Finite Element Method: Plane Stress and Plane Strain

Plane stress and Plane strain

- Finite element in 2-D: Thin plate element required 2 coordinates - Plane stress and plane strain problems

- Constant-strain triangular element

- Equilibrium equation in 2-D

plane stress: The stress state when normal stress, which is perpendicular to the plane x-y, and shear stress are both zero

.

plane strain

: The strain state when normal strain

e

_z , which is perpendicular to the plane

x - y

,

and shear strain g

_xz

, g

_yz

are both zero

.

Stress and strain in 2-D

Stresses in 2-D Principal stress and its direction

{ } s s s t

=

U V|

W|

R S|

T|

x y xy

s s s s s

t s

s s s s s

t s

1

2 2

2

2 2

2 2

2 2

= +

+ F -

HG I

KJ ^{+ =}

= +

- F -

HG I

KJ ^{+ =}

x y x y

xy

x y x y

xy

max

min

tan 2 2

q t

s s

p

xy

x y

= -

Displacement and rotation of plane element x- y

e e u

g u

x y xy

u

x y

u y x

= ¶

¶ = ¶

¶ = ¶

¶ + ¶

¶

{ } e

e e g

=

R S|

T|

U V|

W|

x y xy

{ } s = D [ ]{ } e

Stress-strain matrix(or material composed matrix) of isotropic material for plane stress(

s

_z

= t

_xz

= t

_yz

= 0

)

[ ] D E

= - -

L N MM MM

O Q PP PP

1

1 0

1 0

0 0 1 2 n

2

n

n n

Stress-strain matrix(or material composed matrix) of isotropic material for plane deformation (

e

_z

= g

_xz

= g

_yz

= 0

)

[ ] D ( )( E )

= + -

-

-

-

L N MM MM

O Q PP PP

1 1 2

1 0

1 0

0 0 1 2

2

n n

n n

n n

n

General steps of formulation process for plane triangular element Step 1: Determination of element type

Step 2: Determination of displacement function

Step 3: Definition of relations deformation rate – strain and stress-strain Step 4: Derivation of element stiffness matrix and equation

Step 5: Introduction a combination of element equation and boundary conditions for obtaining entire equations

Step 6: Calculation of nodal displacement

Step 7: Calculation of force(stress) in an element

Step 1: Determination of element type

{ } d

d d d

u u u

i j m

i i j j m m

=

R S|

T|

U V|

W| ⁼ R S ||

| T ||

| U V ||

| W ||

|

u u u

Considering a triangular element, the nodes i j m , , are notated in the anti clock wise direction.

The way to name the nodal numbers in an entire structure must be devised to avoid the

element area comes to be negative.

Step 2: Determination of displacement function

u x y a a x a y x y a a x a y ( , )

( , )

= + +

= + +

1 2 3

4 5 6

u

Linear function gives a guarantee to satisfy the compatibility.

A general displacement function { } y containing function u and u can be expressed as below.

{ } y = + +

+ +

RS T UV W ⁼ L

NM O

QP R S ||

| T ||

| U V ||

| W ||

|

a a x a y a a x a y

x y

x y a a a a a a

1 2 3

4 5 6

1 2 3 4 5 6

1 0 0 0

0 0 0 1

Substitute nodal coordinates to the equation for obtaining the values of a .

Calculation of a a a

₁

,

₂

,

₃

:

u a a x a y u a a x a y u a a x a y

i i i

j j j

m m m

= + +

= + +

= + +

1 2 3

1 2 3

1 2 3

or

u u u

x y x y x y

a a a

i j m

i i

j j

m m

R S|

T|

U V|

W| ⁼ L N MM M

O Q PP P

R S|

T|

U V|

W|

1 1 1

1 2 3

Solving a , { } a = [ ] { } x

^-1

u

Obtaining the inverse matrix of [ ] x , [ ] x

A

i j m

i j m

i j m

-

=

L N MM M

O Q PP

1

1 P

2

a a a

b b b

g g g

where 2 A = x y

_i

(

_j

- y

_m

) + x y

_j

(

_m

- y

_i

) + x

_m

( y

_i

- y

_j

) : 2 times of triangle area

a a a

b b b

g g g

i j m j m j i m i m m i j i j

i j m j m i m i j

i m j j i m m j i

x y y x y x x y x y y x

y y y y y y

x x x x x x

= - = - = -

= - = - = -

= - = - = -

After calculation of

[ ] x

^-1 , the equation

{ } a = [ ] { } x

^-1

u

can be expressed as an extended matrix form.

