All of the parameters in this example are the same as in Ex. 14–4 with the exception that we are using helical gears. Thus, several terms will be the same as Ex. 14–4. The reader should verify that the following terms remain unchanged: K_{o}=1, Y_{P}=0.303, Y_{G}=0.412, m_{G}=3.059, \left(K_{s}\right)_{p}=1.043, \left(K_{s}\right)_{G}=1.052, \left(Y_{N}\right)_{P}=0.977,\left(Y_{N}\right)_{G}=0.996, K_{R}=0.85, K_{T}=1, C_{f}=1, C_{p}=2300 \sqrt{ psi }, \left(S_{t}\right)_{P}=31350 \text { psi }, \left(S_{t}\right)_{G}=28260 psi, \left(S_{c}\right)_{P}= 106 380 psi, \left(S_{c}\right)_{G}=93500 psi, \left(Z_{N}\right)_{P}=0.948, \left(Z_{N}\right)_{G}=0.973, and C_{H}=1.005, For helical gears, the transverse diametral pitch, given by Eq. (13–18), is
P_{n}=\frac{P_{t}}{\cos \psi} (13–18)
P_{t}=P_{n} \cos \psi=10 \cos 30^{\circ}=8.660 \text { teeth/in }
Thus, the pitch diameters are d_{P}=N_{P} / P_{t}=17 / 8.660=1.963 \text { in and } d_{G}=52 / 8.660= 6.005 in. The pitch-line velocity and transmitted force are
V=\frac{\pi d_{P} n_{P}}{12}=\frac{\pi(1.963) 1800}{12}=925 ft / min
W^{t}=\frac{33000 H}{V}=\frac{33000(4)}{925}=142.7 lbf
As in Ex. 14–4, for the dynamic factor, B = 0.8255 and A = 59.77. Thus, Eq. (14–27) gives
K_{v}=\left\{\begin{array}{ll}\left(\frac{A+\sqrt{V}}{A}\right)^{B} & V \text { in } ft / min \\\left(\frac{A+\sqrt{200 V}}{A}\right)^{B} & V \text { in } m / s\end{array}\right. (14–27)
K_{v}=\left(\frac{59.77+\sqrt{925}}{59.77}\right)^{0.8255}=1.404
The geometry factor I for helical gears requires a little work. First, the transverse pressure angle is given by Eq. (13–19)
\cos \psi=\frac{\tan \phi_{n}}{\tan \phi_{t}} (13–19)
\phi_{t}=\tan ^{-1}\left(\frac{\tan \phi_{n}}{\cos \psi}\right)=\tan ^{-1}\left(\frac{\tan 20^{\circ}}{\cos 30^{\circ}}\right)=22.80^{\circ}
The radii of the pinion and gear are r_{P}=1.963 / 2=0.9815 \text { in and } r_{G}=6.004 / 2= 3.002 in, respectively. The addendum is a=1 / P_{n}=1 / 10=0.1, and the base-circle radii of the pinion and gear are given by Eq. (13–6) with \phi=\phi_{t}:
r_{b}=r \cos \phi (13–6)
\left(r_{b}\right)_{P}=r_{P} \cos \phi_{t}=0.9815 \cos 22.80^{\circ}=0.9048 \text { in }
\left(r_{b}\right)_{G}=3.002 \cos 22.80^{\circ}=2.767 \text { in }
From Eq. (14–25), the surface strength geometry factor
Z=\left[\left(r_{P}+a\right)^{2}-r_{b P}^{2}\right]^{1 / 2}+\left[\left(r_{G}+a\right)^{2}-r_{b G}^{2}\right]^{1 / 2}-\left(r_{P}+r_{G}\right) \sin \phi_{t} (14–25)
\begin{aligned}Z=& \sqrt{(0.9815+0.1)^{2}-0.9048^{2}}+\sqrt{(3.004+0.1)^{2}-2.769^{2}} \\&-(0.9815+3.004) \sin 22.80^{\circ} \\=& 0.5924+1.4027-1.5444=0.4507 in\end{aligned}
Since the first two terms are less than 1.544 4, the equation for Z stands. From Eq. (14–24) the normal circular pitch p_{N} is
p_{N}=p_{n} \cos \phi_{n} (14–24)
p_{N}=p_{n} \cos \phi_{n}=\frac{\pi}{P_{n}} \cos 20^{\circ}=\frac{\pi}{10} \cos 20^{\circ}=0.2952 \text { in }
From Eq. (14–21), the load sharing ratio
m_{N}=\frac{p_{N}}{0.95 Z} (14–21)
m_{N}=\frac{p_{N}}{0.95 Z}=\frac{0.2952}{0.95(0.4507)}=0.6895
Substituting in Eq. (14–23), the geometry factor I is
I=\left\{\begin{array}{ll}\frac{\cos \phi_{t} \sin \phi_{t}}{2 m_{N}} \frac{m_{G}}{m_{G}+1} & \text { external gears } \\\frac{\cos \phi_{t} \sin \phi_{t}}{2 m_{N}} \frac{m_{G}}{m_{G}-1} & \text { internal gears }\end{array}\right. (14–23)
I=\frac{\sin 22.80^{\circ} \cos 22.80^{\circ}}{2(0.6895)} \frac{3.06}{3.06+1}=0.195
From Fig. 14–7, geometry factors J_{P}^{\prime}=0.45 \text { and } J_{G}^{\prime}=0.54. Also from Fig. 14–8 the J-factor multipliers are 0.94 and 0.98, correcting J_{P}^{\prime} \text { and } J_{G}^{\prime} to
