Direct Method Truss Analysis Matrice Di Rigidezza Composizione

7/31/2019 Direct Method Truss Analysis Matrice Di Rigidezza Composizione

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CCOORRNNEELLLLU N I V E R S I T Y 3MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (9/1/2011)

Truss analysis

Internal forces in a truss element act along

the member However, displacements at the nodes can

have both components (x- and y-directions,

in local coordinates). This is due to rotation

'( ) '( )

1 2 0e e

y yF F= =


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Truss analysis

To analyze a truss

element in the globalcoordinates xand y, youneed to account for bothcomponents ofdisplacement:

Also note that the crosssection of truss elementscan vary as shown.

( ) ( ) ( ) ( )

1 1 2 2, , ,

e e e e

x y x yu u u u


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Stiffness of a truss element The internal force in the

truss is given (see free body

diagram) as:

Assuming elasticdeformations:

The (small) strain is given as:

Finally:

( ) ( )

2 1

e e e e ep F F A = = =

( ) ( )

2 1

e e e e e ep F F A E = = =

2 1

e ee

e

u u

L

=

ep

( ) ( )

2 1 2 1

2 1

( )

( )

e ee e e e

e

e e e

A EF F u u

L

k u u

= = =

=

( )

1

eF ( )

2

eF

( ) ( )( ) ( )

1 1

( ) ( ) ( ) ( )

2 2

e ee e

e e e e

F uk k

F k k u

=

e e

e

e

A Ek

L=
http://www.amazon.com/First-Course-Finite-Elements/dp/0470035803http://www.amazon.com/First-Course-Finite-Elements/dp/0470035803http://www.amazon.com/First-Course-Finite-Elements/dp/0470035803http://www.amazon.com/First-Course-Finite-Elements/dp/0470035803http://www.amazon.com/First-Course-Finite-Elements/dp/0470035803http://www.amazon.com/First-Course-Finite-Elements/dp/0470035803


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Element stiffness in global coordinates

We need to be able to

transform displacementsfrom the xand yaxes todisplacements along the xand yaxes. We start withthe reverse:

Transformation matrix T(e)

( ) ( )( )

'( ) ( )

1 1

'( ) ( )

1 1

'( ) ( )

2 2

'( ) ( )2 2

{ }{ ' }

cos sin 0 0

sin cos 0 0

0 0 cos sin

0 0 sin cos

e ee

e ee ex x

e ee ey y

e e e e

x x

e ee ey y

T dd

u u

u u

u u

u u

=

The angle is measuredanti-clockwise from

xto x

e


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Coordinate transformation

( ) ( )( )

'( ) ( )

1 1

'( ) ( )

1 1

'( ) ( )

2 2

'( ) ( )

2 2

[ ] { }{ ' }

cos sin 0 0

sin cos 0 0

0 0 cos sin

0 0 sin cos

e ee

e ee ex x

e ee ey y

e e e e

x x

e ee e

y y

T dd

u u

u u

u u

u u

=

{ ' } [ ]{ }

e e e

d T d=

{ } [ ] { ' }e e T ed T d=

:

[ ] [ ] [ ]e T e

Note that

T T I=


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4 4

cos sin 0 0 cos sin 0 0

sin cos 0 0 sin cos 0 0

0 0 cos sin 0 0 cos sin

0 0 sin cos 0 0 sin cos

1 0 0 0

0 1 0 00 0 1 0

0 0 0 1

T ee

e e e e

e e e e

e e e e

e e e e

TT

I

=

Coordinate transformation

Verify that:4 4

[ ] [ ] [ ]e T e

xT T I=


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Stiffness of a truss element

Similarly for the forces:

( )

'( ) ( )

1 1

'( ) ( )

1 1

'( ) ( )

2 2'( ) ( )

