Page 365 - Shigley's Mechanical Engineering Design

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340 Mechanical Engineering Design

EXAMPLE 6–20 A rotating shaft experiences a steady torque T = 1360LN(1, 0.05) lbf · in, and at a
shoulder with a 1.1-in small diameter, a fatigue stress-concentration factor K f =
1.50LN(1, 0.11), K fs = 1.28LN(1, 0.11), and at that location a bending moment of
M = 1260LN(1, 0.05) lbf · in. The material of which the shaft is machined is hot-rolled
1035 with S ut = 86.2LN(1, 0.045) kpsi and S y = 56.0LN(1, 0.077) kpsi. Estimate the
reliability using a stochastic Gerber failure zone.
Solution Establish the endurance strength. From Eqs. (6–70) to (6–72) and Eq. (6–20), p. 288,

S = 0.506(86.2)LN(1, 0.138) = 43.6LN(1, 0.138) kpsi
e
k a = 2.67(86.2) −0.265 LN(1, 0.058) = 0.820LN(1, 0.058)
k b = (1.1/0.30) −0.107 = 0.870
k c = k d = k f = LN(1, 0)

S e = 0.820LN(1, 0.058)0.870(43.6)LN(1, 0.138)
S e = 0.820(0.870)43.6 = 31.1 kpsi
¯
2
2 1/2
C Se = (0.058 + 0.138 ) = 0.150
and so S e = 31.1LN(1, 0.150) kpsi.
Stress (in kpsi):
32(1.50)LN(1, 0.11)1.26LN(1, 0.05)
32K f M a
σ a = 3 = 3
πd π(1.1)
32(1.50)1.26
¯ σ a = 3 = 14.5 kpsi
π(1.1)
2
2 1/2
C σa = (0.11 + 0.05 ) = 0.121
16(1.28)LN(1, 0.11)1.36LN(1, 0.05)
16K fs T m
m = 3 = 3
πd π(1.1)
16(1.28)1.36
¯ τ m = 3 = 6.66 kpsi
π(1.1)
2 1/2
2
C τm = (0.11 + 0.05 ) = 0.121
2 2 1/2 2 2 1/2

¯ σ = ¯σ + 3¯ τ = [14.5 + 3(0) ] = 14.5 kpsi
a a a

2 2 1/2 2 1/2
τ

¯ σ = ¯σ + 3¯ = [0 + 3(6.66) ] = 11.54 kpsi
m m m
¯ σ a 14.5
r = = = 1.26
¯ σ 11.54
m
Strength: From Eqs. (6–80) and (6–81),
⎧ ⎫

2
1.26 86.2 2 ⎨ 2(31.1) 2 ⎬
¯
S a = −1 + 1 + = 28.9 kpsi
1.26(86.2)
2(31.1) ⎩ ⎭