1

MS5019 – FEM 1

MS5019 – FEM 2

5.1. In-plane Problem and Plane ElementThe previous chapters have discussed 1-D elements (line elements) in the form of bar, truss, beam, and frame elements.One order higher than one-dimensional elements are plane elements, or 2-D elements. The simplest two-dimensional element is the membrane element, which can transfer only in-plane loads; there are no out-of-plane loads. Consequently, membrane elements can carry only in-plane stress but cannot transfer bending moments or torsion. Figure 5-1 shows structures which can be modelled by membrane elements.The simplest of membrane elements is so-called constant strain triangle (CST). The assumption for this element is that the element displacement field is linear, which thus yields constant strains within the element. This chapter will discuss the derivation of the CST element equations using POMPE.

2

MS5019 – FEM 3

Figure 5-1 Structures which can be modelled by membrane elements

(b) Spanner under line-pressure load

(a) Plate with hole under in-plane axial load

MS5019 – FEM 4

5.2. Plane Stress and Plane StrainPlane problem can be categorized either as plane stress or plane strain. Consider the in-plane loaded plate shown in Figure 5-2.

x

z y

Figure 5-2 A plate under in-plane loading

The plane stress condition dictates that the stress component in the z-direction is equal to zero. This situation occurs if the thickness of the plate (in z-direction) is very small.The plane strain condition prevails if the strain in z-direction is very small (zero). Thick structures are normally treated as plane strain problems.

3

MS5019 – FEM 5

{ }

)4.2.5(][}{

)3.2.5(}]{[}{

)2.2.5(}]{[}{

)1.2.5()(

Txyyx

xv

yu

yvxu

xy

y

x

D

yvxu

dB

τσσσ

εσ

ε

γεε

ε

=

=

=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

+=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

=

∂∂

∂∂

∂∂∂∂

as written becan components stress thewhere

bygiven is iprelationshstrain -stress Thedirection.- in thent displaceme

theis anddirection - in thent displaceme theis where

or

bygiven is plate for the iprelationshplacement strain/dis The

MS5019 – FEM 6

)7.2.5(}]{][[}{

obtain we(5.2.3), Eq. into (5.2.2) Eq. ngsubstitutiby case,any In

)6.2.5(00

0101

)21)(1(][

:problems For

)5.2.5(00

0101

1][

:problems For strain. planeor stress

plane as dcategorize is problem theon whether depends ][ of valueThe

221

21

2

dBD

ED

ED

D

=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

−

−+=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−=

−

−

σ

νννν

νν

νν

ν

ν

ν

strain plane

stress plane

4

MS5019 – FEM 7

Figure 5-3 Basic triangular element

5.3. Derivation of CST Element Stiffness MatrixStep 1 Select Element TypeConsider the basic triangular element shown in Figure 5-3, with nodes i, j, and m labeled in a CCW manner. Here, all information are based on this CCW system of labelling. It is important that a consistent labelling system for the whole body is used to avoid problems in the calculations such as negative element areas.

),( ii yxi

),( mm yxm

),( jj yxj

Here (xi, yi), (xj, yj), and (xm, ym) are the known nodal coordinates of nodes i, j, and m respectively. The nodal displacement of nodes i, j, and m are (ui, vi),(uj, vj), and (um, vm) respectively.

iuju

mu

jviv

mv

MS5019 – FEM 8

)1.3.5(}{}{}{

}{

bygiven ismatrix nt displaceme nodal The

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

=⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧=

m

m

j

j

i

i

m

j

i

vuvuvu

ddd

d

Step 2 Select Displacement FunctionsWe select a linear displacement function for each element as.

)2.3.5(),(),(

654

321

yaxaayxvyaxaayxu

++=++=

The linear function ensures that compatibility will be satisfied.

