plane stress and plane strain - seoul national university · 2018. 1. 30. · stress-strain...

Mechanics and Design

Finite Element Method: Plane Stress and Plane Strain

SNU School of Mechanical and Aerospace Engineering

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 gxz yz, are both zero.




Stress and strain in 2-D

Stresses in 2-D Principal stress and its direction

{ }ssst

=

UV|W|

RS|

T|x

y

xy

ss s s s

t s

ss s s s

t s

1

22

2

22

2 2

2 2

=+

+-F

HGIKJ + =

=+

--F

HGIKJ + =

x y x yxy

x y x yxy

max

min

tan 22

qt

s spxy

x y

=-




Displacement and rotation of plane element x y-

e e u g ux y xy

ux y

uy x

=¶¶

=¶¶

=¶¶

+¶¶

{ }eeeg

=

RS|T|

UV|W|

x

y

xy




{ } [ ]{ }s e= D

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

stress(s t tz xz yz= = = 0 )

[ ]D E

=- -

L

N

MMMM

O

Q

PPPP1

1 01 0

0 0 12

2n

nn

n

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

(e g gz xz yz= = = 0 )

[ ]( )( )

D E=

+ -

--

-

L

N

MMMM

O

Q

PPPP1 1 2

1 01 0

0 0 1 22

n n

n nn 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

{ }dddd

u

u

u

i

j

m

i

i

j

j

m

m

=RS|T|UV|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 yx y a a x a y

( , )( , )

= + += + +

1 2 3

4 5 6u

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 =+ ++ +RST

UVW =LNM

OQP

R

S

|||

T

|||

U

V

|||

W

|||

a a x a ya a x a y

x yx y

aaaaaa

1 2 3

4 5 6

1

2

3

4

5

6

1 0 0 00 0 0 1

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




Calculation of a a a1 2 3, , :

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

i i i

j j j

m m m

= + += + +

= + +

1 2 3

1 2 3

1 2 3

or

uuu

x yx yx y

aaa

i

j

m

i i

j j

m m

RS|T|UV|W|=L

NMMM

O

QPPP

RS|T|UV|W|

111

1

2

3

Solving a , { } [ ] { }a x u= -1

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

A

i j m

i j m

i j m

- =

L

NMMM

O

QPPP

1 12

a a ab b bg g g

where 2A x y y x y y x y yi j m j m i m i 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 xy y y y y yx x x x x x

= - = - = -

= - = - = -

= - = - = -




After calculation of [ ]x -1, the equation { } [ ] { }a x u= -1

can be expressed as an extended

matrix form.

aaa

A

uuu

i j m

i j m

i j m

i

j

m

1

2

3

12

RS|T|UV|W|=

L

NMMM

O

QPPP

RS|T|UV|W|

a a ab b bg g g

Similarly,

aaa

A

i j m

i j m

i j m

i

j

m

4

5

6

12

RS|T|UV|W|=

L

NMMM

O

QPPP

RS|T|UV|W|

a a ab b bg g g

uuu

Derivation of displacement function u x y( , ) (u can also be derived similarly)

{ } [ ] [ ]u x y

aaa

Ax y

uuu

i j m

i j m

i j m

i

j

m

=RS|T|UV|W|=

L

NMMM

O

QPPP

RS|T|UV|W|

1 12

11

2

3

a a ab b bg 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

( , ) {( ) ( )

( ) }

= + + + + +

+ + +

124

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

= + + + + +

+ + +

124

Simple expression of u and u :

u x y N u N u N ux y N N N

i i j j m m

i i j j m m

( , )( , )

= + +

= + +u u u u where

NA

x y

NA

x y

NA

x y

i i i i

j j j j

m m m m

= + +

= + +

= + +

12

121

2

( )

( )

( )

a b g

a b g

a b g




{ }( , )( , )

yu u u u

u

u

u

=RST

UVW =+ ++ +

RSTUVW=LNM

OQP

R

S

|||

T

|||

U

V

|||

W

|||

u x yx y

N u N u N uN N N

N N NN 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 00 0 0

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

where [ ]NN N N

N N Ni j m

i j m=LNM

OQP

0 0 00 0 0

The displacement function { }y is represented with shape functions N Ni j, , Nm and

nodal displacement { }d .




Review of characteristics of shape function:

Ni＝1, N j＝0, and Nm＝0 at nodes ( , )x yi i

A change of Ni of general elements across the surface x y-




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

Deformation rate:

{ }eeeg

u

u

=

RS|T|

UV|W|=

¶¶¶¶

¶¶

+¶¶

R

S|||

T|||

U

V|||

W|||

x

y

xy

ux

yuy x

Calculation of partial differential terms

¶¶

= =¶¶

+ + = + +ux

uxN u N u N u N u N u N ux i i j j m m i x i j x j m x m, , , ,( )

NA x

x yAi x i i ii

, ( )=¶¶

+ + =1

2 2a b g b

, N AN

Aj xj

m xm

, ,,= =b b2 2

\¶¶

= + +ux A

u u ui i j j m m1

2( )b b b




Likewise,

¶¶

= + +

¶¶

+¶¶

= + + + + +

u g u g u g u

u g b u g b u g b u

y Auy x A

u u u

i i j j m m

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

12

12

( )

( )

Summarizing the deformation rate equation

{ } [ ]{ }eb b b

g g gg b g b g b

u

u

u

=

L

NMMM

O

QPPP

R

S

|||

T

|||

U

V

|||

W

|||

= =RS|T|UV|W|

12

0 0 00 0 0

A

u

u

u

B d B B Bddd

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

AB

AB

Ai

i

i

i i

j

j

j

j j

m

m

m

m m

=L

NMMM

O

QPPP

=

L

NMMM

O

QPPP

=L

NMMM

O

QPPP

12

00 1

2

00 1

2

00

bg

g b

bg

g b

bg

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

sst

eeg

x

y

xy

x

y

xy

DRS|T|

UV|W|=

RS|T|

UV|W|

[ ] ® { } [ ][ ]{ }s = D B d




Step 4: Derivation of element stiffness matrix and equation

Using minimum potential energy principle.

Total potential energy p p u up p i i j m b p su u U= = + + +( , , ,..., ) W W W

Strain energy U dV D dVV

T

V

T= =zzz zzz1

212

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

Potential energy of body force WbV

TX dV= - zzz { } { }y

Potential energy of concentrated load W pTd P= -{ } { }

Potential energy of distributed load (or surface force) WSS

TT dS= -zz { } { }y




\ = -

- -

= -

- -

= -

zzz zzzzz

zzz zzzzz

zzz

p pV

T T

V

T 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

12

12

12

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

{ } { } { } [ ] { }

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

{ } { } { } [ ] { }

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

where

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

T

S

T= + +zzz zz (7.2.46)

Condition having pole values

¶

¶=LNM

OQP

- =zzzp p T

VdB D B dV d f

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

® [ ] [ ][ ] { } { }B D B dV d fT

Vzzz =




So, the element stiffness matrix is [ ] [ ] [ ][ ]k B D B dVT

V

= zzz

Case of an element having constant thickness t : [ ] [ ] [ ][ ]

[ ] [ ][ ]

k t B D B dx dy

tA B D BA

T

T

=

=

zz

Matrix [ ]k is a 6×6 matrix, and the element equation is as below

ffffff

k k kk 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

MMMM

O

Q

PPPP

R

S

|||

T

|||

U

V

|||

W

|||

LL

M M ML

u

u

u




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

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 kTT= = =

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




T

C SS C

C SS C

C SS C

=

-

-

-

L

N

MMMMMMM

O

Q

PPPPPPP

0 0 0 00 0 0 0

0 0 0 00 0 0 00 0 0 00 0 0 0

where

CS==

cossin

qq

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 E= = ´ =1 30 10 0 306in. psi, , .n




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

F TA

F

F

=

= ´

=

1212

1000 1 10

5000

( )( . )psi in in.

lb

The global system of the governing equation is

{ } [ ]{ }F K d= or

FFFFFFFF

RRRR

K

dddddddd

Kdddd

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

50000

50000

0000

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 BT=

· Element 1

- Calculation of matrix [ ]B

[ ]B

A

i j m

i j m

i i j j m m

=

L

NMMM

O

QPPP

12

0 0 00 0 0b b b

g g gg 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 yy yy yx xx xx x

= - = - =

= - = - =

= - = - = -

= - = - = -

= - = - =

= - = - =

10 10 010 0 100 10 100 20 200 0 020 0 20

and

A bh=

= FHGIKJ =

1212

20 10 100( )( ) in.2




Then [ ]B is

[ ]B =-

-- -

L

NMMM

O

QPPP

1200

0 0 10 0 10 00 20 0 0 0 20

20 0 0 10 20 10

- Matrix [ ]D (plane stress)

D E=

- -

L

N

MMMM

O

Q

PPPP=

L

NMMM

O

QPPP( )

( ).

..

.1

1 01 0

0 0 12

30 100 91

1 0 3 00 3 1 00 0 0 35

2

6

n

nn

n

- Calculation of stiffness matrix

i j m

k tA B D BT

= = =

= =

- -- -

- -- -- - -

- - -

L

N

MMMMMMM

O

Q

PPPPPPP

1 3 2

75 0000 91

140 0 0 70 140 700 400 60 0 60 4000 60 100 0 100 60

70 0 0 35 70 35140 60 100 70 240 130

70 400 60 35 130 435

[ ] [ ] [ ][ ] ,.




· Element 2

- Calculation of matrix [ ]B

[ ]B

A

i j m

i j m

i i j j m m

=

L

NMMM

O

QPPP

12

0 0 00 0 0b b b

g g gg 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 yy yy yx xx xx x

= - = - = -

= - = - =

= - = - =

= - = - =

= - = - = -

= - = - =

0 10 1010 0 100 0 020 20 00 20 2020 0 20

and

A bh=

= FHGIKJ =

1212

20 10 100( )( ) in.2




Then matrix [ ]B is

[ ]B =-

-- -

L

NMMM

O

QPPP

1200

10 0 10 0 0 00 0 0 20 0 200 10 20 10 20 0

- Matrix [ ]D (plane stress)

[ ] ( )

.

..

.D =

L

NMMM

O

QPPP

30 100 91

1 0 3 00 3 1 00 0 0 35

6

- Calculation of stiffness matrix

i j m

k

= = =

=

- -- -

- - -- - -- -

- -

L

N

MMMMMMM

O

Q

PPPPPPP

1 4 3

75 0000 91

100 0 100 60 0 600 35 70 35 70 0

100 70 240 130 140 6060 35 130 435 70 400

0 70 140 70 140 060 0 60 400 0 400

[ ] ,.




Element 1:

1 2 3 4

375 0000 91

28 0 28 14 0 14 0 00 80 12 80 12 0 0 0

28 12 48 26 20 14 0 014 80 26 87 12 7 0 00 12 20 12 20 0 0 0

14 0 14 7 0 7 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 0

[ ] ,.

k =

- -- -

- - -- - -- -

- -

L

N

MMMMMMMMMMM

O

Q

PPPPPPPPPPP

Element 2:

1 2 3 4

375 0000 91

20 0 0 0 0 12 20 120 7 0 0 14 0 14 70 0 0 0 0 0 0 00 0 0 0 0 0 0 00 14 0 0 28 0 28 14

12 0 0 0 0 80 12 8020 14 0 0 28 12 48 2612 7 0 0 14 80 26 87

[ ] ,.

k =

- -- -

- -- -- - -

- - -

L

N

MMMMMMMMMMM

O

Q

PPPPPPPPPPP




(3) Calculation of displacement: Superpositioning element stiffness matrix, global system of

stiffness matrix is obtained as below.

1 2 3 4

375 0000 91

48 0 28 14 0 26 20 120 87 12 80 26 0 14 7

28 12 48 26 20 14 0 014 80 26 87 12 7 0 00 26 20 12 48 0 28 14

26 0 14 7 0 87 12 8020 14 0 0 28 12 48 2612 7 0 0 14 80 26 87

[ ] ,.

K =

- - -- - -

- - -- - -- - -

- - -- - -

- - -

L

N

MMMMMMMMMMM

O

Q

PPPPPPPPPPP




Substituting [ ]K to { } [ ]{ }F K d= ,

RRRR

ddd

x

y

x

y

x

y

x

1

1

2

2

3

3

4

50000

50000

375 0000 91

48 0 28 14 0 26 20 120 87 12 80 26 0 14 7

28 12 48 26 20 14 0 014 80 26 87 12 7 0 0

0 26 20 12 48 0 28 1426 0 14 7 0 87 12 8020 14 0 0 28 12 48 2612 7 0 0 14 80 26 87

0000

R

S

|||||

T

|||||

U

V

|||||

W

|||||

L

N

MMMMMMMMMMM

O

Q

PPPPPPPPPPP

=

- - -

- - -

- - -

- - -

- - -

- - -

- - -

- - -

,.

d y4

R

S

|||||

T

|||||

U

V

|||||

W

|||||

Applying given boundary conditions with elimination of columns and rows.

50000

50000

48 0 28 140 87 12 80

28 12 48 2614 80 26 87

375 0000 91

3

3

4

4

RS||

T||

UV||

W||

--

- -- -

L

N

MMMM

O

Q

PPPP

RS||

T||

UV||

W||

=,.

dddd

x

y

x

y




Transposing the displacement matrix to the left side

dddd

x

y

x

y

3

3

4

4

1

60 91375 000

10

48 0 28 140 87 12 80

28 12 48 2614 80 26 87

50000

50000

609 64 2

66371041

RS||

T||

UV||

W||

--

- -- -

L

N

MMMM

O

Q

PPPP

RS||

T||

UV||

W||

RS||

T||

UV||

W||

= = ´

-

-.,

.

.

..

in.

The solution of 1-D beam under tension force is

d = =´

= ´ -PLAE

( , )( )

10 000 2010 30 10

670 1066 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

{ }( )

sn

n

nn

b b b

g g g

g b g b g b

=- -

´

L

N

MMMM

O

Q

PPPPFHIKL

NMMM

O

QPPP

R

S

|||

T

|||

U

V

|||

W

|||

EA

dddddd

x

y

x

y

x

y

1

1 01 0

0 01

2

12

0 0 00 0 02

1 3 2

1 3 2

1 1 3 3 2 2

1

1

3

3

2

2

Calculating,

sst

x

y

xy

RS|T|

UV|W|=RS|T|

UV|W|

1005301

2 4.psi




- Element 2

{ }( )

sn

n

nn

b b b

g g g

g b g b g b

=- -

´

L

N

MMMM

O

Q

PPPPFHIKL

NMMM

O

QPPP

R

S

|||

T

|||

U

V

|||

W

|||

EA

dddddd

x

y

x

y

x

y

1

1 01 0

0 01

2

12

0 0 00 0 02

1 4 3

1 4 4

1 1 4 4 3 3

1

1

4

4

3

3

Calculating,

sst

x

y

xy

RS|T|

UV|W|= -

-

RS|T|UV|W|

995122 4..

psi

plane stress and plane strain - seoul national university · 2018. 1. 30. · stress-strain...

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