Plane Crack Problem
Airy Stress Function
For the plane problem, the equations of equilibrium are
satisfied when the stress components are expressed by
the
Airy stress function through
x
,
,
11
22
xxx
rr










21
,
,
222
r
r
rr
rr

law, the strains can be expressed in terms of .
x
It can be shown that the compatibility equation, when
expressed in terms of the Airy stress function, satisfies
the biharmonic equation:
 
.
11
,0
2
2
22
2
222











rrrr
The boundary conditions for this plane crack problem
22
The boundary conditions for this plane crack problem
are: for

23
Where p and p are harmonic functions of r and θ
(i.e. and
0
2
 p
).0
2
 q
Now consider a separate solution, of
the following form (Williams, 1957):




,




rR
,sincos
21


rArAp 








.
2
sin
sin
2







B
A
r
24






.
2
sin
sin
22








2coscos12
11
 BAr














rx
rr
r
1
25





BA
The admissible cases are: (i) cos λ π = 0,
hence
where Z is an integer including zero, and thus
B
1
= A
1
λ /(λ + 2) or (ii) sin λπ = 0 and hence λ = Z and
B
1
= A
1
. Since the governing equations are linear, any
linear combination of the admissible solutions provides
a solution, hence
λ
can have any satisfying:
2
12


Z



26


r
ij
~
.
~


r
ij
.~
2
1
2


r
ijij

The total strain energy within an annular region,with
inner and outer radii
r
0
and R, respectively, centered at
the crack tip, with unit thickness is
 
.~
2
1
00



Thus the physically admissible values of λ are
,
2
)( ,,2,
2
3
,1,
2
1
,0,
2
1
Z
or


where Z is –1, 0, or a positive integer. Taking the most
dominant singular term (λ = 1/2 and thus B
1
=A
1
/3)
we find that:

29
we find that:
 


 





~
2/10
2
1
1


rrrA
ijij
I
ijij

The higher order terms, with exponents greater than
zero, vanish as r
 0. We write where
K
I
is the stress intensity factor.
Thus we have that:

2/
1 I
KA 
 