Page 344 - Shigley's Mechanical Engineering Design

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Fatigue Failure Resulting from Variable Loading 319

From Table A–20 we find the minimum strengths to be S ut = 440 MPa and S y =
370 MPa. The endurance limit of the rotating-beam specimen is 0.5(440) = 220 MPa.
The surface factor, obtained from Eq. (6–19) and Table 6–2, p. 287, is
−0.265
k a = 4.51S ut = 4.51(440) −0.265 = 0.899
From Eq. (6–20) the size factor is

d 42
−0.107 −0.107
k b = = = 0.833
7.62 7.62
The remaining Marin factors are all unity, so the modified endurance strength S e is
S e = 0.899(0.833)220 = 165 MPa

(a) Theoretical stress-concentration factors are found from Table A–16. Using a/D =
6/42 = 0.143 and d/D = 34/42 = 0.810, and using linear interpolation, we obtain
A = 0.798 and K t = 2.366 for bending; and A = 0.89 and K ts = 1.75 for torsion.
Thus, for bending,

π A 4 4 π(0.798) 4 4 3 3
Z net = (D − d ) = [(42) − (34) ] = 3.31 (10 )mm
32D 32(42)
and for torsion
π A 4 4 π(0.89) 4 4 3 4
J net = (D − d ) = [(42) − (34) ] = 155 (10 )mm
32 32
Next, using Figs. 6–20 and 6–21, pp. 295–296, with a notch radius of 3 mm we find the
notch sensitivities to be 0.78 for bending and 0.81 for torsion. The two corresponding
fatigue stress-concentration factors are obtained from Eq. (6–32) as
K f = 1 + q(K t − 1) = 1 + 0.78(2.366 − 1) = 2.07

K fs = 1 + 0.81(1.75 − 1) = 1.61
The alternating bending stress is now found to be

M 150 6
σ xa = K f = 2.07 −6 = 93.8(10 )Pa = 93.8MPa
Z net 3.31(10 )
and the alternating torsional stress is
−3
TD 120(42)(10 ) 6
τ xya = K fs = 1.61 −9 = 26.2(10 )Pa = 26.2MPa
2J net 2(155)(10 )
The midrange von Mises component σ is zero. The alternating component σ is given

m a
by
1/2
2 2 2 2 1/2

σ = σ + 3τ = [93.8 + 3(26.2 )] = 104.2MPa
a xa xya
Since S e = S a , the fatigue factor of safety n f is
Answer n f = S a = 165 = 1.58
σ a 104.2