fem zabaras fem4linearelasticity

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CCOORRNNEELLLLU N I V E R S I T Y 1

MAE 4700 FE Analysis for Mechanical & Aerospace Design

N. Zabaras (10/29/2009)

MAE4700/5700Finite Element Analysis for

Mechanical and Aerospace DesignCornell University, Fall 2009

Nicholas ZabarasMaterials Process Design and Control Laboratory

Sibley School of Mechanical and Aerospace Engineering101 Rhodes Hall

Cornell UniversityIthaca, NY 14853-3801
http://mpdc.mae.cornell.edu/Courses/MAE4700/MAE4700.htmlhttp://mpdc.mae.cornell.edu/Courses/MAE4700/MAE4700.html


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N. Zabaras (10/29/2009)

Finite element discretization

Consider a problem domain with boundary discretized

with nel number of2D finite elements (triangles or

quadrilaterals).


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N. Zabaras (10/29/2009)


Thexand ycomponents of the displacement field

are generally approximated by the same shape functions. Let n be the number of nodes in the finite element mesh.

There are two degrees-of-freedom per node for each of the

two components of the nodal displacements, so the nodaldisplacement matrix is:

Txx y

y

uu u u

u

= =

1

1

2

2 1 1 2 2 ...

...

x

y

x T

y x y x y xn yn

xn

yn

u

u

u

ud u u u u u u

u

u

= =


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N. Zabaras (10/29/2009)


The finite element approximation of the displacement field

can be expressed by:

The global shape functions Niare C0 continuous (smooth

over element domains but with kinks at the elementboundaries).

Therefore, the integral over in the weak form is computed

as a sum of integrals over element domainse

1

1

( , ) ( , )

( , ) ( , )

nh

x i xii

nh y i yi

i

u x y N x y u

u x y N x y u

=

=

=

=

( ) 01 1 1

T T

e e e

t

nel nel nelTe e e e eS S

e e e

w D u d w td w bd w U = = = = +


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N. Zabaras (10/29/2009)


Here ue is the finite element approximation of thedisplacement field in element e.

The element shape function matrix Ne in the matrix form is

where nen is the number of element nodes.

( ) 01 1 1T T

e e e

t


e e ew D u d w td w bd w U = = = = +

{

int ( , )

( , ) ( , )e e e

Nodalx and ydisplacementsdisplacements

at po x y

u x y N x y d =

1 2

1 2

0 0 ... 0

0 0 ... 0

e e enene

e e enen

N N N

N N N N

=


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N. Zabaras (10/29/2009)

Strain displacement matrix

We need to compute the strains in terms of the element

shape functions and the nodal displacements. Applying the

symmetric gradient operator to Ne gives

where the strain-displacement matrixe

is defined as:

{ } { }[ ]

{ }e

ex

e e e e e y S S

e Bxy

u N d

= = =

[ ] [ ]

1 2

1 2

1 1 2 2

0 0 ... 0

0 0 ... 0

...

ee e

nen

ee e

e e

nen

S

e ee e e e

nen nen

NN N

x x xNN N

B N y y y

N N N N N N

y x y x y x

= =


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N. Zabaras (10/29/2009)

Approximation of the weight functions

The vector of weight functions is approximated by the same

shape functions as the displacement fields:

Similarly applying the symmetric gradient on the vector of

weight functions gives:

We now need to substitute these equations into the weak

form.

{ } { }

{ }( ) { }

int( , )

( , ) ( , ) ( , ) ( , )T T Te e e eT e e e e

NodalVirtualValuesdisplacement

vector at pox y

w x y N x y w w x y N x y w w N = = =

{ }[ ]

{ } { } { }( ) { }( , ) ( , ) ( , ) ( , )e

T T Te e e eT e e e eS S S

B

w x y N x y w w x y B x y w w B = = =


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N. Zabaras (10/29/2009)

Weak formulation

Substituting into the weak form and recalling the local to

global degrees of freedom transformations, de=Led,

weT

=wT

LeT

yields

The arbitrary weight functions w(x,y) has been replaced by

arbitrary parameters wF which is the portion ofw

corresponding to nodes not on an essential boundary.

