Using the Airy Stress Function approach, it was shown that the plane
elasticity formulation with zero body forces reduces to a single governing
biharmonic equation. In Cartesian coordinates it is given by
 
4
x
4
 
4
2
x y
2
 
4
2

y
4
   0
4
and the stresses are related to the stress function by
 
2
x 
y
2
 
2
, y 
x
2
 
2
,  xy  
xy
We now explore solutions to several specific problems in both
Cartesian and Polar coordinate systems
Cartesian Coordinate Solutions
Using Polynomials
In Cartesian coordinates we choose Airy stress function solution of polynomial form

( x , y ) 

 A
m
mn
x y
n
m 0 n0
where Amn are constant coefficients to be determined. This method produces
polynomial stress distributions, and thus would not satisfy general boundary
conditions. However, we can modify such boundary conditions using Saint-Venant’s
principle and replace a non-polynomial condition with a statically equivalent loading.
This formulation is most useful for problems with rectangular domains, and is
commonly based on the inverse solution concept where we assume a polynomial
solution form and then try to find what problem it will solve.
Noted that the three lowest order terms with m + n  1 do not contribute to the
stresses and will therefore be dropped. It should be noted that 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 the biharmonic equation for any
choice of constants Amn. However, for higher order terms, constants Amn will have to
be related in order to have the polynomial satisfy the biharmonic equation.
Example 8.1 Uniaxial Tension of a Beam
y
T
T
2c
x
2l
Stress Field
Boundary Conditions:
Displacement Field (Plane Stress)
u
 x (  l , y )  T ,  y ( x , c )  0
x
 xy (  l , y )   xy ( x ,  c )  0
v
Since the boundary conditions specify constant
stresses on all boundaries, try a second-order
stress function of the form
  A 02 y
2
 x  2 A 02 ,  y   xy  0
The first boundary condition implies that A02 = T/2,
and all other boundary conditions are identically
satisfied. Therefore the stress field solution is
given by
 x  T ,  y   xy  0
y
 ex 
 ey 
u 
T
1
E
1
E
(  x   y ) 
y

v
x  f ( y) , v  
x
 2 e xy 
E
(  y   x )   
E
u
T
T
T
E
y  g ( x)
E
 xy

f ( y )  o y  uo
g ( x)  o x  vo
 0  f ( y )  g ( x )  0
. . . Rigid-Body Motion
“Fixity conditions” needed to determine RBM terms
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
Displacement Field (Plane Stress)
Boundary Conditions:
u
 y ( x ,  c )  0 ,  xy ( x ,  c )   xy (  l , y )  0

c
c
 x (  l , y ) dy  0 ,

c
c
 x  6 A 03 y ,  y   xy  0
3
Moment boundary condition implies that
A03 = -M/4c3, and all other boundary conditions
are identically satisfied. Thus the stress field is
x  
3M
2c
3
v
 x (  l , y ) ydy   M
Expecting a linear bending stress distribution,
try second-order stress function of the form
  A 03 y
x
y ,  y   xy  0
y
u
y

 
3M
2 Ec
 
3M
2 Ec
v
x
3
3
y  u  
3M
2 Ec
y  v 
0 
3M 
4 Ec
3M
2 Ec
3
3
xy  f ( y )
y  g(x)
2
3
x  f ( y )  g ( x )  0
f ( y )  o y  uo
g(x) 
3M
4 Ec
x  o x  vo
2
3
“Fixity conditions” to determine RBM terms:
v (  l , 0 )  0 and u (  l , 0 )  0
u o   o  0 , v o   3 Ml
2
/ 16 Ec
3
Example 8.2 Pure Bending of a Beam
Solution Comparison of Elasticity
with Elementary Mechanics of Materials
y
M
2c
M
x
I  2c / 3
3
2l
Elasticity Solution
x  
u  
Mxy
EI
M
I
, v 
Mechanics of Materials Solution
y ,  y   xy  0
M
8 EI
[4 y  4 x  l ]
2
2
2
Uses Euler-Bernoulli beam theory to
find bending stress and deflection of
beam centerline
x  
M
I
y ,  y   xy  0
v  v ( x ,0 ) 
M
[4 x  l ]
2
2
8 EI
Two solutions are identical, with the exception of the x-displacements
Example 8.3 Bending of a Beam by
Uniform Transverse Loading
w
wl
wl
2c
x
y
2l
Stress Field
Boundary Conditions:
  A 20 x  A 21 x y  A03 y  A 23 x y 
2
 xy ( x ,  c )  0


