6.15 Numerical Examples
201
Radial stress = σ r =
12 × (0.1)
2
(0.15) 2 − (0.1) 2
−
12
(0.15) 2 − (0.1) 2
(0.1)
2
(0.15)
2
(0.15) 2
σ r = 0
Hoop stress = σ θ =
12 × (0.1)
2
(0.15) 2 − (0.1) 2
+
(0.1)
2
(0.15)
2
(0.15) 2
12
(0.15) 2 − (0.1) 2
= 9.6 + 9.6
or σ θ = 19.2 MPa.
Example 6.3 A steel tube, which has an outside diameter of 10 cm and inside diameter of 5 cm, is subjected to an internal pressure of 14 MPa and an external pressure
of 5.5 MPa. Calculate the maximum hoop stress in the tube.
Solution: The maximum hoop stress occurs at r = a.
Therefore,
maximum hoop stress = (σ θ ) max =
p i a
2
− p 0 b
2
b 2 − a 2
+
p i − p 0
b 2 − a 2
a
2 b
2
a 2
=
p i a
2
− p 0 b
2
b 2 − a 2
+
p i − p 0
b 2 − a 2
b
2
=
p i a
2
− p 0 b
2
+ p i b
2
− p 0 b
2
b 2 − a 2
(σ θ ) max =
p i (a
2
+ b
2
) − 2 p 0 b
2
b 2 − a 2
Therefore,
(σ θ ) max =
14
(0.05)
2
+ (0.1)
2
− 2 × 5.5 × (0.1)
2
(0.1) 2 − (0.05) 2
Or (σ θ ) max = 8.67 MPa.
Example 6.4 A steel cylinder which has an inside diameter of 1 m is subjected to
an internal pressure of 8 MPa. Calculate the wall thickness if the maximum shearing
stress is not to exceed 35 MPa.
Solution: The critical point lies on the inner surface of the cylinder, i.e. at r = a.
We have,
radial stress = 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
At r = a and p 0 = 0,