5

MS5019 – FEM 9

)3.3.5(1000

000133

}{

as expressed becan , and functions thestoreswhich ,}{function nt displaceme general The

6

5

4

3

2

1

321

321

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

⎥⎦

⎤⎢⎣

⎡=

⎭⎬⎫

⎩⎨⎧

++++

=Ψ

Ψ

aaaaaa

yxyx

yxaayxaa

vu

MS5019 – FEM 10

)5.3.5(111

as formmatrix in expressed (5.3.4) Eqs. of efirst thre the withbeginning s'for solvecan Weetc. ),,(),,( where

)4.3.5(

yield to(5.3.2) Eq. into dsubstituteare scoordinate nodal thefirstly, (5.3.2), Eq.in toof valuesobtain the To

3

2

1

654

654

654

321

321

321

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

==++=++=++=++=++=++=

aaa

yxyxyx

uuu

ayxuuyxuuyaxaavyaxaavyaxaavyaxaauyaxaauyaxaau

mm

jj

ii

m

j

i

iiiiii

mmm

mjj

iii

mmm

jjj

iii

6

MS5019 – FEM 11

becomes formexpansion on which ],[ oft determinan theis

)8.3.5(111

2

where

)7.3.5(21][

as found becan ][ of inverse themethod,cofactor Using(5.3.5). Eq. of sideright on thematrix 33 theis ][ where

)6.3.5(}{][}{obatin wes,a'for solvingor

1

1

xyxyxyx

A

Ax

xx

uxa

mm

jj

ii

mji

mji

mji

=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

=

×=

−

−

γγγβββααα

MS5019 – FEM 12

obtaincan we(5.3.6), Eqs. of last three theusing Similarly,

)11.3.5(21

as formmatrix in expressed becan (5.3.6) Eq. known, ][With

)10.3.5(

and triangle, theof area theis Here

)9.3.5()()()(2

3

2

1

1-

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

=⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

−=−=−=−=−=−=

−=−=−=

−+−+−=

m

j

i

mji

mji

mji

ijmmijjmi

jimimjmji

jijimimimjmjmji

jimimjmji

uuu

Aaaa

x

xxxxxxyyyyyy

xyyxxyyxxyyx

A

yyxyyxyyxA

γγγβββααα

γγγβββααα

7

MS5019 – FEM 13

obtain we(5.3.13), Eq. into (5.3.11) Eq. ngSubstituti

)13.3.5(]1[}{

have weform,matrix in expressed(5.3.2) Eq. begin with We. and,, ntsdisplaceme nodalunknown

theand ,,,, variablescoordinateknown the, and variablescoordinate theof in term manner) analogousan in derived becan ( }{ of