2 2

[ ]{ ' } { }

cos sin 0 0

sin cos 0 0

0 0 cos sin

0 0 sin cos

ee e

e ee ex x

e ee ey y

e e e e

x xe ee e

y y

TF F

F F

F F

F F

F F

=

{ } [ ] { ' }e e T eF T F=

{ ' } [ ]{ }e e e

F T F=


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Truss element stiffness( ) ( ) ( ) ( )

[ ] [ ] [ ' ][ ]e e T e e

K T K T =

( ) ( )

1 0 1 0

0 0 0 0[ ' ] ,

1 0 1 0

0 0 0 0

e eK k

=

( )

cos sin 0 0

sin cos 0 0[ ]

0 0 cos sin

0 0 sin cos

e e

e e

e

e e

e e

T

=

2 2

2 2

( ) ( )

2 2

2 2

cos sin cos cos sin cos

sin cos sin sin cos sin[ ]

cos sin cos cos sin cossin cos sin sin cos sin

e e e e e e

e e e e e e

e e

e e e e e e

e e e e e e

K k

=


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Truss element stiffness

( ) ( )2 21 1

( ) 2 21 ( )

( ) 2 2

2

2 2( )

2


sin cos sin sin cos sin



e ee e e e e ex x

e e e e e e ey e

e e e e e e e

x

e e e e e ee

y

F u

F ukF

F

=

( )

1

( )

2

( )

2

e

x

e

x

e

y

u

u

Note the 2x2symmetric submatrix structure This implies that you can reverse the numbering of nodes (1 and 2)

without any changes in the element stiffness.

( ) ( )2 22 2

( ) 2 22 ( )

( ) 2 2

1

2 2( )

1





e ee e e e e ex x

e e e e e e ey e

e e e e e e e

x

e e e e e ee

y

F u

F uk

F

F

=

( )2

( )

1

( )

1

ex

e

x

e

y

u

u


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Assembly process

The assembly process is identical to the

one discussed for `spring structures and itwill not be repeated here in its generalform (no need to show at this point

complicated looking matrix operations). We will however provide soon a simple

example demonstrating this assemblyprocess.


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Generalizing the application of essential BCs

You already have seen through an example

how essential boundary conditions areapplied to the global system of eqs:

In essence, we partition the stiffness matrix

in a way that separates known from unknowndegrees of freedom as follows:

[ ]{ } { }K d F=

EE EF E

T

EF F F F

K K fd

K K d f

=

:Ed Known displacements

:Fd Unknown displacements

:Ef Unknown reaction forcescorresponding tonodes/directionswith prescribed displacement:Ff Applied (known) forces


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Generalizing the application of essential BCs

EE EF E

TEF F F F

K K fd

K K d f

=

EE EF F E

TEEF F F F

K d K d f

K d K d f

+ =

+ =

The unknown displacements are obtained from the 2nd eq. as:

1( )

T TE EEF F F F F F F EF

K d K d f d K f K d + = =

With known , we can return to the 1st eq. to compute thereaction forces:

EE EF FEf K d K d= +

Fd

Note that the matrix is symmetric and positive definite,so a solution for always exists!

FK

Fd


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A truss example

Construct the global stiffness matrix and load vector

Partition the matrices and solve for the unknowndisplacements at point B, and displacement in x

direction at point D. Find the stresses in the three bars

Find the reactions at C, D and F

E= 107Pa

Note that point D isfree to move in the x

direction

1 2 3

1,23,4 5,6

7,8

1 210A A

=

22A A=

3A A=


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A truss example: Element 1

(1)

1/ 2 1/ 2 1/ 2 1/ 2

1/ 2 1/ 2 1/ 2 1/ 2[ ]

1/ 2 1/ 2 1/ 2 1/ 22

1/ 2 1/ 2 1/ 2 1/ 2

EAK

=

7 8 1 2

7

8

1

2

(1) 0135 =

Note: Recall that you can number the corresponding global nodes in thesequence 1 2 7 8 without any changes in .(1)[ ]K

