Question 12.3: Determining the Tooth Loads of a Bevel Gearset A set of 20° ......
Determining the Tooth Loads of a Bevel Gearset
A set of 20° pressure angle straight bevel gears is to be used to transmit 20 hp from a pinion operating at 500 rpm to a gear mounted on a shaft that intersects the shaft at an angle of 90° (Figure 12.12a). Calculate:
a. The pitch angles and average radii for the gears.
b. The forces on the gears.
c. The torque produced about the gear shaft axis.
a. Equation (12.13) gives
r_s=\frac{\omega_g}{\omega_p}=\frac{N_p}{N_g}=\frac{d_p}{d_g}=\tan \alpha_p=\cot \alpha_g (12.13)
r_s=\frac{200}{500}=\frac{1}{2.5} \quad \text { or } \quad d_g=2.5 d_p=25 in .
\alpha_p=\tan ^{-1}\left(\frac{1}{2.5}\right)=21.8^{\circ} \quad \text { and } \quad \alpha_g=90^{\circ}-\alpha_p=68.2^{\circ}
Hence,
r_{g, \text { avg }}=r_g-\frac{b}{2} \sin \alpha_g=12.5-(1) \sin 68.2^{\circ}=11.6 \text { in. }
r_{p, \text { avg }}=r_p-\frac{b}{2} \sin \alpha_p=5-(1) \sin 21.8^{\circ}=4.6 in .
b. Through the use of Equation (11.22),
F_t=\frac{33,000 hp }{V} (11.22)
F_t=\frac{33,000 hp }{\pi d_{p, avg } n_p / 12}=\frac{33,000(20)(12)}{\pi(9.2) 500}=548 lb
From Equations (12.17), the pinion forces are
\begin{aligned} & F_a=F_t \tan \phi \sin \alpha \\ & F_r=F_t \tan \phi \cos \alpha \end{aligned} (12.17)
F_a=F_t \tan \phi \sin \alpha_p=548\left(\tan 20^{\circ}\right)\left(\sin 21.8^{\circ}\right)=74 lb
F_r=F_t \tan \phi \cos \alpha_p=548\left(\tan 20^{\circ}\right)\left(\cos 21.8^{\circ}\right)=185 lb
As shown in Figure 12.12(b), the pinion thrust force equals the gear radial force, and the pinion radial force equals the gear thrust force.
c. Torque, T=F_t\left(d_g / 2\right)=548(12.5)=6.85 kips \cdot \text { in } .