Using Airy Stress Function approach, plane elasticity formulation with zero
body forces reduces to a single governing biharmonic equation.
In Cartesian coordinates it is given by
 4
 4
 4
4

2



0
x 4
x 2 y 2 y 4
and the stresses are related to the stress function by
 2
 2
 2
 x  2 ,  y  2 ,  xy  
y
x
xy
We now explore solutions to several specific problems in both
Cartesian and Polar coordinate systems
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
Cartesian Coordinate Solutions
Using Polynomials
In Cartesian coordinates we choose Airy stress function solution of polynomial form
( x, y ) 


 A
mn x
m
y n , Amn  Constants to be Determined
m 0 n 0
Method produces polynomial stress distributions, and thus would not satisfy general
boundary conditions. However, using Saint-Venant’s principle we can replace a nonpolynomial condition with a statically equivalent polynomial loading. This formulation
is most useful for problems with rectangular domains, and is commonly based on
inverse solution concept where we assume a polynomial solution form and then try
to find what problem it will solve.
Notice that the three lowest order terms with m + n  1 do not contribute to the
stresses and will therefore be dropped. Second order terms will produce a constant
stress field, third-order terms will give a linear distribution of stress, and so on for
higher-order polynomials.
Terms with m + n  3 will automatically satisfy biharmonic equation for any choice of
constants Amn. However, for higher order terms, constants Amn will have to be related
in order to have polynomial satisfy biharmonic equation.
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
Example 8.1 Uniaxial Tension of a Beam
y
T
T
2c
x
2l
Stress Field
Boundary Conditions:
 x (  l , y )  T ,  y ( x,  c )  0
 xy (l , y )   xy ( x,c)  0
Since the boundary conditions specify constant
stresses on all boundaries, try a second-order
stress function of the form
  A02 y 2
 x  2 A02 ,  y   xy  0
Displacement Field (Plane Stress)
u
1
T
 e x  (  x   y ) 
x
E
E
v
1
T
 e y  (  y   x )   
y
E
E
u
T
T
x  f ( y ) , v   y  g ( x )
E
E
 xy
u v

 2e xy 
 0  f ( y )  g ( x )  0
y x

f ( y )  o y  uo
The first boundary condition implies that A02 = T/2,
and all other boundary conditions are identically
. . . Rigid-Body Motion
satisfied. Therefore the stress field solution is
g ( x )  o x  vo
given by
“Fixity conditions” needed to determine RBM terms
 x  T ,  y   xy  0
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
u (0,0)  v(0,0)   z (0,0)  0  f ( y )  g ( x)  0
Example 8.2 Pure Bending of a Beam
y
M
2c
M
x
2l
Stress Field
Boundary Conditions:
 y ( x, c )  0 ,  xy ( x, c )   xy ( l , y )  0

c
c
 x ( l , y )dy  0 ,

c
c
 x ( l , y ) ydy   M
Expecting a linear bending stress distribution,
try second-order stress function of the form
  A03 y 3
 x  6 A03 y ,  y   xy  0
Moment boundary condition implies that
A03 = -M/4c3, and all other boundary conditions
are identically satisfied. Thus the stress field is
3M
 x   3 y ,  y   xy  0
2c
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
Displacement Field (Plane Stress)
u
3M
3M

yu
xy  f ( y )
3
x
2 Ec
2 Ec3
v
3M
3M 2

yv
y  g ( x)
3
y
2 Ec
4 Ec3
u v
3M

0
x  f ( y )  g ( x )  0
3
y x
2 Ec
f ( y )  o y  uo
3M 2
x  o x  vo
4 Ec3
“Fixity conditions” to determine RBM terms:
g ( x) 
v( l ,0)  0 and u( l ,0)  0
uo  o  0 , vo  3Ml 2 / 16Ec3
Example 8.2 Pure Bending of a Beam
Solution Comparison of Elasticity
with Elementary Mechanics of Materials
y
M
M
2c
x
2l
Elasticity Solution
M
x  
y ,  y   xy  0
I
Mxy
M
u
,v
[ 4 y 2  4 x 2  l 2 ]
EI
8 EI
I  2c 3 / 3
Mechanics of Materials Solution
Uses Euler-Bernoulli beam theory to
find bending stress and deflection of
beam centerline
M
x  
y ,  y   xy  0
I
M
v  v ( x,0) 
[4 x 2  l 2 ]
8EI
Two solutions are identical, with the exception of the x-displacements
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
Example 8.3 Bending of a Beam by
Uniform Transverse Loading
w
wl
wl
2c
x
y
2l
Stress Field
Boundary Conditions:
  A20 x 2  A21 x 2 y  A03 y 3  A23 x 2 y 3 
 xy ( x, c )  0
2 3
y )
3
 y  2 A20  2 A21 y  2 A23 y 3
 x  6 A03 y  6 A23 ( x 2 y 
 y ( x, c )  0
 y ( x , c )   w