),( function,nt displaceme general thederive ready to are weNow

)12.3.5(21

3

2

1

i

i

6

5

4

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧=

Ψ

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

=⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

aaa

yxu

uuyx

vyxux

vvv

Aaaa

mj

mj

m

j

i

mji

mji

mji

γγαα

γγγβββααα

L

MS5019 – FEM 14

)14.3.5()(

)()(21)},({

as obtained becan ),(function nt displaceme theSimilarly,

)13.3.5()(

)()(21)},({

as written becan grearrangin andtion multiplicaafter which

)12.3.5(]1[21}{

⎭⎬⎫

⎩⎨⎧

+++

+++++=

⎭⎬⎫

⎩⎨⎧

+++

+++++=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

=

mmmm

jjjjiiii

mmmm

jjjjiiii

m

j

i

mji

mji

mji

vyxvyxvyx

Ayxv

yxv

uyxuyxuyx

Ayxu

uuu

yxA

u

γβαγβαγβα

γβαγβαγβα

γγγβββααα

8

MS5019 – FEM 15

)17.3.5(),(),(

}{

)16.3.5(),(),(

)15.3.5(

)(21

)(21

)(21

⎭⎬⎫

⎩⎨⎧

++++

=⎭⎬⎫

⎩⎨⎧

=Ψ

++=++=

++=

++=

++=

mmjjii

mmjjii

mmjjii

mmjjii

mmmm

jjjj

iiii

vNvNvNuNuNuN

yxvyxu

vNvNvNyxvuNuNuNyxu

yxA

N

yxA

N

yxA

N

vu

obtain weform,matrix in (5.3.16) Eqs. Expressing

as (5.3.14) and (5.3.13) Eqs. rewritecan we(5.3.15), Eqs. using Thus,

define weform,simpler in andfor (5.3.14) and (5.3.13) Eq. express To

γβα

γβα

γβα

MS5019 – FEM 16

)20.3.5(000

000][

bygiven is ][ where

)19.3.5(}]{[}{

have weform,matrix dabbreviatein (5.3.18) Eq. expressing Finally,

)18.3.5(000

000}{

or

⎥⎦

⎤⎢⎣

⎡=

=Ψ

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

⎥⎦

⎤⎢⎣

⎡=Ψ

mji

mji

m

m

j

j

i

i

mji

mji

NNNNNN

N

N

dN

vuvuvu

NNNNNN

9

MS5019 – FEM 17

Matrix [N] is the shape function matrix of the CST. This is the same in concept as the shape matrix of the previous 1-D elements. Eq. (5.3.19) express the general displacements as functions of {d} in terms of the shape functions Ni, Nj, and Nm. As the shape functions are linear, the displacement field, within the element, is also linear. A linear displacement field yields a constant strain field in the element.For instance, Ni represents the shape of the variable u when plotted over the surface of the element for ui = 1 and all other d.o.f. Equal to zero, that is, uj = um = vi = vj = vm = 0. In addition, u(xi,yi) must equal to ui. Therefore, we must have Ni = 1, Nj = 0, and Nm = 0 at (xi,yi). Similarly, u(xj,yj) = uj. Therefore, Ni = 0, Nj = 1, and Nm = 0 at (xj,yj).Finally, Ni + Nj + Nm = 1 for all x and y locations on the surface of the element. The proof of this relationship follows that given for the bar element in Section 3.2.

MS5019 – FEM 18

Step 3 Define the Strain/Displacement & Stress/Strain RelationshipsWe express the element strains and stresses in terms of the unknown nodal displacements.

( )

)23.3.5(, or

)22.3.5(,

have wents,displaceme for the (5.3.16) Eqs. Using

)21.3.5()(

}{

bygiven areelement 2D with theassociatedstrain The

,,, mxmjxjixix

mmjjiix

yv

xu

yvxu

xy

y

x

uNuNuNu

uNuNuNx

uxu

++=

++∂∂

==∂∂

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

+=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

=

∂∂

∂∂

∂∂∂∂

γεε

ε

StrainsElement

10

MS5019 – FEM 19

( )

( ) )26.3.5(21

have we(5.3.23), Eq.in (5.3.25) and (5.3.24) Eqs. using Therefore,

)25.3.5(2

and 2

similarly, and,

)24.3.5(22

1expressed becan (5.3.23) Eq.

in functions shape theof sderivative the(5.3.15), Eq. Using.0 and 0 similarly, alue;constant v a is ),( cause-be 0 used have We . e.g. able, that varirespect to

ation withdifferentiindicatesvariableaby followed coma thewhere

,,

,

,,

,,

mmjjii

mxm

jxj

iiiixi

xmxjiii

xiixi

uuuAx

u

AN

AN

Ayx

AN

uuyxuuuxNN

βββ

ββ

βγβα

++=∂∂

==

=++=

====∂∂=

MS5019 – FEM 20

( )

( )

)28.3.5(000000

2A1}{

yields (5.3.21) Eq.in (5.3.27) and (5.3.26) Eq. ngSubstituti

)27.3.5(

21

21

obtain weSimilarly,

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

=

+++++=∂∂

+∂∂

++=∂∂

m

m

j

j

i

i

mmjjii

mji

mji

mmjjiimmjjii

mmjjii

vuvuvu

vvvuuuAy

vxu

vvvAy

v

βγβγβγγγγ

βββε

βββγγγ

γγγ

11

MS5019 – FEM 21

[ ]

(CST). anglestrain triconstant a callediselement thehence, constant; be will(5.3.31) Eq.in strains The

)32.3.5(][ ][ where)31.3.5(}]{[}{

as written becan (5.3.29) Eq. form, simplifiedIn

)30.3.5(][][][

where

)29.3.5(}{}{}{

}{ or

00

2

10

0

2

10

0

2

1

mji

mm

m

m

m

jj

j

j

j

ii

i

i

i

m

j

i

mji

BBBBdB

BBB

ddd

BBB

AAA

==

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧=

ε

ε

βγγ

β

βγγ

β

βγγ

β

MS5019 – FEM 22

)34.3.5(}]{][[}{

)33.3.5(][

dBD

D

D

xy

y

x

xy

y

x

=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

=⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

σ

γεε

τσσ

written becan iprelationshain stress/str the(5.3.33), Eq.in (5.3.31) Eq. ngSubstituti problems.strain planefor 6) (5.2. Eq.by

and problems stress planefor 5) (5.2. Eq.by given is ][ where

bygiven is iprelationshain stress/str plane-in thegeneral,In ipRelationsh ainsStress/Str

