FEM 34

)10.4.4(

12

2

12

2

have webeam,element -one aover acting load ddistributeuniformly afor 1)-4 Tablein 4 case load using(or (4.4.7) - (4.4.5) Eqs.

and ˆ of definition theusing example,For . forces nodal global actualobtain the we(4.4.8), Eq. into Fo forces nodal equivalent and d ments

-displace global thengsubstituti then and (4.4.9) Eq.in dfor solvingOn )9.4.4(

2

2

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

−

−

−

=

=

wL

wL

wL

wL

F

w

o

o

o

fF

KdF

18

MS5019 – FEM 35

forces. nodal local equivalent theare ˆ where

)10.4.4(ˆˆˆˆ as

locally (4.4.8) Eq. applyingby structures of elements individualin ˆ forcenodal localobtain thetobasislocalaon applied becan concept This

o

o

o

f

fdkf

f

−=

4.5. Potential Energy Approach to DeriveBeam Element Equations

We will now derive the beam element equations using the POMPE. The procedure is similar to that used in Section 3.8 in deriving the bar element equations. We use the same notation here as in Section 3.8.

MS5019 – FEM 36

.1

y

2

1

2

1

21

ˆ moments (3) and ;ˆ forces edconcentrat nodal (2) ); surface

over (acting ˆ loading surface e transvers(1) of PE therepresent (4.5.3) Eq.of side hand-right on the termsThe neglected. now are forcesbody where

)3.5.4(ˆˆˆˆˆˆ

bygiven is forces of PE theloads, nodal edconcentrat and ddistributeboth tosubjectedelement beam single afor and

)2.5.4(bygiven is beam a

for energy Ustrain for the expression ldimensiona-one general thewhere)1.5.4(

is beam afor PE totalThe

1

iiy

iii

iiyiyy

xx

p

mPS

T

mdPdSvT

dVU

U

S

V

∑∑∫∫

∫∫∫

==

−−−=Ω

=

Ω+=

φ

εσ

π

19

MS5019 – FEM 37

( )

as econveniencfor here repeated (4.1.10), Eq.iprelationshplacement strain/dis theintofor (4.1.5) Eq. ngSubstituti

)6.5.4(ˆˆˆˆˆˆˆˆ

becomes PE total the(4.5.3), - (4.5.1)Eqs.in (4.5.5) and (4.5.4) Eqs. Usingidth.constant w theis where)5.5.4(ˆ

is acts loading surface theover which area aldifferenti theand)4.5.4(ˆ

as expressed becn then element beam for the volumealdifferenti The. aresection -crossconstant have element to beam heConsider t

12.-4 Figurein shown length ofelement beam for thefunction nt displaceme e transvers theis Again,

2

1021

ˆ∑∫∫ ∫∫=

+−−=

=

=

iiiiyiy

L

yxxp mdPxdvTbxddA

bxdbdS

xddAdV

AL

x Aφεσπ

MS5019 – FEM 38

Figure 4-12 Beam element subjected to surface loading and concentrated nodal forces

{ } [ ]{ } )8.5.4(ˆˆ

as rotations and ntsdisplaceme nodal of in termsstrain theexpress we

)7.5.4(ˆˆˆ

3

2

33

2

32ˆ66ˆ124ˆ66ˆ12

2

2

dy

xdvdy

LLLx

LLx

LLLx

LLx

x

x

−+−−−−=

−=

ε

ε

20

MS5019 – FEM 39

{ } [ ]{ }

[ ] [ ]

{ } [ ]{ }[ ] [ ]

{ } [ ][ ]{ }

{ } { } [ ] { } { } )14.5.4(ˆˆˆˆˆˆ

asnotation matrix in expressed is (4.5.6) Eq.energy potential total theNext,)13.5.4(ˆˆ

obtain we(4.5.11), Eq.in (4.5.9) Eq. Using.elasticity of modulus theis and

)12.5.4(where)11.5.4(

bygiven is iprelationshain stress/str The

)10.5.4(

define wewhere)9.5.4(ˆˆ

or

ˆ 021

2ˆ66ˆ124ˆ66ˆ123

2

33

2

3

∫ ∫∫∫ −−=

−=

==

=

−=

−+−−−

x APdxdvTbxdAd

dBDy

EEDD

B

dBy

TL

Tyx

Txp

x

xx

LLLx

LLx

LLLx

LLx

x

εσπ

σ

εσ

ε

MS5019 – FEM 40

{ } [ ] [ ]{ } { } [ ] { } { }

as formmatrix in written arewhich equations,element four obtain we, minimize tozero toeach term equating

and , ˆ and ,ˆ , ˆ ,ˆ respect to with (4.5.15) Eq.in atingDifferenti

}.ˆ{ offunction a as expressed now is (4.5.15), Eq.In (4.5.15). Eq. of side hand-right on the first term obtain the to