c
c
c
c
c
2
 xy   2 A 21 x  6 A 23 xy
x 
 x (  l , y ) ydy  0
BC’s
3
A 23
y
5
3
y )
3
 y  2 A 20  2 A 21 y  2 A 23 y
 x (  l , y ) dy  0
 xy (  l , y ) dy   wl
2
 x  6 A03 y  6 A 23 ( x y 
 y ( x , c )   w

3
2
 y ( x, c)  0
c
2
3
2
2
3w  l
2
3w
2 3
2
 2  y 
(
x
y

y )
3
4 c  c
5 
4c
3
y  
w
 xy  
3w

2
4c
3w
y
4c
x
3w
4c
3
w
4c
3
xy
2
y
3
5
Example 8.3 Beam Problem
Stress Solution Comparison of Elasticity
with Elementary Mechanics of Materials
w
wl
wl
2c
x
y
2l
Elasticity Solution
x 
w
(l  x ) y 
2
2
2I
w
I
(
y
3
2

c y
3
w  y
2 3
2

 c y  c 
2I  3
3

3
y  
Mechanics of Materials Solution
 xy  
w
x(c  y )
2
2
5
)
x 
My
I

w
(l  x ) y
2
2
2I
y  0
 xy 
VQ
It
 
w
x(c  y )
2
2
2I
2I
Shear stresses are identical, while normal stresses are not
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 the two theories exist
at top and bottom of beam, and actual difference in
stress values is w/5. For most beam problems where
l >> c, the bending stresses will be much greater than
w, and thus the differences between elasticity and
strength of materials will be relatively small.
y/w - Elasticity
y/w - Strength of Materials
Maximum difference between the two theories is w
and this occurs at the top of the 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
3
2
3
w y
c y  3w  1 y
1 y




 x (l , y ) 


3



I  3
5 
2 3 c
5 c 
w
wl
wl
2c
x
y
2l
End stress distribution does not
vanish and is nonlinear but gives
zero resultant force.
 x ( l , y ) / w
Example 8.3 Beam Problem
w
wl
wl
2c
x
y
2l
Displacement Field (Plane Stress)
w
u 
[( l x 
2
2 EI
v  
x
3
) y  x(
3
w
y
[(
2 EI
4
3
2

2c y
3
2

2y
c y
12
2
)  x(
5
2
2c y
)  (l  x )
2
2
3
f ( y)  o y  uo , g ( x) 
y
u 
[( l x 
2
2 EI
x
3
) y  x(
2y
3
3
x 
4
24 EI

3
5
)  x(
y
w
[l  (
2
4 EI
3
3
c y
2
l
2
4

2
2
3
)]  f ( y )
y
4
2

c y
6
8
2
)]  g ( x )
5
  )c ]x   o x  vo
2
2
12 4  c
u o   o  0 , vo 
[1 
(  ) 2]
24 EI
5 5 2 l
5 wl
2c
4
2
3
)]
3
2
 5 wl 4
12 4  c

 [  (  ) c ] x  
[1 
(  ) 2 ]
12
2
5 2
24
EI
5
5 2 l

4
 (
5
4
2
2
3
2
4
2
2
w  y
c y
2c y
y
y
c y
2
2

v  


  [( l  x )


]
2 EI  12
2
3
2
6
5
x
2c
3
2
w
2
2c y
c y
2
2
Choosing Fixity Conditions u ( 0 , y )  v (  l , y )  0
w
3
3
3

y
v ( 0 , 0 )  v max
12 4  c

[1 
(  ) 2]
24 EI
5 5 2 l
5 wl
4
Strength of Materials: v max
2

5 wl
4
24 EI
Good match for beams where l >> c
Cartesian Coordinate Solutions
Using Fourier Methods
A more general solution scheme for the biharmonic equation may be
found using Fourier methods. Such techniques generally use separation of
variables along with Fourier series or Fourier integrals.
 
