Question 11.8: Design of a Speed Reducer for Wear by the AGMA Method Determ......

Allowable contact stress is estimated from Equation (11.44) as

\sigma_{c, \text { all }}=\frac{S_C C_L C_H}{K_T K_R}         (11.44)

\sigma_{c, \text { all }}=\frac{S_c C_L C_H}{K_T K_R}       (a)

In the preceding,

S_c =127.5 ksi (from Table 11.11, for average strength)

C_L =1.0 (from Figure 11.19, for indefinite life)

C_H=1.0+A\left(\frac{N_g}{N_p}-1.0\right)=1.0    (by Equation 11.46 )

C_H=1.0+A\left(\frac{N_g}{N_p}-1.0\right) \quad\left(\text { for } \frac{H_{B p}}{H_{B g}}<1.70\right)         (11.46)

K_T =1.0 and K_R =1.25 (both from Example 11.7)

Hence,

\sigma_{c, \text { all }}=\frac{127,500(1.0)(1.0)}{(1.0)(1.25)}=102  ksi

The maximum allowable transmitted load is now determined, from Equation (11.42), setting \sigma_{c, \text { all }}=\sigma_c, \text { as }

\sigma_c=C_p\left(F_t K_o K_{ v } \frac{K_s}{b d} \frac{K_m C_f}{I}\right)^{1 / 2}      (11.42)

F_t=\left(\frac{102,000}{C_p}\right)^2 \frac{1.0}{K_o K_{ v }} \frac{b d}{K_s} \frac{I}{K_m C_f}

where

C_p=2300 \sqrt{ psi } (by Table 11.10)

b=1.5 \text { in., } d_p=1.8  in \text {. }

K_{ \upsilon }=1.55, K_o=1.75, K_s=1, K_m=1.6   (all from Example 11.7)

C_f = 1.0 (for smooth surface finish)

I=\frac{\sin \phi \cos \phi}{2 m_N} \frac{m_G}{m_G+1}=0.107       (using Equation 11.43b)

I=\frac{\sin \phi \cos \phi}{2 m_N} \frac{m_G}{m_G+1}          (11.43b)

Therefore,

F_t=\left(\frac{102,000}{2300}\right)^2 \frac{1.0}{(1.75)(1.55)} \frac{1.5(1.8)}{1.0} \frac{0.107}{1.6(1.0)}=131  lb

This value applies to both mating gear tooth surfaces. The corresponding power, using Equation (11.22) with V=754 fpm (from Example 11.7), is

F_t=\frac{33,000 hp }{V}        (11.22)

hp =\frac{F_t V}{33,000}=\frac{131(754)}{33,000}=2.99