)16.5.4(inertia ofmoment theof definition theused have wewhere

)15.5.4(ˆˆˆˆˆˆˆ2

as formmatrix in (4.5.14), Eq.PE, total theexpress wedirection, in the length)unit per (load load line theas ˆ defining and (4.5.13), Eq. and (4.5.12), (4.5.9), Eqs. Using

2211

2

00

p

yy

p

TL

TTL

TT

p

y

dd

d

dAyI

PdxdNdwxddBBdEI

Tbw

A

πφφ

π

π

∫∫

∫∫

=

−−=

=

21

MS5019 – FEM 41

[ ] [ ] { } [ ] { } { }

{ } [ ] { }

[ ] [ ] )19.5.4(ˆ]ˆ[

(4.5.17) Eq. from have we},d̂]{ˆ[}ˆ{Since (4.1.13). Eq. toidentical then are (4.5.17) Eq. evaluatingexplicitlyby given equationselement four the, (4.5.18) Eq. Using

)18.5.4(ˆˆˆ

have weloading, edconcentrat and loading ddistribute from resultingforces nodal thoseof sum theasmatrix force nodal thengRepresenti

)17.5.4(0ˆˆˆˆ

0

0

00

∫

∫

∫∫

=

=

+=

=−−

LT

LT

LT

LT

xdBBEIk

kf

PxdwNf

PxdwNdxdBBEI

MS5019 – FEM 42

.previously developped (4.1.14) Eq. toidentical is (4.5.20) Eq. expected, Aselement. beam afor matrix stiffness local therepresents (4.5.20)Equation

)20.5.4(

4626612612

2646612612

]ˆ[

as formexplicit in evaluatedis ]ˆ[matrix stiffness g,integratin and (4.5.19) Eq.in (4.5.10) Eq. Using

22

22

3

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−−−

−−

=

LLLLLL

LLLLLL

LEIk

k

Actually, we can also use the Galerkin’s method to derive the stiffness matrix of a beam element. This method is not intended to be studied in this course. Interested students are encouraged to read the reference [1-3] to explore the method.

22

MS5019 – FEM 43

4.6. Other Type of Line ElementMany structures, such as building and bridges, are composed of

frames and/or grids element. The stiffness matrix of this kind of element is more complex then the beam stiffness matrix that has been explained previously.

The most complex line element is a space frame , which has 2 nodes at each element and has 6 d.o.f (3 translations and 3 rotations) per nodes. So it’s stiffness matrix will be of size (12 x 12). The stiffness matrix of this type of element is given in Eq. (4.6.1).

The students who are interested to explore more about the other line element are encouraged to read the ref. [1-3] and/or other FE books.

MS5019 – FEM 44

z

y

x

zd

yd

xd

z

y

x

zd

yd

xd

LEI

LEI

LGJ

L

EI

L

EIL

EI

LEI

LAE

LEI

LEI

LEI

LEI

L

EILEI

LGJ

LGJ

L

EI

L

EI

L

EI

L

EILEI

LEI

LEI

LEI

LAE

LAE

zyxzdydxdzyxzdydxd

z

y

yy

yz

zzz

yyy

yyyy

zzzz

k

2ˆ

2ˆ

2ˆ

2ˆ

2ˆ

2ˆ

1ˆ

1ˆ

1ˆ

1ˆ

1ˆ

1ˆ

4

4

612

612

264

264

612612

612612

2ˆ

2ˆ

2ˆ

2ˆ

2ˆ

2ˆ

1ˆ

1ˆ

1ˆ

1ˆ

1ˆ

1ˆ

symetry

]ˆ[

0

00

00

000

00000

0000

00000

0000000

000000

0000000

0000000000

23

23

2

2

2323

2323

φ

φ

φ

φ

φ

φ

φφφφφφ

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡ −

=

−

−

−

−−−

−−

23

MS5019 – FEM 45

MS5019 – FEM 46

Reference:1. Logan, D.L., 1992, A First Course in the Finite Element Method,

PWS-KENT Publishing Co., Boston.

2. Imbert, J.F.,1984, Analyse des Structures par Elements Finis, 2nd Ed., Cepadues.

3. Zienkiewics, O.C., 1977, The Finite Eelement Method, 3rd ed., McGraw-Hill, London.

4. Finlayson, B.A., 1972, The Method of Weighted Residuals and Variational Principles, Academic Press, New York.