4
 ( x , y )  X ( x )Y ( y )
x
4
Choosing X  e  x , Y  e  y
 
4
2
x y
2
 
4

2
y
4
0
   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  C 0  C 1 x  C 2 x  C 3 x
2
3
   0  C 4 y  C 5 y  C 6 y  C 7 xy  C 8 x y  C 9 xy
2
3
2
2
Example 8.4 Beam with Sinusoidal Loading
qosinπx/l
y
qol/
qol/
2c
x
l
Stress Field
A   D (  c tanh  c  1)
Boundary Conditions:
 x (0, y )   x (l , y )  0
 xy ( x ,  c )  0
 y ( x , c )  0
 y ( x , c )   q o sin(  x / l )


c
c
c
c
 xy ( 0 , y ) dy   q o l / 
 xy ( l , y ) dy  q o l / 
B   C (  c coth  c  1)
  sin  x [( A  C  y ) sinh  y  ( B  D  y ) cosh  y ]
 x   sin  x [( A sinh  y  C (  y sinh  y  2 cosh  y )
2
C 
 B cosh  y  D (  y cosh  y  2 sinh  y )]
 y    sin  x [( A  C  y ) sinh  y  ( B  D  y ) cosh  y ]
2
 xy    cos  x [( A cosh  y  C (  y cosh  y  2 sinh  y )
2
 B sinh  y  D (  y sinh  y  2 cosh  y )]
l
2
  c
c
c 
2 2
 sinh
cosh
l  l
l
l 
 q o sinh
D 
  c
 sinh
2
l  l
2
2
c
 q o sinh
 
c
l
c
l

l
cosh
c 
l 
Example 8.4 Beam Problem
qosinπx/l
y
qol/
Bending Stress
qol/
2c
x
 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
 q o sinh
C 
  c
 sinh
2
l  l
2
2
l
c
cosh
l
c 
l 
q o cosh
, D 
  c
 sinh
2
l  l
2
2
A   D (  c tanh  c  1) , B   C (  c coth  c  1) ,  
l
c
l
c
cosh
l
c 
l 

l

y
y 
c
y

 y cosh
 2 l 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

x  l/2
c
y 


   c coth
 l  cosh

l
l
l
l




c
c

 c  l sinh
cosh

l
l

For the case l  c :
 y sinh
D 
3qol
 2 l cosh
y
5
4c 
3
y
5
, C  0 , A  D , B  0
3qol  y
3qol
y
y 
x
x
cosh
 sinh

y sin

 sin
3 3
3 2
4c   l
l
l 
l
2c 
l
3
x  
2
qol
Strength
of Materials
Theory :  x  
My
I
 
2
2
sin
x
l
3
2c / 3
y

3qol
2
2c 
2
3
y sin
x
l
Example 8.4 Beam Problem
qosinπx/l
y
qol/
qol/
2c
x
l
Displacement Field (Plane Stress)
u 

v
cos  x{ A (1   ) sinh  y  B (1   ) cosh  y

sin  x{ A (1   ) cosh  y  B (1   ) sinh  y
E
E
 C [(1   )  y sinh  y  2 cosh  y ]
 C [(1   )  y cosh  y  (1   ) sinh  y ]
 D [(1   )  y cosh  y  2 sinh  y ]}   o y  u o
 D [(1   )  y sinh  y  (1   ) cosh  y ]}   o y  v o
v ( x ,0 ) 
D

 o  vo  0 , u o 
u ( 0 ,0 )  v ( 0 ,0 )  v ( l ,0 )  0
[ B (1   )  2 C ]
E
sin  x [ 2  (1   ) c tanh  c ]
E
For the case l >> c
D  
3q o l
5
4c 
3
5
v ( x ,0 )  
Strength of Materials v ( x ,0 )  
3q o l
3q o l
2c  E
3
4
2c  E
3
4
4
sin
4
x
l
sin
x
l
[1 
1   c
2
l
tanh
c
l
]
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.
b

 cos 

n
a
a
x [ B n cosh  n y  C n  n y sinh  n y ]
x
n 1

 cos 

m
y [ F m cosh  m x  G m  m x sinh  m x ]  C 0 x
2
b
m 1
p(x)

