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
