The pinion of Examples 14–1 and 14–2 is to be mated with a 50-tooth gear manufactured of ASTM No. 50 cast iron. Using the tangential load of 382 lbf, estimate the factor of safety of the drive based on the possibility of a surface fatigue failure.

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From Table A–5 we find the elastic constants to be [latex]E_{P}=30 Mpsi , v_{P}=0.292, E_{G}=[/latex] 14.5 Mpsi, [latex]v_{G}=0.211[/latex]. We substitute these in Eq. (14–13) to get the elastic coefficient as

[latex]C_{p}=\left[\frac{1}{\pi\left(\frac{1-v_{P}^{2}}{E_{P}}+\frac{1-v_{G}^{2}}{E_{G}} ight)} ight]^{1 / 2}[/latex] (14–13)

[latex]C_{p}=\left\{\frac{1}{\pi\left[\frac{1-(0.292)^{2}}{30\left(10^{6} ight)}+\frac{1-(0.211)^{2}}{14.5\left(10^{6} ight)} ight]} ight\}^{1 / 2}=1817[/latex]

From Example 14–1, the pinion pitch diameter is [latex]d_{P}=2[/latex] in. The value for the gear is [latex]d_{G}=50 / 8=6.25[/latex] in. Then Eq. (14–12) is used to obtain the radii of curvature at the pitch points. Thus

[latex]r_{1}=\frac{d_{p} \sin \phi}{2} \quad r_{2}=\frac{d_{G} \sin \phi}{2}[/latex] (14–12)

[latex]r_{1}=\frac{2 \sin 20^{\circ}}{2}=0.342 ext { in } \quad r_{2}=\frac{6.25 \sin 20^{\circ}}{2}=1.069 ext { in }[/latex]

The face width is given as F = 1.5 in. Use [latex]K_{v}=1.52[/latex] from Example 14–1. Substituting all these values in Eq. (14–14) with [latex]\phi=20^{\circ}[/latex] gives the contact stress as

[latex]\sigma_{C}=-C_{p}\left[\frac{K_{v} W^{t}}{F \cos \phi}\left(\frac{1}{r_{1}}+\frac{1}{r_{2}} ight) ight]^{1 / 2}[/latex] (14–14)

[latex]\sigma_{C}=-1817\left[\frac{1.52(380)}{1.5 \cos 20^{\circ}}\left(\frac{1}{0.342}+\frac{1}{1.069} ight) ight]^{1 / 2}=-72400 psi[/latex]

The surface endurance strength of cast iron can be estimated from

[latex]S_{C}=0.32 H_{B} kpsi[/latex]

for [latex]10^{8}[/latex] cycles, where [latex]S_{C}[/latex] is in kpsi. Table A–24 gives [latex]H_{B}=262[/latex] for ASTM No. 50 cast iron. Therefore [latex]S_{C}=0.32(262)=83.8 kpsi[/latex]. Contact stress is not linear with transmitted load [see Eq. (14–14)]. If the factor of safety is defined as the loss-of-function load divided by the imposed load, then the ratio of loads is the ratio of stresses squared. In other words,

[latex]n=\frac{ ext { loss-of-function load }}{ ext { imposed load }}=\frac{S_{C}^{2}}{\sigma_{C}^{2}}=\left(\frac{83.8}{72.4} ight)^{2}=1.3[/latex]

One is free to define factor of safety as [latex]S_{C} / \sigma_{C}[/latex]. Awkwardness comes when one compares the factor of safety in bending fatigue with the factor of safety in surface fatigue for a particular gear. Suppose the factor of safety of this gear in bending fatigue is 1.20 and the factor of safety in surface fatigue is 1.34 as above. The threat, since 1.34 is greater than 1.20, is in bending fatigue since both numbers are based on load ratios. If the factor of safety in surface fatigue is based on [latex]S_{C} / \sigma_{C}=\sqrt{1.34}=1.16[/latex], then 1.20 is greater than 1.16, but the threat is not from surface fatigue. The surface fatigue factor of safety can be defined either way. One way has the burden of requiring a squared number before numbers that instinctively seem comparable can be compared. In

Question 14.3: The pinion of Examples 14–1 and 14–2 is to be mated with a 5...

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