Local node 1

Local node 2


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(2)

0 0 0 0

0 2 2 0 2 2

[ ] 0 0 0 02

0 2 2 0 2 2

EA

K

=

7 8 3 4

7

8

3

4

090 =

22A A=

3,4

Local node 1

Local node 2


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(3)

1/ 2 1/ 2 1/ 2 1/ 2

1/ 2 1/ 2 1/ 2 1/ 2[ ]

1/ 2 1/ 2 1/ 2 1/ 221/ 2 1/ 2 1/ 2 1/ 2

EAK

=

7 8 5 6

7

8

5

6

045 =

3A A=

Local node 1

Local node 2


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A truss example: Assembly (element 1)

1/ 2 1/ 2 0 0 0 0 1/ 2 1/ 2

1/ 2 1/ 2 0 0 0 0 1/ 2 1/ 2

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0[ ]

0 0 0 0 0 0 0 02

0 0 0 0 0 0 0 0

1/ 2 1/ 2 0 0 0 0 1/ 2 1/ 2

1/ 2 1/ 2 0 0 0 0 1/ 2 1/ 2

EAK

=

1

2

3

4

5

6

7

8

1 2 3 4 5 6 7 8

(1)

1/ 2 1/ 2 1/ 2 1/ 2

1/ 2 1/ 2 1/ 2 1/ 2[ ]1/ 2 1/ 2 1/ 2 1/ 22

1/ 2 1/ 2 1/ 2 1/ 2

EAK

=

7 8 1 2

7

8

1

2


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1/ 2 1/ 2 0 0 0 0 1/ 2 1/ 21/ 2 1/ 2 0 0 0 0 1/ 2 1/ 2

0 0 0 0 0 0 0 0

0 0 0 2 2 0 0 0 2 2[ ]

0 0 0 0 0 0 0 02

0 0 0 0 0 0 0 0

1/ 2 1/ 2 0 0 0 0 1/ 2 0 1/ 2 0

1/ 2 1/ 2 0 2 2 0 0 1/ 2 1/ 2 2 2

EAK

=

+ + +

1 2 3 4 5 6 7 8

1

2

3

4

5

6

7

8

( 2)

0 0 0 0

0 2 2 0 2 2[ ]0 0 0 02

0 2 2 0 2 2

EAK

=

7 8 3 4

7

8

3

4


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1/ 2 1 / 2 0 0 0 0 1/ 2 1/ 21/ 2 1/ 2 0 0 0 0 1 / 2 1/ 2

0 0 0 0 0 0 0 0

0 0 0 2 2 0 0 0 2 2[ ]

0 0 0 0 1/ 2 1/ 2 1/ 2 1/ 22

0 0 0 0 1/ 2 1/ 2 1/ 2 1/ 2

1/ 2 1/ 2 0 0 1/ 2 1/ 2 1/ 2 0 1/ 2 1/ 2 0 1/ 2

1/ 2 1 / 2 0 2 2 1 / 2 1/ 2 1/ 2 1/ 2 1/ 2 2 2 1 / 2

EAK

=

+ + + + + + +

1 2 3 4 5 6 7 81

2

3

4

5

6

7

8

(3)

1/ 2 1/ 2 1/ 2 1/ 2

1/ 2 1/ 2 1/ 2 1/ 2

[ ] 1/ 2 1 / 2 1/ 2 1/ 22

1/ 2 1 / 2 1/ 2 1/ 2

EAK

=

7 8 5 6

7

8

5

6


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A truss example: Assembly

1/ 2 1/ 2 0 0 0 0 1/ 2 1/ 21/ 2 1/ 2 0 0 0 0 1/ 2 1/ 2

0 0 0 0 0 0 0 0

0 0 0 2 2 0 0 0 2 2[ ]