12

MS5019 – FEM 23

Step 4 Derive the Element Stiffness Matrix and Equations

Using the POMPE, we can generate the equations for a typical CSTelement. The total PE of the element is a function of the nodal displacements ui, vi, uj, ..., vm such that

)38.3.5(}}{{}{21

have we), (5.3.33 Eq. usingor

)37.3.5(}{}{21

bygiven isenergy strain thewhere)36.3.5(

bygiven is PE totalThe)35.3.5(),,,(

∫∫∫

∫∫∫

=

=

Ω+Ω+Ω+=

=

V

V

dVDU

dVU

U

vuvu

T

T

spbp

mjiipp

εε

σε

π

ππ L

MS5019 – FEM 24

act. }{ tractions theover which surface therepresents and )lb/inor N/min (typically tractionssurface therepresents }{ where

)41.3.5(}{}{bygiven is ) tractionssurface(or loads ddistribute theof PE The

loads. nodal edconcentrat the}{ and ntsdisplaceme nodal therepresents }{ where)40.3.5(}{}{

bygiven is loads edconcentrat theof PE The

).lb/inor N/min (typicallydensity or weight t volumeweight/unibody theis }{ and function,nt displaceme general again the is }{ where

)39.3.5(}{}{bygiven isforcesbody theof PE The

22

33

TST

dST

PdPd

X

dVX

S

Tp

Tp

Tb

V

∫∫

∫∫∫

Ψ−=Ω

−=Ω

Ψ

Ψ−=Ω

13

MS5019 – FEM 25

[ ]

is that element;an on }{ system load totaltherepresent (5.3.43) Eq. of termslast three that theseecan we(5.3.41), - (5.3.39) Eqs. From

)43.3.5(}{][}{}{}{

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

Therefore, (5.3.42). Eqs. of integral theofout taken becan }{ so s,coordinate general oft independen are }{nt displaceme nodal The

)42.3.5(}{][}{}{}{

}{][}{}]{][[][}{have we(5.3.41), Eq.

- (5.3.38) Eq.in strain for the (5.3.31) Eq. and }{for (5.3.19) Eq. g Usin

21

21

f

dSTNdPd

dVXNdddVBDBd

dx-yd

dSTNdPd

dVXNddVdBDBd

S

TTT

TTTTp

S

TTT

TTTTp

VV

VV

⎥⎦

⎤⎢⎣

⎡−−

⎥⎥⎦

⎤

⎢⎢⎣

⎡−

⎥⎥⎦

⎤

⎢⎢⎣

⎡=

−−

−=

Ψ

∫∫

∫∫∫∫∫∫

∫∫

∫∫∫∫∫∫

π

π

MS5019 – FEM 26

)46.3.5(0}{}{]][[][}{

obtain we4), and3 Chaptersin elements beam andbar for the done wasas ),( since

ntsdisplaceme nodal therespect to with of derivative partial theTaking

)45.3.5(}{}{}{]][[][}{21

obtain we(5.3.43), Eq.in (5.3.44) Eq. Usingly.respectivetractions,surface theand forces, nodal edconcentrat theforces,body epresent th

-re (5.3.44) Eq. of sideright on the terms thirdand second, first, thewhere

)44.3.5(}{][}{}{][}{

p

pp

p

p

=−⎥⎥⎦

⎤

⎢⎢⎣

⎡=

∂

∂

=

−⎥⎥⎦

⎤

⎢⎢⎣

⎡=

++=

∫∫∫

∫∫∫

∫∫∫∫∫

fddVBDBd

fdddVBDBd

dSTNPdVXNf

V

V

V

T

TTT

S

TT

π

πππ

π

d

14

MS5019 – FEM 27

(5.3.6).or (5.3.5) Eq.bygiven is ][ and (5.3.32), Eq.by given is ][ (5.3.9), Eq.by given is where

