For β = −0.056 for a through-hardened steel, grade 1, continue Ex. 14–6 for wear.
Question 14.7: For β = −0.056 for a through-hardened steel, grade 1, contin...
From Fig. 14–5,
\left(S_{c}\right)_{P}=322(300)+29100=125700 psi
From Eq. (14–45),
\left(S_{c}\right)_{G}=\left(S_{c}\right)_{P} m_{G}^{\beta} (14–45)
\left(S_{c}\right)_{G}=\left(S_{c}\right)_{P}\left(\frac{64}{18}\right)^{-0.056}=125700\left(\frac{64}{18}\right)^{-0.056}=117100 psi
\left(H_{B}\right)_{G}=\frac{117100-29200}{322}=273 \text { Brinell }
which is slightly less than the pinion hardness of 300 Brinell.
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