x 

 n cos  n x [ B n cosh  n y  C n (  n y sinh  n y  2 cosh  n y )]
2
n 1



2
m
cos  m y [ F m cosh  m x  G m  m x sinh  m x ]
m 1

 y     cos  n x [ B n cosh  n y  C n  n y sinh  n y ]  2 C 0
2
n
n 1



2
m
cos  m y [ F m cosh  m x  G m (  m x sinh  m x  2 cosh  m x )]
m 1

 xy 

2
n
sin  n x [ B n sinh  n y  C n (  n y cosh  n y  sinh  n y )]
2
m
sin  m y [ F m sinh  m x  G m (  m x cosh  m x  sinh  m x )]
n 1



m 1
Boundary Conditions
 x (a, y )  0
 xy (  a , y )  0
 xy ( x ,  b )  0
 y ( x , b )   p ( x )
Use Fourier series theory to handle general
boundary conditions, and this 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
r 
1 
r r
1  
2

r
2

 
2
 
r
 r  
Biharmonic Governing Equation
2
2
 2
1 
1 
    2 
 2
2

r
r

r
r 

4
2
  1  


r  r  

   0

S
r
r
2
  2
1 
1 
  2 
 2
2

r
r

r
r 

R
y
Traction Boundary Conditions
Tr  f r ( r , ) , T  f  ( r , )


r

x
Polar Coordinate Formulation
Plane Elasticity Problem
Strain-Displacement
er 
e 
e r 
u r
r
u  
1
 ur 

r
 
u 
u 
1  1 u r

 

2  r 
r
r 
Hooke’s Law
Plane
Strain
 r   (er  e )  2 er
    (er  e )  2 e
 z   (er  e )   ( r    )
 r   2  e r  ,   z   rz  0
Plane
er 
1
E
ez  
e r 
Stress
(  r    ) , e  

E
( r    )  
1 
E
1
E

1 
 r  , e  z  e rz  0
(     r )
(er  e )
General Solutions in Polar Coordinates
Michell Solution
( r ,  )  f ( r ) e
2
 2
1 
1 
    2 
 2
2

r
r

r
r 

b
4
2
f  
f  
1  2b
r
r
2
2
f  
1  2b
r
2
3
2
  2
1 
1 
  2 
 2
2

r
r

r
r 

b (4  b )
2
f 

   0

2
r
4
f 0
Choosing the case where b = in, n = integer gives the general Michell solution
  a 0  a 1 log r  a 2 r  a 3 r log r
2
2
 ( a 4  a 5 log r  a 6 r  a 7 r log r ) 
2
2
 ( a 11 r  a 12 r log r 
a 13
 ( b11 r  b12 r log r 
b13


 (a
r  an2r
n
n1
2n
r
r
 a 14 r  a 15 r   a 16 r  log r ) cos 
3
 b14 r  b15 r   b16 r  log r ) sin 
3
 a n3r
n
 an4r
2n
) cos n 
n2


 (b
n2
r  bn 2 r
n
n1
2n
 bn 3 r
n
 bn 4 r
2n
) sin n 
We will use various
terms from this general
solution to solve
several plane problems
in polar coordinates
Axisymmetric Solutions
Stress Function Approach: =(r)
Navier Equation Approach: u=ur(r)er
(Plane Stress or Plane Strain)
  a 0  a 1 log r  a 2 r  a 3 r log r
2
 r  2 a 3 log r 
   2 a 3 log r 
a1
r
2
a1
r
2
2
 a3  2a2
Displacements - Plane Stress Case
2

1 du r
1

r dr
r
2
ur  0
u r  C1r  C 2
1
r
Gives Stress Forms
r 
A
r
2
 B ,   
A
r
2
 B ,  r  0
1  (1   )


a

2
(
1


)
a
r
log
r

(
1


)
a
r

2
a
(
1


)
r
1
3
3
2

E 
r

 A sin   B cos 
u 
dr
 3a 3  2 a 2
 r  0
ur 
2
d ur
4 r
E
a 3  A cos   B sin   Cr
•
•
Underlined terms represent
rigid-body motion
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 8-14
Example 8.6 Thick-Walled Cylinder
Under Uniform Boundary Pressure
p2
r1
p1
General Axisymmetric
Stress Solution
r 
A
r
  
