Question 11.6: Power Transmitted by a Gear Based on Bending Strength and Us......

We have Y =0.402 for 25 teeth (Table 11.2) and \sigma _o =172 MPa (Table 11.3). The pitch diameter is d=m N=2(25)=50  mm \text { and } V=\pi d n=\pi(0.05)(15)=2.356  m / s =463.7  fpm

a. Using Equation (11.33) with 1/P=m, we have

F_b=\frac{\sigma_o b}{K_f} \frac{Y}{P}          (11.33)

F_b=\frac{\sigma_o b Y m}{K_f}=\frac{1}{1.5}(172 \times 0.402 \times 2)=4.149  kN

b. From Equation (11.24a), the dynamic load is

F_d=\frac{600+V}{600} F_t \quad(\text { for } 0<V \leq 2000  fpm )          (11.24a)

F_d=\frac{600+463.7}{600} F_t=1.77 F_t

The limiting value of the transmitted load, applying Equation (11.34), is

F_b \geq F_d       (11.34)

4.149=1.77 F_t \quad \text { or } \quad F_t=2.344  kN

The corresponding gear power, by Equation (1.15), is

kW =\frac{F V}{1000}=\frac{T n}{9549}        (1.15)

\begin{aligned} kW & =\frac{F_t \pi d n}{60} \\ & =\frac{(2.344) \pi(0.05) 900}{60}=5.52 \end{aligned}