Question 6.8: A 1015 hot-rolled steel bar has been machined to a diameter ...

$k_{b} = 1$               (6–21)

$S_{e} = k_{a}k_{b}k_{c}k_{d}k_{e}k_{f} S′_{e}$ (6–18)

$k_{c}=\begin{cases}1 & bending \\bending0.85 &axial\\axial0.59 &torsion^{17}\end{cases}$

$S_{e} = k_{a}k_{b}k_{c}k_{d}k_{e}k_{f} S′_{e}$
= 0.963(1)(0.85)(1)(0.814)(1)24.5 = 16.3 kpsi

For the fatigue strength at 70 000 cycles we need to construct the S-N equation. From p. 277, since $S_{ut}$ = 49 < 70 kpsi, then f 0.9. From Eq. (6–14)

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

$a =\frac {[0.9(49)^{2}]}{16.3} = 119.3  kpsi$

and Eq. (6–15)

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

$b= −\frac {1}{3}log \left [\frac {0.9 (49)}{16.3}\right]= −0.1441$

Finally, for the fatigue strength at 70 000 cycles, Eq. (6–13) gives

$S_{f} = a N^{b}$           (6–13)

$= 119.3 (70 000)^{−0.1441 }=23.9   kpsi$

17    Use this only for pure torsional fatigue loading. When torsion is combined with other stresses, such as bending, $k_{c}$ = 1 and the combined loading is managed by using the effective von Mises stress as in Sec. 5–5. Note: For pure torsion, the distortion energy predicts that  $(k_{c})_{torsion} = 0.577$.