FEM: Bars and Trusses

FEM: Bars and Trusses

Mohammad Tawfik #WikiCourses

http://WikiCourses.WikiSpaces.com

Introduction to the Finite

Element Method

Bars and Trusses




Bar Example (Ex. 4.5.2, p. 187)

• Consider the bar shown in the above figure.

• It is composed of two different parts. One steel tapered part, and uniform Aluminum part.

• Calculate the displacement field using finite element method.




Bar Example

• The bar may be represented by two

elements.

• The stiffness matrices of the two elements

may be obtained using the following

integration:

2

1

2

122

2221

2

1

11

11x

x

ee

ee

x

x

e dx

hh

hhxEAdx

dx

d

dx

d

dx

ddx

d

xEAK




Bar Example

• For the Aluminum bar: E=107 psi, and A=1

in2. we get:

• For the Steel bar: E=38107 psi, and

A=(1.5-0.5x/96) in2. we get:

11

11

120

10

11

11

120

10 7

2

7 2

1

x

x

Al dxK

11

11

96

10.75.4

11

11

96

5.05.1

96

10.3 7

2

7 2

1

x

x

Fe dxx

K




Bar Example

• Assembling the Stiffness matrix and

utilizing the external forces, we get:

• The boundary conditions may be applied

and the system of equations solved.

0

0

10

10.2

0

33.833.80

33.88.575.49

05.495.49

105

5

3

2

1

4

R

u

u

u




Bar Example

• Solving, we get:

• For the secondary

variables:

inu

u

181.0

061.0

3

2

lbR 30000




Trusses

• A truss is a set of bars that are connected

at frictionless joints.

• The Truss bars are generally oriented in

the plain.




Trusses

• Now, the problem lies in the

transformation of the local displacements

of the bar, which are always in the

direction of the bar, to the global degrees

of freedom that are generally oriented in

the plain.




Equation of Motion

0

0

0000

0101

0000

0101

2

1

2

2

1

1

F

F

v

u

v

u

h

EA




Transformation Matrix

DOF

dTransformeDOFLocal

v

u

v

u

CosSin

SinCos

CosSin

SinCos

v

u

v

u

2

2

1

1

2

2

1

1

00

00

00

00

DOF

dTransformeDOFLocal T




The Equation of Motion Becomes

• Substituting into the

FEM:

• Transforming the

forces:

• Finally:

FTK

FTTKTTT

FK




Recall

TKTKT

CosSin

SinCos

CosSin

SinCos

T

00

00

00

00

Where:

0000

0101

0000

0101

h

EAK




Element Stiffness Matrix in Global

Coordinates

CosSin

SinCos

CosSin

SinCos

CosSin

SinCos

CosSin

SinCos

h

EAK

00

00

00

00

0000

0101

0000

0101

00

00

00

00




Element Stiffness Matrix in Global

Coordinates

22

22

22

22

22

12

2

1

22

12

2

1

22

12

2

1

22

12

2

1

SinSinSinSin

SinCosSinCos

SinSinSinSin

SinCosSinCos

h

EAK




Example: 4.6.1 pp. 196-201

• Use the finite element analysis to find the

displacements of node C.




Element Equations

0000

0101

0000

0101

1

L

EAK

1010

0000

1010

0000

2

L

EAK

3536.03536.03536.03536.0

3536.03536.03536.03536.0

3536.03536.03536.03536.0

3536.03536.03536.03536.0

3

L

EAK




Assembly Procedure

3536.13536.0103536.03536.0

3536.03536.0003536.03536.0

101000

000101

3536.03536.0003536.03536.0

3536.03536.0013536.03536.1

L

EAK




Global Force Vector

P

P

F

F

F

F

F

F

F

F

F

F

Fy

x

y

x

y

x

y

x

y

x

2

2

2

1

1

3

3

2

2

1

1

Remember!

NO distributed load

is applied to a truss




Boundary Conditions

02211 VUVU

Remove the corresponding rows and columns

P

P

V

U

L

EA

23536.13536.0

3536.03536.0

3

3

Continue! (as before)




Results

EA

PLV

EA

PLU

3 ,828.5 33

PFF

PFPF

yx

yx

3 ,0

, ,

22

11




Postcomputation

e

e

e

e

e

A

P

A

P 21

e

e

eee

e

u

u

L

EA

P

P

2

1

2

1

11

11

2

2

1

1

2

2

1

1

00

00

00

00

v

u

v

u

CosSin

SinCos

CosSin

SinCos

v

u

v

u




Postcomputation

A

P

A

P2 ,

3 ,0 )3()2()1(




Summary

• In this lecture we learned how to apply the

finite element modeling technique to bar

problems with general orientation in a

plain.

Date post:	15-Jul-2015
Category:	Education
Upload:	mohammad-tawfik
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FEM: Bars and Trusses

Education