0 0 0 0 1/ 2 1/ 2 1/ 2 1/ 22

0 0 0 0 1/ 2 1/ 2 1/ 2 1/ 2

1/ 2 1/ 2 0 0 1/ 2 1/ 2 1 0

1/ 2 1/ 2 0 2 2 1/ 2 1/ 2 0 1 2 2

EAK

=

+

1 2 3 4 5 6 7 8

1

2

3

4

5

6

7

8


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A l P i i d BC


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1

2

3

4

6

5

7

8

1/ 2 1/ 2 0 0 0 0 1/ 2 1/ 2 0

1/ 2 1/ 2 0 0 0 0 1/ 2 1/ 2 0

0 0 0 0 0 0 0 0 0

00 0 0 2 2 0 0 0 2 2

00 0 0 0 1/ 2 1/ 2 1/ 2 1/ 22

0 0 0 0 1/ 2 1/ 2 1/ 2 1/ 2

1/ 2 1/ 2 0 0 1/ 2 1/ 2 1 0

1/ 2 1/ 2 0 2 2 1/ 2 1/ 2 0 1 2 2

d

d

d

dEA

d

d

d

d

= =

=

= =

+

1

2

3

4

6

3

0

10

0

r

r

r

r

r

N

=

A truss example: Partition and BCs

[ ]E

K

[ ]F

K

[ ]EF

K

[ ]EF

TKF

f

Ef

0Ed =

Fd


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A l R i l l i


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1

2

3

4

6

5

7

8

1/ 2 1/ 2 0 0 0 0 1/ 2 1/ 2 0

1/ 2 1/ 2 0 0 0 0 1/ 2 1/ 2 0

0 0 0 0 0 0 0 0 0

00 0 0 2 2 0 0 0 2 2

00 0 0 0 1/ 2 1/ 2 1/ 2 1/ 22

0 0 0 0 1/ 2 1/ 2 1/ 2 1/ 2

1/ 2 1/ 2 0 0 1/ 2 1/ 2 1 0

1/ 2 1/ 2 0 2 2 1/ 2 1/ 2 0 1 2 2

d

d

d

dEA

d

d

d

d

= =

=

= =

+

1

2

3

4

6

3

0

10

0

r

r

r

r

r

N

=

A truss example: Reaction calculation

[ ]E

K

[ ]F

K

[ ]EF

K

[ ]EF

TKF

f

Ef

0Ed =

Fd

EE EF FEf K d K d= +

A t l R ti l l ti


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1

52

3 7

4 8

6

0 1/ 2 1/ 2

0 1/ 2 1/ 2

0 0 02

0 0 2 2

1/ 2 1/ 2 1/ 2 F

EEF

d

f K

r

drEA

r d

r d

r

=

A truss example: Reaction calculation

1

2

3

4

6

10001000

0

1000

0

Fx

Fy

Cx

Cy

Dy

Rr N

Rr N

r R

Nr R

rR

=

A t l C t th t


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cos sin cos sin { }e

e e e e e e

e

Ed

L =

A truss example: Compute the stresses

' ' ' '

2 1 2 1

e e e ee e ex x x x

e e

u u u uE

L L

= =

Combining the 2 Eqs gives:{ }

ed

'( ) ( )

1 1

'( ) ( )

1 1

'( ) ( )

2 2

'( ) ( )

2 2

cos sin 0 0

sin cos 0 0

0 0 cos sin

0 0 sin cos

e ee ex x

e ee ey y

e e e e

x x

e ee e

y y

u u

u u

u u

u u

=

However:

A t l C t th t


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cos sin cos sin { }e

e e e e e e

e

Ed

L =

A truss example: Compute the stresses

Applying this to each element, we have:

(1)

0.033284

0.0052 2 2 2

141.42102 2 2 22

0

= =

m

mE

kPa

[ ](2 )

0.033284

0.0050 1 0 1 50

0

0

= =

m

mE kPa

(3)

0.033284

0.0052 2 2 20

0.0382842 2 2 22

0

= =

m

mEkPa

m


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Th di i l ( ) t t t


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Three-dimensional (space) truss structures

'( ) '( )( ) ( )

1 1

'( ) ( ) ( ) '( )

2 2

e ee e

x x

e e e e

x x

F uk kF k k u

=

A unit vector along the direction x of a 3D truss

element has the components (direction cosines ofthe axes between x and x,y,z, respectively):

where are the nodalcoordinates in the (x,y,z) system.