( , )eT eT eT w x y w N =( , )eT eT eT

Sw x y w B =

( ) 01 1 1

T T

e e e

t


e e e

w D u d w td w bd w U = = = = +

e e e e

Su B d= =

01

0T T T T

e e e

t

nelT e e e e e e e

Fe

w L B D B d L d N td N bd w U =

=


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N. Zabaras (10/29/2009)

Element stiffness matrix and load vector

From the above weak form we can conclude that:

01

0T T T T

e e e

t

nelT e e e e e e e

Fe

w L B D B d L d N td N bd w U =

=

T

e

e e e eK B D B d

=

T T

e e

t

e

e

t

e e e

ffbodyboundaryforce vectorforce vector

f N td N bd

= +


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N. Zabaras (10/29/2009)


We can now re-write this weak form as:

This is simplified as:

01 0

T T T T

e e e

t

nelT e e e e e e e

Few L B D B d L d N td N bd w U =

=

01 1

0T Tnel nel

T e e e e eF

e e

K f

w L K L d L f w U = =

=

00T

F

rresidual

w Kd f w U

=

00ET T

E F F

F

rw w w U

r

=


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N. Zabaras (10/29/2009)


Since wE=0 and wF=0 is arbitrary, it follows that rF=0.

Partioning the finite element nodes in Eand Fnodes, gives:

This equation is solved using the two-step partition

approach discussed in an earlier lecture.

00

ET T

E F F F

r

w w w U r

=

E EF E E E

TF F EF F

K K d f r

d fK K

+ =


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N. Zabaras (10/29/2009)

Piecewise-linear interpolation on triangles

Consider that h consists ofE3-node triangular elements

and that we have linear interpolation of thex-

and y-displacements on each element e.

This is a constant strain element.

We already have seen how tocompute the coefficients and

eventually the basis functions,e.g.

1 2 3

4 5 6

e

x

e

y

u x y

u x y

= + +

= + +

1 1 1 2 1 3 1

2 2 1 2 2 3 2

3 3 1 2 3 3 3

( , ) ,

( , ) ,

( , ) ,

ex

e

x

e

x

u x y x y

u x y x y

u x y x y

= + +

= + +

= + +

( , )e

xu x y

( , )xu x y


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N. Zabaras (10/29/2009)


We determine the 3 constants as follows:

Solving this system of equations leads to:

31 2

31 2

1 2

1 1 2 3 3 2 2 3 1 1 3 3 1 2 2 1

2 1 2 3 2 3 1 3 1 2

3 1 3 2 2 1 3

1 ( ) ( ) ( )2

1

( ) ( ) ( )2

1( ) ( )

2

e e e

x x x

e aa a

e e e

x x xe bb b

e e

x x

e c c

u x y x y u x y x y u x y x yA

u y y u y y u y yA

u x x u x x

A

= + +

= + +

= +

3

3 2 1( )

e

x

c

u x x +

1 1 1 1 1 1

2 2 2 2 2 2

3 3 3 3 3 3

( , ) 1

( , ) 1( , ) 1

e e

x x

e e

x x

e e

x x

u u x y x y

u u x y x yu u x y x y

=

1 2 3

e

xu x y = + +


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N. Zabaras (10/29/2009)


Substituting the coefficients in the first approximation:

where:

1 1 2 2 3 3( ) ( , ) ( , ) ( , )e e e e e e e

x x x xu x u N x y u N x y u N x y= + +

[ ] [ ]

[ ] [ ]

[ ] [ ]

1 1 1 1 2 3 3 2 2 3 3 2

2 2 2 2 3 1 1 3 3 1 1 3

3 3 3 3 1 2 2 1 1 2 2 1

1 1( , ) ( ) ( ) ( )

2 2

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

1 1( , ) ( ) ( ) ( )

