Question 6.68: A 1 1/2 -in-diameter hot-rolled steel bar has a 3/16 -in dia...

Given:    S_{u t}=58  kpsi .

Eq. (6-70):      S _{e}^{\prime}=0.506(76) L N (1,0.138)=38.5 L N (1,0.138)  kpsi

Table 6-13:      k _{a}=14.5(76)^{-0.719} L N (1,0.11)=0.644 L N (1,0.11)

Eq. (6-24):      d_{e}=0.370(1.5)=0.555 \text { in }

Eq. (6-20):     k_{b}=(0.555 / 0.3)^{-0.107}=0.936

Eq. (6-70):

Table A-16:  d / D=0, a / D=(3 / 16) / 1.5=0.125, A=0.80 \therefore K_{t}=2.20 .

From Eqs. (6-78) and (6-79) and Table 6-15

Table A-16:

Eq. (5-43), p. 250:  z=-\frac{\ln \left[(23.2 / 10.4) \sqrt{\left(1+0.10^{2}\right) /\left(1+0.176^{2}\right)}\right]}{\sqrt{\ln \left[\left(1+0.176^{2}\right)\left(1+0.10^{2}\right)\right]}}=-3.94

Table A-10:      p_{f}=0.0000415 \Rightarrow R=1-p_{f}=1-0.0000415=0.99996

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Eq. (6-20): k_{b}= \begin{cases}(d / 0.3)^{-0.107}=0.879 d^{-0.107} & 0.11 \leq d \leq 2 \text { in } \\ 0.91 d^{-0.157} & 2<d \leq 10 \text { in } \\ (d / 7.62)^{-0.107}=1.24 d^{-0.107} & 2.79 \leq d \leq 51 mm \\ 1.51 d^{-0.157} & 51<d \leq 254 mm \end{cases}

Eq. (6-24): d_{e}=0.370 d

Eq. (6-70): S _{e}^{\prime}= \begin{cases}0.506 \bar{S}_{u t} L N (1,0.138) kpsi \text { or } MPa & \bar{S}_{u t} \leq 212 kpsi (1460 MPa ) \\ 107 L N (1,0.139) kpsi & \bar{S}_{u t}>212 kpsi \\ 740 L N (1,0.139) MPa & \bar{S}_{u t}>1460 MPa \end{cases}

Eq. (6-78) : \bar{K}_{f}=\frac{K_{t}}{1+\frac{2\left(K_{t}-1\right)}{K_{t}} \frac{\sqrt{a}}{\sqrt{r}}}

Eq. (6-79) : K _{f}=\bar{K}_{f} L N \left(1, C_{K_{f}}\right)

Eq. (5-43) :  z=-\frac{\mu_{\ln S}-\mu_{\ln \sigma}}{\left(\hat{\sigma}_{\ln S}^{2}+\hat{\sigma}_{\ln \sigma}^{2}\right)^{1 / 2}}=-\frac{\ln \left(\frac{\mu_{S}}{\mu_{\sigma}} \sqrt{\frac{1+C_{\sigma}^{2}}{1+C_{S}^{2}}}\right)}{\sqrt{\ln \left[\left(1+C_{S}^{2}\right)\left(1+C_{\sigma}^{2}\right)\right]}}