Boundary Conditions
 r ( r1 )   p 1 ,  r ( r2 )   p 2
 B
2
A
r
2
2
A
r1 r2 ( p 2  p 1 ) 1
2
  
ur 
r
2

2
2
2
r
r [( 1  2  ) B 
2
2
2
2
2
r1 p 1  r2 p 2
2
r2  r1
r2  r1
r2  r1
r1 r2 ( p 2  p 1 ) 1
1 
E
2
2
r1 p 1  r2 p 2
2
r2  r1
2
r1 p 1  r2 p 2
2
2
Using Strain Displacement
Relations and Hooke’s Law
for plane strain gives the
radial displacement
2
B 
2
2
r2  r1
 B
r2
r 
r1 r2 ( p 2  p 1 )
2

2
A
r
2
2
r2  r1
2
2
]
2 2
2
2
r1 p 1  r2 p 2 
1    r1 r2 ( p 2  p 1 ) 1

 (1  2  )
r

2
2
2
2
E 
r2  r1
r
r2  r1

Example 8.6 Cylinder Problem Results
Internal Pressure Only
Dimensionless Stress
r1/r2 = 0.5
r1
p
r2
θ /p
r /p
r/r
2
Dimensionless Distance,
r/r2
(   ) max  ( r1  r2 ) /( r2  r1 ) p  ( 5 / 3 ) p
2
Thin-Walled Tube Case:
t  r2  r1  1
ro  ( r1  r2 ) / 2
 
2
pr o
t
2
2
Matches with Strength
of Materials Theory
Special Cases of Example 8-6
Pressurized Hole in an Infinite Medium
p 2  0 and r2  
Stress Free Hole in an Infinite Medium
Under Equal Biaxial Loading at Infinity
p 1  0 , p 2   T , r2  
T
r1
p
r1
2
 r   p1
r1
r
2
2
r1
,    p1
r
1   p 1 r1
2
ur 
E
r
2
, z  0
2
2


r1 
r1 
 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
y
a
T
 r ( , ) 
T
  ( , ) 
T
T
x
(1  cos 2  )
2
(1  cos 2  )
2
 r (  ,  )  
T
sin 2 
2
Try Stress Function
  a 0  a 1 log r  a 2 r  a 3 r log r
2
 ( a 21 r  a 22 r  a 23 r
2
 r  a 3 (1  2 log r )  2 a 2 
   a 3 ( 3  2 log r )  2 a 2 
 r   ( 2 a 21  6 a 22 r 
2
6 a 23
r
4
a1
r
 ( 2 a 21 
2
a1
r

2
6 a 23
r
4

4 a 24
r
 ( 2 a 21  12 a 22 r 
4
2 a 24
r
2
) sin 2 
2
) cos 2 
6 a 23
r
4
) cos 2 
4
2
2
 a 24 ) cos 2 
r 
2
4
2
T 
a  T 
3a
4a
1  2   1  4  2
2 
r  2 
r
r
 
2
4
T 
a  T 
3a
1  2   1  4
2 
r  2 
r
 r
4
2
T 
3a
2a
   1  4  2
2
r
r

 cos 2 



 cos 2 



 sin 2 


Example 8.7 Stress Results
2
4
2
T 
a  T 
3a
4a



 r  1  2   1  4  2
2
r  2
r
r
y
a
T
 
x
2
4
T 
a  T 
3a
1  2   1  4
2 
r  2 
r
 r  
90
4
2
T 
3a
2a
1  4  2
2 
r
r

 cos 2 


 max    ( a ,   / 2 )  3T

 sin 2 


3
120
60
2
  ( a , ) / T
150
30
, /T
T

 cos 2 


   ( a , ) / T
1
180
0
(
210
r 
, )/T
a 2
330
240
300
270
  ( a ,  )  T (1  2 cos 2  )
  ( a , 0 )   T ,   ( a , 30 )  0
o
r/a
Superposition of Example 8.7
Biaxial Loading Cases
T2
T1
=
+
T1
T2
Equal Biaxial Tension Case
T1 = T2 = T
2
2


r1 
r1 
 r  T  1  2  ,    T  1  2 
r 
r 


 max  (   ) max    ( r1 )  2 T
Tension/Compression Case
T1 = T , T2 = -T
4
2

3a
4a
 r  T  1  4  2
r
r

4

3a
    T  1  4
r

 r

 cos 2 


 cos 2 

4
2

3a
2a
  T  1  4  2
r
r


 sin 2 

  ( a , 0 )    ( a ,  )   4 T ,   ( a ,  / 2 )    ( a ,3  / 2 )  4 T
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
K=4
T
Stress Concentration Around
Stress Free Elliptical Hole – Chapter 10
Maximum Stress Field
 