2 2

e

e e

e

e e

e

e e

N x y a b x c y x y x y y y x x x yA A



= + + = + +

= + + = + +

= + + = + +

1 2 3

e

xu x y = + +

1 1 1 2 2 3 3

2 1 1 2 2 3 3

3 1 1 2 2 3 3

1

2

1

2

1

2

e e e

x x x

e

e e e

x x x

e

e e e

x x x

e

u a u a u aA

u b u b u bA

u c u c u c

A

= + +

= + +

= + +


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N. Zabaras (10/29/2009)

Linear shape functions over a triangular element

Please note that:

( , )i

e

j j ij N x y =

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

e e e e e e e

x x x xu x u N x y u N x y u N x y= + +

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

e e e e e e e

y y y yu x u N x y u N x y u N x y= + +

[ ]

[ ]

[ ]

1 1 1 1

2 2 2 2

3 3 3 3

1( , )

2

1( , )2

1( , )

2

e

e

e

e

e

e

N x y a b x c yA

N x y a b x c yA

N x y a b x c yA

= + +

= + +

= + +

2( , )

eN x y

1( , )

eN x y

3 ( , )e

N x y

( , )

e

xu x y


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N. Zabaras (10/29/2009)

3-node (constant strain) triangular element

We have 6 DOF per element:

The Ne

matrix for the interpolation ofdisplacements is:

The Be matrix for computing strains in terms ofde is:

1 1 2 2 3 3

Te e e e e e e

x y x y x yd u u u u u u =

1 2 3

1 2 3

0 0 0

0 0 0

e e e ex e

e e e ey

u N N Nd

u N N N

=

1, 2, 3,

1, 2, 3,

1, 1, 2, 2, 3, 3,

0 0 0

0 0 0

e e e e x x x x

e e e e e y y y y

e e e e e e e

xy y x y x y x

N N N

N N N d

N N N N N N

=

,

,

1,2,3

ee

ii x i

ee ii y i

NN bx

NN c

y

i

= =

= =

=

where


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N. Zabaras (10/29/2009)


We can compute the derivatives of the shape functions

analytically:

where

[ ]

1 2 3

1 2 3

1 1 2 2 3 3

0 0 01

0 0 02

e

ex

e e

y e

exy

B

b b b

c c c d Ac b c b c b

=

1 2 3 2 3 1 3 1 2

1 3 2 2 1 3 3 2 1

, ,

, ,

e e e e e e

e e e e e e

b y y b y y b y y

c x x c x x c x x

= = =

= = =

1 1

2 2

3 3

11

det 12

1

e e

e e e

e e

x y

x y

x y

=


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N. Zabaras (10/29/2009)


As expected, the Be matrix is not a function ofxory, i.e.,this is a constant strain element.

We will not list the stiffness matrix (6 x 6) explicitly. The

required matrix multiplications are performed in thecomputer.

T

e

e e e eK B D B d

= ( )T

e e e e eK A B D B unit thickness=


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N. Zabaras (10/29/2009)


For plane stress, the element stiffness

For variable thickness within the element, you need to

evaluate using Gauss quadrature

T

e

e e e eK B D B d

=

1 1

1 11 2 3

2 21 2 32

2 21 1 2 2 3 3

3 3

3 3

0

01 0 0 0 0

01 11 0 0 0 0

02 1 2(1 )0 00 2

0

e

e e

e e

b c

c bv b b b

b c EK v c c c t dA

c b A v Av c b c b c bb c

c b

=

( )

1 1

1 11 2 3

2 21 2 32

2 21 1 2 2 3 3

3 3

3 3

0

01 0 0 0 0

01 0 0 0 0

04 1(1 )

0 002

0

e ee

e

b c

c bv b b b

b ct EK v c c c

c bv Av c b c b c b

b cc b

=

e

e

A

t dA


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N. Zabaras (10/29/2009)

