1 Finite Element Method FEM FOR TRUSSES for readers of all backgrounds G. R. Liu and S. S. Quek CHAPTER 4:

1

FFinite Element Methodinite Element Method

FEM FOR TRUSSES

for readers of all backgroundsfor readers of all backgrounds

G. R. Liu and S. S. Quek

CHAPTER 4:

2Finite Element Method by G. R. Liu and S. S. Quek

CONTENTSCONTENTS INTRODUCTION FEM EQUATIONS

– Shape functions construction– Strain matrix– Element matrices in local coordinate system– Element matrices in global coordinate system– Boundary conditions– Recovering stress and strain

EXAMPLE– Remarks

HIGHER ORDER ELEMENTS


INTRODUCTIONINTRODUCTION

Truss members are for the analysis of skeletal type systems – planar trusses and space trusses.

A truss element is a straight bar of an arbitrary cross-section, which can deform only in its axis direction when it is subjected to axial forces.

Truss elements are also termed as bar elements. In planar trusses, there are two components in the x and y

directions for the displacement as well as forces at a node. For space trusses, there will be three components in the x,

y and z directions for both displacement and forces at a node.


INTRODUCTIONINTRODUCTION

In trusses, the truss or bar members are joined together by pins or hinges (not by welding), so that there are only forces (not moments) transmitted between bars.

It is assumed that the element has a uniform cross-section.


Example of a truss structureExample of a truss structure


FEM EQUATIONSFEM EQUATIONS

Shape functions constructionStrain matrixElement matrices in local coordinate systemElement matrices in global coordinate

systemBoundary conditionsRecovering stress and strain


Shape functions constructionShape functions construction

Consider a truss element

D3i - 1

D3i - 2

D3i

D3j - 1

D3j - 2

D3j

le

x

u1

u2

u(x)

fs1

fx

global node j local node 2

global node i local node 1

fs2

X

Y

Z

o

0



00 1

1

( ) 1h T

T

u x x x

p α

p α

Let

Note: Number of terms of basis function, xn determined by n = nd - 1

At x = 0, u(x=0) = u1

At x = le, u(x=le) = u2

1 0

2 1

1 0

1 e

u

lu

0 1

1 2

1 0

1 1

e e

u

ul l

1 2

1 1

2 2

( ) ( )

1 0

( ) 1 1 ( )1 1

( )

h Te

e ee e N x N x

e

u ux xu x x x

u ul ll l

x

P α N d

dN



)()()( 21 xNxNx N

1

2

( ) 1

( )

e

e

xN x

l

xN x

l

N1 N2

x

le 0

1 1

1 2

2 11 1 2 2 1( ) ( ) ( )

e

u uu x N x u N x u u x

l

(Linear element)


Strain matrixStrain matrix

2 11 1 2 2 1( ) ( ) ( )

e

u uu x N x u N x u u x

l

2 1x

e

u uu

x l

or

eex Lx

uBdNd

1 11

e e e e

x xL

x l l l l

B Nwhere


Element Matrices in the Local Coordinate Element Matrices in the Local Coordinate SystemSystem

0

1

1 11 1d d

1 1 1e

e

l eTe

e e eV

e

l AEV A E x

l l l

l

k B cB

Note: ke is symmetrical

Proof: BcB][BcBB]cB TTTTTTT [


Element Matrices in the Local Coordinate Element Matrices in the Local Coordinate SystemSystem

1 1 1 2

02 1 2 2

2 1d d

1 26

e

e

lT ee e

V

N N N N A lV A l x

N N N N

m N N

Note: me is symmetrical too

111

022

1

2d d d

2

e

e e

x esl sT T

e b s xs x eV S

s

f lffN

f V f S f xfN f l

f

f N N


Element matrices in global coordinate Element matrices in global coordinate systemsystem

Perform coordinate transformation

Truss in space (spatial truss) and truss in plane (planar truss)



Spatial truss

ee TDd (Relationship between local DOFs and global DOFs)

whereeijijij

ijijij

nml

nml

000

000T

j

j

j

i

i

i

e

D

D

D

D

D

D

3

13

23

3

13

23

D,

cos( , )

cos( , )

cos( , )

j iij

e

j iij

e

j iij

e

X Xl x X

l

Y Ym x Y

l

Z Zn x Z

l

Direction cosines

(2x1)

(6x1)



Spatial truss (Cont’d)

