A 1035 steel has a tensile strength of 70 kpsi and is to be used for a part that sees 450°F in service. Estimate the Marin temperature modification factor and \left(S_{e}\right)_{450^{\circ}} if
(a) The room-temperature endurance limit by test is \left(S_{e}^{\prime}\right)_{70^{\circ}}=39.0 kpsi \text { . }
(b) Only the tensile strength at room temperature is known.
Question : A 1035 steel has a tensile strength of 70 kpsi and is to be ...
(a) First, from Eq. (6–27),
\begin{aligned}k_{d}=& 0.975+0.432\left(10^{-3}\right)T_{F}-0.115\left(10^{-5}\right) T_{F}^{2} \\&+0.104\left(10^{-8}\right)T_{F}^{3}-0.595\left(10^{-12}\right) T_{F}^{4}\end{aligned} (6–27)
\begin{aligned}k_{d}=& 0.975+0.432\left(10^{-3}\right)(450)-0.115\left(10^{-5}\right)\left(450^{2}\right)\\&+0.104\left(10^{-8}\right)\left(450^{3}\right)-0.595\left(10^{-12}\right)\left(450^{4}\right)=1.007\end{aligned}
Thus,
\left(S_{e}\right)_{450^{\circ}}=k_{d}\left(S_{e}^{\prime}\right)_{70^{\circ}}=1.007(39.0)=39.3 kpsi
(b) Interpolating from Table 6–4 gives
\left(S_{T} / S_{R T}\right)_{450^{\circ}}=1.018+(0.995-1.018) \frac{450-400}{500-400}=1.007
Thus, the tensile strength at 450°F is estimated as
\left(S_{ut}\right)_{450^{\circ}}=\left(S_{T} / S_{RT}\right)_{450^{\circ}}\left(S_{ut}\right)_{70^{\circ}}=1.007(70)=70.5 kpsi
From Eq. (6–8) then,
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)
\left(S_{e}\right)_{450^{\circ}}=0.5\left(S_{ut}\right)_{450^{\circ}}=0.5(70.5)=35.2 kpsi
Part a gives the better estimate due to actual testing of the particular material.