Elasticity
c
c
c
c
c
c
 xy  2 A21 x  6 A23 xy 2
 x ( l , y )dy  0
 x ( l , y ) ydy  0
 xy ( l , y )dy   wl
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
A23 5
y
5
BC’s
3w  l 2 2 
3w
2
 2   y  3 ( x 2 y  y 3 )
x 
4c  c
5
4c
3
w 3w
w
y   
y  3 y3
2 4c
4c
3w
3w
 xy  
x  3 xy 2
4c
4c
Example 8.3 Beam Problem
Stress Solution Comparison of Elasticity
with Elementary Mechanics of Materials
w
wl
wl
2c
x
y
2l
Elasticity Solution
w 2
w y3 c2 y
2
x 
(l  x ) y  ( 
)
2I
I 3
5
w  y3
2 
 y   
 c 2 y  c 3 
2I  3
3 
 xy  
w
x(c 2  y 2 )
2I
Mechanics of Materials Solution
My w 2

(l  x 2 ) y
I
2I
y  0
x 
 xy 
VQ
w

x(c 2  y 2 )
It
2I
Shear stresses are identical, while normal stresses are not
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
Example 8.3 Beam Problem
Normal Stress Comparisons of Elasticity
with Elementary Mechanics of Materials
x – Stress at x=0
y - Stress
l/c = 2
l/c = 3
l/c = 4
x/w - Elasticity
x/w - Strength of Materials
Maximum differences between two theories exist at top
and bottom of beam, difference in stress is w/5. For most
beam problems (l >> c), bending stresses will be much
greater than w, and differences between elasticity and
strength of materials will be relatively small.
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
y/w - Elasticity
y/w - Strength of Materials
Maximum difference between two theories is w and
occurs at top of beam. Again this difference will be
negligibly small for most beam problems where l >> c.
These results are generally true for beam problems
with other transverse loadings.
Example 8.3 Beam Problem
Normal Stress Distribution on Beam Ends
w  y 3 c 2 y  3w  1 y 3 1 y 
 
 3 

 x (l , y )   
I 3
5  2 3 c
5c
w
wl
wl
2c
x
y
2l
End stress distribution does not
vanish and is nonlinear but gives
zero resultant force.
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
 x (l , y) / w
Example 8.3 Beam Problem
w
wl
wl
2c
x
y
2l
Displacement Field (Plane Stress)
w
x3
2 y 3 2c 2 y
y3
2c 3
2
2
u
[(l x  ) y  x (

)  x (  c y 
)]  f ( y )
2 EI
3
3
5
3
3
2
w
y 4 c 2 y 2 2c 3 y
y 4 c2 y 2
2
2 y
v
[( 

)  ( l  x )
 ( 
)]  g ( x )
2 EI 12
2
3
2
6
5
w 4
w 2 8
f ( y )  o y  uo , g ( x) 
x 
[l  (  )c 2 ]x 2  o x  vo
24 EI
4 EI
5
Choosing Fixity Conditions u(0, y )  v( l , y )  0
w
x3
2 y 3 2c 2 y
y3
2c 3
[(l 2 x  ) y  x (

)  x (  c 2 y 
)]
2 EI
3
3
5
3
3
w  y 4 c 2 y 2 2c 3 y
y 2 y 4 c2 y 2

v


 [(l 2  x 2 )


]
2 EI  12
2
3
2
6
5
5wl 4
12 4  c 2
uo  o  0 , vo 
[1  (  ) 2 ]
24 EI
5 5 2 l
u
x4
l2
4  2 2  5wl 4
12 4  c 2

 [  (  )c ]x  
[1  (  ) 2 ]
12
2
5 2
5 5 2 l
 24EI
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
5wl 4
12 4  c 2
[1  (  ) 2 ]
24EI
5 5 2 l
5wl 4
v

Strength of Materials: max
24EI
v(0,0)  vmax 
Good match for beams where l >> c
Cartesian Coordinate Solutions
Using Fourier Methods
Fourier methods provides a more general solution scheme for biharmonic
equation. Such techniques generally use separation of variables along
with Fourier series or Fourier integrals.
( x, y )  X ( x)Y ( y )
 4
 4
 4
2 2 2  4 0
x 4
x y
y
Choosing X  e x , Y  ey
  i
  sin x[( A  Cy ) sinh y  ( B  Dy ) coshy ]
 cosx[( A  C y ) sinh y  ( B   D y ) coshy ]
 sin y[(E  Gx ) sinh x  ( F  Hx ) cosh x ]
 cos y[(E   G x ) sinh x  ( F   H x ) cosh x ]
 0  0
0  C0  C1 x  C2 x 2  C3 x 3
0  C4 y  C5 y 2  C6 y 3  C7 xy  C8 x 2 y  C9 xy 2
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
Example 8.4 Beam with Sinusoidal Loading
qosinπx/l
y
qol/
qol/
2c
x
l
Stress Field
Boundary Conditions:
 x (0, y )   x (l , y )  0
  sin x[( A  Cy ) sinh y  ( B  Dy ) cosh y ]
 xy ( x, c )  0
 x   sin x[( A sinh y  C (y sinh y  2 coshy )
 y ( x , c )  0
 y ( x, c )   qo sin(x / l )


c
c
c
c
 xy (0, y )dy   qo l / 
 xy (l , y )dy  qo l / 
2
 B coshy  D(y coshy  2 sinh y )]
 y   2 sin x[( A  Cy ) sinh y  ( B  Dy ) coshy ]
 xy   2 cosx[( A coshy  C (y coshy  2 sinh y )
 B sinh y  D(y sinh y  2 coshy )]
A   D(c tanhc  1)
B  C (c cothc  1)
c
l
C
 2  c
c
c 
2 2   sinh cosh 
l  l
l
l 
 qo sinh
c
l
D
2
  c
c
c 
2 2   sinh cosh 
l  l
l
l 
 qo sinh

Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island

l
Example 8.4 Beam Problem
Bending Stress
 x   sin x[( A sinh y  C (y sinh y  2 cosh y )
2
 B cosh y  D(y cosh y  2 sinh y )]
c
c
 qo sinh
qo cosh
l
l
C
, D
2
2
  c
c
c 
  c
c
c 
2 2   sinh cosh 
2 2   sinh cosh 
l  l
l
l 
l  l
l
l 

A   D(c tanh c  1) , B  C (c coth c  1) ,  
l

y
y 
c 
y
y cosh
 2l sinh
  c tanh
 l  sinh

q
c
x
l
l 
l
l


 x   o sinh sin
c
c
2
l
l 
c  l sinh cosh

l
l
y sinh

y
y 
c 
y 
 2l cosh
  c coth
 l  cosh 
l
l 
l
l


c
c

c  l sinh cosh

l
l
3qo l 5
, C  0 , A  D , B  0
4c 3 5
3q l 3  y
3q l 2
y
y 
x
x
 x   3o 3  cosh
 sinh  sin
  3o 2 y sin
l
l 
l
l
4c   l
2c 
For the case l  c : D  
qo l 2
x
sin
y
2
3q l 2
My
x
Strength of Materials Theory :  x  
  3 l
  3o 2 y sin
I
l
2c / 3
2c 
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
qosinπx/l
y
qol/
qol/
2c
x
l
x l/2
Example 8.4 Beam Problem
qosinπx/l
y
qol/
qol/
2c
x
l
Displacement Field (Plane Stress)

cos x{ A(1  ) sinh y  B(1  ) cosh y
E
 C[(1  )y sinh y  2 cosh y ]
u
 D[(1  )y cosh y  2 sinh y ]}  o y  uo
u(0,0)  v(0,0)  v(l ,0)  0

sin x{ A(1  ) cosh y  B(1  ) sinh y
E
 C[(1  )y cosh y  (1  ) sinh y ]
v
 D[(1  )y sinh y  (1  ) cosh y ]}  o y  vo
o  vo  0 , u o 

[ B(1  )  2C ]
E
D
sin x[2  (1  )c tanh c ]
E
3qo l 4
x
1   c
c
3qo l 5
For the case l >> c
v( x,0)   3 4 sin [1 
tanh ]
D 3 5
4c 
2c  E
l
2 l
l
4
3q l
x
Strength of Materials v( x,0)   3 o 4 sin
2c  E
l
v ( x,0) 
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
Example 8.5 Rectangular Domain with
Arbitrary Boundary Loading
y
p(x)
Must use series representation for Airy stress
function to handle general boundary loading.

   cos n x[ Bn cosh n y  Cn n y sinh  n y ]
b
a
a
x
n 1

  cos m y[ Fm cosh  m x  Gm m x sinh m x ]  C0 x 2
b
m 1

 x    2n cos n x[ Bn cosh  n y  C n ( n y sinh  n y  2 cosh  n y )]
n 1

   2m cos m y[ Fm cosh  m x  Gm  m x sinh  m x]
m 1

 y    cos n x[ Bn cosh  n y  C n  n y sinh  n y ]  2C 0
n 1
2
n
p(x)
Boundary Conditions
 x ( a, y )  0 ,  xy ( a, y )  0
 xy ( x,b)  0 ,  y ( x,b)   p ( x)

   2m cos m y[ Fm cosh  m x  G m ( m x sinh  m x  2 cosh  m x)]
m 1

 xy    2n sin  n x[ Bn sinh  n y  C n ( n y cosh  n y  sinh  n y )]
n 1

   2m sin  m y[ Fm sinh  m x  Gm ( m x cosh  m x  sinh  m x)]
m 1
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
Using Fourier series theory to handle
general boundary conditions, generates a
doubly infinite set of equations to solve for
unknown constants in stress function form.
See text for details
Polar Coordinate Formulation
Airy Stress Function Approach  = (r,θ)
Airy Representation
1  1  2 
r 

r r r 2  2
 2
  2
r
  1  
 r   

r  r  
Biharmonic Governing Equation
 2 1 
1  2   2 1 
1 2 
    2 
 2 2  2 
 2 2   0
r r r   r
r r r  
 r
4
S
r
r
R
Traction Boundary Conditions
y
Tr  f r (r, ) , T  f  (r, )


r

x
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
Polar Coordinate Formulation
Plane Elasticity Problem
Strain-Displacement
ur
r
u 
1
e   u r   
r
 