)50.3.5(]][[][][ yield tointegral theof

outcan taken it element, CSTfor y and x offunction anot is integrand theAs

)49.3.5(]][[][][ becomes (5.3.48) Eq. , hicknessconstant t ofelement an For

)48.3.5(]][[][][ seen thatbeen can it (5.3.47) Eq. from },{}]{[ that Knowing

)47.3.5(}{}{]][[][

have we(5.3.46), Eq. Rewriting

DBABDBAtk

dydxBDBtkt

dVBDBkfdk

fddVBDB

T

A

T

T

T

V

V

=

=

=

=

=⎥⎥⎦

⎤

⎢⎢⎣

⎡

∫∫

∫∫∫

∫∫∫

MS5019 – FEM 28

)51.3.5(

bygiven isequation element The

them).offunction a is ][ (as and properties mechanical theof ands)coordinate nodal of in terms defined are ][ and (because scoordinate nodal

theoffunction aisandmatrix 66ais][element, CST ofmatrix stiffness The

3

3

2

2

1

1

666564636261

565554535251

464544434241

363534333231

262524232221

161514131211

3

3

2

2

1

1

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

=

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

×

vuvuvu

kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk

ffffff

DEBA

k

y

x

y

x

y

x

ν

15

MS5019 – FEM 29

Step 5 Assembling of Element Stiffness Matrix and Introduce BC

We obtain the global structure stiffness matrix and equations by using the direct stiffness method as

(5.3.44). Eq. usingly consistentby or loads) nodal edconcentrat(including nodesproper at the loads ddistribute and forcesbody

lumpingby obtained forces nodal global equivalent of vector theis

)54.3.5(}{}{

and system, scoordinate global theof termsin defined are matrices stiffnesselement all (5.3.52), Eq.in where,

)53.3.5(}]{[}{and

)52.3.5(][][

1

)(

1

)(

∑

∑

=

=

=

=

=

N

e

e

N

e

e

fF

x-y

dKF

kK

MS5019 – FEM 30

In the formulation of the element stiffness matrix Eq. (5.3.50), the matrix has been derived for a general orientation in global coordinates. All element matrices are expressed in the global-coordinates orientation. Therefore, no transformation from local to global equations is necessary.

If the local axes for the triangular element are not parallel to the global axes, we must apply matrix transformation similar to those introduced in Chapter 3. We illustrate the transformation of axes for the triangular element shown in Figure 5-4. Local nodal forces are shown in the figure.

The transformation from local to global equations follows the procedure outlined in Section 3.4. We have the same general expressions, Eqs. (3.4.14), (3.4.16), and (3.4.22), to relate local to global displacements, forces, stiffness matrices, respectively; that is.

)55.3.5(ˆˆˆ TkTkTffTdd T===bemust (5.3.55) Eqs.in used matrix ation transformfor the (3.4.15) Eq. where T

16

MS5019 – FEM 31

Figure 5-4 Triangular element with local axes not parallel to the global axes

y

xi

m

j

myf

xyθ

mxf

jxf

jyf

ixfiyf

4.-5 Figurein shown is and,sin,cos where

)51.3.5(

toexpanded is (3.4.15) Thuselement. CST in thepresent are d.o.f. additional twobecause expanded

00000000

0000000000000000

θθθ ==

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=

−

−

−

SCCSSC

CSSC

CSSC

T

MS5019 – FEM 32

Step 6 Solve for the Nodal Displacements

We determine the unknown global structure nodal displacement by solving the system of algebric equations given by Eq. (5.3.53).

Step 7 Solve for the Elements Forces (Stresses)

Having solved for the nodal displacements, we obtain the strains and stresses in the global x and y directions in the elements by using Eqs. (5.3.31) and (5.3.34). Finally, we determine the maximum and minimum in-plane principal stresses σ1 and σ2 by using Mohr circle formula, where these stresses are usually assumed to act at the centroid of the element.

EXAMPLE 7.1 and 7.2

17

MS5019 – FEM 33

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.