Ch 2 Stiffness Method 09

8/3/2019 Ch 2 Stiffness Method 09

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MS5019 FEM 1

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2.1. Definition of the Stiffness Matrix

element.singleaof

forcelocal

nt todisplacemenodal)(coordinate-localrelateswhere

such thatmatrixaismatrix,Stiffness

f

k

dkf

k

z,y,x

=

element.for theconvenientbeto

upsetsystemcoordinate-localatoreferredquantitiesdenotes

symboltheandvector,aormatrixadenotesnotationThe bold

structure.ormediumwholetheofforceglobalto

ntdisplacemenodal)(coordinate-globalrelatesmatrix

stiffnessaelement,withstructureormediumcontinuousaFor

Fd

K x, y, z

n


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MS5019 FEM 3

2.2. Stiffness Matrix for a Spring Element Consider a linear spring element shown in Figure 2-1. References point 1 and

2 (nodes) are located at the ends of the element.

We want to develop a relationship between nodal force and nodaldisplacement for a spring element. This relationship will be the stiffnessmatrix as follows.

k1 2

Lxx df 11

, xx df 22,

x

Figure 2-1 Linear spring element with positive nodal displacement and force conventions

)1.2.2(

2

1

2221

1211

2

1

=

x

x

x

x

d

d

kk

kk

f

f

.determinedbetoare(2.1)Eq.inmatrixtheofelementthewhere kijk

MS5019 FEM 4

Step 1 Select Element Type

Consider a linear spring subjected to resulting nodal forces Tdirected a longthe spring axial direction as shown in Figure 2-2, so as to be inequilibrium.

The spring is represented by labeling nodes at each end. The original distancebetween nodes before deformation is denoted byL.

x

xd1

xd2

k1 2 xTT

L

1 2k

Figure 2-2 Linear spring subjected to tensile forces


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MS5019 FEM 5

Step 2 Select a Displacement Functions

A displacement function is assumed. Here a linear displacement

varation along the axis of the spring is assumed because a linear

function with specified endpoints has a unique path. Therefore

In general, the total number of coefficients a is equal to the total

number of degrees of freedom (dof) associated with the element. Here

the total number of dof is two an axial displacement at each nodes of

the element. In matrix form Eq. (2.2.2) becomes

)2.2.2( 21 xaau +=

[ ] )3.2.2(12

1

=a

axu

MS5019 FEM 6

We now want to express as a function of the nodal displacement. We

achieve this by evaluating at each node and solving for a1 and a2from Eq. (2.2.2) as follows

)6.2.2(

,for(2.5)Eq.solvingor,

)5.2.2()(

)4.2.2()0(

122

2

122

11

L

dda

a

dLadLu

adu

xx

xx

x

=

+==

==

On substituting Eqs. (2.2.4) and (2.2.6) into Eq. (2.2.2), we have

)7.2.2(

1

12x

xx dxL

ddu +

=


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MS5019 FEM 7

In matrix form, we express Eq. (2.2.7) as

[ ]

)10.2.2(

and

1Here

)9.2.2(

or

)8.2.2(

1

21

2

121

2

1

L

xN

L

xN

d

dNNu

d

d

L

x

L

xu

x

x

x

x

==

=

=

are called theshape function because theNis express the shape of the

assumed displacement function over the domain of the element when

the ith element degree of freedom has unit value and all other dof are

zero.

MS5019 FEM 8

In the case,N1 danN2 are linear functions that have the properties that

N1 = 1 at node 1 and N1 = 0 at node 2, whereas N2 = 1 at node 2 and

N2 = 0 at node 1. Also,N1 +N2 = 1 for any axial coordinate along the

bar.

In addition, theNis are often called interpolation function because

we are interpolating to find the value of a function between given

nodal value. The interpolation function may be different from the

actual function except at the endpoints or nodes where the

interpolation function and actual function must be equal to specified

nodal values.


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MS5019 FEM 9

Step 3 Define the Strain-displacement and Stress-strainRelationships

The tensile forces Tproduce a total elongation (deformation)of the

spring. For the linear spring, Tand are related through Hookes law

by.