2 1 2 1 2 1

2 1 2 1 2 1

2 2 2

, ,

( ) ( ) ( )

s s s

e e e e e e

e e e

e e e

e e e e e e e

x x y y z zl m n

L L L

L x x y y z z

= = =

= + +

1 1 1 22 2( , , ) ( , , )

e e e e e ex y z and x y z

Three dimensional (space) tr ss str ct res


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'( ) '( )( ) ( )

1 1

'( ) ( ) ( ) '( )

2 2

e ee e

x x

e e e e

x x

F uk k

F k k u

=

The displacement transformation then takes the form:

Similar transformation is applied for the forces:

( )

1

( )

1

'( ) ( )

1 1

'( ) ( )

2 2

( )[ ] 2

( )

2

{ }

0 0 0[ ]{ }

0 0 0

s s s

s s s

e

e

e

x

e

y

e e ee e

x z e e

e e e e e

x x

eT y

e

z

d

uu

l m nu uT d

u l m n u

u

u

=

'( )

1

'( )

2

[ ]{ }

e

x e e

e

x

FT F

F

=

Three dimensional (space) truss structures


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'( ) '( )( ) ( )

1 1

'( ) ( ) ( ) '( )

2 2

'

=

e

e ee e

x x

e e e e

x x

K

F uk k

F k k u

Similarly to the derivation for 2D trusses, the stiffness

matrix in global coordinates is then:

( )

6 6 6 2 2 62 2

[ ] [ ] ' [ ] e e T e e

x x xx

K T K T

Stiffness of a space element


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Stiffness of a space element

2 2

2 2

2 2

( )

2 2

2 2 2

[ ]

s s s s s s s s s s

s s s s s s s s s s

s s s s s s s s s s

s s s s s s s s s s

s s s s s s s s s s

s s s

e e e e e e e e e e

e e e e e e e e e e

e e e e e e e e e ee e

e

e e e e e e e e e e e

e e e e e e e e e e

e e e

l m l n l l m l n l

m l m m n m l m m n

n l m n n n l m n nE AK

L l m l n l l m l n l

m l m m n m l m m nn l m

=

2 2

s s s s s s s

e e e e e e en n n l m n n

2 1 2 1 2 1

2 1 2 1 2 1

2 2 2

: , ,

( ) ( ) ( )

s s s

e e e e e e

e e ee e e

e e e e e e e

x x y y z zwhere l m n

L L L

L x x y y z z

= = =

= + +

Computing the stresses in a space truss element


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{ }s s s s s s

ee e e e e e e e

e

El m n l m n d

L =

Computing the stresses in a space truss element

' '

2 1

' '' '2 1

2 1( )

e ee x x

e

e e ee e e ex x

x xe e

u u

L

u u EE u u

L L

=

= =

Combining the 2 Eqs gives:

( )

1

( )

1

'( ) ( )

1 1

'( ) ( )

2 2

( )

2( )

2

0 0 0[ ]{ }

0 0 0

s s s

s s s

e

x

e

y

e e ee e

x z e e

e e e e e

x x

e

ye

z

u

ul m nu u

T du l m n u

u

u

=

However:

Accounting for thermal effects in truss analysis


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Accounting for thermal effects in truss analysis

Consider a truss structure that is heated. We need toaccount for thermal expansion effects. Note:

Hookes law is now modified as follows (using the xcoordinate system):

2 1( ) ( )

e ee e e e e e e e e

elastic thermal e

total strain

u uE E E T

L

= = =

0

( ) ( ) 2 1

2 1

2 1 0

( )

