A 17-tooth 20◦ normal pitch-angle helical pinion with a right-hand helix angle of 30◦ rotates at 1800 rev/min when transmitting 4 hp to a 52-tooth helical gear. The normal diametral pitch is 10 teeth/in, the face width is 1.5 in, and the set has a quality number of 6. The gears are straddle-mounted

Question

A 17-tooth 20◦ normal pitch-angle helical pinion with a right-hand helix angle of 30◦ rotates at 1800 rev/min when transmitting 4 hp to a 52-tooth helical gear. The normal diametral pitch is 10 teeth/in, the face width is 1.5 in, and the set has a quality number of 6. The gears are straddle-mounted with bearings immediately adjacent. The pinion and gear are made from a through-hardened steel with surface and core hardnesses of 240 Brinell on the pinion and surface and core hardnesses of 200 Brinell on the gear. The transmission is smooth, connecting an electric motor and a centrifugal pump. Assume a pinion life of $10^{8}$ cycles and a reliability of 0.9 and use the upper curves in Figs. 14–14 and 14–15.

(a) Find the factors of safety of the gears in bending.

(b) Find the factors of safety of the gears in wear.

(c) By examining the factors of safety identify the threat to each gear and to the mesh.

(a) Find the factors of safety of the gears in bending.

(b) Find the factors of safety of the gears in wear.

(c) By examining the factors of safety identify the threat to each gear and to the mesh.

Accepted Answer

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: [latex]K_{o}=1, Y_{P}=0.303, Y_{G}=0.412[/latex], [latex]m_{G}=3.059[/latex], [latex]\left(K_{s} ight)_{p}=1.043[/latex], [latex]\left(K_{s} ight)_{G}=1.052[/latex], [latex]\left(Y_{N} ight)_{P}=0.977[/latex],[latex]\left(Y_{N} ight)_{G}=0.996[/latex], [latex]K_{R}=0.85[/latex], [latex]K_{T}=1[/latex], [latex]C_{f}=1[/latex], [latex]C_{p}=2300 \sqrt{ psi }[/latex], [latex]\left(S_{t} ight)_{P}=31350 ext { psi }[/latex], [latex]\left(S_{t} ight)_{G}=28260 psi[/latex], [latex]\left(S_{c} ight)_{P}=[/latex] 106 380 psi, [latex]\left(S_{c} ight)_{G}=93500 psi[/latex], [latex]\left(Z_{N} ight)_{P}=0.948[/latex], [latex]\left(Z_{N} ight)_{G}=0.973[/latex], and [latex]C_{H}=1.005[/latex], For helical gears, the transverse diametral pitch, given by Eq. (13–18), is

[latex]P_{n}=\frac{P_{t}}{\cos \psi}[/latex] (13–18)

[latex]P_{t}=P_{n} \cos \psi=10 \cos 30^{\circ}=8.660 ext { teeth/in }[/latex]

Thus, the pitch diameters are [latex]d_{P}=N_{P} / P_{t}=17 / 8.660=1.963 ext { in and } d_{G}=52 / 8.660=[/latex] 6.005 in. The pitch-line velocity and transmitted force are

[latex]V=\frac{\pi d_{P} n_{P}}{12}=\frac{\pi(1.963) 1800}{12}=925 ft / min[/latex]

[latex]W^{t}=\frac{33000 H}{V}=\frac{33000(4)}{925}=142.7 lbf[/latex]

As in Ex. 14–4, for the dynamic factor, B = 0.8255 and A = 59.77. Thus, Eq. (14–27) gives

[latex]K_{v}=\left\{\begin{array}{ll}\left(\frac{A+\sqrt{V}}{A} ight)^{B} & V ext { in } ft / min \\left(\frac{A+\sqrt{200 V}}{A} ight)^{B} & V ext { in } m / s\end{array} ight.[/latex] (14–27)

[latex]K_{v}=\left(\frac{59.77+\sqrt{925}}{59.77} ight)^{0.8255}=1.404[/latex]

The geometry factor I for helical gears requires a little work. First, the transverse pressure angle is given by Eq. (13–19)

[latex]\cos \psi=\frac{ an \phi_{n}}{ an \phi_{t}}[/latex] (13–19)

[latex]\phi_{t}= an ^{-1}\left(\frac{ an \phi_{n}}{\cos \psi} ight)= an ^{-1}\left(\frac{ an 20^{\circ}}{\cos 30^{\circ}} ight)=22.80^{\circ}[/latex]

