Question 14.4: A 17-tooth 20° pressure angle spur pinion rotates at 1800 re...

There will be many terms to obtain so use Figs. 14–17 and 14–18 as guides to what is needed.

d_{P}=N_{P} / P_{d}=17 / 10=1.7 \text { in } \quad d_{G}=52 / 10=5.2 \text { in }

 

V=\frac{\pi d_{P} n_{P}}{12}=\frac{\pi(1.7) 1800}{12}=801.1 ft / min

 

W^{t}=\frac{33000 H}{V}=\frac{33000(4)}{801.1}=164.8 lbf

 

Assuming uniform loading, K_{o}=1. To evaluate K_{v}, from Eq. (14–28) with a quality number O_{n}=6,

 

\begin{array}{l}A=50+56(1-B) \\B=0.25\left(12-Q_{v}\right)^{2 / 3}\end{array} (14–28)

 

\begin{array}{l}B=0.25(12-6)^{2 / 3}=0.8255 \\A=50+56(1-0.8255)=59.77\end{array}

 

Then from Eq. (14–27) the dynamic factor is

 

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{801.1}}{59.77}\right)^{0.8255}=1.377

 

To determine the size factor, Ks , the Lewis form factor is needed. From Table 14–2, with N_{P}=17 \text { teeth } teeth, Y_{P}=0.303. Interpolation for the gear with N_{G}=52 teeth yields Y_{G}=0.412. Thus from Eq. (a) of Sec. 14–10, with F = 1.5 in,

 

 

\sigma=\frac{M}{I / c}=\frac{6 W^{t} l}{F t^{2}} (a)

 

\left(K_{s}\right)_{P}=1.192\left(\frac{1.5 \sqrt{0.303}}{10}\right)^{0.0535}=1.043

 

\left(K_{s}\right)_{G}=1.192\left(\frac{1.5 \sqrt{0.412}}{10}\right)^{0.0535}=1.052

 

The load distribution factor K_{m} is determined from Eq. (14–30), where five terms are needed. They are, where F = 1.5 in when needed:

 

K_{m}=C_{m f}=1+C_{m c}\left(C_{p f} C_{p m}+C_{m a} C_{e}\right) (14–30)

 

Uncrowned, Eq. (14–30): C_{m c}=1,

 

K_{m}=C_{m f}=1+C_{m c}\left(C_{p f} C_{p m}+C_{m a} C_{e}\right) (14–30)

 

Eq. (14–32): C_{p f}=1.5 /[10(1.7)]-0.0375+0.0125(1.5)=0.0695

 

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)

 

Bearings immediately adjacent, Eq. (14–33): C_{p m}=1

 

C_{p m}=\left\{\begin{array}{ll}1 & \text { for straddle-mounted pinion with } S_{1} / S<0.175 \\1.1 & \text { for straddle-mounted pinion with } S_{1} / S \geq 0.175\end{array}\right. (14–33)

 

Commercial enclosed gear units (Fig. 14–11): C_{m a}=0.15

Eq. (14–35): C_{e}=1

 

C_{e}=\left\{\begin{array}{cl}0.8 & \text { for gearing adjusted at assembly, or compatibility } \\1 & \text { is improved by lapping, or both }\end{array}\right. (14–35)

 

Thus,

 

K_{m}=1+C_{m c}\left(C_{p f} C_{p m}+C_{m a} C_{e}\right)=1+(1)[0.0695(1)+0.15(1)]=1.22

 

Assuming constant thickness gears, the rim-thickness factor K_{B}=1. The speed ratio is m_{G}=N_{G} / N_{P}=52 / 17=3.059. The load cycle factors given in the problem statement, with N \text { (pinion) }=10^{8} cycles and N(\text { gear })=10^{8} / m_{G}=10^{8} / 3.059 cycles, are

 

\left(Y_{N}\right)_{P}=1.3558\left(10^{8}\right)^{-0.0178}=0.977

 

\left(Y_{N}\right)_{G}=1.3558\left(10^{8} / 3.059\right)^{-0.0178}=0.996

 

From Table 14.10, with a reliability of 0.9, K_{R}=0.85. From Fig. 14–18, the temperature and surface condition factors are K_{T}=1 \text { and } C_{f}=1. From Eq. (14–23), with m_{N}=1 for spur gears,

 

 

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{\cos 20^{\circ} \sin 20^{\circ}}{2} \frac{3.059}{3.059+1}=0.121

 

From Table 14–8, C_{p}=2300 \sqrt{ ps }.

