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1/6

1

2D Truss

60o 60o

2.5m

25kN 1

2 3

A=6cm2

200GPa

'

'

1,

Uu

'

'1

1

,

V

v

1

'

2

2,

Uu

'

'2

2

,V

v

X

Y

'

X

' Y

Where

' 1

1 2

,U U are forces along the member (along local co-ordinate direction, 'X )' 1

1 2,V V are the forces perpendicular to the member (along local co-ordinate direction,'

Y )' ' ' '

1 2 1 2, , ,u u v v are the corresponding displacements

Let 1 1 1 1 2 2 2 2, , , , , , ,U V u v U V u v be the corresponding global values

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2

' ' '

1 1 2

' ' '

2 2 1

( )

( )

AEU u u

l

AEU u u

l

' '1 2 0v v

' '

1 1

' '

1 1

' '

2 2

' '

2 2

1 0 1 0

0 0 0 0

1 0 1 0

0 0 0 0

u U

v VAE

l u U

v V

' '

1 1

' '

' ' '1 1

' '

2 2

' '

2 2

1 0 1 0

0 0 0 0, ,

1 0 1 00 0 0 0

e e

u U

v VAEk d F

l u Uv V

but

' ' '

e ek d F

'e ed T d 'F T F

'e ek T d T F Multiplying both sides with

1T

1 1'

e eT k T d T T F

1 '

e e

e e

k d F

where k T k T

Solutioncos( ) sin( ) 0 0

sin( ) cos( ) 0 0

0 0 cos( ) sin( )

0 0 sin( ) cos( )

T

7/28/2019 2D Truss and Direct Stiffness

3/6

3

1

cos( ) sin( ) 0 0

sin( ) cos( ) 0 0

0 0 cos( ) sin( )

0 0 sin( ) cos( )

TT T

Hence

cos( ) sin( ) 0 0 1 0 1 0

sin( ) cos( ) 0 0 0 0 0 0

0 0 cos( ) sin( ) 1 0 1 0

0 0 sin( ) cos( ) 0 0 0 0

cos( ) sin( ) 0 0

sin( ) cos( ) 0 0

0 0 cos( ) sin( )0 0 sin( ) cos( )

e

AEk

l

2 2

2 2

2 2

2 2

e

C C S C C S

C S S C S S AEk

l C C S C C S

C S S C S S

2 2

2 2

1 11 2 2

1

2 2

e

C C S C C S

C S S C S S A E

k l C C S C C S

C S S C S S

1 11

1

0.25 0.433 0.25 0.433

0.433 0.75 0.433 0.75

0.25 0.433 0.25 0.433

0.433 0.75 0.433 0.75

e

A Ek

l

2 22

2

0.25 0.433 0.25 0.433

0.433 0.75 0.433 0.75

0.25 0.433 0.25 0.433

0.433 0.75 0.433 0.75

e

A Ek

l

AssemblyNumber of rows=columns=no. of nodes X degrees of freedom=3X2=6

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4

Applying Boundary conditions ( ) 4.8 [. .] =.

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5

Numerical example: Direct stiffness matrix method

Calculate the nodal displacements and element forces for the composite bar shown

in figure:

30cm 20cm 10cm

10cm

dia

5cm

dia

Steel

Aluminium

4P

P

P

P=5kN

Esteel=210GPa

EAl=70GPa

(1) (2) (3)

1 2 3 4Nodes

Elements

Element 1

Element 2

Element 3

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6

Global stiffness matrix

9

5.495 5.495 0 0

5.495 5.495 2.06 2.06 010

0 2.06 2.06 1.37375 1.37375

0 0 1.37375 1.37375

K

Applying the point loads and reaction at nodes

1

3

29

33

4

5.495 5.495 0 0

5.495 5.495 2.06 2.06 0 2 5 1010

0 2.06 2.06 1.37375 1.37375 00 0 1.37375 1.37375 4 5 10

au R

u

u

u

1

1

29 4

3

4

0

1 0 0 0 0

0 7.555 2.06 0 110 10

0 2.06 3.43375 1.37375 0

0 0 1.37375 1.37375 2

u

u

u

u

u

1

62

5

3

5

4

0

1.819 10

1.153 10

2.609 10

u

u

u

u