The radii of the pinion and gear are [latex]r_{P}=1.963 / 2=0.9815 ext { in and } r_{G}=6.004 / 2=[/latex] 3.002 in, respectively. The addendum is [latex]a=1 / P_{n}=1 / 10=0.1[/latex], and the base-circle radii of the pinion and gear are given by Eq. (13–6) with [latex]\phi=\phi_{t}[/latex]:

[latex]r_{b}=r \cos \phi[/latex] (13–6)

[latex]\left(r_{b} ight)_{P}=r_{P} \cos \phi_{t}=0.9815 \cos 22.80^{\circ}=0.9048 ext { in }[/latex]

[latex]\left(r_{b} ight)_{G}=3.002 \cos 22.80^{\circ}=2.767 ext { in }[/latex]

From Eq. (14–25), the surface strength geometry factor

[latex]Z=\left[\left(r_{P}+a ight)^{2}-r_{b P}^{2} ight]^{1 / 2}+\left[\left(r_{G}+a ight)^{2}-r_{b G}^{2} ight]^{1 / 2}-\left(r_{P}+r_{G} ight) \sin \phi_{t}[/latex] (14–25)

[latex]\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}[/latex]

Since the first two terms are less than 1.544 4, the equation for Z stands. From Eq. (14–24) the normal circular pitch [latex]p_{N}[/latex] is

[latex]p_{N}=p_{n} \cos \phi_{n}[/latex] (14–24)

[latex]p_{N}=p_{n} \cos \phi_{n}=\frac{\pi}{P_{n}} \cos 20^{\circ}=\frac{\pi}{10} \cos 20^{\circ}=0.2952 ext { in }[/latex]

From Eq. (14–21), the load sharing ratio

[latex]m_{N}=\frac{p_{N}}{0.95 Z}[/latex] (14–21)

[latex]m_{N}=\frac{p_{N}}{0.95 Z}=\frac{0.2952}{0.95(0.4507)}=0.6895[/latex]

Substituting in Eq. (14–23), the geometry factor I is

[latex]I=\left\{\begin{array}{ll}\frac{\cos \phi_{t} \sin \phi_{t}}{2 m_{N}} \frac{m_{G}}{m_{G}+1} & ext { external gears } \\frac{\cos \phi_{t} \sin \phi_{t}}{2 m_{N}} \frac{m_{G}}{m_{G}-1} & ext { internal gears }\end{array} ight.[/latex] (14–23)

[latex]I=\frac{\sin 22.80^{\circ} \cos 22.80^{\circ}}{2(0.6895)} \frac{3.06}{3.06+1}=0.195[/latex]

From Fig. 14–7, geometry factors [latex]J_{P}^{\prime}=0.45 ext { and } J_{G}^{\prime}=0.54[/latex]. Also from Fig. 14–8 the J-factor multipliers are 0.94 and 0.98, correcting [latex]J_{P}^{\prime} ext { and } J_{G}^{\prime}[/latex] to

[latex]J_{P}=0.45(0.94)=0.423[/latex]

[latex]J_{G}=0.54(0.98)=0.529[/latex]

The load-distribution factor [latex]K_{m}[/latex] is estimated from Eq. (14–32):

[latex]C_{p f}=\left\{\begin{array}{ll}\frac{F}{10 d}-0.025 & F \leq 1 ext { in } \\frac{F}{10 d}-0.0375+0.0125 F & 1

[latex]C_{p f}=\frac{1.5}{10(1.963)}-0.0375+0.0125(1.5)=0.0577[/latex]

with [latex]C_{m c}=1, C_{p m}=1, C_{m a}=0.15[/latex] from Fig. 14–11, and [latex]C_{e}=1[/latex]. Therefore, from Eq. (14–30),

[latex]K_{m}=C_{m f}=1+C_{m c}\left(C_{p f} C_{p m}+C_{m a} C_{e} ight)[/latex] (14–30)

[latex]K_{m}=1+(1)[0.0577(1)+0.15(1)]=1.208[/latex]

(a) Pinion tooth bending. Substituting the appropriate terms into Eq. (14–15) using [latex]P_{t}[/latex] gives

