Civil Engineering - UPSC Civil Service Exam Questions - Discussion
Discussion Forum : UPSC Civil Service Exam Questions - Section 6 (Q.No. 10)
10.
A train is hauled by 2-8-2 locomotive with 22.5 tonnes load on each driving axle. Assuming the coefficient of rail-wheel friction to be 0.25, what would be the hauling capacity of the locomotive ?
Discussion:
4 comments Page 1 of 1.
Dhanalakshmi said:
8 years ago
Hauling capacity- f*n*w.
Where f- coefficient of friction
n- no of DRIVING AXLES
W- total load on each axle.
In 2-8-2 locomotive 1st 2 refers to leading wheels , and 8 refers to driving wheels, and 2nd 2 refers to trailing wheels. So no of driving axles are 8/2 = 4.
Hence hauling capacity= 0.25 *4 *22.5.
= 22.5 tonnes.
Where f- coefficient of friction
n- no of DRIVING AXLES
W- total load on each axle.
In 2-8-2 locomotive 1st 2 refers to leading wheels , and 8 refers to driving wheels, and 2nd 2 refers to trailing wheels. So no of driving axles are 8/2 = 4.
Hence hauling capacity= 0.25 *4 *22.5.
= 22.5 tonnes.
(1)
Ishfaq Hussain wani said:
2 months ago
Axle load = (2 + 8 + 2)/2 = 6.
Number of driving axis = 8/4 = 4,
Hauling capacity = (0.25 * 4 * 22.5) = 22.5 tonnes.
Number of driving axis = 8/4 = 4,
Hauling capacity = (0.25 * 4 * 22.5) = 22.5 tonnes.
Kusumanjali said:
5 years ago
In 8/2, that 2 indicates the driving wheel number or trailing wheel? Explain, please.
Md.Faysal Kabir said:
4 years ago
2-8-2 means 2 leading wheel, 8 driving wheel (axle no 4), 2 trailing wheel.
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