Element nodal body force for 3-node triangular

To evaluate this integral, we start by interpolating b with a

linear function using area coordinates. At the end, the

needed integration is performed analytically. We interpolate the body force in the element by the linear

shape functions in the triangular coordinates as

T

e

e e f N bd

=

3

3 1

31

1

e ee e i xi

x xi ieie e

i e e y yii yi

i

N bb b

b N

b b N b

=

==

= = =


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N. Zabaras (10/29/2009)


Substitution and performing the

integration gives:

Note that computing integrals like this, you can use the

following:

T

e

e e f N bd =

3

1

e e x xie

ie ei y yi

b b

b Nb b=

= =

1 1 2 3

1 1 2 3

3

1 2 32

1 1 2 32

1 2 33

1 2 33

0 2

0 2

20

2120

202

0

e

e

x x xe

y y y

ee e e

xi x x xe ei ee i y y yyi

e x x x

e y y y

N b b b

N b b b

b b b bNA t f N d

b b bbN

b b bNb b b

N

=

+ + + + + +

= = + + + + + +

( ) ( ) ( )( )

1 2 3

! ! !2

2 !ee e e e

A

N N N dA A =+ + +


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N. Zabaras (10/29/2009)


Assume that the body force is constant in the element in

each direction with components bx

and by

, then:

1 2 3

1 2 3

1 2 3

1 2 3

1 2 3

1 2 3

2

2

2

212

2

2

x x x

y y y

e e x x xe

y y y

x x x

y y y

b b b

b b b

b b bA tf

b b b

b b b

b b b

+ +

+ + + +

= + +

+ +

+ +

3

x

y

e exe

y

x

y

b

b

bA tf

b

b

b

=

In this case of constant body force in each direction,

the force is distributed equally in the three nodes.


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N. Zabaras (10/29/2009)

Nodal boundary force vector for 3-node triangular

We can now substitute in the expression for the boundary

force:

T

e

t

e e

f N td =

1 2

1 2

1 211 2

1 1 21 2

1(1 ) 0

22

10 (1 ) 22 1 1

(1 ) (1 )1 22 2(1 ) 0

2 1 1 2 6(1 ) (1 )1 2 2 00 (1 )2

00 0

0 0

x x

y y

ex xx xe e

y yx x

t t

t t

t tt t L Lt

f t d

t tt t

+ + + + +

+= = + + +

+


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N. Zabaras (10/29/2009)

Nodal boundary force vector for 3-node triangular

For a constant traction over the segment 12,

T

e

t

e e

f N td =

1 2

1 2

1 2

1 2

2

2

2

6 2

00

00

x x x

y y y

e exx xe

yy y

t t t

t t t

t t t Lt Lt

f t t t

+ + +

= = +

1 2 xx xt t t= =

1 2 yy yt t t= =

In this case of constant traction components,

the force (traction times the area of the

segment where it is applied), it is distributed

equally at the 2 nodes of the segment.

E l f i b d di i


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N. Zabaras (10/29/2009)

Example of traction boundary conditions

Considera unit square of thickness tunder plane stress

and tractions conditions as shown. Assume that it is

discretized with 2 linear (triangular elements). Compute the

form of the assembled nodal force vector due to applied

tractions.

1 2

34

1

2

3

4

We denote the 4 boundary sides as 1,2,3,4

Traction is applied on sides 2 and 3.

1

2

0

01

0

121

0

1

tf

=

Node 2

Node 3

Node 4

Node 1

M l b d di i


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N. Zabaras (10/29/2009)

More general boundary conditions

Since this is a 2D problem, on a given boundary we may

have different conditions (displacement/traction) applied on

the 2 directions along and normal to the boundary segment.

On a given boundary we cannot describe on the same

direction both displacement and traction (recall from earlierlectures you cannot specify on the same part of the

boundary work conjugate variables).