1  1 ur u u 
er  

 
2  r 
r
r 
er 
Hooke’s Law
Plane Strain
 r  (er  e )  2er
   (er  e )  2e
 z  (er  e )  ( r    )
 r  2er , z   rz  0
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
Plane Stress
1
1
er  ( r    ) , e  (    r )
E
E


e z   ( r    )  
(er  e )
E
1 
1 
er 
 r , ez  erz  0
E
General Solutions in Polar Coordinates
Michell Solution
(r, )  f (r)e
b
 2 1 
1  2   2 1 
1 2 
    2 
 2 2  2 
 2 2   0

r
r

r
r



r
r

r
r  


4
2
1  2b 2
1  2b 2
b 2 (4  b 2 )
f   f  
f  
f 
f 0
r
r2
r3
r4
Choosing the case where b = in, n = integer gives the general Michell solution
  a0  a1 log r  a 2 r 2  a3 r 2 log r
 ( a 4  a5 log r  a6 r 2  a7 r 2 log r )
a13
 a14 r 3  a15 r  a16 r log r ) cos 
r
b
 (b11r  b12 r log r  13  b14 r 3  b15 r  b16 r log r ) sin 
r
 ( a11r  a12 r log r 

  ( a n1r n  a n 2 r 2n  a n 3 r n  a n 4 r 2n ) cos n
n 2

  (bn1r n  bn 2 r 2n  bn 3 r n  bn 4 r 2n ) sin n
n 2
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
Will use various terms
from this general
solution to solve
several plane problems
in polar coordinates
Axisymmetric Solutions
Stress Function Approach: =(r)
  a0  a1 log r  a2 r  a3r log r
2
2
a1
 a 3  2a 2
r2
a
   2a 3 log r  21  3a 3  2a 2
r
 r  0
 r  2a 3 log r 
Navier Equation Approach: u=ur(r)er
(Plane Stress or Plane Strain)
d 2 u r 1 dur 1

 ur  0
r dr r 2
dr 2
1
u r  C1r  C 2
r
Gives Stress Forms
A
A
 r  2  B ,     2  B ,  r  0
r
r
Displacements - Plane Stress Case
1  (1  )

ur  
a1  2(1  )a3 r log r  (1  )a3 r  2a2 (1  )r 
E
r

 A sin   B cos 
Underlined terms represent
4 r
rigid-body motion
u 
a3  A cos   B sin   Cr
E
•
•
Elasticity
a3 term leads to multivalued behavior, and is not found following the
displacement formulation approach
Could also have an axisymmetric elasticity problem using  = a4
which gives r =  = 0 and r = a4/r  0, see Exercise 8-15
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
Example 8.6 Thick-Walled Cylinder Under
Uniform Boundary Pressure
p2
General Axisymmetric
Stress Solution
r1
p1
A
r  2  B
r
A
   2  B
r
r2
r 
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
r (r1 )   p1 , r (r2 )   p2
r12 r22 ( p2  p1 )
A
r22  r12
r12 p1  r22 p2
B
r22  r12
r12 r22 ( p2  p1 ) 1 r12 p1  r22 p2

r22  r12
r2
r22  r12
  
Using Strain Displacement
Relations and Hooke’s Law
for plane strain gives the
radial displacement
Boundary Conditions
r12 r22 ( p2  p1 ) 1 r12 p1  r22 p2

r22  r12
r2
r22  r12
1 
A
r[(1  2) B  2 ]
E
r
r12 p1  r22 p2
1    r12 r22 ( p2  p1 ) 1


 (1  2)
E 
r22  r12
r
r22  r12
ur 

r

Example 8.6 Cylinder Problem Results
Internal Pressure Only
r1
p
r2
Dimensionless Stress
r1/r2 = 0.5
θ /p
r /p
r/r
2
Dimensionless Distance,
r/r2
( ) max  (r12  r22 ) /(r22  r12 ) p  (5 / 3) p
Thin-Walled Tube Case:
t  r2  r1  1 ro  (r1  r2 ) / 2
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
 
pro
t
Matches with Strength
of Materials Theory
Special Cases of Example 8-6
Pressurized Hole in an Infinite Medium
p2  0 and r2  
Stress Free Hole in an Infinite Medium
Under Equal Biaxial Loading at Infinity
p1  0 , p2  T , r2  
T
r1
p
r1
r12
r12
 r   p1 2 ,    p1 2 ,  z  0
r
r
1   p1r12
ur 
E
r
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
 r12 
 r12 
 r  T 1  2  ,    T 1  2 
 r 
 r 
 max  ( ) max   (r1 )  2T
T
Example 8.7 Infinite Medium with a Stress
Free Hole Under Uniform Far Field Loading
Boundary Conditions
 r (a, )   r (a, )  0
T
(1  cos 2)
2
T
  (, )  (1  cos 2)
2
T
 r (, )   sin 2
2
 r (, ) 
y
T
a
T
x
Try Stress Function
  a0  a1 log r  a 2 r 2  a3 r 2 log r
 (a 21 r 2  a22 r 4  a23 r 2  a24 ) cos 2
6a
a
4a
 r  a3 (1  2 log r )  2a 2  12  (2a 21  423  224 ) cos 2
r
r
r
6a
a
   a3 (3  2 log r )  2a 2  12  (2a 21  12a 22 r 4  423 ) cos 2
r
r
6a
2a
 r  (2a 21  6a 22 r 2  423  224 ) sin 2
r
r
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
T  a2
 r  1  2
2 r
 T  3a 4 4a 2 
  1  4  2  cos 2
r
r 
 2
T  a 2  T  3a 4 
   1  2   1  4  cos 2
2 r  2
r 
T  3a 4 2a 2 
 r   1  4  2  sin 2
2
r
r 
Example 8.7 Stress Results
T  a2
 r  1  2
2 r
y
a
T  a 2  T  3a 4 
   1  2   1  4  cos 2
2 r  2
r 
T
x
 r  
90

