Question 14.8: Design a 4:1 spur-gear reduction for a 100-hp, three-phase s...

Make the a priori decisions:

Function: 100 hp, 1120 rev/min, R=0.95, N=10^{9} \text { cycles, } K_{o}=1

• Design factor for unquantifiable exingencies: n_{d}=2

• Tooth system: \phi_{n}=20^{\circ}

• Tooth count: N_{P}=18 \text { teeth, } N_{G}=72 teeth (no interference)

• Quality number: Q_{v}=6, use grade 1 material

• Assume m_{B} \geq 1.2 \text { in Eq. (14-40), } K_{B}=1

Pitch: Select a trial diametral pitch of P_{d}=4 teeth/in. Thus, d_{p}=18 / 4=4.5 in and d_{G}=72 / 4=18 in. From Table 14–2, Y_{P}=0.309, Y_{G}=0.4324 (interpolated). From Fig. 14–6, J_{P}=0.32, J_{G}=0.415

 

 

 

V=\frac{\pi d_{P} n_{P}}{12}=\frac{\pi(4.5) 1120}{12}=1319 ft / min

 

W^{t}=\frac{33000 H}{V}=\frac{33000(100)}{1319}=2502 lbf

 

From Eqs. (14–28) and (14–27),

 

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

 

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)

 

B=0.25\left(12-Q_{v}\right)^{2 / 3}=0.25(12-6)^{2 / 3}=0.8255

 

A=50+56(1-0.8255)=59.77

 

K_{v}=\left(\frac{59.77+\sqrt{1319}}{59.77}\right)^{0.8255}=1.480

 

From Eq. (14–38), K_{R}=0.658-0.0759 \ln (1-0.95)=0.885. From Fig. 14–14,

 

K_{R}=\left\{\begin{array}{ll}0.658-0.0759 \ln (1-R) & 0.5<R<0.99 \\0.50-0.109 \ln (1-R) & 0.99 \leq R \leq 0.9999\end{array}\right. (14–38)

 

\left(Y_{N}\right)_{p}=1.3558\left(10^{9}\right)^{-0.0178}=0.938

 

\left(Y_{N}\right)_{G}=1.3558\left(10^{9} / 4\right)^{-0.0178}=0.961

 

From Fig. 14–15,

 

\left(Z_{N}\right)_{P}=1.4488\left(10^{9}\right)^{-0.023}=0.900

 

\left(Z_{N}\right)_{G}=1.4488\left(10^{9} / 4\right)^{-0.023}=0.929

 

From the recommendation after Eq. (14–8), 3p ≤ F ≤ 5p. Try F = 4p = 4π/P = 4π/4 = 3.14 in. From Eq. (a), Sec. 14–10,

 

\sigma=\frac{K_{v} W^{t}}{F m Y} (14–8)

 

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

 

K_{s}=1.192\left(\frac{F \sqrt{Y}}{P}\right)^{0.0535}=1.192\left(\frac{3.14 \sqrt{0.309}}{4}\right)^{0.0535}=1.140

 

From Eqs. (14–31), (14–33) and (14–35), C_{m c}=C_{p m}=C_{e}=1. From Fig. 14–11, C_{m a}=0.175 for commercial enclosed gear units. From Eq. (14–32), F /\left(10 d_{P}\right)=3.14 / [10(4.5)] = 0.0698. Thus,

 

C_{m c}=\left\{\begin{array}{ll}1 & \text { for uncrowned teeth } \\0.8 & \text { for crowned teeth }\end{array}\right. (14–31)

 

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)

 

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

 

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)

 

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)[0.0715(1)+0.175(1)]=1.247

 

From Table 14–8, for steel gears, C_{p}=2300 \sqrt{ psi }. From Eq. (14–23), with m_{G}=4 and m_{N}=1

 

 

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{4}{4+1}=0.1286

 

Pinion tooth bending. With the above estimates of K_{s} \text { and } K_{m} from the trial diametral pitch, we check to see if the mesh width F is controlled by bending or wear considerations. Equating Eqs. (14–15) and (14–17), substituting n_{d} W^{t} \text { for } W^{t} , and solving for the face width (F)_{\text {bend }} necessary to resist bending fatigue, we obtain

 

\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)

 

\sigma_{\text {all }}=\left\{\begin{array}{ll}\frac{S_{t}}{S_{F}} \frac{Y_{N}}{K_{T} K_{R}} & \text { (U.S. customary units) } \\\frac{S_{t}}{S_{F}} \frac{Y_{N}}{Y_{\theta} Y_{Z}} & \text { (SI units) }\end{array}\right. (14–17)

 

(F)_{\text {bend }}=n_{d} W^{t} K_{o} K_{v} K_{s} P_{d} \frac{K_{m} K_{B}}{J_{P}} \frac{K_{T} K_{R}}{S_{t} Y_{N}} (1)

 

