Finite Element Method - WordPress.com · Finite Element Method FEM FOR TRUSSES . 2 CONTENTS ... Consider a truss element D 3i - 1 D 3i - 2 D 3i D 3j - 1 D 3j - 2 D 3j l e x u 1 u

1

Finite Element Method

FEM FOR

TRUSSES

2

CONTENTS

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

3

INTRODUCTION

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.

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INTRODUCTION

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.

5

Example of a truss structure

6

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

7


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

8


0

0 1

1

( ) 1h T

T

u x x x

p α

pα

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 T

e

e e

e eN x N x

e

u ux xu x x x

u ul ll l

x

P α N d

d

N

9


)()()( 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)

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Strain 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

11

Element Matrices in the Local Coordinate

System

0

1

1 11 1d d

1 1 1

e

e

l eT

e

e e eV

e

l AEV A E x

l l l

l

k B cB

Note: ke is symmetrical

Proof: BcB][BcBB]cBTTTTTTT [

12

Element Matrices in the Local Coordinate

System

1 1 1 2

02 1 2 2

2 1d d

1 26

e

e

lT e

e 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 es

l sT T

e b s x

s x eV Ss

f lf

fNf V f S f x

fN f lf

f N N

13


system

Perform coordinate transformation

Truss in space (spatial truss) and truss in

plane (planar truss)

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system

Spatial truss

ee TDd (Relationship between local

DOFs and global DOFs)

where

eijijij

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 i

ij

e

j i

ij

e

j i

ij

e

X Xl x X

l

Y Ym x Y

l

Z Zn x Z

l

Direction cosines

(2x1)

(6x1)

15


system

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

16


system


ee TDd

eeeee fdmdk eeeee fDTmTDk

e

T

ee

T

ee

TfTDTmTDTkT )()(

eeeee FDMDK

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system


2 2

2 2

2 2

2 2

2 2

T

e 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 2

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

18


system


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

T

e 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

19


system


1

1

1

1

1

1

( )2

( )2

( )2

( )2

( )2

( )2

x es ij

x es ij

x es ij

T

e ey e

s ij

y e

s ij

y e

s 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

e

x es

f lf

f lf

f

20


system

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)

21


system

Planar truss (Cont’d)

2 2

2 2

2 2

2 2

ij ij ij ij ij ij

ij ij ij ij ij ijT

e 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

22


system

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

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Boundary conditions

Singular K matrix rigid body movement

Constrained by supports

Impose boundary conditions cancellation

of rows and columns in stiffness matrix,

hence K becomes SPD

Recovering stress and strain

x e eE E Bd BTD (Hooke’s law)

x

24

EXAMPLE

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

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EXAMPLE

Exact solution of 2

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

26

Remarks

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

27

HIGHER ORDER ELEMENTS

1 2 3 4 1 2 3

Quadratic element Cubic element

1

2

3

1( ) (1 )

2

1( ) (1 )

2

( ) (1 )(1 )

N

N

N

2

1

2

2

2

3

2

4

1( ) (1 )(1 9 )

16

1( ) (1 )(1 9 )

16

9( ) (1 3 )(1 )

16

9( ) (1 3 )(1 )

16

N

N

N

N

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Finite Element Method - WordPress.com · Finite Element Method FEM FOR TRUSSES . 2 CONTENTS ... Consider a truss element D 3i - 1 D 3i - 2 D 3i D 3j - 1 D 3j - 2 D 3j l e x u 1 u