 cos 2

T  3a 4 2a 2
1  4  2
2 
r
r
 max    (a,  / 2)  3T

 sin 2

3
120
60
2
 (a, ) / T
150
30
  (a, ) / T
1
180
, /T
T
 T  3a 4 4a 2
  1  4  2
r
r
 2
0
r 
 ( , ) / T
a 2
210
330
240
300
270
  (a, )  T (1  2 cos 2)
  (a,0)  T ,   (a,30o )  0
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
r/a
Superposition of Example 8.7
Biaxial Loading Cases
T2
T1
=
+
T1
T2
Equal Biaxial Tension Case
T1 = T2 = T
 r12 
 r12 
 r  T 1  2  ,    T 1  2 
 r 
 r 
 max  (  ) max    (r1 )  2T
Tension/Compression Case
T1 = T , T2 = -T
 3a 4 4a 2 
 r  T 1  4  2  cos 2
r
r 

 3a 4 
   T 1  4  cos 2
r 

 3a 4 2a 2 
 r  T 1  4  2  sin 2
r
r 

  (a,0)    (a, )   4T ,   (a,  / 2)    (a,3 / 2)  4T
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
Review Stress Concentration Factors
Around Stress Free Holes
T
y
r1
T
a
T
T
x
K=2
K=3
T
T
T
T
T
45o
=
T
T
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
K=4
T
Stress Concentration Around
Stress Free Elliptical Hole – Chapter 10
Maximum Stress Field
b

 S 1  2 
a

25
x
Stress Concentration Factor
b
 max
x  S
y
a
 
20
15
()max/S
10
5
Circular Case
0
0
1
2
3
4
5
6
7
Eccentricity Parameter, b/a
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
8
9
10
Stress Concentration Around Stress Free
Hole in Orthotropic Material – Chapter 11
x(0,y)/S
y
S
S
Orthotropic Case Carbon/Epoxy
x
Isotropic Case
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
2-D Thermoelastic Stress Concentration
Problem Uniform Heat Flow Around
Stress Free Insulated Hole – Chapter 12
q
y
a
x
Stress Field
1 Eqa  a a 3 
  3  sin 
r  
2 k r r 
1 Eqa  a a 3 
   sin 
  
2 k  r r 3 
1 Eqa  a a 3 
   cos 
 r 
2 k  r r 3 
Eqa
 max    (a, )  
sin 
k
 max  (a,  / 2)   Eqa / k
Maximum compressive stress on hot side of hole    / 2
Maximum tensile stress on cold side    / 2
Steel Plate: E = 30Mpsi (200GPa) and = 6.5in/in/oF (11.7m/m/oC),
qa/k = 100oF (37.7oC), the maximum stress becomes 19.5ksi (88.2MPa)
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
Nonhomogeneous Stress Concentration Around Stress
Free Hole in a Plane Under Uniform Biaxial Loading
with Radial Gradation of Young’s Modulus – Chapter 14
3.5
n
b/a = 20
 = 0.25
n = 0 (homogeneous case)
n = 0.2
n = 0.4
n = 0.6
b/a = 20
 = 0.25
3
Stress Concentration Factor, K
n = -0.2
r
E (r )  Eo  
a
2.5
homogeneous case
2
1.5
1
-0.4
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
-0.3
-0.2
0.1
0
-0.1
Power Law Exponent, n
0.2
0.3
0.4
Three Dimensional Stress Concentration
Problem – Chapter 13
S
Normal Stress on the x,y-plane (z = 0)
z

4  5 a 3
9
a5 

 z (r ,0)  S 1 

3
5 
2
(
7

5

)
r
2
(
7

5

)
r


y
x
a
 z (a,0)  ( z ) max 
27  15
S
2(7  5)
  0.3 
( z ) max
 2.04
S
S
3
2.15
Stress Concentration Factor
2.2
Normalized Stress in Loading Direction
3.5
2.5
Two Dimensional Case: (r,/2)/S
2
1.5
1
0.5
2.1
2.05
2
1.95
Three Dimensional Case: z(r,0)/S ,  = 0.3
1.9
0
1
2
3
4
Dimensionless Distance, r/a
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
5
0
0.1
0.2
0.3
Poisson's Ratio
0.4
0.5
Wedge Domain Problems
y
Use general stress function solution to include
terms that are bounded at origin and give
uniform stresses on the boundaries
  r 2 (a2  a6   a21 cos2  b21 sin 2)
 r  2a2  2a6   2a21 cos 2  2b21 sin 2
r

   2a 2  2a6   2a 21 cos 2  2b21 sin 2


x
 r  a6  2b21 cos 2  2a 21 sin 2
Quarter Plane Example ( = 0 and  = /2)
y
  (r ,  / 2)  0
 r (r ,  / 2)  S
S
r

