Question 11.CS.1: 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

$V=\frac{\pi d n}{12}$      (11.20)

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

The allowable bending stress is estimated from Equation (11.36):

$\sigma_{\text {all }}=\frac{S_t K_L}{K_T K_R}$       (a)

where

$S_t=41.5  ksi \text { (from Table 11.6, for average strength) }$

$K_L=1.0 \text { (from Table 11.7, for indefinite life) }$

$K_T=1 \text { (oil temperature should be }<160^{\circ} F \text { ) }$

$K_R=1.25 \text { (by Table } 11.8 \text {, for } 99.9 \% \text { reliability) }$

Carrying the foregoing values into Equation (a) results in

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

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

$\begin{aligned} & \sigma=F_t K_o K_{ \upsilon } \frac{P}{b} \frac{K_s K_m}{J} \quad \text { (US customary units) } \\ & \sigma=F_t K_o K_{ \upsilon } \frac{1.0}{b m} \frac{K_s K_m}{J} \quad \text { (SI units) } \end{aligned}$     (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 11.16(a), 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)

$hp =\frac{F_t V}{33,000}=\frac{270(754)}{33,000}=6.2$