( ) ,

e

e ee e e e e e e e e

e

e ee e e e e e e

e

u uF F p A A E T

L

A Ek u u A E k

L

= = = = =

= =

( ) ( )( ) ( )

1 1

0( ) ( ) ( ) ( )

2 2

l n

1

1

e ee e

e e e

e e e e

Therma odal vector

F uk kA E

F k k u

+ =

e e e

elastic thermal

total strain

= +

Element equations with thermal effects
http://www.amazon.com/First-Course-Finite-Elements/dp/0470035803


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Expanding these equations to include nodal displacementsin the y axis gives:

We need to transform this to xand ydisplacements (ourdegrees of freedom for this element)

( )( ) ( )

'( ) '( )

1 1

'( ) '( )

1 1( )

0'( ) '( )

2 2'( ) '( )

2 2

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

1 1 0 1 0

0 0 0 0 0

1 1 0 1 0

0 0 0 0 0

e ethermale e

e e

x x

e e

y ye e e e

e e

x xe e

y y

F KF d

F u

F uA E k

F u

F u

+ =

{ ' } [ ]{ }e e ed T d= { ' } [ ]{ }e e eF T F=



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We can transform these element equations as follows:

From these equations, we conclude that:

( )( ) ( )

'( ) '( )

1 1

'( ) '( )

1 1( )

0'( ) '( )

2 2

'( ) '( )

2 2

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

1 1 0 1 0

0 0 0 0 0

1 1 0 1 0

0 0 0 0 0

e ethermale e

e e

x x

e e

y ye e e e

e e

x x

e e

y y

F KF d

F u

F u

A E kF u

F u

+ =

{ ' } [ ]{ }e e ed T d=

{ ' } [ ]{ }e e eF T F=

( )[ ]{ } { ' } [ ' ][ ]{ }

e e e e e e

thermalT F F K T d + =

( )

( )

[ ]{ }

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

eethermal

e e T e e T e e e

thermal

KF

F T F T K T d + =



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( )

( )

[ ]{ }

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

ee

thermal

e e T e e T e e e

thermal

KF

F T F T K T d + =

( )

( ) 2 21

( ) 2 21 ( )

0( ) 2

2

( )

2

{ }{ }

cos cos sin cos cos sin cos

sin sin cos sin sin cos sin

cos cos sin cos

sin

eethermal

e e e e e e e ex

e e e e e e e ey

e e e ee e e e

x

ee

y

FF

F

F

A E kF

F

+ =

( ) ( )

( )

1

( )

1

2 ( )

2

2 2 ( )

2

[ ]

cos sin cos


e e

e

x

e

y

e e e e e

x

e e e e e e e

y

K d

u

u

u

u

( )

cos sin 0 0

sin cos 0 0[ ]

0 0 cos sin

0 0 sin cos

e e

e e

e

e e

e e

T

=

Use:

Finally we obtain:

(for 2D trusses)



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What do you need to do to account for thermal effects in truss analysis?

For each truss element that is heated, simply add to the element force, thefollowing extra term

You will need to define at which truss elements thermal effects take place andfor each of them read the values and

0 0

{ }

cos

sin,

cos

sin

ethermal

e

e

e e e e e e

e

e

F

A E where T

=

e .eT

Principle of minimum potential energy


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An alternative equivalent approach to solving

many structural problems is the principle ofminimum potential energy.

From all the possiblecompatible

displacements of a structure, the one thatminimizes the total potential energyis theexact solution.

Potential energyfor given

displacements=

Strain energyfor these givendisplacements

-

Work done by externalloads on these given

displacements



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Let us see this principle applied to the trussproblems discussed earlier.