  (r ,0)   r (r ,0)  0
Elasticity
S 

(  2  cos 2  sin 2)
2 2
2
S 

   (  2  cos 2  sin 2)
2 2
2
S

 r  (1  cos 2  sin 2)
2
2
r 
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
x
Half-Space Examples
Uniform Normal Stress Over x  0
Boundary Conditions
  (r ,0)   r (r ,0)  0
T
x
 r (r , )  0 ,   (r , )  T
Try Airy Stress Function

r
  a6 r 2   b21r 2 sin 2
   2a6   2b21 sin 2
 r  a6  2b21 cos 2
y
Use BC’s To Determine Stress Solution
T
(sin 2  2)
2
T
 
(sin 2  2)
2
T
 r 
(1  cos 2)
2
r  
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
Half-Space Under Concentrated Surface
Force System (Flamant Problem)
Y
Boundary Conditions
X
x
  (r ,0)   r (r ,0)  0

 r (r , )  0 ,   (r , )  0
r
C
 Forces   Xe
1
 Ye 2 
C
Try Airy Stress Function
  (a12 r log r  a15 r) cos 
 (b12 r log r  b15 r) sin 
y
Use BC’s To Determine Stress Solution
2
[ X cos   Y sin ]
r
    r  0
r  
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
Flamant Solution Stress Results
Normal Force Case
2Yx 2 y
 x   r cos   
( x 2  y 2 ) 2
2
2Y
sin 
r
    r  0
r  
or in Cartesian
components
2Yy 3
 y   r sin   
( x 2  y 2 ) 2
2
 xy
Y
2Yxy 2
  r sin  cos   
( x 2  y 2 ) 2
x
y=a
y
xy/(Y/a)
Dimensionless Stress
r = constant
y/(Y/a)
 y  2Y / a
Dimensionless Distance, x/a
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
Flamant Solution Displacement Results
Normal Force Case
u r 1
2Y
 (  r   )  
sin 
r
E
Er
u
1 u 1
2 Y
  r 
 (    r ) 
sin 
r r  E
Er
1 u r u u 1
 r 


  r  0
r 
r
r 
r 
Y

[(1  )(  ) cos   2 log r sin ]
E
2
Y

u 
[ (1  )(  ) sin   2 log r cos   (1  ) cos ]
E
2
ur 
Note unpleasant feature of 2-D model that
displacements become unbounded as r  
0.1
Y
0
On Free Surface y = 0
Y
u r ( r,0)  u r ( r, )  
(1  )
2E
Y
u ( r,0)  u ( r, )  
[(1  )  2 log r ]
E
-0.1
-0.2
-0.3
-0.4
-0.5
-0.6
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
-0.5
0
0.5
Comparison of Flamant Results with
3-D Theory - Boussinesq’s Problem
Cartesian Solution
P
u
x
Px  z 1  2 
Py  z 1  2 
P 
z2 



,
v


,
w

2
(
1


)





4R  R 2 R  z 
4R  R 2 R  z 
4R 
R 2 
 3x 2 z
z
R
x 2 (2 R  z ) 


 3  (1  2) 
2 
 R R  z R( R  z ) 
 R
z
P 3y 2 z
R
y 2 (2 R  z ) 


y  
 (1  2) 


2 
R
R

z
2R 2  R 3
R
(
R

z
)


P
x  
2R 2
y
z
z  
Free Surface Displacements
u z ( R ,0 ) 
P(1  )
2R
Corresponding 2-D Results
u (r ,0)  
P
[(1  )  2 log r ]
E
3-D Solution eliminates the
unbounded far-field behavior
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
3Pz 3
P  3 xyz (1  2)(2 R  z ) xy 
,  xy  



5
2R
2R 2  R 3
R( R  z ) 2

3Pyz 2
3Pxz 2
 yz  
,



xz
2R 5
2R 5
Cylindrical Solution
2
P  rz (1  2 )r   r  P  3r z  (1  2) R 
ur 

Rz 
2R 2  R 3
4R  R 2
R  z 
(1  2) P  z
R 
P 
z 2   

2

uz 
R R  z 
2R
2(1  )  2 

4R 
R 
3P rz 2
3Pz 3
z  
,  rz  
u  0
2R 5
2R 5
Half-Space Under Uniform Normal
Loading Over –a  x  a
p
x
2
a 1
a
2Y
sin  cos2 
r
2Y
 y   r sin 2   
sin 3 
r
2Y
 xy   r sin  cos   
sin 2  cos 
r
 x   r cos2   

2p
cos2  d

2p 2
d y  
sin  d

2p
d xy  
sin  cos d

d x  
dx
y
r
d
dY = pdx = prd /sin

2 p 2
p
2
cos

d



[2( 2  1 )  (sin 22  sin 21 )]
 1
2
2 p 2 2
p
y  
sin  d   [2(2  1 )  (sin 22  sin 21 )]

 1
2

2p 2
p
 xy  
sin  cos  d 
[cos22  cos 21 ]