 max

x  S
y
b
25
x
S t re s s C onc e nt ra tion Fa c tor
a
b 

 S 1  2 
a 

20
()max/S
15
10
5
Circular Case
0
0
1
2
3
4
5
6
7
E ccentr icity P ar ameter , b/a
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
2-D Thermoelastic Stress Concentration
Problem Uniform Heat Flow Around
Stress Free Insulation Hole – Chapter 12
q
y
a
x
Stress Field
3
1 E  qa  a a 


r  
 r  r 3  sin 
2
k


3
1 E  qa  a a 
  3  sin 
  
r
2
k
r 

3
1 E  qa  a a 
  3  cos 
 r 
r
2
k
r 

 max    ( a ,  )  
E  qa
sin 
k
 max   ( a ,   / 2 )   E  qa / k
Maximum compressive stress on hot side of hole 
Maximum tensile stress on cold side     / 2
 /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)
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 = 0 (homogeneous case)
n = 0.2
n = 0.4
n = 0.6
n
b/a = 20
 = 0.25
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
-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
3
5

4  5 a
9
a 

 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 
2 (7  5 )
  0 .3 
S
(  z ) max
 2 . 04
S
S
2.2
3
S tre s s C o n c e n tra tio n Fa c to r
Normalized Stress in Loading Direction
3.5
2.5
Two Dimensional Case: (r,/2)/S
2
1.5
1
0.5
2.15
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
5
0
0.1
0.2
0.3
P o isso n 's R a tio
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 ( a 2  a 6   a 21 cos 2   b 21 sin 2  )
2
 r  2 a 2  2 a 6   2 a 21 cos 2   2 b 21 sin 2 
r

   2 a 2  2 a 6   2 a 21 cos 2   2 b 21 sin 2 


x
 r    a 6  2 b 21 cos 2   2 a 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
x
r 
S 

(  2   cos 2   sin 2  )
2 2
2
 
S 

(  2   cos 2   sin 2  )
2 2
2
 r 
S
2
(1  cos 2  

2
sin 2  )
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
  a 6 r   b 21 r sin 2 
2
2
   2 a 6   2 b 21 sin 2 
 r    a 6  2 b 21 cos 2 
y
Use BC’s To Determine Stress Solution
r  
 
 r 
T
2
T
2
T
2
(sin 2   2  )
(sin 2   2  )
(1  cos 2  )
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
 Forces
C
   X e 1  Ye 2 
C
Try Airy Stress Function
  ( a 12 r log r  a 15 r  ) cos 
 ( b12 r log r  b15 r  ) sin 
y
Use BC’s To Determine Stress Solution
r  
2
r
[ X cos   Y sin  ]
    r  0
Flamant Solution Stress Results
Normal Force Case
2
 x   r cos   
2
r  
2Y
sin 
r
    r  0
or in Cartesian
components
 y   r sin
2
 
2 Yx y
( x  y )
2 Yy
2
3
( x  y )
2
2
 xy   r sin  cos   
Y
2
2
2
2 Yxy
2
( x  y )
2
2
x
r = constant
Dimensionless Stress
y=a
xy/(Y/a)
y/(Y/a)
y
 y  2Y /  a
Dimensionless Distance, x/a
2
Flamant Solution Displacement Results
Normal Force Case
r 
 
 r 
u r

r
ur
r
1
E

(  r    )  
1 u 
r 
1 u r
r 


u 
r
1
E

2Y
 Er
sin 
(     r ) 
u

r
1

2Y
 Er
ur 
sin 
u 
 r  0
Y
E
Y
E
[( 1   )(  

) cos   2 log r sin  ]
2
[  (1   )(  

) sin   2 log r cos   (1   ) cos  ]
2
Note unpleasant feature of 2-D model that
displacements become unbounded as r  
0.1
Y
0
On Free Surface y = 0
u r ( r ,0 )  u r ( r ,  )  
Y
(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
-0.5
0
0.5
Comparison of Flamant Results with
3-D Theory - Boussinesq’s Problem
Cartesian Solution
P
x
u 
1  2 
Py  z
1  2 
P
 z
 2 
 , v 
 2 
 , w 
4  R  R
R z 
4  R  R
R z 
4  R
Px
x  
y
z
Free Surface Displacements
u z ( R ,0 ) 
y  
z  
P (1   )
P
2  R
E
[(1   )  2 log r ]
3-D Solution eliminates the
unbounded far-field behavior
2 R
2
P
2 R
2
3 Pz
3
2 R
5
2
3x 2 z
 z
R
x (2 R  z ) 