{ }

'( ) '( ) '( ) '( )

1 1 2 2

( / )

,

1( )

2

min

e

e e e e

d e

e e e e e e e e

x x x xExternal Work

Elastic strain energy densitywork volume

PE PE U W

PE dV F u F u

=

= +

Assemblyprocess



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Lets apply this principle to one truss

element. We need to minimize with respectto the nodal displacements (localcoordinates) Recall that:'( ) '( )1 2, .

e e

x xu u

'

2 1'

,

e ee e e ex x

e

u uEL

= =

2 '( ) '( ) '( ) '( )

1 1 2 2

'

2 '( ) '( ) '( ) '( )2 1

1 1 2 2

'

1

2

'1( )

2

e

ee

e e e e e e e e

x x x x

e ee e e e e ex x

x x x xe

A dx

PE E dV F u F u

u uE dV F u F u

L

= =



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' '1 2 1 2

'

2 '( ) '( ) '( ) '( )2 1

1 1 2 2

, ' , '

'1( )

2

min mine e e ex x x x

e ee e e e e e e ex x

x x x xe

u u u u

u uPE E A L F u F u

L

=

Take partial derivatives of wrt :ePE'

1 2, '

e e

x xu u

' '( )

1 2 1

1

0 ( ' ) 0

'

e e ee e e

x x xe e

x

PE E Au u F

u L

= =

' '( )

2 1 2

2

0 ( ' ) 0'

e e ee e e

x x xe e

x

PE E Au u F

u L

= =

'( ) '( )( ) ( )

1 1

'( ) ( ) ( ) '( )

2 2

e ee e

x x

e e e e

x x

F uk k

F k k u

=

These are the same Eqs as those obtained with the direct method!



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In general (not just for mechanics

problems!), the principle of minimumpotential energy takes the following form:

or after assembly:

Note that the mimimization gives us thefamiliar solution:


( ) ( ) ( )

{ } { }

1{ } [ ]{ } { } { }

2

min mine e T e e T

d de e

PE d K d d F

=

{ } { }

1{ } [ ]{ } { } { }

2min min

T T

d d

d K d d F

PE

=

[ ]{ } { }K d F=



48/55

CCOORRNNEELLLLU N I V E R S I T Y 48MAE 4700 FE Analysis for Mechanical & Aerospace Design

N. Zabaras (9/1/2011)


We will not use this method to repeat the trusscalculations.

However, it will be our starting point for computingthe stiffness of beam elements (lecture 4).

The method of minimum potential energy can beapplied to many problems not related to mechanics

however there are many problems where thistechnique is not applicable.

After discussing beam bending problems, we willneed to look for more powerful (`unfortunately alsomore mathematical) methods (weak (Galerkin)formulations).

Revisiting the 2-node truss element


49/55


N. Zabaras (9/1/2011)

Revisiting the 2-node truss element

Up to now we used the direct method to express the nodal loads vs.nodal displacements for the 2-node truss element.

Let us linearly interpolate the displacement of any point xin theelement in terms of the nodal displacements:

( ) ( )1 2

'( )

'( ) '( ) '( ) ( ) ( )1

1 2 '( )

2

' ' ' '(1 ) [1 , ] [ ] { }

e e

e

e e e e ex

x x xe e e e e

x basis functionsmatrixN N

Nodaldisplacementselement basis

functions

ux x x xu u u N d

L L L L u

= + = =

The strain in this 2-node element can nowbe computed as follows:

'

'

ee xdu

dx =

' ( ) ( ) ( )'( ) ( ) ( ) ( ) ( ) ( )[ ]{ }

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

e e e ee e e e e e e

xx

du d N d dN u N d d B d dx dx dx= = = =

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

'( )

2

1 1 1 1[ ] [ , ] [ , ] [ , ]

' '

e e ee ee e x x x

e e e e ee

xGradient ofbasis functions

matrix

u u udN dN B

dx dx L L L L Lu

= = = =

This is exactly the sameapproximation we usedbefore (constant strain

element)

'( )e

xu


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Truss analysis with displacement constraints


51/55


N. Zabaras (9/1/2011)


Up to this point, we imposed essential boundary conditionsin terms of prescribed x- or y- nodal displacements. How

about if the support is inclined as in the figure below:

Here, we dont know the displacements at node 1 but we

know the relation between u1xand u1y. In general we writethese constrains on our nodal degrees of freedom as:Cd=q.