a a

a A

u u u

i j m

i j m

i j m

i j m 1

2

3

1 2

R S|

T|

U V|

W| ⁼ L N MM M

O Q PP P

R S|

T|

U V|

W|

a a a

b b b

g g g

Similarly,

a a

a A

i j m

i j m

i j m

i j m 4

5

6

1 2

R S|

T|

U V|

W| ⁼ L N MM M

O Q PP P

R S|

T|

U V|

W|

a a a

b b b

g g g

u u u

Derivation of displacement function

u x y ( , )

(

u

can also be derived similarly)

{ } u [ x y ] [ ]

a a

a A x y

u u u

i j m

i j m

i j m

i j m

=

R S|

T|

U V|

W| ⁼

L N MM M

O Q PP P

R S|

T|

U V|

1 1 W|

2 1

1

2

3

a a a

b b b

g g g

Arranging by depolyment:

u x y x y u x y u

x y u

i i i i j j j j

m m m m

( , ) {( ) ( )

( ) }

= + + + + +

+ + +

1

24 a b g a b g

a b g

As the same way

u a b g u a b g u

a b g u

( , ) {( ) ( )

( ) }

x y x y x y

x y

i i i i j j j j

m m m m

= + + + + +

+ + +

1 24

Simple expression of u ^and u ^:

u x y N u N u N u

x y N N N

i i j j m m

i i j j m m

( , ) ( , )

= + +

= + +

u u u u where

N A x y

N A x y

N A x y

i i i i

j j j j

m m m m

= + +

= + +

= + +

1 2

1 2

1 2

( )

( )

( )

a b g

a b g

a b g

{ } ( , ) ( , )

y u u u u

u

u

u

= RS T UV W ⁼ RS T ^{+ + + +} UV W ⁼ L

NM O

QP R S ||

| T ||

| U V ||

| W ||

|

u x y x y

N u N u N u

N N N

N N N

N N N

u u u

i i j j m m

i i j j m m

i j m

i j m

i i j j m m

0 0 0

0 0 0

Making the equation be simple in a form of matrix, { } y = N d [ ]{ }

where [ N ] N N N

N N N

i j m

i j m

= L

NM O

QP

0 0 0

0 0 0

The displacement function { } y is represented with shape functions N N

_i

,

_j

_, N

_m

and

nodal displacement { } d .

Review of characteristics of shape function:

N

_i

＝1, N

_j

_{＝0, and} N

_m

＝0 at nodes ( , x y

_{i i}

)

A change of N

_i

of general elements across the surface x - y

Step 3: Definition of relations deformation rate – strain and stress-strain

Deformation rate:

{ } e

e e g

u u

=

R S|

T|

U V|

W| ⁼

¶

¶

¶

¶

¶

¶ + ¶

¶

R S ||

| T ||

|

U V ||

| W ||

|

x y xy

u x y u

y x

Calculation of partial differential terms

¶

¶ = = ¶

¶ + + = + +

u x u

x N u N u N u N u N u N u

x i i j j m m i x i j x j m x m

,

( )

, , ,

N A x x y

i x i i i

A

i

,

= ¶ ( )

¶ + + =

1

2 a b g 2 b

, N

A N

j x

A

j

m x

m

,

= b ,

,

= b

2 2

\ ¶

¶ u = + +

x 1 A

_i

u

_{i j}

u

_{j m m}

u

2 ( b b b )

Likewise,

¶

¶ = + +

¶

¶ + ¶

¶ = + + + + +

u g u g u g u

u g b u g b u g b u

y A

u

y x A u u u

i i j j m m

i i i i j j j j m m m m

1 2

1 2

( )

( )

Summarizing the deformation rate equation

{ } e [ ]{ }

b b b

g g g

g b g b g b

u u u

=

L N MM M

O Q PP P

R S ||

| T ||

| U V ||

| W ||

|

= =

R S|

T|

U V|

W|

1 2

0 0 0

0 0 0

A

u u u

B d B B B

d d d

i j m

i j m

i i j j m m

i i j j m m

i j m

i j m

where

[ B ] [ ] [ ]

A B

A B

i

A

i i

i i

j

j j

j j

m

m

m

m m

=

L N MM M

O Q PP

P ⁼

L N MM M

O Q PP

P ⁼

L N MM M

O Q PP P

1 2

0

0 1

2

0

0 1

2

0 0

b g g b

b g g b

b g

g b

Strain is constant in an element, for matrix B regardless of x and y coordinates, and is influenced by only nodal coordinates in an element.

