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6 Two-Dimensional Problems in Elasticity …
The solution of this equation is the same as in Eq. (6.22)
u = C 1 r + C 2 /r.
where C 1 and C 2 are constants of integration. Therefore, σ θ and σ r are given by
σ θ =
E
(1 − 2ν)(1 + ν)
C 1 + (1 − 2ν)
C 2
r 2
σ r =
E
(1 − 2ν)(1 + ν)
C 1 − (1 − 2ν)
C 2
r 2
Applying the boundary conditions,
σ r = −p i when r = a
σ r = −p 0 when r = b
Therefore,
E
(1 − 2ν)(1 + ν)
C 1 − (1 − 2ν)
C 2
a 2
= −p i
E
(1 − 2ν)(1 + ν)
C 1 − (1 − 2ν)
C 2
b 2
= −p o
Solving, we get
C 1 =
(1 − 2ν)(1 + ν)
E
p 0 b
2
− p i a
2
a 2 − b 2
and
C 2 =
(1 + ν)
E
( p 0 − p i )a
2 b
2
a 2 − b 2
Substituting these, the stress components become
σ r =
p i a
2
− p 0 b
2
b 2 − a 2
−
p i − p 0
b 2 − a 2
a
2 b
2
r 2
(6.32)
σ θ =
p i a
2
− p 0 b
2
b 2 − a 2
+
p i − p 0
b 2 − a 2
a
2 b
2
r 2
(6.33)
σ z = 2ν
p 0 a
2
− p i b
2
b 2 − a 2
(6.34)