 1
2
x  
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
Half-Space Under Uniform Normal
Loading - Results
0.5
0.45
Dimensionless Maximum Shear Stress
Dimensionless Stress
xy /p
y/p
0.4
Concentrated Loading
max/(Y/a)
0.35
0.3
0.25
Distributed Loading
max/p
0.2
0.15
0.1
0.05
Dimensionless Distance, x/a
max - Contours
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
0
0
1
2
3
4
5
6
7
Dimensionless Distance, y / a
8
9
10
Generalized Superposition Method
Half-Space Loading Problems
p(s)
x
a
a
t(s)
y
2y
x  


2 y3
y  

 xy  
Elasticity
p ( s )( x  s ) 2
2
ds 
2
2
2
 a [( x  s )  y ]

2y2

a
t ( s )( x  s ) 3
ds
 a [( x  s ) 2  y 2 ]2

a
p( s)
2y2
ds 
 a [( x  s ) 2  y 2 ]2

a

p ( s )( x  s )
2y
s

 a [( x  s ) 2  y 2 ]2


a
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
t ( s )( x  s )
ds
 a [( x  s ) 2  y 2 ]2

a
t ( s )( x  s ) 2
ds
 a [( x  s ) 2  y 2 ]2

a
Photoelastic Contact Stress Fields
Elasticity
(Point Loading)
(Uniform Loading)
(Flat Punch Loading)
(Cylinder Contact Loading)
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
Notch/Crack Problem
y
r



x
Stress Free Faces
 = 2 - 

Try Stress Function:   r [ A sin   B cos   C sin(  2)  D cos(  2)]
   (  1)r  2 [ A sin   B cos  C sin(  2)  D cos(  2)]
 r  (  1)r  2 [ A cos  B sin   C (  2) cos(  2)  D(  2) sin(  2)]
Boundary Conditions:  (r,0)   r (r,0)   (r,2)   r (r,2)  0 
n
sin 2(  1)  0     1 , n  0,1, 2, 
2
At Crack Tip r  0: Stress  O(r   2 ) , Displacement  O(r  1 )
Finite Displacements and Singular Stresses at Crack Tip  1<  <2   = 3/2
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
Notch/Crack Problem Results
y
3 1 
3

3
5
 
A
(sin


5
sin
)

B
(cos


cos
)
4 r 
2
2
2
3
2 
3 1 
3

3
 
 
A
(sin


3
sin
)

B
(cos


cos
)
4 r 
2
2
2
2 
r  
r



x
Stress Free Faces
 = 2 - 
 r  
3
2
3
  
2
3
 r  
2
r  
Transform to  Variable
•
•
•
•
Elasticity
3 1 
3

3
1
 
A
(cos


cos
)

B
(sin


sin
)
4 r 
2
2
2
3
2 
A

B

cos (3  cos ) 
sin (1  3 cos )
2
2
r
2 r
A

3B

cos (1  cos ) 
sin (1  cos )
2
2
r
2 r
A

B

sin (1  cos ) 
cos (1  3 cos )
2
2
r
2 r
Note special singular behavior of stress field O(1/r)
A and B coefficients are related to stress intensity factors and are useful in fracture
mechanics theory
A terms give symmetric stress fields – Opening or Mode I behavior
B terms give antisymmetric stress fields – Shearing or Mode II behavior
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
Crack Problem Results
Contours of Maximum Shear Stress
Mode I (Maximum shear stress contours)
Mode II (Maximum shear stress contours)
Experimental Photoelastic Isochromatics
Courtesy of URI Dynamic Photomechanics Laboratory
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
Mode III Crack Problem – Exercise 8-41

y
z
r
Contours for Mode III Crack Problem
●

x
Anti-Plane Strain Case
u  v  0 , w  w( x, y)
 2 w 1 w 1  2 w
 w 2 
 2
0
2
r r r 
r
2

A

A

w  A r sin ,  z 
cos ,  zr 
sin
2
2
2
2 r
2 r
Stresses Again Or 1 / 2 
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
z - Stress Contours
Curved Beam Under End Moments
r
M
2
a
M
 r (a)   r (b)  0
 r (a)   r (b)  0
b
  dr  0
a

b
  rdr  M
a
Dimensionless Stress, a /M
b

  a0  a1 log r  a2 r 2  a3r 2 log r
4M a 2b2
b
r
a
r  
[ 2 log( )  b 2 log( )  a 2 log( )]
N
r
a
b
r
2 2
4M a b
b
r
a
  
[ 2 log( )  b 2 log( )  a 2 log( )  b 2  a 2 ]
N
r
a
b
r
 r  0
Elasticity
b/a = 4
Theory of Elasticity
Strength of Materials
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
Dimensionless Distance, r/a
Curved Cantilever Beam
P
r
a
b
 r ( a, )   r (b, )  0
 r ( a, )   r (b, )  0

b

b

b

b

b
a
a
a
a
a
 = /2 b/a = 4
Theory of Elasticity
Strength of Materials
 r ( r,0)dr  P
b
  ( r,0)dr     ( r,0) rdr  0
Dimensionless Distance, r/a
a
  ( r,  / 2)dr   P
  ( r,  / 2)rdr  P( a  b) / 2
 r ( r,  / 2)dr  0
  ( Ar 3 
Elasticity
Dimensionless Stress, a/P