For this problem, the constraint is thatthere is no normal displacement

at the support 1



52/55


N. Zabaras (9/1/2011)


Note that at node 1 we dont have essential boundaryconditions we have a displacement constraint.

To solve this problem we use the principle of minimumpotential energy with the constraint Cd=q:

Here, we enforce the constraint using Lagrange multipliers.

Find d and such that

{ }int

inint

1min { } [ ]{ } { } { } { } ([ ]{ } { })

2

TT T

d Lagrange ConstramultiplierPotential energyenforcingof unconstra edthe constrasystem

L d K d d F C d q= +



53/55


N. Zabaras (9/1/2011)


is the Lagrange multiplier that enforces the constraint itis nothing else but the reaction force at node 1 (normal to

the support!)

Minimization is now performed with respect to both d and .

Find d and such that

0

T d FK C

qC

=

{ }

int

inint

1min { } [ ]{ } { } { } { } ([ ]{ } { })

2

TT T

d

Lagrange ConstramultiplierPotential energyenforcingof unconstra edthe constrasystem

L d K d d F C d q= +

Apply essential boundary conditionsand then solve for {dF} and

(here, reaction force at node 1)

Displacement constraints: Implementation


54/55


N. Zabaras (9/1/2011)

Displacement constraints: Implementation

How do you implement this in the MatLab

libraries of HW1?

Introduce the constraints in the InputData.m and thenmodify the stiffness and load vectors in the NodalSoln.m.

Apply the essential boundary conditions first before youaugment the reduced global equations (Kf) with theLagrange multiplier.

% Read information for constraints

C = zeros(1,neq-length(debc)); %The dimension of C is neq minus the% prescribed DOF via essential BCs% Here there is only one constraint

C(1) = sin(pi/6); C(2) = cos(pi/6);q = 0;

InputData.m

1 1

sin 30 cos 30 0x y

u u+ =

Displacement constraints: NodalSoln.m


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MAE 4700 FE Analysis for Mechanical & Aerospace Design

Displacement constraints: NodalSoln.mfunction [d, rf, lambda] = NodalSoln(K, R, debc, ebcVals, C, q)

% K=global stiffness, R=global force, debc=degrees of freedom with specified values, ebcVals=specified displacements

dof = length(R); % Extract the total degreess-of-freedom

df = setdiff(1:dof, debc); % Sets the difference between two sets of indices, i.e. the global degrees of freedom minus the% degrees of freedom with prescribed essential boundary conditions

Kf = K(df, df); % Remove eqs. corresponding to prescribed displacements

Rf = R(df) - K(df, debc)*ebcVals; % Modify the remaining load vector to account for the essential boundary conditions

[m n] = size(C); % Extract number of constraints

Kf = [Kf C'; % Augment global equations with the Lagrange multipliersC zeros(m)];

Rf = [Rf;q]; % Augment load vector

dfVals = Kf\Rf; % Solve the linear system of equations. Here for simplicity, we use Gauss elimination.

d = zeros(dof,1); % Restore the solution vector (i.e. include back the nodes with prescribed displacements).d(debc) = ebcVals; % Use the originally established ordering of the nodes.d(df) = dfVals(1:(length(dfVals)-m));

rf = K(debc,:)*d - R(debc); % Calculate the reaction vector at nodes with prescribed displacements

lambda = dfVals((length(dfVals)-m+1):length(dfVals)); % Calculate Langrange multipliers (reactions at nodes with constraints)

0

T d FK C

qC =

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Direct Method Truss Analysis Matrice Di Rigidezza Composizione