Page 140 - Shigley's Mechanical Engineering Design

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Load and Stress Analysis 115

EXAMPLE 3–14 An aluminum-alloy pressure vessel is made of tubing having an outside diameter of 8 in
1
and a wall thickness of in.
4
(a) What pressure can the cylinder carry if the permissible tangential stress is
12 kpsi and the theory for thin-walled vessels is assumed to apply?
(b) On the basis of the pressure found in part (a), compute the stress components
using the theory for thick-walled cylinders.
Solution (a) Here d i = 8 − 2(0.25) = 7.5 in, r i = 7.5/2 = 3.75 in, and r o = 8/2 = 4 in. Then
t/r i = 0.25/3.75 = 0.067. Since this ratio is less than 0.1, the theory for thin-walled
vessels should yield safe results.
We first solve Eq. (3–53) to obtain the allowable pressure. This gives

Answer p = 2t(σ t ) max = 2(0.25)(12)(10) 3 = 774 psi
d i + t 7.5 + 0.25
(b) The maximum tangential stress will occur at the inside radius, and so we use
r = r i in the first equation of Eq. (3–50). This gives

2 r 2 r + r 2 4 + 3.75 2
2
2
i
Answer (σ t ) max = r p i 1 + o = p i o i = 774 = 12 000 psi
2
2
2
r − r 2 r 2 r − r 2 4 − 3.75 2
o i i o i
Similarly, the maximum radial stress is found, from the second equation of Eq. (3–50)
to be
Answer σ r =−p i =−774 psi
The stresses σ t and σ r are principal stresses, since there is no shear on these surfaces.
Note that there is no significant difference in the stresses in parts (a) and (b), and so the
thin-wall theory can be considered satisfactory for this problem.

3–15 Stresses in Rotating Rings

Many rotating elements, such as flywheels and blowers, can be simplified to a rotating
ring to determine the stresses. When this is done it is found that the same tangential and
radial stresses exist as in the theory for thick-walled cylinders except that they are
caused by inertial forces acting on all the particles of the ring. The tangential and radial
stresses so found are subject to the following restrictions:
• The outside radius of the ring, or disk, is large compared with the thickness r o ≥ 10t.
• The thickness of the ring or disk is constant.
• The stresses are constant over the thickness.
The stresses are 11
3 + ν 2 2 r r 1 + 3ν 2
2 2
i o
2
σ t = ρω r + r + − r
o
i
8 r 2 3 + ν
(3–55)
3 + ν 2 2 r r 2
2 2
2
i o
σ r = ρω r + r − − r
o
i
8 r 2
11 Ibid, pp. 348–357.