A 1015 hot-rolled steel bar has been machined to a diameter of 1 in. It is to be placed in reversed axial loading for 70 000 cycles to failure in an operating environment of 550°F. Using ASTM minimum properties, and a reliability of 99 percent, estimate the endurance limit and fatigue strength at 70 000 cycles.
Question : A 1015 hot-rolled steel bar has been machined to a diameter ...
From Table A–20, S_{u t}=50 kpsi \text { at } 70^{\circ} F \text { . } Since the rotating-beam specimen endurance limit is not known at room temperature, we determine the ultimate strength at the elevated temperature first, using Table 6–4. From Table 6–4,
\left(\frac{S_{T}}{S_{RT}}\right)_{550^{\circ}}=\frac{0.995+0.963}{2}=0.979
The ultimate strength at 550°F is then
\left(S_{ut}\right)_{550^{\circ}}=\left(S_{T} / S_{RT}\right)_{550^{\circ}}\left(S_{u t}\right)_{70^{\circ}}=0.979(50)=49.0 kpsi
The rotating-beam specimen endurance limit at 550°F is then estimated from Eq. (6–8) as
S_{e}^{\prime}=\left\{\begin{array}{ll}0.5 S_{u t} & S_{u t} \leq 200 kpsi (1400 MPa ) \\100 kpsi & S_{u t}> kpsi \\700 MPa & S_{u t}>1400 MPa\end{array}\right. (6–8)
S_{e}^{\prime}=0.5(49)=24.5 kpsi
Next, we determine the Marin factors. For the machined surface, Eq. (6–19) with Table 6–2 gives
k_{a}=a S_{u t}^{b} (6–19)
k_{a}=a S_{ut}^{b}=2.70\left(49^{-0.265}\right)=0.963
For axial loading, from Eq. (6–21), the size factor k_{b}=1 \text { , } and from Eq. (6–26) the loading factor is k_{c}=0.85. The temperature factor k_{d}=1 \text { , } since we accounted for the temperature in modifying the ultimate strength and consequently the endurance limit. For 99 percent reliability, from Table 6–5, k_{e}=0.814 .Finally, since no other conditions were given, the miscellaneous factor is k_{f}=1. The endurance limit for the part is estimated by Eq. (6–18) as
k_{b}=1 (6–21)
S_{e}=k_{a} k_{b} k_{c} k_{d} k_{e} k_{f} S_{e}^{\prime} (6–18)
k_{c}=\left\{\begin{array}{ll}1 & \text { bending } \\0.85 & \text { axial } \\0.59 & \text { torsion }\end{array}\right. (6–26)
S_{e}=k_{a} k_{b} k_{c} k_{d} k_{e} k_{f} S_{e}^{\prime}
=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. since S_{u t}=49<70 kpsi \text { , then } f=0.9 From Eq. (6–14)
a=\frac{\left(f S_{u t}\right)^{2}}{S_{e}} (6–14)
a=\frac{\left(f S_{u t}\right)^{2}}{S_{e}}=\frac{[0.9(49)]^{2}}{16.3}=119.3 kpsi
and Eq. (6–15)
b=-\frac{1}{3} \log \left(\frac{f S_{u t}}{S_{e}}\right) (6–15)
b=-\frac{1}{3} \log \left(\frac{f S_{u t}}{S_{e}}\right)=-\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)
S_{f}=aN^{b}=119.3(70000)^{-0.1441}=23.9 kpsi