à CST: constant-strain triangle

Relation of stress-strain

s s t

e e g

x y xy

x y xy

D

R S|

T|

U V|

W| ⁼

R S|

T|

U V|

W|

[ ] ® { } s = D B d [ ][ ]{ }

Using minimum potential energy principle.

Total potential energy

p

_p

= p

_p

( , u

_i

u

_i

, u

_j

,..., u

_m

) = U + W

_b

+ W

_p

+ W

_s

Strain energy

U dV D dV

V T

V T

= 2 ¹ zzz = zzz

1

{ } { } e s 2 { } [ ]{ } e e

Potential energy of body force

W

_b

V

T

X dV

= - zzz ^{{ } { } y}

Potential energy of concentrated load

W

_p

= -{ } { } d

^T

P

Potential energy of distributed load (or surface force)

W

_S

S

T

T dS

= - zz ^{{ } { } y}

\ = -

- -

= -

- -

= -

zzz zzz

zzz zz zzz

zzz zz

p

_p

V

T

V

T

T

S

T T

T T

V

T V

T

T T

S T

T T

V

T

d B D B d dV d N X dV

d P d N T dS

d B D B dV d d N X dV

d P d N T dS

d B D B dV d d f

1 2

1 2

1 2

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

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

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

where

{ } f [ N ] { } X dV { } P [ N ] { } T dS

V

T

S

= zzz + + zz

T ^(7.2.46)

Condition having pole values

¶

¶ = L

NM O

QP ^{- =}

zzz

p

_{p T}

d

V

B D B dV d f

{ } [ ] [ ][ ] { } { } 0

®

[ ] [ ][ ] B

^T

D B dV d { } { } f

zzz

V

⁼

So, the element stiffness matrix is

[ ] k [ ] [ ][ ] B

^T

D B dV

V

= zzz

Case of an element having constant thickness

t

:

[ ] [ ] [ ][ ] [ ] [ ][ ]

k t B D B dx dy tA B D B

A T

T

=

=

zz

Matrix

[ ] k

is a 6×6 matrix, and the element equation is as below

f f f f f f

k k k

k k k

k k k

u u u

x y x y x y 1

1

2

2

3

3

11 12 16

21 22 26

61 62 66

1

1

2

2

3

3

R S ||

||

T ||

||

U V ||

||

W ||

||

=

L N MM MM

O Q PP PP

R S ||

| T ||

| U V ||

| W ||

|

L L

M M M

L

u

u

u

a global coordinate system of equation.

[ ] K [ k

^{( )e}

]

e N

=

=

å

1

and

{ } F { f

^{( )e}

}

e N

=

=

å

1

{ } F = [ ]{ } K d

Step 6: Calculation of nodal displacement

Step 7: Calculation of force(stress) in an element

Transformation from the global coordinate system to the local coordinate system: (See Ch. 3)

$ $ $

d = Td f = T f k = T k T

^T

Constant-strain triangle(CST) has 6 degrees of freedom.

T

C S S C

C S S C

C S S C

= -

-

-

L

N MM MM MM M

O

Q PP PP PP P

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

where

C S

=

= cos

sin q

q

A triangular element with local coordinates system not along to the global coordinate system

Finite Element Method in a plane stress problem

Find nodal displacements and element stresses in the case of the thin plate(see below figure) under surface force.

thickness

t = 1 in. , E = 30 10 ´

⁶

psi , n = 0 30 .

(1) Discretization: Surface tension force is replaced by the following nodal loads.

