6.8 Thick-Walled Cylinder Subjected to Internal and External Pressures
177
Now, from Hooke’s law,
ε r =
1
E
σ r − ν(σ θ + σ z )
ε θ =
1
E
σ θ − ν(σ r + σ z )
ε z =
1
E
σ z − ν(σ r + σ θ )
Since ε z = 0, then
0 =
1
E
σ z − ν(σ r + σ θ )
σ z = v(σ r + σ θ )
Hence,
ε r =
(1 + ν)
E
[(1 − ν)σ r − νσ θ ]
ε θ =
(1 + ν)
E
[(1 − ν)σ θ − νσ r ]
Solving for σ θ and σ r,
σ θ =
E
(1 − 2ν)(1 + ν)
[νε r + (1 − ν)ε θ ]
σ r =
E
(1 − 2ν)(1 + ν)
[(1 − ν)ε r + νε θ ]
Substituting the values of ε r and ε θ , the above expressions for σ θ and σ r can be
written as
σ θ =
E
(1 − 2ν)(1 + ν)
ν
du
dr
+ (1 − ν)
u
r
σ r =
E
(1 − 2ν)(1 + ν)
(1 − ν)
du
dr
+
vu
r
Substituting these in the equation of equilibrium (6.21), we get
d
dr
(1 − ν) r
du
dr
+ νu
− ν
du
dr
− (1 − ν)
u
r
= 0
or
du
dr
+ r
d
2 u
dr 2 −
u
r
= 0
d
2 u
dr 2 +
1
r
du
dr
−
u
r 2 = 0