[latex]\sigma=\left\{\begin{array}{ll}W^{t} K_{o} K_{v} K_{s} \frac{P_{d}}{F} \frac{K_{m} K_{B}}{J} & ext { (U.S. customary units) } \W^{t} K_{o} K_{v} K_{s} \frac{1}{b m_{t}} \frac{K_{H} K_{B}}{Y_{J}} & ext { (SI units) }\end{array} ight.[/latex] (14–15)

[latex]\begin{aligned}(\sigma)_{P} &=\left(W^{t} K_{o} K_{v} K_{s} \frac{P_{t}}{F} \frac{K_{m} K_{B}}{J} ight)_{P}=142.7(1) 1.404(1.043) \frac{8.66}{1.5} \frac{1.208(1)}{0.423} \&=3445 psi\end{aligned}[/latex]

Substituting the appropriate terms for the pinion into Eq. (14–41) gives

[latex]S_{F}=\frac{S_{t} Y_{N} /\left(K_{T} K_{R} ight)}{\sigma}=\frac{ ext { fully corrected bending strength }}{ ext { bending stress }}[/latex] (14–41)

[latex]\left(S_{F} ight)_{P}=\left(\frac{S_{t} Y_{N} /\left(K_{T} K_{R} ight)}{\sigma} ight)_{P}=\frac{31350(0.977) /[1(0.85)]}{3445}=10.5[/latex]

Gear tooth bending. Substituting the appropriate terms for the gear into Eq. (14–15) gives

[latex](\sigma)_{G}=142.7(1) 1.404(1.052) \frac{8.66}{1.5} \frac{1.208(1)}{0.529}=2779 psi[/latex]

Substituting the appropriate terms for the gear into Eq. (14–41) gives

[latex]\left(S_{F} ight)_{G}=\frac{28260(0.996) /[1(0.85)]}{2779}=11.9[/latex]

(b) Pinion tooth wear. Substituting the appropriate terms for the pinion into Eq. (14–16) gives

[latex]\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}} & ext { (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}}} & ext { (SI units) }\end{array} ight.[/latex] (14–16)

[latex]\begin{aligned}\left(\sigma_{c} ight) P &=C_{p}\left(W^{t} K_{o} K_{v} K_{s} \frac{K_{m}}{d_{p} F} \frac{C_{f}}{I} ight)_{p}^{1 / 2} \&=2300\left[142.7(1) 1.404(1.043) \frac{1.208}{1.963(1.5)} \frac{1}{0.195} ight]^{1 / 2}=48230 psi\end{aligned}[/latex]

Substituting the appropriate terms for the pinion into Eq. (14–42) gives

[latex]S_{H}=\frac{S_{c} Z_{N} C_{H} /\left(K_{T} K_{R} ight)}{\sigma_{c}}=\frac{ ext { fully corrected contact strength }}{ ext { contact stress }}[/latex] (14–42)

[latex]\left(S_{H} ight)_{P}=\left(\frac{S_{c} Z_{N} /\left(K_{T} K_{R} ight)}{\sigma_{c}} ight)_{P}=\frac{106400(0.948) /[1(0.85)]}{48230}=2.46[/latex]

Gear tooth wear. The only term in Eq. (14–16) that changes for the gear is [latex]K_{s}[/latex]. Thus,

[latex]\left(\sigma_{c} ight)_{G}=\left[\frac{\left(K_{s} ight)_{G}}{\left(K_{s} ight)_{P}} ight]^{1 / 2}\left(\sigma_{c} ight)_{P}=\left(\frac{1.052}{1.043} ight)^{1 / 2} 48230=48440 psi[/latex]

Substituting the appropriate terms for the gear into Eq. (14–42) with [latex]C_{H}=1.005[/latex] gives

[latex]\left(S_{H} ight)_{G}=\frac{93500(0.973) 1.005 /[1(0.85)]}{48440}=2.22[/latex]

(c) For the pinion we compare [latex]S_{F} ext { with } S_{H}^{2}[/latex], or 10.5 with [latex]2.46^{2}=6.05[/latex], so the threat in the pinion is from wear. For the gear we compare [latex]S_{F} ext { with } S_{H}^{2}[/latex], or 11.9 with [latex]2.22^{2}=4.93[/latex], so the threat is also from wear in the gear. For the meshing gearset wear controls.

Question 14.5: A 17-tooth 20◦ normal pitch-angle helical pinion with a righ...

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