)13.2.2(

becomes(2.2.11)Eq.(2.2.7),Eq.ofuseMaking

)12.2.2()0()(

havewespring,theofndeformatiotheisbecausewhere,

)11.2.2(

12 xx dd

uLu

kT

=

=

=

MS5019 FEM 10

Step 4 Derive the Element Matrix and Equations

)16.2.2()(

)(

obtainwe(2.2.15),Eq.rewritingor

)15.2.2()(

)(

havewe(2.2.14),and(2.2.13),(2.2.11),Eqs.Using

)14.2.2(and

haveweforces,nodalforconventionsignBy the

122

211

122

121

21

xxx

xxx

xxx

xxx

xx

ddkf

ddkf

ddkfT

ddkfT

TfTf

=

=

==

==

==


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MS5019 FEM 11

matrix.squaresymmetrica

isthatobserveWeelement.springlinearaformatrixstiffnesstheas

)18.2.2(

obtainwe

(2.2.17),Eq.toapplied(2.2.1)Eq.ofuseandmatrixstiffnessaofdefinition

basicourForaxis.thealongspringfor theholdsiprelationshThis

)17.2.2(

yieldsequationsmatrixsingleain(2.2.16)Eq.expressingNow

2

1

2

1

k

k

=

=

kk

kk

x

d

d

kk

kk

f

f

x

x

x

x

MS5019 FEM 12

Step 5 Assemble the Element Equations to Obtain the Total/

Global Equations and Introduce Boundary Equations

The total stiffness matrix and total force vector are assembled using

nodal force equilibrium equations, force/deformation and compatibility

equations from Section 2.2, and the direct stiffness method described

in Section 2.4. This step applies for structures composed of more than

one element such that

[ ] { }

frame.referenceglobalain

expressedmatricesforceandstiffnesselementnowareandwhere

and

fk

FKN

e

eN

e

e )19.2.2(1

)(

1

)( ==

==== fFkK

The sign used in this context means that all element matrices mustbe assembled properly according to the direct stiffness method

described in Section 2.4.


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Step 6Solve for the Nodal Displacements

The displacements are then determined by imposing boundary

conditions and solving a system of equations, F = K d, simultaneously.

Step 7Solve for the Element Forces

The element forces are determined by back-substitution, applied to

each element, into equation similar to Eq. (2.2.16).

Step 8 Interpret the results

Finally, the element forces resulted can be analysed to check if all

element forces are below its allowable values.

MS5019 FEM 14

2.3. Example of a Spring Assemblage

We will consider the specific example of the two-spring assemblage

shown in Figure 2-3. Here we fix node 1, and apply axial forces at node

2 and node 3. The stiffness of spring elements 1 and 2 are k1 and k2respectively. The x axis is the global axis of the assemblage. In this

case, the local axis of each element coincides with the global axis of the

assemblage.

1 2

1k

31 2

xF3 xF2

x

Figure 2-3 Two-spring assemblage


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MS5019 FEM 15

Using Eq. (2.17) the element stiffness matrix for each element cam beexpressed

)2.3.2(

2,elementforand

)1.3.2(

1elementfor

2

3

22

22

2

3

3

1

11

11

3

1

=

=

x

x

x

x

x

x

x

x

d

d

kk

kk

f

f

d

d

kk

kk

f

f

Furthermore, element 1 and 2 must remain connected at common node3 throught the displacement. This is called thecontinuity or

compatibility requirement. The compatibility requirement yields

(superscript refers to the element number.)

)3.3.2(3)2(

3

)1(

3 xxx ddd ==

MS5019 FEM 16

Based on the sign convention for element nodal forces given in Figure 2-1,

we can write nodal equilibrium equations at node 3, 2, and 1 as

)6.3.2(

)5.3.2(

)4.3.2(

)1(

11

)2(

22

)2(

3

)1(

33

xx

xx

xxx

fF

fF

ffF

=

=

+=

Where F1x results from the reaction at the fixed support. To furtherclarify the resulting Eqs. (2.3.4) (2.3.6), free-body diagrams of each

element and node (using the established sign conventions for element

nodal forces) are shown in Figure 2-4.

