Question 6.18: Estimate Kf and q for the steel shaft given in Ex. 6–6, p. 2...

\sqrt {a} =\frac {139}{S_{ut}} =\frac {139}{690} = 0.2014\sqrt {mm}

From Eq. (6–78),

\overline {K}_{f} =\frac {K_{t}}{1 +\frac {2(K_{t} − 1)}{K_{t}} \frac {\sqrt {a}}{\sqrt {r}}}                   (6–78)

\overline {K}_{f} =\frac {K_{t}}{1 +\frac {2(K_{t} − 1)}{K_{t}} \frac {\sqrt {a}}{\sqrt {r}}}=\frac {1.65}{1 +\frac {2(1.65− 1)}{1.65} \frac {\sqrt {0.2014}}{\sqrt {3}}}=1.51

which is 2.5 percent lower than what was found in Ex. 6–6.

From Table 6–15, C_{Kf} = 0.11. Thus from Eq. (6–79),

K_{f} =\overline{ K}_{f} LN (1,C_{Kf})                (6–79)

K_{f}= 1.51 LN(1, 0.11)

From Eq. (6–77), with K_{t}= 1.65

\begin{aligned} \bar{q} &=\frac{\bar{K}_{f}-1}{K_{t}-1} \\ \hat{\sigma}_{q} &=\frac{C \bar{K}_{f}}{K_{t}-1} \\ C_{q} &=\frac{C \bar{K}_{f}}{\bar{K}_{f}-1} \end{aligned}           (6-77)

\overline {q} =\frac {1.51 − 1}{1.65 − 1} = 0.785

C_{q} =\frac {C_{Kf} \overline{K}_{f}}{\overline{K}_{f}− 1} =\frac {0.11(1.51)}{1.51 − 1} = 0.326

\hat{σ}_{q} = C_{q} \overline {q} = 0.326(0.785) = 0.256

So,

q = LN(0.785, 0.256)