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Chapter 9

Q. 9-1

A pair of spur gears with a 20° pressure angle, full depth, involute teeth transmit 30 hp. The pinion is mounted on a jack shaft assembly that is directly coupled to the shaft of the electric motor operating at 600 rpm as shown in Figure 9–2(a) and (b). The pinion has 36 teeth and a diametral pitch of 5. The gear has 60 teeth and is mounted on a jack shaft assembly that is directly coupled to the output roller.
The output roller is part of the driven machinery. Compute the following:
a. The velocity ratio and the gear ratio for the gear pair.
b. The rotational speed of the gear.
c. The pitch diameter of the pinion and the gear.
d. The center distance of the shafts carrying the pinion and the gear.
e. The pitch line speed for both the pinion and the gear.
f. The torque on the pinion shaft and gear shaft.
g. The tangential force and on the teeth of each gear.
h. The radial force acting on the teeth of each gear.
i. The normal force acting on the teeth of each gear.


Verified Solution

a. From Equation (9–3) (VR=\frac {ω_P}{ω_G}=\frac {n_P}{n_G}=\frac {R_G}{R_P}=\frac {D_G}{D_P}=\frac {N_G}{N_P}), with the pinion as the driver and the gear being driven, the velocity ratio is:
Velocity ratio = VR=\frac {ω_p}{ω_g}=\frac {N_g}{N_p}= \frac {60}{36}= 1.667

c. The pitch diameters of the pinion and gear are:
D_p = \frac{N_p}{P_d}= \frac{36}{5} = 7.200 in     D_g = \frac{N_g}{P_d}= \frac{60}{5} = 12.000 in
d. The center distance for the pinion and gear:

CD=\frac{D_p}{D_g}=\frac{7.200 in + 12.000 in}{2}= 9.600 in

e. The pitch line speed can be solved using either the pinion or the gear:

\begin{array}{l}v_{t}=\frac{D}{2} \cdot \omega \\\\v_{t p}=\frac{D_{p}}{2} \cdot \omega_{p}=\frac{7.200 \mathrm{in}}{2} \cdot \frac{600 \mathrm{rev}}{\mathrm{min}} \cdot \frac{2 \pi \mathrm{rad}}{1 \mathrm{rev}} \cdot \frac{1 \mathrm{ft}}{12 \mathrm{in}}=1130 \mathrm{fpm} \\\\v_{\mathrm{tg}}=\frac{D_{g}}{2} \cdot \omega_{g}=\frac{12.000 \mathrm{in}}{2} \cdot \frac{360\mathrm{rev}}{\mathrm{min}} \cdot \frac{2 \pi \mathrm{rad}}{1 \mathrm{rev}} \cdot \frac{1 \mathrm{ft}}{12 \mathrm{in}}=1130 \mathrm{fpm}\end{array}
f. The torque on the pinion shaft and gear shaft can be determined by solving the power equation in terms of torque
\begin{array}{l}T=\frac{P}{\omega}\\\\T_{p}=\frac{P}{\omega_{p}}=\frac{30 \mathrm{hp}}{600 \mathrm{rpm}} \cdot \frac{33000 \frac{\mathrm{lb} \cdot \mathrm{ft}}{\mathrm{min}}}{1 \mathrm{hp}} \cdot \frac{1 \mathrm{rev}}{2 \pi \mathrm{rad}} \cdot \frac{12 \mathrm{in}}{1 \mathrm{ft}}=3151.3 \mathrm{lb} \cdot \mathrm{in} \\\\T_{g}=\frac{P}{\omega_{g}}=\frac{30 \mathrm{hp}}{360 \mathrm{rpm}} \cdot \frac{33000 \frac{\mathrm{b} \cdot \mathrm{ft}}{\mathrm{min}}}{1 \mathrm{hp}} \cdot \frac{1 \mathrm{rev}}{2 \pi \mathrm{rad}} \cdot \frac{12 \mathrm{in}}{1 \mathrm{ft}}=5252.1 \mathrm{lb} \cdot \mathrm{in}\end{array}
You will notice the ratio of the torque on the gear shaft to the torque on the pinion shaft is the same as the gear ratio: The power through each shaft is 30 \mathrm{hp}. If the rotational speed of the shaft decreases, the torque will increase proportionally keeping the power constant.

m_g=\frac{T_g}{T_p}= \frac{5252.1}{3151.3}= 1.67

g. The tangential force is solved using the pinion torque and the pitch circle radius or the gear torque and gear pitch circle radius. This is the drive force the pinion tooth applies to the gear tooth and equal and opposite reaction force the gear tooth applies to the pinion tooth.

W_t=\frac{T_p}{(\frac{D_p}{2})}=\frac{3151.3 lb . in}{(\frac{7.200 in}{2})} = 875.4 lb

W_t=\frac{T_g}{(\frac{D_g}{2})}=\frac{5252.1 lb . in}{(\frac{12.000 in}{2})} = 875.4 lb

h. The radial force is calculated using the tangential force and the pressure angle. As you can see the direction of the force is toward the centerline of the shaft and tends to separate the gear set.

W_t=W_t .tan(\varphi)=875.4 lb . tan(20°) = 318.6 lb

i. The normal force to the tooth profile is solved using Pythagorean theorem. This force is along the line of action and is normal to the tooth profile as shown in Figure 9–4.

W_n= \sqrt{(W_t)^2+(W_r)^2}= \sqrt{(875.4 lb)^2+(318.6 lb)^2}= 931.5 lb