Figure 2-4 Nodal forces consistent with element forces sign convention

232

xF2

11

xF3xF1

)2(

2xf)2(

3xf)1(

3xf)1(

1xf)1(

1xf

Based on the sign convention for element nodal forces given in Figure 2-1,

we can write nodal equilibrium equations at node 3, 2, and 1 as

)6.3.2(

)5.3.2(

)4.3.2(

)1(

11

)2(

22

)2(

3

)1(

33

xx

xx

xxx

fF

fF

ffF

=

=

+=


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MS5019 FEM 17

Application of Newtons third law for each node and element gives

xxx

xxx

xxxxx

dkdkF

dkdkF

dkdkdkdkF

31111

22322

223231113

)7.3.2(

=

+=

++=

In matrix form, Eqs. (2.3.7) are expressed by

)8.3.2(

0

0

1

2

3

11

22

1221

1

2

3

+

=

x

x

x

x

x

x

d

d

d

kk

kk

kkkk

F

F

F

Or rearranging, Eqs. (2.3.8) in numerically increasing order of the nodal

dof, we have

)9.3.2(0

0

3

2

1

2121

22

11

3

2

1

+

=

x

x

x

x

x

x

d

d

d

kkkk

kk

kk

F

F

F

MS5019 FEM 18

Eq. (2.3.9) is now written as the single matrix equation

.orthecalledis

)11.3.2(0

0

and,the

calledis,thecallediswhere

)10.3.2(

2121

22

11

3

2

1

3

2

1

matrixstiffnessglobaltotal

kkkk

kk

kk

vectorntdisplacemenodalglobal

d

d

d

vectorforcenodalglobal

F

F

F

x

x

x

x

x

x

+

=

=

K

dKF


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MS5019 FEM 19

2.4.Assembling the Total Stiffmess Matrix bySuperposition (Direct Stiffness Method)

element.eachwithassociatedfreedomof

degreestheindicates'in thecolumntheabovewrittens'theHere

)1.4.2(

22

22)2(

11

11)1(

2331

k

kk

ix

xxxx

dkk

kk

kk

kk

dddd

=

=

2.4.Assembling the Total Stiffmess Matrix bySuperposition (Direct Stiffness Method)

This method is based on proper superposition of the individualelement matrices making up a structure.

Referring to the-spring assemblage of Section 2.3, the element stiffnessmatrices are given in Eqs. (2.3.1) and (2.3.2) as

MS5019 FEM 20

To superpose the element matrices, all of them must be expended to the

order of the total structure (spring assemblage) stiffness matrix so that

each element matrix is associated with all the dof of the structure.

To expand each element stiffness matrix to the order of total stiffness

matrix, we simply add rows and columns of zero for the those

displacement not associated with that particular element.

The expanded form of each element equation can be expressed as

)2.4.2(

101

000

101

1elementfor

)1(

3

)1(

2

)1(

1

)1(

3

)1(

2

)1(

1

1

=

x

x

x

x

x

x

f

f

f

d

d

d

k


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MS5019 FEM 21

)5.4.2(

110

110

000

101

000

101

obtainwe(2.4.4),Eq.in(2.4.3)and(2.4.2)Eqs.Using

)4.4.2(

0

0

inresultsnodeeachatmequilibriuforcegconsiderinNow

)3.4.2(

110

110

0001elementfor

3

2

1

)2(

3

)2(

2

)2(

1

2

)1(

3

)1(

2

)1(

1

1

3

2

1

)2(

3

)2(

2

)1(

3

)1(

1

)2(

3

)2(

2

)2(

1

)2(

3

)2(

2

)2(

1

2

=

+

=

+

=

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

F

F

F

d

d

d

k

d

d

d

k

F

F

F

f

f

f

f

f

f

f

d

d

d

k

MS5019 FEM 22

(2.3.9).Eq.toidenticalision,superpositthrough

obtained(2.4.6),Equation.assemblagetotaltheofntdisplaceme

3nodethe,(2.3.3)Eq.byand,,reallyis,reallyisbecausedroppedbeenhaventsdisplacemenodalthe

withassociatednumberselementtheindicatingtssuperscriptheHere

)6.4.2(0

0

inresults(2.4.5)Eq.gSimplifyin

3

)2(

3

)1(

32

)2(

2

1

)1(

1

3

2

1

3

2

1

2121

22

11

xxxxx

xx

x

x

x

x

x

x

ddddddd

F

F

F

d

d

d

kkkk

kk

kk

==

=

+

The method of directly assembling individual element stiffness matrices

to form the total structure stiffness matrix and the the total set of

stiffness equations is calleddirect stiffness method. It is the most

important step in the FEM.