The most general boundary conditions are the following:

,

,

x x tx

y y ty

x x ux

y y uy

n t on

n t on

u u on

u u on

=

= =

=

i

i

where:

0

0

tx ux

ty uy

tx ux

ty uy

=

= =

=

E l bl


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N. Zabaras (10/29/2009)

Example problem

Construct the weak form corresponding to the plane stress

problem shown in the figure. Consider a one-element

triangular mesh. The boundary conditions are as follows.

edge BC is constrained in yand traction freex,

edge AB is constrained inxand traction free in y.

edge AC is subject to prescribed normal traction

Compute:

(a) Construct the stiffness matrix

(b) Calculate the nodal force vector

(c) The unknown displacement vector(d) The stress at (1.5, 1.5).

6

310 , 0.3 E Pa v= =

21 3 3 4.52

eA m= =

1

2

3

E l bl Sh f ti


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N. Zabaras (10/29/2009)

Example problem: Shape functions

The displacement vector is:

The shape functions are:

6310 , 0.3 E Pa v= =

213 3 4.5

2

eA m= =

1

2

3

Te e e e e e e Ax Ay Bx By Cx Cyd u u u u u u =

( ) ( )1 2 3 3 2 2 3 3 21 1

( ) ( ) 0 0 0 39 32

e

e

y

N x y x y y y x x x y yA= + + = + + =

( ) ( )2 3 1 1 3 3 1 1 31 1

( ) ( ) 9 0 3 3 19 3 32

e

e

x yN x y x y y y x x x y x y= + + = =

( ) ( )3 1 2 2 1 1 2 2 11 1

( ) ( ) 0 0 3 09 32

e e

xN x y x y y y x x x y x= + + = + + =

E l bl Be d D t i


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N. Zabaras (10/29/2009)

Example problem: Be and D matrices

The non-zero derivatives of the

shape functions are:

The Be matrix is computed next:

The elastic property matrix is:

6310 , 0.3 E Pa v= =

213 3 4.5

2

eA m= =

1

2

3

1, 2, 3,

1, 2, 3,

1, 1, 2, 2, 3, 3,

0 0 0 0 0 1 0 1 01

0 0 0 0 1 0 1 0 0

3 1 0 1 1 0 1

e

e e e e x x x x

e e e e e e y y y y

e e e e e e e xy y x y x y x

B

N N N

N N N d d

N N N N N N

= =

31 2 21 1 1 1, , ,3 3 3 3

ee e e N N N N

y x y x

= = = =

6

2

1 0 32.967 9.89 0

1 0 10 9.89 32.967 01

(1 ) 0 0 11.53850 0

2

vE

D vv

v

= =

E l bl Stiff


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N. Zabaras (10/29/2009)

Example problem: Stiffness

The element stiffness for unit

thickness is given as:

With substitution:

Finally:

6310 , 0.3 E Pa v= =

213 3 4.5

2

eA m= =

1

2

3

Te e e e eK A B D B=

6

0 0 1

0 1 032.967 9.89 0 0 0 1 0 1 0

1 0 11 14.5 10 9.89 32.967 0 0 1 0 1 0 0

0 1 13 30 0 11.5385 1 0 1 1 0 1

1 0 0

0 0 1

eK

=

6

5.769 0 5.769 5.769 0 5.769

0 16.48 4.945 16.48 4.945 0

5.769 4.945 22.25 10.71 16.48 5.76910

5.769 16.48 10.71 22.25 4.945 5.769

0 4.945 16.48 4.945 16.48 0

5.769 0 5.769 5.769 0 5.769

eK

=

E l bl B d l d


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N. Zabaras (10/29/2009)

Example problem: Boundary load

The boundary load is given as:

With substitution:

where:

6310 , 0.3 E Pa v= =

213 3 4.5

2

eA m= =

1

2

3

T

e

t

e e f N td

=

1 3

1 31 31

11 3

1 3

1 3

1(1 ) 0

22

10 (1 ) 2

2 1 1(1 ) (1 )0 0 02 2

0 0 1 1 02 6(1 ) (1 )

2 21 2(1 ) 0

2 21

0 (1 )2

x x

y yx x

e

y y

x x

y y

t t

t tt tL L

f d

t tt t

t t

+ + + + = = + + + +

+ +

1 3

1 3

3 2

215cos45 15 /

2

215sin 45 15 /

2

ox x

oy y

L m

t t N m

t t N m

=

= = =

= = =

E l bl B d d b d f


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N. Zabaras (10/29/2009)

