Page 341 - Shigley's Mechanical Engineering Design

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316 Mechanical Engineering Design
Solution Some preparatory work is needed. From Table A–24, S ut = 31 kpsi, S uc = 109 kpsi,

k a k b S = 14 kpsi. Since k c for axial loading is 0.9, then S e = (k a k b S )k c = 14(0.9) =
e e
2
12.6 kpsi. From Table A–15–1, A = t(w − d) = 0.375(1 − 0.25) = 0.281 in , d/w =
0.25/1 = 0.25, and K t = 2.45. The notch sensitivity for cast iron is 0.20 (see p. 296),
so
K f = 1 + q(K t − 1) = 1 + 0.20(2.45 − 1) = 1.29

K f F a 1.29(0) K f F m 1.29(1000) −3
(a) σ a = = = 0 σ m = = (10 ) = 4.59 kpsi
A 0.281 A 0.281
and
31.0
S ut
Answer n = = = 6.75
σ m 4.59
F 1000
(b) F a = F m = = = 500 lbf
2 2

K f F a 1.29(500) −3
σ a = σ m = = (10 ) = 2.30 kpsi
A 0.281
σ a
r = = 1
σ m
From Eq. (6–52),

(1)31 + 12.6 4(1)31(12.6)
S a = −1 + 1 + = 7.63 kpsi
2 [(1)31 + 12.6] 2
7.63
S a
Answer n = = = 3.32
σ a 2.30
1 1.29(650) −3
(c) F a = |300 − (−1000)|= 650 lbf σ a = (10 ) = 2.98 kpsi
2 0.281
1 1.29(−350) −3
F m = [300 + (−1000)] =−350 lbf σ m = (10 ) =−1.61 kpsi
2 0.281
3.0
σ a
r = = =−1.86
σ m −1.61
From Eq. (6–53), S a = S e + (S e /S ut − 1)S m and S m = S a /r. It follows that
12.6
S e
S a = = = 18.5 kpsi
1 S e 1 12.6
1 − − 1 1 − − 1
r S ut −1.86 31
18.5
S a
Answer n = = = 6.20
σ a 2.98
Figure 6–31b shows the portion of the designer’s fatigue diagram that was constructed.