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MS5019 FEM 23

2.5.Boundary Conditions

We must specify boundary (or support) conditions for structure models

such as the spring assemblage of Figure 2-4, or K will be singular; that

is the determinant ofK will be zero and, therefore, its inverse will not

exist. Without specifying adequate kinematic constraints or support

conditions, the structure will be free to move as a rigid body.

Boundary conditions (BC) are of two general type:

1. Homogeneous BC the most common occur at locations that are

completely prevented from movement,2. Non- homogeneous BC occur where finite non-zero values of

displacement are specified, such as the settlement of a support.

MS5019 FEM 24

In general, specified support conditions are treated mathematically by

partitioning the global equilibrium equations as follows:

.ofiondeterminatfor theallowingthussingular,longer

noisthatassumewe(2.5.2),Eq.In the(2.5.2).Eq.fromtermining

-deafter(2.5.3)Eq.fromfoundisnodes.ntdisplacemespecifiedat the

forcesnodalunknowntheareandforcesnodalknownthearewhere

)3.5.2(and

)2.5.2(

havewe(2.5.1),From.nodes.ntsdisplacemespecifiedthe

thebeandntdisplacemefreeornedunconstraithebeletwewhere

)1.5.2(

inresults(2.4.5)Eq.gSimplifyin

1

111

2

21

2221212

2121111

21

2

1

2

1

2221

1211

d

Kd

F

FF

dKdKF

dKFdK

dd

F

F

d

d

KK

KK

+=

=

=


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MS5019 FEM 25

Homogenoue BCTo illustrate the two general types of BC, let us consider Eq. (2.4.6),

derived for the spring assemblage of Figure 2-4. We will first consider

the case of homogenous BC. Hence, all BC are such that the

displacements are zero at certain nodes.Here we have d1x = 0 because

node 1 is fixed. Therefore, Eq. (2.4.6) can be written as.

xxx

xxx

xxx

x

x

x

x

x

Fdkkdkk

Fdkdk

Fdkdk

F

F

F

d

d

kkkk

kk

kk

3321221

23222

13121

3

2

1

3

2

2121

22

11

)()0(

)5.5.2()0(0

)0()0(

becomesfromexpandedinwritten(2.5.4),Equation

)4.5.2(

0

0

0

=++

=+

=+

=

+

MS5019 FEM 26

We have now effectively partitioned off the first column and row ofK

and the first row ofd and F to arrive at Eq. (2.5.6).

For homogenous BC, Eq. (2.5.6) could have been obtained directly by

deleting the row and column of Eq. (2.5.4) corresponding to the zero-

displacement degree of freedom. Here row 1 and column 1 are deleted

because d1x = 0. However, F1x is not necessary zero and must be

determined as follows.

)6.5.2(

havewe

form,matrixinwritten(2.5.5),Eqs.ofthirdandsecondheConsider t

3

2

3

2

212

22

=

+

x

x

x

x

F

F

d

d

kkk

kk


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MS5019 FEM 27

)9.5.2(

isresultsThe(2.5.8).Eq.intostituted

-subfor(2.5.7)Eq.usingbyandforcesnodalappliedofterms

inforce)(the1nodeatforcenodalunknowntheexpresscanWe

)8.5.2(

asreactionobtain thewe(2.5.5),Eqs.offirstin the(2.5.7)Eq.Using

)7.5.2(11

11

2

1havewe,andfor(2.5.6)Eq.solvingAfter

321

332

311

1

3

2

11

11

3

2

1

212

22

3

2

32

xxx

xxx

xx

x

x

x

x

x

x

x

xx

FFF

dFF

reaction

dkF

F

F

F

kk

kkk

F

F

kkk

kk

d

d

dd

=

=

+

=

+

=

MS5019 FEM 28

Note for homogenoue BC.

For all homogenous BC, we can delete the rows and culumns

corresponding to the zero-displacement dof from the original set of

equatons and then solve for the unknown displacements.

This procedure is useful for hand calculations. (More practical,

computer-assisted scheme for solving the system of simultaneousequations will be discussed in another chapter.)

Students are encouraged to review the numerical method to solve the

system of simultaneous equations, i.e. Gauss or Gauss-Seidel method or

other.


