Question 11.7: Design of a Speed Reducer for Bending Strength by the AGMA M...

The pinion pitch diameter and number of teeth of the gear are

d_p=\frac{N_p}{P}=\frac{18}{10}=1.8  \text { in., } \quad N_g=N_p\left(\frac{1}{r_s}\right)=18(2)=36

The pitch-line velocity, using Equation 11.20, is

V=\frac{\pi d_p n_p}{12}=\frac{\pi(1.8)(1600)}{12}=754  fpm

The allowable bending stress is estimated from Equation 11.36:

\sigma_{\text {all }}=\frac{S_{ t } K_L}{K_T K_R}                 (a)

where

S_{t} = 41.5 ksi (from Table 11.6, for average strength)

K_{L} = 1.0 (from Table 11.7, for indefinite life)

K_{T} = 1 (oil temperature should be <160°F)

K_{R} = 1.25 (by Table 11.8, for 99.9% reliability)

Carrying the foregoing values into Equation (a) results in

\sigma_{\text {all }}=\frac{41.5(1)}{1(1.25)}=33.2  ksi

The maximum allowable transmitted load is now obtained, from Equation 11.35 by setting \sigma _{all} = \sigma, as

\sigma=F_t K_o K_{\upsilon} \frac{P}{b} \frac{K_s K_m}{J}       (11.35)

F_t=\frac{33,200}{K_o K_{\upsilon}} \frac{b}{P} \frac{J}{K_s K_m}         (b)

In the foregoing, we have

P = 10

b = 1.5 in.

K_{\upsilon} = 1.55 (from curve C of Figure 11.15)

J = 0.235 (from Figure 11.16a, the load acts at the tip of the tooth, N_{p} = 18)

K_{o} = 1.75 (by Table 11.4)

K_{s} = 1.0 (for standard gears)

K_{m} = 1.6 (from Table 11.5)

Equation (b) yields

F_t=\frac{33,200(1.5)(0.235)}{(1.75)(1.55)(10)(1.0)(1.6)}=270  lb

The allowable power is then, by Equation 11.22,

F_t=\frac{33,000 hp }{V}     (11.22)

hp =\frac{F_{ t } V}{33,000}=\frac{270(754)}{33,000}=6.2