 (1  2  )  


3
2 
R
R

z
R
R
(
R

z
)



2
3y 2z
 z
R
y (2 R  z ) 

 (1  2  )  


3
2
R z
R ( R  z )  
R
 R
,  xy  
 yz  
P
2 R
3 Pyz
2 R
2
 3 xyz
(1  2  )( 2 R  z ) xy 
 3 

2
R(R  z)
 R

2
5
,  xz  
2
3 Pxz
2 R
5
Cylindrical Solution
Corresponding 2-D Results
u  ( r ,0 )  
P
2

z 
 2 (1   )  2 

R 

(1  2  ) r 
 rz
ur 

2
4  R  R
R  z 
r 
2

z 
uz 
 2 (1   )  2 
4  R 
R 
 
P
P
u  0
P
2 R
2
 3 r 2 z (1  2  ) R 



3
R z 
R

(1  2  ) P  z
R 

2

R  z 
2 R
R
z  
3 Pz
3
2 R
5
,  rz  
3 P rz
2 R
2
5
Half-Space Under Uniform Normal
Loading Over –a  x  a
p
x
2
a 1
a
 x   r cos   
2Y
 y   r sin
2Y
2
2
 
sin  cos 
2
r
r
sin
 xy   r sin  cos   
y
dx

r
d
x  
2p
y  
2p
 xy  
2p





2
1
2
1

2
1
dY = pdx = prd /sin

cos  d   
2
sin
2
 d  
p
2
p
2
sin  cos  d  
3
2Y
r

sin
2
 cos 
d x  
2p
d y  
2p
d  xy  
2p



cos
2
 d
sin
2
 d
sin  cos  d 
[ 2 (  2   1 )  (sin 2  2  sin 2  1 )]
[ 2 (  2   1 )  (sin 2  2  sin 2  1 )]
p
2
[cos 2  2  cos 2  1 ]
Half-Space Under Uniform Normal
Loading - Results
0.5
Dimensionless Stress
xy /p
y/p
Dimensionless Maximum Shear Stress
0.45
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
0
Dimensionless Distance, x/a
max - Contours
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
x  
2y
y  
2y
 xy  
2y


p ( s )( x  s )
a
[( x  s )  y ]
3



a
2
a
a
2

2
2
2
p(s)
[( x  s )  y ]
2
2
ds 
p ( s )( x  s )
a
[( x  s )  y ]
2
2

ds 
2
a
2
2
2
s

a
t ( s )( x  s )
a
[( x  s )  y ]
2y
2
2


2y


a
a
3
2
2
ds
t ( s )( x  s )
[( x  s )  y ]
2
2
a
t ( s )( x  s )
a
[( x  s )  y ]
2
2
2
2
2
ds
ds
Photoelastic Contact Stress Fields
(Point Loading)
(Uniform Loading)
(Flat Punch Loading)
(Cylinder Contact Loading)
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 
sin 2  (   1)  0   
At Crack Tip r  0:
n
 1 , n  0 , 1, 2 , 
2
Stress  O ( r
2
) , Displaceme
nt  O ( r
 1
)
Finite Displacements and Singular Stresses at Crack Tip  1<  <2   = 3/2
Notch/Crack Problem Results
y
r  
r


 

x
Stress Free Faces
 = 2 - 
 
5
3

3
3 1 
  5 sin )  B (cos   cos )
A (sin
2 
3
2
2
2
4 r 
 
3

3
3 1 
  3 sin )  B (cos   cos )
A (sin
2 
2
2
2
4 r 
 r  
 
1
3

3
3 1 
  sin )
A (cos   cos )  B (sin
2 
3
2
2
2
4 r 
r  
Transform to  Variable
  
 r  
•
•
•
•
3 A
2
2
cos
r
( 3  cos  ) 

sin

2
B
sin
(1  cos  ) 
3B
sin
B
2 r
(1  3 cos  )

(1  cos  )
2
2 r
(1  cos  ) 

