Page 396 - Shigley's Mechanical Engineering Design

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Shafts and Shaft Components 371
Table 6–6: k e = 0.814

S e = 0.787(0.870)0.814(52.5) = 29.3 kpsi
For a rotating shaft, the constant bending moment will create a completely reversed
bending stress.

M a = 1260 lbf · in T m = 1100 lbf · in M m = T a = 0
Applying Eq. (7–7) for the DE-Goodman criteria gives
2 1/2 2 1/2

1 16 4 (1.58 · 1260) 3 (1.37 · 1100)
= + = 0.615
n π(1.1) 3 29 300 105 000
Answer n = 1.63 DE-Goodman
Similarly, applying Eqs. (7–9), (7–11), and (7–13) for the other failure criteria,

Answer n = 1.87 DE-Gerber

Answer n = 1.88 DE-ASME Elliptic

Answer n = 1.56 DE-Soderberg

For comparison, consider an equivalent approach of calculating the stresses and apply-
ing the fatigue failure criteria directly. From Eqs. (7–5) and (7–6),
1/2

32 · 1.58 · 1260
2

σ = 3 = 15 235 psi
a
π (1.1)
1/2

16 · 1.37 · 1100
2

σ = 3 3 = 9988 psi
m
π (1.1)
Taking, for example, the Goodman failure critera, application of Eq. (6–46)
gives
1 σ a σ m 15 235 9988
= + = + = 0.615
n S e S ut 29 300 105 000
n = 1.63
which is identical with the previous result. The same process could be used for the other
failure criteria.
(b) For the yielding factor of safety, determine an equivalent von Mises maximum
stress using Eq. (7–15).
1/2

32(1.58) (1260) 16(1.37) (1100)
2 2

σ max = 3 + 3 3 = 18 220 psi
π (1.1) π (1.1)
Answer n y = S y = 82 000 = 4.50
σ 18 220
max