Question 6.15: Given a part with Sut = 151 kpsi and at the critical locatio...

From Fig. 6–18, p. 277, for S_{ut} = 151 kpsi, f = 0.795. From Eq. (6–14),

a =\frac{( f S_{ut} )^{2}}{S_{e}}                  (6–14)

a =\frac{( f S_{ut} )^{2}}{S_{e}} =\frac {[0.795(151)]^{2}}{67.5 }= 213.5  kpsi

From Eq. (6–15),

b = −\frac {1}{3} log \left ( \frac {f S_{ut}}{S_{e}}\right)                   (6–15)

b = −\frac {1}{3} log \left ( \frac {f S_{ut}}{S_{e}}\right)= −\frac {1}{3} log \left [ \frac {0.795 (151)}{67.5}\right]= −0.0833

So,

S_{f} = 213.5N^{−0.0833}                      N =\left ( \frac {S_{f}}{213.5}\right)^{−1/0.0833}                        (1), (2)

We prepare to add two columns to the previous table. Using the Gerber fatigue criterion, Eq. (6–47), p. 298, with S_{e} = S_{f} , and n = 1, we can write

Gerber     \frac{nσ_{a}}{S_{e}} + \left( \frac {nσ_{m}}{S_{ut}}\right)^{2}= 1                     (6–47)

S_{f}= \begin{cases} \frac {σ_{a}}{1 − (σ_{m}/S_{ut} )^{2}} & σ_{m} > 0 \\ S_{e} &σ_{m} ≤ 0  \end{cases}                   (3)

Cycle 1: r = σ_{a}/σ_{m} = 70/10 = 7, and the strength amplitude from Table 6–7, p. 299, is

Table 6–7  Amplitude and Steady Coordinates of Strength and Important Intersections in First Quadrant for Gerber and Langer Failure Criteria

S_{a} =\frac {7^{2}151^{2}}{2 (67.5)} \left\{-1+\sqrt {1+\left [\frac {2(67.5)}{7(151)}\right]^{2}}\right\}= 67.2     kpsi

Since σ_{a} > S_{a} , that is, 70 > 67.2, life is reduced. From Eq. (3),

S_{f} =\frac {70}{1 − (10/151)^{2}} = 70.3  kpsi

and from Eq. (2)

N =\left( \frac {70.3}{213.5} \right)^{−1/0.0833}= 619(10^{3}) cycles

Cycle 2: r = 10/50 = 0.2, and the strength amplitude is

S_{a} =\frac {0.2^{2}151^{2}}{2 (67.5)} \left\{-1+\sqrt {1+\left [\frac {2(67.5)}{0.2(151)}\right]^{2}}\right\}= 24.2      kpsi

Since σ _{a} < S_{a} , that is 10 < 24.2, then S_{f} = S_{e} and indefinite life follows. Thus, N→ ∞.

Cycle 3: r = 10/−30 = −0.333, and since σ_{m} < 0, S_{f} = S_{e} , indefinite life follows and N→ ∞