F TA F

F

=

= ´

= 1 2 1

2 1000 1 10 5000

( psi )( in . in. ) lb

The global system of the governing equation is

{ } F = [ ]{ } K d

or

F F F F F F F F

R R R

R K

d d d d d d d d

K d d d d

x y x y x y x y

x y x y

x y x y x y x y

x y x y 1

1

2

2

3

3

4

4

1

1

2

2

1

1

2

2

3

3

4

4

3

3

4

4

5000 0 5000

0

0 0 0 0

R S

|| ||

| T

|| ||

| U V

|| ||

| W

|| ||

|

=

R S

|| ||

| T

|| ||

|

U V

|| ||

| W

|| ||

|

=

R S

|| ||

| T

|| ||

| U V

|| ||

| W

|| ||

|

=

R S

|| ||

| T

|| ||

| U

[ ] [ ] V

|| ||

| W

|| ||

|

where

[ ] K

is a 8×8 matrix.

(2) A combination of stiffness matrix:

[ ] k = tA B [ ] [ ][ ] D B

· Element 1

- Calculation of matrix

[ ] B

[ ] B A

i j m

i j m

i i j j m m

=

L N MM M

O Q PP P

1 2

0 0 0

0 0 0

b b b

g g g

g b g b g b

where

b b b g g g

i j m

j m i

m i j

i m j

j i m

m j i

y y y y

y y x x x x x x

= - = - =

= - = - =

= - = - = -

= - = - = -

= - = - =

= - = - =

10 10 0 10 0 10

0 10 10 0 20 20 0 0 0 20 0 20

and

A = bh

= F HG I

KJ ⁼

1 2 1

2 ( 20 10 )( ) 100 in.

²

Then is

[ ] B =

- -

- -

L N MM M

O Q PP P

1 200

0 0 10 0 10 0

0 20 0 0 0 20

20 0 0 10 20 10

- Matrix

[ ] D

(plane stress)

D E

= - -

L N MM MM

O Q PP PP ⁼

L N MM M

O Q PP P

( )

( ) .

. . 1 .

1 0

1 0

0 0 1 2

30 10 0 91

1 0 3 0 0 3 1 0 0 0 0 35

2

6

n

n

n n

- Calculation of stiffness matrix

i j m

k tA B

^T

D B

= = =

= =

- -

- -

- -

- -

- - -

- - -

L

N MM MM MM M

O

Q PP PP PP P

1 3 2

75 000 0 91

140 0 0 70 140 70

0 400 60 0 60 400

0 60 100 0 100 60

70 0 0 35 70 35

140 60 100 70 240 130

70 400 60 35 130 435

[ ] [ ] [ ][ ] ,

.

- Calculation of matrix

[ ] B

[ ] B A

i j m

i j m

i i j j m m

=

L N MM M

O Q PP P

1 2

0 0 0

0 0 0

b b b

g g g

g b g b g b

where

b b b g g g

i j m

j m i

m i j

i m j

j i m

m j i

y y y y

y y x x x x x x

= - = - = -

= - = - =

= - = - =

= - = - =

= - = - = -

= - = - =

0 10 10 10 0 10 0 0 0 20 20 0 0 20 20 20 0 20

and

A = bh

= F HG I

KJ ⁼

1 2 1

2 ( 20 10 )( ) 100 in.

²

Then matrix is

[ ] B =

-

-

- -

L N MM M

O Q PP P

1 200

10 0 10 0 0 0

0 0 0 20 0 20

0 10 20 10 20 0

- Matrix

[ ] D

(plane stress)

[ ] ( ) .

. .

. D =

L N MM M

O Q PP P

30 10 0 91

1 0 3 0 0 3 1 0 0 0 0 35

6

- Calculation of stiffness matrix

i j m

k

= = =

=

- -

- -

- - -

- - -

- -

L

N MM MM MM M

O

Q PP PP PP P

1 4 3

75 000 0 91

100 0 100 60 0 60

0 35 70 35 70 0

100 70 240 130 140 60

60 35 130 435 70 400

0 70 140 70 140 0

[ ] ,

.

Element 1:

1 2 3 4

375 000 0 91

28 0 28 14 0 14 0 0

0 80 12 80 12 0 0 0

28 12 48 26 20 14 0 0

14 80 26 87 12 7 0 0

0 12 20 12 20 0 0 0

14 0 14 7 0 7 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

[ ] ,

k = .

- -

- -

- - -

- - -

- -

- -

L

N MM MM MM MM MM M

O

Q PP PP PP PP PP P

Element 2:

1 2 3 4

375 000 0 91

20 0 0 0 0 12 20 12

0 7 0 0 14 0 14 7

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 14 0 0 28 0 28 14

12 0 0 0 0 80 12 80

20 14 0 0 28 12 48 26

12 7 0 0 14 80 26 87

[ ] ,

k = .

- -

- -

- -

- -

- - -

- - -

L

N MM MM MM MM MM M

O

Q PP PP PP PP PP P

(3) Calculation of displacement: Superpositioning element stiffness matrix, global system of stiffness matrix is obtained as below.

1 2 3 4

375 000 0 91

48 0 28 14 0 26 20 12

0 87 12 80 26 0 14 7

28 12 48 26 20 14 0 0

14 80 26 87 12 7 0 0

0 26 20 12 48 0 28 14

26 0 14 7 0 87 12 80

20 14 0 0 28 12 48 26

12 7 0 0 14 80 26 87

[ ] ,

K = .

- - -

- - -

- - -

- - -

- - -

- - -

- - -

- - -

L

N MM MM MM MM MM M

O

Q

PP PP

PP PP

PP P

Substituting

[ ] K

_to

{ } F = [ ]{ } K d

, R

R R R

d d d

x y x y

x y x 1

1 2 2

3 3 4

5000 0 5000

0

375 000 0 91

48 0 28 14 0 26 20 12

0 87 12 80 26 0 14 7

28 12 48 26 20 14 0 0

14 80 26 87 12 7 0 0

0 26 20 12 48 0 28 14

26 0 14 7 0 87 12 80

20 14 0 0 28 12 48 26

12 7 0 0 14 80 26 87

0 0 0 0

R S

|| ||

| T

|| ||

| U V

|| ||

| W

|| ||

|

L

N MM MM MM MM MM M

O

Q PP PP PP PP PP P

=

- - -

- - -

- - -

- - -

- - -

- - -

- - -

- - -

, .

d_4y

R S

|| ||

| T

|| ||

| U V

|| ||

| W

|| ||

|

Applying given boundary conditions with elimination of columns and rows.

5000 0 5000

0

48 0 28 14

0 87 12 80

28 12 48 26

14 80 26 87

375 000 0 91

3

3

4

4

R S ||

T ||

U V ||

W ||

-

-

- -

- -

L N MM MM

O Q PP PP

R S ||

T ||

U V ||

W ||

= ,

.

d d d d

x y x y

Transposing the displacement matrix to the left side

d d d d

x y x y 3

3

4

4

1

0 91

6

375 000 10

48 0 28 14

0 87 12 80

28 12 48 26

14 80 26 87

5000 0 5000

0

609 6 4 2 663 7 104 1

R S ||

T ||

U V ||

W ||

-

-

- -

- -

L N MM MM

O Q PP PP

R S ||

T ||

U V ||

W ||

R S ||

T ||

U V ||

W ||

= = ´

-

.

-

,

. . . .

in.

The solution of 1-D beam under tension force is

d = =

´ = ´

^-

PL AE

( , )

( )

10 000 20

10 30 10

₆

670 10

⁶

in.

Therefore, x-component of the displacement at nodes in the equation (7.5.27) of 2-D plane is quite accurate when the grid is considered as coarse grid.

(4) Stresses at each node:

{ } s = D B d [ ][ ]{ }

- Element 1

{ } s ( ) n

n n

n

b b b

g g g

g b g b g b

= - -

´

L N MM MM

O Q PP PP F

H I

K L N MM M

O Q PP P

R S ||

| T ||

| U V ||

| W ||

|

E

A

d d d d d d

x y x y x y

1

1 0

1 0

0 0 1 2

1 2

0 0 0

0 0 0

2

1 3 2

1 3 2

1 1 3 3 2 2

1 1 3 3 2 2

Calculating,

s s t

x y xy

R S|

T|

U V|

W| ⁼ R S|

T|

U V|

W|

1005 301

2 4 .

psi

- Element 2

{ } s ( ) n

n n

n

b b b

g g g

g b g b g b

= - -

´

L N MM MM

O Q PP PP F

H I

K L N MM M

O Q PP P

R S ||

| T ||

| U V ||

| W ||

|

E

A

d d d d d d

x y x y x y

1

1 0

1 0

0 0 1 2

1 2

0 0 0

0 0 0

2

1 4 3

1 4 4

1 1 4 4 3 3

1 1 4 4 3 3

Calculating,

s s t

x y xy

R S|

T|

U V|

W| ^{= - -} R S|

T|

U V|

W|

995 1 2 2 4 . .

psi