B
 Cr  Dr log r ) sin 
r
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
P
a 2b2 a 2  b2
 r  (r  3 
) sin 
N
r
r
P
a 2b2 a 2  b2
   (3r  3 
) sin 
N
r
r
P
a 2b2 a 2  b2
 r   ( r  3 
) cos 
N
r
r
Disk Under Diametrical Compression
P
D
=
P
Flamant Solution (1)
+
+
Flamant Solution (2)
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
Radial Tension Solution (3)
Disk Problem – Superposition of Stresses
y
P
 (x1)  
2P
cos 1 sin 2 1
r1
 (y1)  
2P
cos 3 1
r1
(xy1)  
2P
cos 2 1 sin 1
r1
1
x
2
r2
 (x2 )  
2P
cos  2 sin 2  2
r2
 (y2 )  
2P
cos 3  2
r2
2P  ( R  y) x 2 ( R  y) x 2 1 
x  

 
 
r14
r24
D
(xy2 )  
2P
cos 2  2 sin  2
r2
2P  ( R  y)3 ( R  y)3 1 
y  

 
  r14
r24
D
 (x3)   (y3) 
2P
,  (xy3)  0
D
P
 xy
2P  ( R  y)2 x ( R  y)2 x 



 
r14
r24

r1, 2 
Elasticity
r1
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
x 2  (R  y)2
Disk Problem – Results
y
P
1
x-axis (y = 0)
r1
x
2
r2
2P  D 2  4 x 2 
 x ( x,0) 
D  D 2  4 x 2 
y-axis (x = 0)
2
2P
 Constant
D
2P  2
2
1
 y (0, y )  

 

 D  2y D  2y D
 x (0, y ) 

2P 
4D 4
 y ( x,0)  
 1

2
2 2
D  ( D  4 x )

 xy ( x,0)  0
 xy (0, y )  0
P
(Theoretical max Contours)
(Photoelastic Contours)
(Courtesy of URI Dynamic Photomechanics Lab)
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
Applications to Granular Media Modeling
Contact Load Transfer Between Idealized Grains
P
P
P
P
Four-Contact Grain
(Courtesy of URI Dynamic Photomechanics Lab)
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
Contact Between Two Elastic Solids
Generates:
- Contact Area (w)
- Interface Tractions (pc)
- Local Stresses in Each Body
pc
w
Creates Complicated Nonlinear Boundary Condition:
Boundary Condition Changing With Deformation; i.e. w
and pc Depend on Deformation, Load, Elastic Moduli,
Interfacial Friction Characteristics
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
2-D Elastic Half-Space Subjected
to a Rigid Indenter
Rigid Indenter
a
a
Local stresses and deformation
determined from Flamant solution
See Section 8.4.9 and Exercise 8.38
x
uy
y
a
1   x
  2 a t ( s ) log x  s ds  a
p
(
s
)
ds

p
(
s
)
ds
1
 E -a
x
2 E  -a
a
2 a
1   x
a
uy  
p ( s ) log x  s ds 
t
(
s
)
ds

t
(
s
)
ds
 2
x
E -a
2 E  -a
a1 and a2 are rigid body motion constants
Frictionless Case (t = 0)
Elasticity
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island

ux  




du x
1 

p( x)
dx
E
du y
2 a p( s)

ds
dx
E -a x  s

2-D Elastic Half-Space Subjected
Frictionless Flat Rigid Indenter
Rigid Indenter
a
a
 a2  x2
x
2 Py
x   2

Elasticity
2 Py 2
 xy   2

y
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
 a2  x2
a
( x  s) 2
a
a  s [( x  s )  y ]

2 Py 3
y   2

Max Shear
Stress Contours
P
1 
P sin 1 ( x / a) , x  a
E
1/ 2
 x  x2
 
2
uy  
log   2  1   u yo , x  a
E
 a  a
 
P
Frictionless Rigid Punch Loading on a Half-Space
p ( x) 
ux  
y
Unbounded Stresses
at Edges of Indenter
p( s)
ds  0
xs
a
-a
Solution
x
uy
p ( x) 

u y  u yo  constant
P
2
2
2
2 2
ds

a
1
a
a  s [( x  s )  y ]

a
( x  s)
a
a  s [( x  s )  y ]
2
2
2
2
2
2
2 2
2 2
ds
ds
2-D Elastic Half-Space Subjected
Frictionless Cylindrical Rigid Indenter
Rigid Indenter
P
u y proportional to  x 2 / 2 R
R
a
x
a
Solution
uy
p( x) 
y
p( x) 
2P
a 2
a2  x2
2P
a 2
a2  x2
4 Py
x   2 2
 a
a
-a
Elliptical Distributed Normal Loading on a Half-Space
x
4 Py 3
y   2 2
 a
4 Py 2
 xy   2 2
 a
Max Shear
Stress Contours
y
Elasticity
p( s)
E
ds

x
-a x  s
2R
a
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
a

a
a2 
4 PR
E
a 2  s 2 ( x  s) 2
ds
[( x  s ) 2  y 2 ]2
a2  s2
ds
 a [( x  s ) 2  y 2 ]2
a

a 2  s 2 ( x  s)
ds
 a [( x  s ) 2  y 2 ]2

a