Question

A pair of spur gears with 20 degree pressure angle, full-depth, involute teeth transmits 65 hp. The pinion is mounted on a shaft connected to a 4 cylinder diesel engine that operates at 1250 rpm in the clockwise direction. The pinion has 26 teeth and a diameteral pitch of 6. The gear has 48 teeth and drives a concrete mixer. The gears are manufactured to an AGMA quality number of A11. The face width of the gears is 3 inches and are machined from solid bar stock (blanks). The gears are enclosed in a commercial gear box.
1. What is the pitch line speed of the gear train in feet per minute?
2. What is the center distance of the gear pair in inches?
3. What is the torque on the pinion in Lbin?
4. What is the torque on the gear in lbin?
5. What is the output horsepower of the gearbox?
6. What is the tangential force acting on the gear teeth (Wt) in pounds?
7. What is the radial force acting on the gear teeth in pounds?
8. What is the normal force acting on the gear teeth in pounds?
9. What is the bending stress for the pinion teeth (St) in PSI ?

Answer 1

Answer:

See explaination

Explanation:

a) The pitch circle diameter of pinion in inches is given by

Dp=Np/P

Where

Np= No. of teeth in pinion = 26

P =diametral pitch= 6

Hence

Dp= 26/6 = 4.333 in

Pinion angular speed\omega _{p} =1250 rpm = 130.9 rad/s

Therefore pitch line speed

V=Dp/2\omega _{p} = 4.333/2x130.9

= 283.62 in/s

V= 23.63 ft/s

V=1418 ft/min

b) The pitch circle diameter og gear is given by

Dg= Ng/P= 48/6 = 8 in

The center distance is given by

C=(Dp+Dg)/2

= (4.333+8)/2

C= 6.167 in

c) The torque on the pinion is given by

Tp= P/\omega _{p}

Where

P = transmitted power, =65 hp = 65x550= 35750 lt-lb/s

Tp= 35750/130.9

= 273.1 ft-lb

d) Speed ratio is given as

R=Ng/Np= 48/26 = 1.8461

Hence speed of gear is

\omega _{g}=\frac{\omega _{p}}{R}

= 130.9/1.8461

= 70.9 rad/s

Therefore torque on gear is given as

Tg= P/\omega _{g} = 35750/70.9

= 504.2 ft-lb

e) Assuming transmission eficiency of 100%

Output hp=input hp= 65 hp

f) Tangential force on gear teeth is given by

Fgt= Tg/(Dg/2)

= 504.2x2/8

= 126.05 lb

g) Radial force on ger teeth is given as

Fgr= Fgt tan\phi

Where

\phi is pressure angle = 200

Hence

Fgr= 126.05tan200

= 45.88 lb

h) The normal force on gear teeth is given as

F=Fgt/cos\phi

= 126.05/cos200

= 134.14 lb