Question 11.9: 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}             (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)

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^{\prime}} 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 in., d_{p} = 1.8 in

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)

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

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