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MS5019 FEM 29

Non-homogeneous BC

Next, we consider the case ofnon-homogenous BC. Hence, some of the

specified displacements are non-zero. For simplicitys sake, let d1x = ,

where is a known displacement, in the Eq. (2.4.6). We now have

xxx

xxx

xxx

x

x

x

x

x

Fdkkdkk

Fdkdk

Fdkdk

F

F

F

d

d

kkkk

kk

kk

3321221

23222

13121

3

2

1

3

2

2121

22

11

)(

)11.5.2(0

0becomesformexpandedinwritten(2.5.10)Equation

)10.5.2(0

0

=++

=+

=+

=

+

MS5019 FEM 30

)14.5.2(

haveweform,matrixin(2.5.13)Eqs.Rewritting

)13.5.2()(

inresults(2.5.12)Eqs.ofsiderightthetoknownthengTransformi

)12.5.2()(

0

andforcesnodalside-rightknown

havetheybecause(2.5.11)Eqs.ofthirdandsecondthegConsiderin

31

2

3

2

212

22

3132122

23222

3321221

23222

32

+=

+

+=++=

=++

=+

x

x

x

x

xxx

xxx

xxx

xxx

xx

Fk

F

d

d

kkk

kk

Fkdkkdk

Fdkdk

Fdkkdkk

Fdkdk

FF


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MS5019 FEM 31

Note for homogenoue BC.

When dealing with non-homogeneous BC, one cannot initially delete

row 1 and column 1 of Eq. (2.5.1)), corresponding to the non-

homogeneous BC, as indicated by the resulting Eq. (2.5.14). Had we

done so, the k1 term in Eq. (2.5.14) would have been neglected,resulting in an error in the solution for the displacement.

For non-homogeneous BC, we must, in general, transform the terms

associated with the known displacement to the right-side force matrix

before solving for the unknown nodal displacements. This was illustrated

by transforming the k1 term of the second row of Eqs. (2.5.12) to theright-side of the second row of Eqs. (2.5.13).

Finally, we could solve for the displacement in Eq. (2.5.14) in a manner

similar to that used to solve Eq. (2.5.6).

MS5019 FEM 32

Some properties of the stiffness matrix (that are also applicable to the

generalization of the finite element method).

1. K is symetric, as is each of the element stiffness matrices.

2. K is singular and thus no inverse exists until sufficient BC are

imposed.

3. The main diagonal terms ofK are always positive. Otherwise, apositive nodal force Fi could produce a negative displacement di a

behavior contrary to the physical behavior of any actual structure.

Example 2.1


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MS5019 FEM 33

2.6.Potential Energy ApproachOne of the alternative methods often used to derive the element

equations and the stiffness matrix for an element is based on the

principle of minimum potential energy (POMPE).

Adantages of the POMPE:

1. More general than previous method (equilibrium equations of nodal and

element).

2. More adaptable for the determination of element equations for

complicated elements (those with large numbers of dof) such as plane

stress/strain element, the axisymetric stress element, the bending plate

element, and the 3D solid stress element).

3. Only applicable for elastic materials (principle of virtual work is

applicable for any material behavior).

4. Has lack of physical insight.

MS5019 FEM 34

The total potential energy (TPE), p of a structure is expressed in

term of displacements. In the FE formulation, these will generally be

nodal displacements such that p = p (d1, d2, d3, ..., dn).

When p is minimized with respect to these displacements,

equilibrium equations result. For the spring element, we will show

that the same nodal equilibrium equations result as previously

derived in Section 2.2. The POMPE is stated as follows:

Of all the displacements that satisfy the given boundary conditions of

a structure, those that safisfy the equations of equilibrium are

distinguishable by the stationary value of the potential energy. If the

stationary value is a minimum, the equilibrium state is stable.

To explain this principle, we must first explain the concepts of PE

and of a stationary value of a function.


18/22

MS5019 FEM 35

Total potential energy (PTE) is defined as the sum of the internal

strain energy Uand the potential energy of the external force ;thus is

Strain energy is the capacity of internal forces (or stresses) to do

work through deformation (strains) in the structure.

Potential energy (PE) of the external force , is the capacity ofexternal forces such body force, surface traction forces, and applied

nodal forces to do work through deformation of the structure.

)1.6.2(+=Up

PE of external forceStrain energy

MS5019 FEM 36

The differential internal work (or strain energy) dUin the spring is

the internal force multiplied by the change in displacement through

which the force moves, given by.