 

 

Next, we need the terms for the gear endurance strength equations. From Table 14–3, for grade 1 steel with H_{B P}=240 \text { and } H_{B G}=200, we use Fig. 14–2, which gives

 

 

Notes: See ANSI/AGMA 2001-D04 for references cited in notes 1–7.

{ }^{1} \text { Hardness } to be equivalent to that at the root diameter in the center of the tooth space and face width.

{ }^{2} \text { See } tables 7 through 10 for major metallurgical factors for each stress grade of steel gears.

{ }^{3} \text { The } steel selected must be compatible with the heat treatment process selected and hardness required.

{ }^{3} \text { The } allowable stress numbers indicated may be used with the case depths prescribed in 16.1.

{ }^{5} See figure 12 for type A and type B hardness patterns.

{ }^{6} \text { If } bainite and microcracks are limited to grade 3 levels, 70 000 psi may be used.

{ }^{3} \text { The } overload capacity of nitrided gears is low. Since the shape of the effective S-N curve is flat, the sensitivity to shock should be investigated before proceeding with the design. [7]

*Tables 8 and 9 of ANSI/AGMA 2001-D04 are comprehensive tabulations of the major metallurgical factors affecting S_{t} \text { and } S_{c} of flame-hardened and induction-hardened (Table 8) and carburized and hardened (Table  9) steel gears.4

Similarly, from Table 14–6, we use Fig. 14–5, which gives

 

 

Notes: See ANSI/AGMA 2001-D04 for references cited in notes 1–5.

{ }^{1} \text { Hardness } to be equivalent to that at the start of active profile in the center of the face width.

{ }^{2} \text { See } Tables 7 through 10 for major metallurgical factors for each stress grade of steel gears.

{ }^{3} \text { The } steel selected must be compatible with the heat treatment process selected and hardness required.

{ }^{4} \text { These } materials must be annealed or normalized as a minimum.

{ }^{5} \text { The } allowable stress numbers indicated may be used with the case depths prescribed in 16.1.

*Table 9 of ANSI/AGMA 2001-D04 is a comprehensive tabulation of the major metallurgical factors affecting S_{t} \text { and } S_{c} of carburized and hardened steel gears.

 

\left(S_{c}\right)_{P}=322(240)+29100=106400 psi

 

\left(S_{c}\right)_{G}=322(200)+29100=93500 psi

 

From Fig. 14–15,

 

\left(Z_{N}\right)_{P}=1.4488\left(10^{8}\right)^{-0.023}=0.948

 

\left(Z_{N}\right)_{G}=1.4488\left(10^{8} / 3.059\right)^{-0.023}=0.973

 

For the hardness ratio factor C_{H}, the hardness ratio is H_{B P} / H_{B G}=240 / 200=1.2. Then, from Sec. 14–12,

 

\begin{aligned}A^{\prime} &=8.98\left(10^{-3}\right)\left(H_{B P} / H_{B G}\right)-8.29\left(10^{-3}\right) \\&=8.98\left(10^{-3}\right)(1.2)-8.29\left(10^{-3}\right)=0.00249\end{aligned}

 

Thus, from Eq. (14–36),

 

C_{H}=1.0+A^{\prime}\left(m_{G}-1.0\right) (14–36)

 

C_{H}=1+0.00249(3.059-1)=1.005

 

(a) Pinion tooth bending. Substituting the appropriate terms for the pinion into Eq. (14–15) 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_{d}}{F} \frac{K_{m} K_{B}}{J}\right)_{P}=164.8(1) 1.377(1.043) \frac{10}{1.5} \frac{1.22(1)}{0.30} \\&=6417 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)]}{6417}=5.62

 

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

 

(\sigma)_{G}=164.8(1) 1.377(1.052) \frac{10}{1.5} \frac{1.22(1)}{0.40}=4854 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)]}{4854}=6.82

 

(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[164.8(1) 1.377(1.043) \frac{1.22}{1.7(1.5)} \frac{1}{0.121}\right]^{1 / 2}=70360 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)]}{70360}=1.69

 

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} 70360=70660 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)]}{70660}=1.52

 

(c) For the pinion, we compare \left(S_{F}\right) P \text { with }\left(S_{H}\right)_{P}^{2} , or 5.73 with 1.69^{2}=2.86, so the threat in the pinion is from wear. For the gear, we compare \left(S_{F}\right)_{G} \text { with }\left(S_{H}\right)_{G}^{2}, or 6.96 with 1.52^{2}=2.31, so the threat in the gear is also from wear.