Question 6.1: A 1/4 -in drill rod was heat-treated and ground. The measure......

Eq. (2-17):      S_{u t}=0.495(490)=242.6\mathrm{~kpsi}\gt 212\mathrm{~kpsi}

S_{u}=\left\{\begin{array}{l l}{{0.495H_{B}}}&{{\mathrm{kpsi}}}\\ {{3.41H_{B}}}&{{\ \ \mathrm{MPa}}}\end{array}\right.                              (2-17)

Eq. (6-8):        S_{e}^{\prime}=100\,\mathrm{kpsi}

S_{e}^{\prime}=\left\{\begin{array}{l l}{{0.5\,S_{u t}}}&{{S_{u t}\leq200\mathrm{~kpsi~(1400~MPa)}}}\\ {{100\ \mathrm{kpsi}}}&{{S_{u t}\gt 200\ \mathrm{kpsi}}}\\ {{700\ \mathrm{MPa}}}&{{S_{u t}\gt 1400\ \mathrm{MPa}}}\end{array}\right.                                   (6-8)

Table 6-2:          a = 1.34, b = −0.085

Eq. (6-19):           k_{a}=1.34(242.6)^{-0.085}=0.840

k_{a}=a S_{u t}^{b}                                        (6-19)

Eq. (6-20):            k_{b}=\left({\frac{1/4}{0.3}}\right)^{-0.107}=1.02

k_{b}={\left\{\begin{array}{l l}{(d/0.3)^{-0.107}=0.879d^{-0.107}}&{0.11\leq d\leq2\operatorname*{in}}\\ {0.91d^{-0.157}}&{2\lt d\leq10\ in}\\ {(d/7.62)^{-0.107}=1.24d^{-0.107}}&{2.79\leq d\leq51\ \operatorname*{mm}}\\ {1.51d^{-0.157}}&{51\lt d\leq254\operatorname*{mm}}\end{array}\right.}                             (6-20)

Eq. (6-18):          S_{e}=k_{a}k_{b}S_{e}^{\prime}=0.840(1.02)(100)=85.7\,\mathrm{kpsi}

S_{e}=k_{a}k_{b}k_{c}k_{d}k_{e}k_{f}S_{e}^{\prime}                              (6-18)