Page 322 - Shigley's Mechanical Engineering Design

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Fatigue Failure Resulting from Variable Loading 297
Solution From Fig. A–15–9, using D/d = 38/32 = 1.1875, r/d = 3/32 = 0.093 75, we read
.
the graph to find K t = 1.65.
.
(a) From Fig. 6–20, for S ut = 690 MPa and r = 3 mm, q = 0.84. Thus, from Eq. (6–32)
.
Answer K f = 1 + q(K t − 1) = 1 + 0.84(1.65 − 1) = 1.55
√ √ √
(b) From Eq. (6–35a) with S ut = 690 MPa = 100 kpsi, a = 0.0622 in = 0.313 mm.
Substituting this into Eq. (6–33) with r = 3 mm gives

K t − 1 . 1.65 − 1
Answer K f = 1 + √ = 1 + = 1.55
1 + a/r 0.313
1 + √
3

Some designers use 1/K f as a Marin factor to reduce S e. For simple loading, infi-
nite life problems, it makes no difference whether S e is reduced by dividing it by K f or
the nominal stress is multiplied by K f. However, for finite life, since the S-N diagram
is nonlinear, the two approaches yield differing results. There is no clear evidence
pointing to which method is better. Furthermore, in Sec. 6–14, when we consider com-
bining loads, there generally are multiple fatigue stress-concentration factors occurring
at a point (e.g. K f for bending and K fs for torsion). Here, it is only practical to modify
the nominal stresses. To be consistent in this text, we will exclusively use the fatigue
stress-concentration factor as a multiplier of the nominal stress.

EXAMPLE 6–7 For the step-shaft of Ex. 6–6, it is determined that the fully corrected endurance limit is
S e = 280 MPa. Consider the shaft undergoes a fully reversing nominal stress in the fil-
let of (σ rev ) nom = 260 MPa. Estimate the number of cycles to failure.

Solution From Ex. 6–6, K f = 1.55, and the ultimate strength is S ut = 690 MPa = 100 kpsi. The
maximum reversing stress is

(σ rev ) max = K f (σ rev ) nom = 1.55(260) = 403 MPa
From Fig. 6–18, f = 0.845. From Eqs. (6–14), (6–15), and (6–16)
( fS ut ) 2 [0.845(690)] 2
a = = = 1214 MPa
S e 280
1 fS ut 1 0.845(690)
b =− log =− log =−0.1062
3 S e 3 280

1/−0.1062

1/b

403
Answer N = σ rev = = 32.3 10 3 cycles

a 1214
Up to this point, examples illustrated each factor in Marin’s equation and stress
concentrations alone. Let us consider a number of factors occurring simultaneously.