J_{P}=0.45(0.94)=0.423
J_{G}=0.54(0.98)=0.529
The load-distribution factor K_{m} is estimated from Eq. (14–32):
C_{p f}=\left\{\begin{array}{ll}\frac{F}{10 d}-0.025 & F \leq 1 \text { in } \\\frac{F}{10 d}-0.0375+0.0125 F & 1<F \leq 17 \text { in } \\\frac{F}{10 d}-0.1109+0.0207 F-0.000228 F^{2} & 17<F \leq 40 \text { in }\end{array}\right. (14–32)
C_{p f}=\frac{1.5}{10(1.963)}-0.0375+0.0125(1.5)=0.0577
with C_{m c}=1, C_{p m}=1, C_{m a}=0.15 from Fig. 14–11, and C_{e}=1. Therefore, from Eq. (14–30),
K_{m}=C_{m f}=1+C_{m c}\left(C_{p f} C_{p m}+C_{m a} C_{e}\right) (14–30)
K_{m}=1+(1)[0.0577(1)+0.15(1)]=1.208
(a) Pinion tooth bending. Substituting the appropriate terms into Eq. (14–15) using P_{t} gives
\sigma=\left\{\begin{array}{ll}W^{t} K_{o} K_{v} K_{s} \frac{P_{d}}{F} \frac{K_{m} K_{B}}{J} & \text { (U.S. customary units) } \\W^{t} K_{o} K_{v} K_{s} \frac{1}{b m_{t}} \frac{K_{H} K_{B}}{Y_{J}} & \text { (SI units) }\end{array}\right. (14–15)
\begin{aligned}(\sigma)_{P} &=\left(W^{t} K_{o} K_{v} K_{s} \frac{P_{t}}{F} \frac{K_{m} K_{B}}{J}\right)_{P}=142.7(1) 1.404(1.043) \frac{8.66}{1.5} \frac{1.208(1)}{0.423} \\&=3445 psi\end{aligned}
Substituting the appropriate terms for the pinion into Eq. (14–41) gives
S_{F}=\frac{S_{t} Y_{N} /\left(K_{T} K_{R}\right)}{\sigma}=\frac{\text { fully corrected bending strength }}{\text { bending stress }} (14–41)
\left(S_{F}\right)_{P}=\left(\frac{S_{t} Y_{N} /\left(K_{T} K_{R}\right)}{\sigma}\right)_{P}=\frac{31350(0.977) /[1(0.85)]}{3445}=10.5
Gear tooth bending. Substituting the appropriate terms for the gear into Eq. (14–15) gives
(\sigma)_{G}=142.7(1) 1.404(1.052) \frac{8.66}{1.5} \frac{1.208(1)}{0.529}=2779 psi
Substituting the appropriate terms for the gear into Eq. (14–41) gives
\left(S_{F}\right)_{G}=\frac{28260(0.996) /[1(0.85)]}{2779}=11.9
(b) Pinion tooth wear. Substituting the appropriate terms for the pinion into Eq. (14–16) gives
\sigma_{c}=\left\{\begin{array}{ll}C_{p} \sqrt{W^{t} K_{o} K_{v} K_{s} \frac{K_{m}}{d_{P} F} \frac{C_{f}}{I}} & \text { (U.S. customary units) } \\Z_{E} \sqrt{W^{t} K_{o} K_{v} K_{s} \frac{K_{H}}{d_{w 1} b} \frac{Z_{R}}{Z_{I}}} & \text { (SI units) }\end{array}\right. (14–16)
\begin{aligned}\left(\sigma_{c}\right) P &=C_{p}\left(W^{t} K_{o} K_{v} K_{s} \frac{K_{m}}{d_{p} F} \frac{C_{f}}{I}\right)_{p}^{1 / 2} \\&=2300\left[142.7(1) 1.404(1.043) \frac{1.208}{1.963(1.5)} \frac{1}{0.195}\right]^{1 / 2}=48230 psi\end{aligned}
Substituting the appropriate terms for the pinion into Eq. (14–42) gives
S_{H}=\frac{S_{c} Z_{N} C_{H} /\left(K_{T} K_{R}\right)}{\sigma_{c}}=\frac{\text { fully corrected contact strength }}{\text { contact stress }} (14–42)
\left(S_{H}\right)_{P}=\left(\frac{S_{c} Z_{N} /\left(K_{T} K_{R}\right)}{\sigma_{c}}\right)_{P}=\frac{106400(0.948) /[1(0.85)]}{48230}=2.46
Gear tooth wear. The only term in Eq. (14–16) that changes for the gear is K_{s}. Thus,
\left(\sigma_{c}\right)_{G}=\left[\frac{\left(K_{s}\right)_{G}}{\left(K_{s}\right)_{P}}\right]^{1 / 2}\left(\sigma_{c}\right)_{P}=\left(\frac{1.052}{1.043}\right)^{1 / 2} 48230=48440 psi
Substituting the appropriate terms for the gear into Eq. (14–42) with C_{H}=1.005 gives
\left(S_{H}\right)_{G}=\frac{93500(0.973) 1.005 /[1(0.85)]}{48440}=2.22
(c) For the pinion we compare S_{F} \text { with } S_{H}^{2}, or 10.5 with 2.46^{2}=6.05, so the threat in the pinion is from wear. For the gear we compare S_{F} \text { with } S_{H}^{2}, or 11.9 with 2.22^{2}=4.93, so the threat is also from wear in the gear. For the meshing gearset wear controls.
