This set of Farm Machinery Multiple Choice Questions on “Mechanical Power Transmission – Gear Drive”.
1. A single reduction gear of 120 KW with a pinion of 150 mm pitch circle diameter and speed 550 rpm is supported in bearings on either side. Calculate the total load due to power transmission.
a) 28.20 KN
b) 35.21 KN
c) 40 KN
d) 45 KN
Answer: a
Explanation: P=(frac{2πNT}{60} )
T=(frac{120*1000*60}{2*π*550}) = 2083.48 Nm
Tangential load on pinion Ft = (frac{T}{r}=frac{2083.48}{0.075})=27779.73 N
Total load due to power transmitted = (frac{Ft}{cosθ} = frac{27779.73}{cos10}) = 28.20 KN.
2. The number of teeth in driving spur gear is 60. The circular pitch of the teeth is 15 mm. the tangential force transmitted is 140 kg. Calculate horsepower transmitted to the driven gear at 180 rpm of driving gear.
a) 2.5
b) 5.04
c) 6
d) 7.55
Answer: b
Explanation: Pitch=(frac{πD}{T} )
D1=(frac{PT}{π}=frac{0.015*60}{π}) = 0.287 m
Peripheral speed, v=( frac{πDN}{60} = frac{(π*0.287*18)}{60}) = 2.70 m/s
H.P = ( frac{F*v}{75}=frac{140*2.70}{75} )
H.P = 5.04.
3. In a simple planetary gear drive, the sun gear is used to give drive a plant carrier mounted with 3 planetary gears, while ring gear is held stationary. If sun gear and all other planetary gears have 20 teeth each and ring gear has 60 teeth, the gear ratio will be __________
a) 4:1
b) 4.5:1
c) 5:1
d) 5.5:1
Answer: a
Explanation: (frac{ns}{nc}=1+frac{60}{20})=4
Gear ratio = 4:1.
4. To obtain 200 kg m of torque at the transmission output shaft, what gear ratio would be necessary? (Assume transmission efficiency 90% and input torque as 20 kg m)
a) 1:9
b) 1:10
c) 1:11
d) 1:12
Answer: c
Explanation: Torque of transmission input shaft = i/p torque * efficiency = 20*(frac{90}{100}) = 18 kg m
Gear ratio = (frac{18}{200}) = 1:11.
5. A tractor has 1.6 m rear wheel diameter. If the e gear ratio of final drive and differential gear is 5:1 and 4:1 respectively calculate forward speed of tractor in kilometre per hour.
a) 5.53
b) 7.48
c) 6.53
d) 8.48
Answer: a
Explanation: (frac{Ne}{Nr} ) = Gear box reduction*Gear ratio of final drive*Gear ratio of differential gear
(frac{1100}{Nr}=frac{3}{1}*frac{5}{1}*frac{4}{1} )
Nr=(frac{1100}{3*5*4}) = 18.33 rpm
Forward speed of tractor = (frac{π*1.6*18.33}{60}) = 5.53 kmph.
6. In an epicyclic gear train, the sun gear on the input shaft is 20 teeth external gear. The planet gear is a 40 teeth external gear and the ring gear is a 100 teeth internal gear. The ring gear is fixed and the sun gear is rotating at 60 rpm counter clockwise. The arm attached to output shaft will rotate at ________
a) 5 rpm
b) 10 rpm
c) 20 rpm
d) 30 rpm
Answer: b
Explanation: (frac{Ns-Na}{Nr-Na}=frac{-Tr}{Ts} )
(frac{60-Na}{0-Na}=frac{-100}{20} )
6Na = 60
Na = 10 rpm.
7. The speed reduction in 1st low gear of tractor box and differential with final drive are 5:1 and 45:1 for a tractor developing 30 KW at engine at an rpm of 2500 with overall efficiency of 85%. The torque is KN-m available at wheel axle?
a) 15
b) 21.50
c) 24.55
d) 36.50
Answer: d
Explanation: Naxle = 1500*(frac{1}{5}*frac{1}{45}) = 6.67 rpm
T=(frac{P*60}{2πN}=frac{25500*60}{2π*6.67}) = 36.50 KN-m.
8. A tractor has 8 forward speeds. The speed ratio varies in exact geometrical progressions. If the speed ratio in highest and lowest gear are 14.9:1 and 108:1, the geometrical constant will be __________
a) 0.328
b) 1.328
c) 2.328
d) 3.328