Page 311 - Shigley's Mechanical Engineering Design

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286 Mechanical Engineering Design

EXAMPLE 6–2 Given a 1050 HR steel, estimate
6
(a) the rotating-beam endurance limit at 10 cycles.
(b) the endurance strength of a polished rotating-beam specimen corresponding to 10 4
cycles to failure
(c) the expected life of a polished rotating-beam specimen under a completely reversed
stress of 55 kpsi.
Solution (a) From Table A–20, S ut = 90 kpsi. From Eq. (6–8),

Answer S = 0.5(90) = 45 kpsi

e
.
(b) From Fig. 6–18, for S ut = 90 kpsi, f = 0.86. From Eq. (6–14),
[0.86(90)] 2
a = = 133.1 kpsi
45
From Eq. (6–15),
1 0.86(90)
b =− log =−0.0785
3 45
Thus, Eq. (6–13) is

S = 133.1 N −0.0785
f
Answer For 10 cycles to failure, S = 133.1(10 ) = 64.6 kpsi
4 −0.0785
4

f
(c) From Eq. (6–16), with σ rev = 55 kpsi,
1/−0.0785
55
Answer N = = 77 500 = 7.75(10 ) cycles
4
133.1
Keep in mind that these are only estimates. So expressing the answers using three-place
accuracy is a little misleading.

6–9 Endurance Limit Modifying Factors
We have seen that the rotating-beam specimen used in the laboratory to determine
endurance limits is prepared very carefully and tested under closely controlled condi-
tions. It is unrealistic to expect the endurance limit of a mechanical or structural mem-
ber to match the values obtained in the laboratory. Some differences include
• Material: composition, basis of failure, variability
• Manufacturing: method, heat treatment, fretting corrosion, surface condition, stress
concentration
• Environment: corrosion, temperature, stress state, relaxation times
• Design: size, shape, life, stress state, speed, fretting, galling