Example problem: Boundary and body forces

With substitution:

The body force is zero

6310 , 0.3 E Pa v= =

213 3 4.5

2

eA m= =

1

2

3

1 3

1 3

1 3

1 3

215

22 22.52

152 22.52

0 0 03 2 3 23

0 0 06 6

2 22.5215

22.52

215

2

x x

y y

e

x x

y y

t t

t t

f

t t

t t

+

+

= = = + +

0.ef =

E l bl E ti l BC


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N. Zabaras (10/29/2009)

Example problem:Essential BC

Based on the essential BC, the

reaction forces are:

The displacement and nodal force

vector are summarized below:

6310 , 0.3 E Pa v= =

213 3 4.5

2

eA m= =

1

2

3

1 1 2 2 3 3

1 2 2 30 0

Te

x y x y x y

T x x y y

r r r r r r r

r r r r

= =

=

11

1

2 2

2 2

3

3 3

022.5

22.5

0,0

22.5

22.5 0

xx

y

x xe e

y y

x

y y

ur

u

ruf d

r u

u

r u

=+ =

= = =

+ =

Example problem Solution step


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N. Zabaras (10/29/2009)

Example problem: Solution step

We are ready to solve for

displacements once we partition the

load/stiffness equations:

We thus solve:

6310 , 0.3 E Pa v= =

213 3 4.5

2

eA m= =

1

2

31

1

26

2

05.769 0 5.769 5.769 0 5.769

0 16.48 4.945 16.48 4.945 0

05.769 4.945 22.25 10.71 16.48 5.76910

5.769 16.48 10.71 22.25 4.945 5.769

0 4.945 16.48 4.945 16.48 0

5.769 0 5.769 5.769 0 5.769

x

y

x

u

u

u

u

=

=

1

2

2

3

33

22.5

22.5

0

22.5

22.50

x

x

yy

x

yy

r

r

r

u

ru

+

= =

+=

1 16 6

3 3

16.48 4.945 22.5 1.0510 10

4.945 16.48 22.5 1.05

y y

x x

u um

u u

= =

Example problem: Stress calculation


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Example problem: Stress calculation

The constant strain in this triangle is

given as:

Using the elastic moduli, we can

now compute the constant stress in

this element:

6310 , 0.3 E Pa v= =

213 3 4.5

2

eA m= =

1

2

3

6

6

6

0

1.05100 0 1 0 1 0 0.35

01 0 1 0 1 0 0 0.35 1003

1 0 1 1 0 1 01.0510

0

e

ex

ey

exy

B

= =

6 6

2

1 0 32.967 9.89 0 0.35 15

1 0 10 9.89 32.967 0 0.35 10 151

(1 ) 0 0 11.5385 0 00 0

2

e ex x

e ey y

e e xy xy

vE

v Pav

v

= = =

Stress recovery


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N. Zabaras (10/29/2009)

Stress recovery

How do we compute the stress at any point inside the

domain?

Since the finite element approximated displacement field

, the stress field , i.e. the stresses arediscontinuous in inter-element boundaries.

We compute the stresses at the Gauss points where they

are more accurate and then extrapolate them at the nodesusing a least squares approximation.

[ ]{ }

ex

e e ey

exy

B d

=

2

1 0

1 01

(1 )0 0 2

e ex x

e ey y

e e xy xy

vE

vv

v

=

0

,u v C

1

C

Example problem: Element equs from min potential energy


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N. Zabaras (10/29/2009)

Example problem: Element equs from min potential energy

The general expression for the strain energy is as follows:

For plane stress, we can write this explicitly as:

This can be further simplified using:

{ } [ ]{ }1 1 1

2 2 2ijkl kl

kl

Tij ij ij ijkl kl

ij ij klV V V

C

U dV C dV C dV

= = =

( )1

2 xx xx yy yy xy xyVU dV= + +

2

1 0

1 01

(1 )0 0

2

xx

yy

xyxy

v

E vv

v

=

Example problem: Element eqs from min potential energy


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N. Zabaras (10/29/2009)

Example problem: Element eqs. from min potential energy

The strain energy for plane stress becomes:

This can be written explicitly as:

We can now introduce our approximations for strain withineach finite element

2

2

1( ) ( ) 22(1 )

xx yy xx xx yy yy xyV

EU dV

= + + + +

{ } [ ]{ }1

2

T

V

U C dV = { }2

1 0

, [ ] 1 01

(1 )0 0

2

xx

yy

xy

vE

C vv

v

= =

where:



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N. Zabaras (10/29/2009)


Thus the strain energy in element e is as follows:

The external work done is similarly computed as:

Finally:

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

1 1

2 2ee T e T e e

TT e e e

eVd B B d

U C dV C dV = =

1{ } [ ] [ ]{ }

2

e e e e T e T e e eU t A d B C B d =

{ }[ ]

{ }[ ]{ } { }

e e

tT T

e T e e T e

xxe e e eext x y x y

e e yy

d N d N

b tW u u u u d

b t

= +

{ } { }e e

t

T T xxe T e e T eext

e e yy

b tW d N d d N d

b t

= +



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N. Zabaras (10/29/2009)


Minimization of this with respect to the nodal displacements

leads to the same algebraic equations as those derived

earlier via the weak form (principle of virtual work):

1

{ } [ ] [ ]{ }2

{ } { }e e

t

e e e T e T e e eext

e

T T xxe T e e T e

e e yy

P U W t A d B C B d

b td N d d N d

b t

= =

+

[ ]{ }

[ ] [ ]{ } e et

e

T T xxe e e T e e e e e

e e e yyK

F

b t

t A B C B d N d N d b t

= +

Example problem


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N. Zabaras (10/29/2009)

Consider the linear elasticity problem shown. The vertical

left edge is fixed. The bottom and the right vertical edgesare traction free. Traction N/m is applied on the top

horizontal edge. E = 3107Pa and = 0.3 . Plane stress

conditions are considered. Discretize the domain using onequadrilateral element.

Example problem

20yt =

0x yt t= =

Example problem


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N. Zabaras (10/29/2009)

Example problem

The elasticity matrix is:

The coordinate matrix is:

The shape functions are:

7

2

1 0 1 0.3 0

1 0 3.310 0.3 1 0

1 (1 ) 0 0 0.350 0

2

vE

D v

v v

= =

1 1

2 2

3 3

4 4

0 1

0 0

2 0.5

2 1

e e

e ee e

e e

e e

x y

x yx y

x y

x y

= =

1 2

3 4

1 1(1 )(1 ), (1 )(1 )

4 4

1 1(1 )(1 ), (1 )(1 )

4 4

N N

N N

= = +

= + + = +

Example problem: Jacobian


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N. Zabaras (10/29/2009)

Example problem: Jacobian

The Jacobian matrix is:

The inverse and determinant of the Jacobian are:

1 11 2 3 4

2 2

1 2 3 4 3 3

4 4

0 1

1 1 1 1 0 0 0 0.125 0.3751

1 1 1 1 2 0.5 1 0.1254

2 1

e e

e ee

e e

e e

x y x y N N N N

x yJ

x y x y N N N N

x y

= = =

+ = +

0.125

+

1

11

3, | | 0.125 0.375

80

3

e eJ J

+ = = +

Example problem: Be matrix


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N. Zabaras (10/29/2009)


The Be matrix is:

To calculate the stiffness matrix we use 4 Gauss points (2in each direction):

The stiffness is:

