Page 395 - Shigley's Mechanical Engineering Design

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370 Mechanical Engineering Design
throughout its entire range. This is evident by noting that it crosses the yield line in
Fig. 6–27. The Gerber and modified Goodman criteria do not guard against yielding,
requiring a separate check for yielding. A von Mises maximum stress is calculated for this
purpose.
2

σ max = (σ m + σ a ) + 3 (τ m + τ a ) 2 1/2
1/2

2 2
32K f (M m + M a ) 16K fs (T m + T a )
= + 3
πd 3 πd 3
(7–15)
To check for yielding, this von Mises maximum stress is compared to the yield
strength, as usual.
S y
n y = (7–16)
σ max

For a quick, conservative check, an estimate for σ max can be obtained by simply

adding σ and σ . (σ + σ ) will always be greater than or equal to σ max , and will

m
a
m
a
therefore be conservative.
EXAMPLE 7–1 At a machined shaft shoulder the small diameter d is 1.100 in, the large diameter D is
1.65 in, and the fillet radius is 0.11 in. The bending moment is 1260 lbf · in and the
steady torsion moment is 1100 lbf · in. The heat-treated steel shaft has an ultimate
strength of S ut = 105 kpsi and a yield strength of S y = 82 kpsi. The reliability goal
is 0.99.
(a) Determine the fatigue factor of safety of the design using each of the fatigue failure
criteria described in this section.
(b) Determine the yielding factor of safety.
Solution (a) D/d = 1.65/1.100 = 1.50, r/d = 0.11/1.100 = 0.10, K t = 1.68 (Fig. A–15–9),
K ts = 1.42 (Fig. A–15–8), q = 0.85 (Fig. 6–20), q shear = 0.88 (Fig. 6–21).

From Eq. (6–32),
K f = 1 + 0.85(1.68 − 1) = 1.58
K fs = 1 + 0.88(1.42 − 1) = 1.37

Eq. (6–8): S = 0.5(105) = 52.5 kpsi
e
Eq. (6–19): k a = 2.70(105) −0.265 = 0.787
−0.107
1.100

Eq. (6–20): k b = = 0.870
0.30
k c = k d = k f = 1