Page 323 - Shigley's Mechanical Engineering Design

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298 Mechanical Engineering Design

EXAMPLE 6–8 A 1015 hot-rolled steel bar has been machined to a diameter of 1 in. It is to be placed
in reversed axial loading for 70 000 cycles to failure in an operating environment of
550°F. Using ASTM minimum properties, and a reliability of 99 percent, estimate the
endurance limit and fatigue strength at 70 000 cycles.
Solution From Table A–20, S ut = 50 kpsi at 70°F. Since the rotating-beam specimen endurance
limit is not known at room temperature, we determine the ultimate strength at the ele-
vated temperature first, using Table 6–4. From Table 6–4,

S T 0.995 + 0.963
= = 0.979
S RT 2
550 ◦
The ultimate strength at 550°F is then
(S ut ) 550 ◦ = (S T /S RT ) 550 ◦ (S ut ) 70 ◦ = 0.979(50) = 49.0 kpsi

The rotating-beam specimen endurance limit at 550°F is then estimated from Eq. (6–8)
as

S = 0.5(49) = 24.5 kpsi

e
Next, we determine the Marin factors. For the machined surface, Eq. (6–19) with
Table 6–2 gives
b
k a = aS = 2.70(49 −0.265 ) = 0.963
ut
For axial loading, from Eq. (6–21), the size factor k b = 1, and from Eq. (6–26) the load-
ing factor is k c = 0.85. The temperature factor k d = 1, since we accounted for the tem-
perature in modifying the ultimate strength and consequently the endurance limit. For
99 percent reliability, from Table 6–5, k e = 0.814. Finally, since no other conditions
were given, the miscellaneous factor is k f = 1. The endurance limit for the part is esti-
mated by Eq. (6–18) as

S e = k a k b k c k d k e k f S e
Answer
= 0.963(1)(0.85)(1)(0.814)(1)24.5 = 16.3 kpsi

For the fatigue strength at 70 000 cycles we need to construct the S-N equation. From
p. 285, since S ut = 49 < 70 kpsi, then f 0.9. From Eq. (6–14)

( fS ut ) 2 [0.9(49)] 2
a = = = 119.3 kpsi
S e 16.3
and Eq. (6–15)
1 fS ut 1 0.9(49)
b =− log =− log =−0.1441
3 S e 3 16.3
Finally, for the fatigue strength at 70 000 cycles, Eq. (6–13) gives

Answer S f = aN = 119.3(70 000) −0.1441 = 23.9 kpsi
b