Equating Eqs. (14–16) and (14–18), substituting n_{d} W^{t} \text { for } W^{t}, and solving for the face width (F)_{\text {wear }} necessary to resist wear fatigue, we obtain

 

\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)

 

\sigma_{c, \text { all }}=\left\{\begin{array}{ll}\frac{S_{c}}{S_{H}} \frac{Z_{N} C_{H}}{K_{T} K_{R}} & \text { (U.S. customary units) } \\\frac{S_{c}}{S_{H}} \frac{Z_{N} Z_{W}}{Y_{\theta} Y_{Z}} & \text { (SI units) }\end{array}\right. (14–18)

 

(F)_{\text {wear }}=\left(\frac{C_{p} Z_{N}}{S_{c} K_{T} K_{R}}\right)^{2} n_{d} W^{t} K_{o} K_{v} K_{s} \frac{K_{m} C_{f}}{d_{P} I} (2)

 

From Table 14–5 the hardness range of Nitralloy 135M is Rockwell C32–36 (302–335 Brinell). Choosing a midrange hardness as attainable, using 320 Brinell. From Fig. 14–4,

 

 

S_{t}=86.2(320)+12730=40310 psi

 

Inserting the numerical value of S_{t} in Eq. (1) to estimate the face width gives

 

(F)_{\text {bend }}=2(2502)(1) 1.48(1.14) 4 \frac{1.247(1)(1) 0.885}{0.32(40310) 0.938}=3.08 \text { in }

 

From Table 14–6 for Nitralloy 135M, S_{c}=170000 psi. Inserting this in Eq. (2), we find

 

 

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.

 

(F)_{\text {wear }}=\left(\frac{2300(0.900)}{170000(1) 0.885}\right)^{2} 2(2502) 1(1.48) 1.14 \frac{1.247(1)}{4.5(0.1286)}=3.44 \text { in }

 

Make face width 3.50 in. Correct K_{s} \text { and } K_{m}:

 

K_{s}=1.192\left(\frac{3.50 \sqrt{0.309}}{4}\right)^{0.0535}=1.147

 

\frac{F}{10 d_{p}}=\frac{3.50}{10(4.5)}=0.0778

 

C_{p f}=0.0778-0.0375+0.0125(3.50)=0.0841

 

K_{m}=1+(1)[0.0841(1)+0.175(1)]=1.259

 

The bending stress induced by W^{t} in bending, from Eq. (14–15), is

 

(\sigma)_{P}=2502(1) 1.48(1.147) \frac{4}{3.50} \frac{1.259(1)}{0.32}=19100 psi

 

The AGMA factor of safety in bending of the pinion, from Eq. (14–41), is

 

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}=\frac{40310(0.938) /[1(0.885)]}{19100}=2.24

 

Gear tooth bending. Use cast gear blank because of the 18-in pitch diameter. Use the same material, heat treatment, and nitriding. The load-induced bending stress is in the ratio of J_{P} / J_{G}. Then

 

(\sigma)_{G}=19100 \frac{0.32}{0.415}=14730 psi

 

The factor of safety of the gear in bending is

 

\left(S_{F}\right)_{G}=\frac{40310(0.961) /[1(0.885)]}{14730}=2.97

 

Pinion tooth wear. The contact stress, given by Eq. (14–16), is

 

\left(\sigma_{c}\right)_{P}=2300\left[2502(1) 1.48(1.147) \frac{1.259}{4.5(3.5)} \frac{1}{0.129}\right]^{1 / 2}=118000 psi

 

The factor of safety from Eq. (14–42), is

 

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}=\frac{170000(0.900) /[1(0.885)]}{118000}=1.465

 

By our definition of factor of safety, pinion bending is \left(S_{F}\right)_{P}=2.24, and wear is \left(S_{H}\right)_{P}^{2}=(1.465)^{2}=2.15.

Gear tooth wear. The hardness of the gear and pinion are the same. Thus, from Fig. 14–12, C_{H}=1, the contact stress on the gear is the same as the pinion, \left(\sigma_{c}\right)_{G}= 118 000 psi. The wear strength is also the same, S_{c}=170000 psi. The factor of safety of the gear in wear is

 

\left(S_{H}\right)_{G}=\frac{170000(0.929) /[1(0.885)]}{118000}=1.51

 

So, for the gear in bending, \left(S_{F}\right)_{G}=2.97, and wear \left(S_{H}\right)_{G}^{2}=(1.51)^{2}=2.29.

Rim. Keep m_{B} \geq 1.2. The whole depth is h_{t}=\text { addendum }+\text { dedendum }=1 / P_{d}+ 1.25 / P_{d}=2.25 / \bar{P}_{d}=2.25 / 4=0.5625 in. The rim thickness t_{R} is

 

t_{R} \geq m_{B} h_{t}=1.2(0.5625)=0.675 \text { in }

 

In the design of the gear blank, be sure the rim thickness exceeds 0.675 in; if it does not, review and modify this mesh design.