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Question 14.7: For β = −0.056 for a through-hardened steel, grade 1, contin...

For β = −0.056 for a through-hardened steel, grade 1, continue Ex. 14–6 for wear.

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From Fig. 14–5,

(S_{c})_{P} = 322(300) + 29 100 = 125 700 psi

From Eq. (14–45),

(S_{c})_{G} = (S_{c})_{P}m^{β}_{G}                 (14–45)

(S_{c})_{G} = (S_{c})_{P} \left(\frac{64}{18}\right)^{−0.056}= 125 700 \left(\frac{64}{18}\right)^{−0.056}= 117 100 psi

(H_{B})_{G} =\frac{117 100 − 29 200}{322} = 273 Brinell

which is slightly less than the pinion hardness of 300 Brinell.

14.5

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