31 2 4

31 2 4

3 31 1 2 2 4 4

0 0 0 0

0 0 0 0e

N N N N

x x x x

N N N N B

y y y y

N N N N N N N N

y x y x y x y x

=

1 1( , ) ( , ), 1

3 3i jW W= = =

{ }

1 11

1 12 2

( , )1 1

1in

| |

| |

T T

T

i j

e e e e e e e

e e e ei j

i jThese are the weights from D integration

the and

directions

K B D B d B D B J d d

W W B D B J

= =

= = =

=



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N. Zabaras (10/29/2009)


The Be contribution from the Gauss

point can be computed as follows:

The stiffness from this Gauss point is then:

1 1

1 1

1 2 3 431 2 4

1

( , )31 2 4 1 2 3 4

( , )

0.44 0.06 0.12 0.38

0.88 0.88 0.24 0.24

e

N N N N N N N N

x x x xJ

N N N N N N N N y y y y

= =

1 1( , ) ( , ),

3 3=

1 1

31 2 4

31 2 4( , )

3 31 1 2 2 4 4

0 0 0 0

0.44 0 0.06 0 0.12 0 0.38 0

0 0 0 0 0 0.88 0 0.88 0 0.24 0 0.24

0.88 0.44 0.88 0.06 0.24 0.12 0.24 0.38

e

N N N N

x x x x

N N N N B y y y y

N N N N N N N N

y x y x y x y x

= =

{ }1 1

11 1

( , )| |

T

e e e eK W W B D B J

Final stiffness matrix


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N. Zabaras (10/29/2009)

Final stiffness matrix

The final stiffness is:

7

1.49 0.74 0.66 0.16 0.98 0.65 0.15 0.08

2.75 0.24 2.46 0.66 1.68 0.16 1.39

1.08 0.33 0.15 0.16 0.56 0.41

2.6 0.08 1.39 0.41 1.5310

2 0.82 1.18 0.25

3.82 0.33 3.53

1.59 0.25

3.67

K

=

Symmetric

Load vector


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N. Zabaras (10/29/2009)

Load vector

There is no body force:

The load contribution from the

applied traction is:

Accounting for reaction forces,

the assembled load vector is:

0.e

f =

14

1

1

1

1

1

1

1

1,1

1

4

1

1

4

1

0

1 0 00

0 1 20

0 0 0 0 0

0 0 0 0 0 0 0|

0 0 20 0 0 20 0

0 0 0 0 0

1 0 00

0 1 20

0

e T T

N d

N d

f N td N t d

N d

N d

=

= = = = =

1

1

2

23

3

4

4

0

20 0

0

0

0

0

0

20

x

y

x

yex

y

x

y

r

r

r

r f and d u

u

u

u

= =

1

2

3

4

0x yt t= =



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N. Zabaras (10/29/2009)


73

3

4

4

01.49 0.74 0.66 0.16 0.98 0.65 0.15 0.08

02.75 0.24 2.46 0.66 1.68 0.16 1.39

01.08 0.33 0.15 0.16 0.56 0.41

02.6 0.08 1.39 0.41 1.5310

2 0.82 1.18 0.25

3.82 0.33 3.53

1.59 0.25

3.67

x

y

x

y

u

u

u

u

1

1

2

2

20

0

0

0

20

x

y

x

y

r

r

r

r

=

3

3 6

4

4

0

0

1.17 0

9.67 0102.67 1.17

9.94 9.67

2.67

9.94

x

y

x

y

u

u du

u

= =

Computing the stresses


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CCOORRNNEELLLLMAE 4700 FE Analysis for Mechanical & Aerospace Design

Computing the stresses

To compute the stress at e.g. the Gauss point1 1

( , ) ( , ),3 3

=

1 1

1 1 1 1 1 1

( , )

7

( , ) ( , ) ( , )

0

0

01 0.3 0 0.44 0 0.06 0 0.12 0 0.38 0

03.310 0.3 1 0 0 0.88 0 0.88 0 0.24 0 0.24

1.170 0 0.35 0.88 0.44 0.88 0.06 0.24 0.12 0.24 0.389.67

2

xxe e e e e e

yy

xy

D D B d

= = = =

.67

9.94

12.5

5.64

45.5

=

fem zabaras fem4linearelasticity

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