F

x

k

Figure 2-5 Force-deformation curve for linear spring

k

F

x

Recall that a linear spring has force related to deformation by F= kx,

where kis the spring constant andx is the deformation of the spring

(Figure 2-5).

)2.6.2(dxFdU=


19/22

MS5019 FEM 37

curve.ndeformatio-force

underareatheisenergystrainthatindicates(2.6.7)Equation

)7.6.2()(

havewe(2.6.6),Eq.in(2.6.3)Eq.Using

)6.6.2(

obtainwe(2.6.5),Eq.ofnintegratioexplicitUpon

)5.6.2(

bygiventhenisenergystraintotalThe

)4.6.2(

becomesenergystrainaldifferentithe(2.6.2),Eq.in(2.6.3)Eq.Using

)3.6.2(asexpressweNow

21

21

2

2

1

0

FxxkxU

kxU

dxkxU

dxkxdU

kxFF

x

==

=

=

=

=

MS5019 FEM 38

The PE of external force, being opposite in sign from the external

work expression because the PE of external force is lost when the

work is done by the external force, is given by

)9.6.2(

becomesTPEthe

(2.6.1),into(2.6.8)and(2.6.6)Eqs.ngsubstitutiTherefore

)8.6.2(

2

21 Fxkx

Fx

p =

=

To apply the POMPE that is, to minimize ofp we take the

variation of p defined in general as

)10.6.2(22

1

1

n

n

ppp

p dd

dd

dd

++

+

= L


20/22

MS5019 FEM 39

The principles states that equilibrium exists when the di define astructure state such that p = 0 for arbitrary admissible variations

di from the equilibrium state. An admissible variation is one inwhich the displacement field still satisfies the BC and inter-element

continuity.

{ }

structure.theofstatemequilibriustatic

definethatdiofvaluesfor thesolvedbemustequationswhere

)11.6.2(0or),,3,2,1(0

Thustly.independenzerobemustthe

withassociatedtscoefficienall),0(for0satisfyTo

nn

dni

d

d

d

p

i

p

i

ip

=

==

=

L

MS5019 FEM 40

Equation (2.6.11) shows that for our purposes throughout this text,

we can interpret the variation ofp (p) as a compact notation

equivalent to differentiation ofp with respect to the unknown nodal

displacements for which p is expressed.

We now derive the spring element equations and stiffness matrix

using the POMPE by analyzing a linear single-dof spring as shown

in Figure 2-6.

k

xf2

xf1

1 2

L

Figure 2-6 Linear spring subjected to nodal forces


21/22

MS5019 FEM 41

( )

( )

( )

( ))14.6.2(

022

022

thatrequires

ntdisplacemenodaleachrespect towithofonMinimizati

)13.6.2(2

obtainwe(2.6.12),Eq.gSimplifyin

(2.6.9).Eq.inspringtheofndeformatiotheiswhere

)12.6.2(becomesTPEthe(2.6.9),Eq.Using

21221

2

11221

1

2211

2

112

2

221

12

2211

2

1221

==

=+=

+=

=

xxx

x

p

xxx

x

p

p

xxxxxxxxp

xx

xxxxxxp

fddkd

fddkd

dfdfddddk

dd

dfdfddk

MS5019 FEM 42

(2.2.18).Eq.2.2,Sectioninobtained

matrixstiffnessthetoidenticalis(2.6.17)Eq.expected,As

)17.6.2(

:(2.6.16)Eq.fromobtainedelement

springfor thematrixstiffnessthehavewe,Since

)16.6.2(

as(2.6.15)Eq.expressweform,matrixIn

)15.6.2()(

)(

havewe(2.6.14),Eq.gSimplifyin

2

1

2

1

212

112

=

=

=

=

=+

kk

kk

f

f

d

d

kk

kk

fddk

fddk

x

x

x

x

xxx

xxx

k

dkf


22/22

MS5019 FEM 43

We have developed the FE spring element equations by minimizing

the TPE with respect to the nodal displacements. Now, we show that

the TPE of an entire structure (here an assemblage of spring

elements) can be minimized with respect to each nodal dof and this

minimization results in the same FE equations used for the solution

as those obtained by the direct stiffness method explained in Section

2.4.

Example 2.5

MS5019 FEM 44

Reference:

1. Logan, D.L., 1992, A First Course in the Finite ElementMethod, PWS-KENT Publishing Co., Boston.

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

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

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Ch 2 Stiffness Method 09