2 2 2( ) ( ) ( )e j i j i j il X X Y Y Z Z

Transformation applies to force vector as well:

ee TFf where

j

j

j

i

i

i

e

F

F

F

F

F

F

3

13

23

3

13

23

F




ee TDd

eeeee fdmdk eeeee fDTmTDk

eT

eeT

eeT fTDTmTDTkT )()(

eeeee FDMDK




2 2

2 2

2 2

2 2

2 2

Te e

ij ij ij ij ij ij ij ij ij ij



ij ij ij ij ij ij ij ij ij ije


ij

l l m l n l l m l n

l m m m n l m m m n

l n m n n l n m n nAE

l l m l n l l m l nl

l m m m n l m m m n

l n

K T k T

2 2ij ij ij ij ij ij ij ij ijm n n l n m n n




2 2

2 2

2 2

2 2

2 2

2 2 2

2 2 2

2 2 2

2 2 26

2 2 2

Te e



ij ij ij ij ij ij ij ij ij ije



ij

l l m l n l l m l n

l m m m n l m m m n

l n m n n l n m n nA l

l l m l n l l m l n

l m m m n l m m m n

l n

M T m T

2 22 2 2ij ij ij ij ij ij ij ij ijm n n l n m n n




1

1

1

1

1

1

( )2

( )2

( )2

( )2

( )2

( )2

x es ij

x es ij

x es ij

Te e

y es ij

y es ij

y es ij

f lf l

f lf m

f lf n

f lf l

f lf m

f lf n

F T f Note:1

1

2

2

x es

ex e

s

f lf

f lf

f



Planar truss

ee TDd

where

ijij

ijij

ml

ml

00

00T ,

j

j

i

i

e

D

D

D

D

2

12

2

12

D

j

j

i

i

e

F

F

F

F

2

12

2

12

FSimilarly (4x1)



Planar truss (Cont’d)

2 2

2 2

2 2

2 2

ij ij ij ij ij ij

ij ij ij ij ij ijTe e

ij ij ij ij ij ije

ij ij ij ij ij ij

l l m l l m

l m m l m mAE

l l m l l ml

l m m l m m

K T k T



Planar truss (Cont’d)

2 2

2 2

2 2

2 2

2 2

2 2

2 26

2 2

ij ij ij ij ij ij

ij ij ij ij ij ijT ee e

ij ij ij ij ij ij

ij ij ij ij ij ij

l l m l l m

l m m l m mA l

l l m l l m

l m m l m m

M T m T


Boundary conditionsBoundary conditionsSingular K matrix rigid body movementConstrained by supportsImpose boundary conditions cancellation

of rows and columns in stiffness matrix, hence K becomes SPD

Recovering stress and strainRecovering stress and strain

x e eE E Bd BTD (Hooke’s law)

x


EXAMPLEEXAMPLE

Consider a bar of uniform cross-sectional area shown in the figure. The bar is fixed at one end and is subjected to a horizontal load of P at the free end. The dimensions of the bar are shown in the figure and the beam is made of an isotropic material with Young’s modulus E.

P

l


EXAMPLEEXAMPLE

Exact solution of2

20x

uE f

x

( )

Pu x x

EA , stress: x

P

A :

FEM:

(1 truss element)1 1

1 1e

AE

l

K = k

1 1

2 2

?1 1

1 1

u FAE

u F Pl

1 1

2 2

?1 1

1 1

u FAE

u F Pl

2

Plu

AE 1

2

0( ) ( ) 1 1e

ux x x x Pu x x xPl

ul l l l EAEA

N d

2

01 1x e

PE E

ul l A

Bd


RemarksRemarks

FE approximation = exact solution in example Exact solution for axial deformation is a first order

polynomial (same as shape functions used) Hamilton’s principle – best possible solution Reproduction property


HIGHER ORDER ELEMENTSHIGHER ORDER ELEMENTS

1 2 3 4 1 2 3

Quadratic element Cubic element

1

2

3

1( ) (1 )

21

( ) (1 )2

( ) (1 )(1 )

N

N

N

21

22

23

24

1( ) (1 )(1 9 )

161

( ) (1 )(1 9 )16

9( ) (1 3 )(1 )

169

( ) (1 3 )(1 )16

N

N

N

N

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1 Finite Element Method FEM FOR TRUSSES for readers of all backgrounds G. R. Liu and S. S. Quek CHAPTER 4: