#### Transcript Plane Stress Formulation

```Formulation of Two-Dimensional
Elasticity Problems
Simplified Elasticity Formulations
The General System of Elasticity Field Equations
of 15 Equations for 15 Unknowns Is Very Difficult
to Solve for Most Meaningful Problems, and So
Modified Formulations Have Been Developed.
Displacement Formulation
Stress Formulation
Eliminate the stresses and strains
from the general system of equations.
This generates a system of three
equations for the three unknown
displacement components.
Eliminate the displacements and
strains from the general system of
equations. This generates a system of
six equations and for the six unknown
stress components.
Solution to Elasticity Problems
x
F(z)
G(x,y)
z
y
Even Using Displacement and Stress Formulations
Three-Dimensional Problems Are Difficult to Solve!
So Most Solutions Are Developed for Two-Dimensional Problems
Two and Three Dimensional Problems
Three-Dimensional
Two-Dimensional
x
y
y
z
z
z
Spherical Cavity
y
x
x
Two-Dimensional Formulation
Plane Stress
Plane Strain
y
y
2h << other dimensions
R
z
x
x
 x   x ( x, y )
R
 y   y ( x, y )
 xy   xy ( x , y )
z
 z   xz   yz  0
u  u( x, y ) , v  v( x, y ) , w  0
Examples of Plane Strain Problems
y
P
z
x
x
y
z
Long Cylinders
Semi-Infinite Regions
Examples of Plane Stress Problems
Thin Plate With
Central Hole
Circular Plate Under
Plane Strain Formulation
u  u( x, y ) , v  v( x, y ) , w  0
Strain-Displacement
ex 
u
x
, ey 
v
y
, e xy 
1  u v 



2  y
x 
e z  e xz  e yz  0
Hooke’s Law
 x   (e x  e y )  2e x
 y   (e x  e y )  2e y
 z   (e x  e y )   ( x   y )
 xy  2  e xy ,  xz   yz  0
Plane Strain Formulation
Stress Formulation
Displacement Formulation
 x
  u v 
 u  (   )


  F x  0
x  x
y 
2
x
  xy
  u v 
 v  (   )


  F y  0
y  x
y 
2
x


  xy
y
 y
y
 ( x   y )  
2
Si
y
R
S = Si + So
So
x
 Fx  0
 Fy  0
F y
 Fx


1    x
y
1



Plane Strain Example
Given the Following
u 
1 
2
ox , v 
E
Displaceme
 (1   )
E
nts , Determine
the Strains
and Stresses
o y , w  0
Strains :
ex 
u
x

1 
2
E
o , ey 
v
y

 (1   )
E
 o , e xy  e yz  e zx  e z  0
Stresses :
 x   (e x  e y )  2e x

 1 2
E

 (1   )( 1  2  )



o   2


o 

E
)


1
(
2
E
(1   )( 1  2  ) 



E
 o
 y   (e x  e y )  2e y  0
 z   ( e x  e y )   (  x   y )    o
 xy   xz   yz  0
Plane Stress Formulation
 x   x ( x , y ) ,  y   y ( x , y ) ,  xy   xy ( x , y ) ,  z   xz   yz  0
Hooke’s Law
ex 
ey 
1
E
1
E
ez  
e xy 
Strain-Displacement
(  x   y )
ex 
(  y   x )

E
( x   y )  
1 
E
u
, ey 
x
e xy 

1 
(ex  e y )
v
y
, ez 
1  u v 



2  y x 
e yz 
1  v w 

  0

2  z
y 
e xz 
1  u w 


0
2  z
x 
 xy , e xz  e yz  0
Note plane stress theory normally neglects some of the
strain-displacement and compatibility equations.
w
z
Plane Stress Formulation
Displacement Formulation
Stress Formulation
  u v 


  F x  0
2 (1   )  x   x
y 
 x
  u v 
 v 


  F y  0
2 (1   )  y   x
y 
  xy
 u 
2
2
E
x
E
x

  xy
y
 y
y
 Fx  0
 Fy  0
F y
 Fx

 (  x   y )   (1   ) 


x
y

2
Si
y

R
S = Si + So
So
x



Correspondence Between Plane Problems
Plane Strain
Plane Stress
 u  (   )
  u v 


  F x  0
x  x
y 
 u 
  u v 


  F x  0
2 (1   )  x   x
y 
 v  (   )
  u v 


  F y  0
y  x
y 
 v 
  u v 


  F y  0
2 (1   )  y   x
y 
2
2
 x
x
  xy
x


  xy
2
y
 y
1
E
x
  xy
 Fy  0
y
E
 x
 Fx  0
F y
 Fx

 ( x   y )  

1    x
y
2
2
x





  xy
y
 y
y
 Fx  0
 Fy  0
F y
 Fx

 (  x   y )   (1   ) 


x
y

2



Elastic Moduli Transformation Relations for Conversion
Between Plane Stress and Plane Strain Problems
Plane Strain
Plane Stress
E
v

E
Plane Stress to Plane Strain
Plane Strain to Plane Stress
1 
2
1 
E (1  2  )
(1   )
2

1 
Therefore the solution to one plane problem also yields the solution
to the other plane problem through this simple transformation
Airy Stress Function Method
Plane Problems with No Body Forces
Stress Formulation
 x
x
  xy
x


  xy
y
 
2
 y
y
Airy Representation
 0
x 
 0
y
2
2
, y 
 ( x   y )  0
2
Biharmonic Governing Equation
 
4
x
4
 
4
2
x y
2
 
4
2

y
4
 
   0
4
(Single Equation with Single Unknown)
x
2
 
2
,  xy  
xy
Polar Coordinate Formulation
  r
 r 
x2
 
 


d
d
r 
r
Strain-Displacement
dr
er 
 r 
rd
 r
r
r
  r
r


r 
( r    )
r
1  

r 
2  r
r
u  
1
 ur 

r
 
u 
u 
1  1 u r

 

2  r 
r
r 
Plane Stress
Plane Strain
 r   ( e r  e )  2 e r
    ( e r  e )  2 e
er 
1
E
(  r    )
 z   ( e r  e  )   (  r    ) e  1 (    )


r
E
 r   2  e r  ,   z   rz  0


ez  
( r    )  
(er  e )
E
1 
Equilibrium Equations
1   r
, e 
Hooke’s Law



r
e r 
dr
dr
 r
r
  r
x1
 r
u r
F
Fr
d
 r
 Fr  0
e r 
 F  0
1 
E
 r  , e  z  e rz  0
Airy Representation
r 
1 
r r
1  
2

r
2

 
2
 
r
 r  
2
  1  


r  r  
2
2
 2
1 
1 
    2 
 2
2
r r
r 
 r
4
2
  2
1 
1 
  2 
 2
2
r r
r 
 r

   0

Solutions to Plane Problems
Cartesian Coordinates
Airy Representation
 
2
x 
y
2
 
2
, y 
x
2
 
2
,  xy  
xy
Biharmonic Governing Equation
 
 
4
x
4
4
2
x y
2
 
4
2

y
4
   0
4
Traction Boundary Conditions
y
S
R
x
Tx  f x ( x, y ) , T y  f y ( x, y )
Solutions to Plane Problems
Polar Coordinates
Airy Representation
r 
1 
r r
1  
2

r
2

2
 
2
,  
r
2
,  r  
  1  


r  r  
Biharmonic Governing Equation
2
 2
1 
1 
    2 
 2
2
r r
r 
 r
4
2
  2
1 
1 
  2 
 2
2
r r
r 
 r

   0

Traction Boundary Conditions
S
R
y

r

x
Tr  f r ( r , ) , T  f  ( r , )
Cartesian Coordinate Solutions
Using Polynomial Stress Functions
 
 
4
x
4
4
2
x y
2

( x, y ) 
4
2

y
 
2
 
0
4
x 
y
2
 
2
, y 
x
2
 
2
,  xy  
xy

 A
x y  A00  A10 x  A01 y  A 20 x  A11 xy  A02 y  
m
mn
n
2
2
m 0 n0
m  n  1 terms do not contribute to the stresses and are therefore dropped
m  n  3 terms will automatically satisfy the biharmonic equation
m  n  3 terms require constants Amn to be related in order to satisfy biharmonic equation
Solution method limited to problems where boundary traction conditions
can be represented by polynomials or where more complicated boundary
Stress Function Example
Consider
 
the Following
F
d
2
 
the Stresses :
2
y
2
 
 
6F
d
3
x(d  y )
2
y 
x
:
xy ( 3 d  2 y )
3
Determine
x 
Stress Function
2
 
2
 0 ,  xy  
xy

6F
d
3
y (d  y )
Appears to Solve the Beam Problem:
y
F
d
x
```