2
2 r
2
r
3 A

2
r
3 A
2
cos
cos

(1  3 cos  )
2
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
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
Mode III Crack Problem – Exercise 8-32

y
z
r
Contours for Mode III Crack Problem
●

x
Anti-Plane Strain Case
u  v  0 , w  w( x, y )
 w
2
 w
2
w  A r sin
r

2
2

1 w
r r
,  z 
1  w
2

A
2 r
Stresses Again
r
2
cos


2
O r
2
0
,  zr 
1 / 2

A
2 r
sin

2
z - Stress Contours
Curved Beam Under End Moments
Dimensionless Stress, a /M
b
2
r
a
M
M
 r ( a )   r (b )  0
 r ( a )   r ( b )  0


b
a
b
a
  dr  0
  rdr   M
  a 0  a 1 log r  a 2 r  a 3 r log r
2
r  
4M
  
4M
 r  0
2
[
a b
N
N
r
2
log(
2
b
)  b log(
2
a
2
[
a b
r
2
b/a = 4
Theory of Elasticity
Strength of Materials
2
log(
2
r
)  a log(
2
b
b
a
)  b log(
2
a
Dimensionless Distance, r/a
)]
r
r
b
)  a log(
2
a
r
)b a ]
2
2
Curved Cantilever Beam
P
r
a
b
 r (a , )   r (b, )  0
Dimensionless Stress, a/P

 = /2 b/a = 4
Theory of Elasticity
Strength of Materials
 r ( a ,  )   r ( b ,  )  0





b
a
b
a
b
a
b
a
b
a
 r  ( r , 0 ) dr  P
Dimensionless Distance, r/a
b
  ( r , 0 ) dr     ( r , 0 ) rdr  0
a
  ( r ,  / 2 ) dr   P
  ( r ,  / 2 ) rdr  P ( a  b ) / 2
 r  ( r ,  / 2 ) dr  0
  ( Ar 
3
B
r
 Cr  Dr log r ) sin 
r 
 
P
2
(r 
a b
2

r
P
a b
( 3r 
 r  
r
P
N
2

) sin 
2
) sin 
r
a b
r
a b
2
3
2
(r 
2
r
2
N
2
3
N
a b
3
2
a b
2

r
2
) cos 
Disk Under Diametrical Compression
P
D
=
P
Flamant Solution (1)
+
+
Flamant Solution (2)
Radial Tension Solution (3)
Disk Problem – Superposition of Stresses
y
P

(1 )
x
2P

 r1
(1 )
(1 )
x
(2)
2P
 r1
y  
 xy  
(2)
x
(3)
3
x
cos  1 sin  1
2P
 r2
2P
 r2
2P
 r2
 y
r1
cos  1
 r1

(2)
1
2P
y  
 xy  
cos  1 sin  1
2
(3)
2
2
cos  2 sin  2
r2
2
P
cos  2
3
x  
cos  2 sin  2
2

2P
D
,  xy  0
2
2
2P (R  y)x
(R  y)x
1




4
4
 
r1
r2
D
y
3
3
2P (R  y)
(R  y)
1
 




4
4
 
r1
r2
D
 xy
2
2
2P (R  y) x (R  y) x 




4
4
 
r1
r2

( 3)
r1 , 2 
x  (R  y)
2
2
Disk Problem – Results
y
P
1
x-axis (y = 0)
r1
x
2
r2
2
2
2P  D  4x 
 x ( x ,0 ) 
 2
2 
D  D  4 x 
y-axis (x = 0)
2
 x (0, y ) 
4

2P 
4D
 y ( x ,0 )  

1


2
2 2
D  ( D  4 x )

 xy ( x , 0 )  0
2P
D
 y (0, y )  
 Constant
2P 
2
2
1 




 D  2y D  2y D
 xy ( 0 , y )  0
P
(Theoretical max Contours)
(Photoelastic Contours)
(Courtesy of URI Dynamic Photomechanics Lab)
Applications to Granular Media Modeling
Contact Load Transfer Between Idealized Grains
P
P
P
P
Four-Contact Grain
(Courtesy of URI Dynamic Photomechanics Lab)