Track Design MCQ Quiz in मराठी - Objective Question with Answer for Track Design - मोफत PDF डाउनलोड करा

Concepts:

A brief description of the fastenings used to fix rails to wooden sleepers is given below:

1. Dog Spikes:

• It has a 16 mm square cross-section.
• It  can be used on both straight and curved tracks
• It is called a dog spike because the head of this spike looks like the ear of a dog

2. Round spikes:

• It consists of an 18 mm round section.
• They have become obsolete now.

3. Screw spikes  

• Indian Railways has developed two sizes of screw spikes with diameters of 20 mm and 22 mm to be used on high-speed and main routes in order to increase the lifespan of wooden sleepers.
• Screw spikes with a diameter of 20 mm are called 'plate screws' and are used for fixing rails to sleepers with the help of bearing plates.
• Screw spikes with a diameter of 22 mm are called 'rail screws' and are used to directly fasten the rails to the wooden sleepers with or without the use of bearing plates.

Track Modulus: It is defined as the weight in kg per cm length of rail that is required to produce a depression of 1 cm in the track.

The value of track modulus varies with gauge, the type, and spacing of sleepers, nature of ballast, weight of the rail section, etc and it varies between 28 to 140.

For the purpose of calculation and in the absence of actual values of the track modulus, it may be taken as:

70 TO 84 → For Broad Gauge Track

42 to 49 → For Metre Gauge Track

Concept:

Versine of Curve (V)

V = \(\frac{{{C^2}}}{{8 \times R}}\) in meter

where, C is chord in meter, R is radius in meters

V = \(\frac{{{{100}^2}}}{{8 \times 500}}\) = 2.5 m = 250 cm

Explanation:

For BG track length of one rail = 12.8 m

Number of sleepers per rail length = 12.8 + 6 = 18.8 = 19

Number of rail length for 1024 m long track

\(= \frac{{1024}}{{12.8}} = 80\)

Concepts:

The hauling capacity of a railway locomotive is given as:

H = μ × W × N

H = Hauling power

N = number of pairs of driving wheels

W = weight exerted on the driving wheels

μ = coefficient of friction 

Calculation:

Given: μ = 0.3, W = 23 tonnes

For Locomotive 1-4-1:

Nos of driving wheels = 4

Nos. of pair of driving wheels are, N = 4/2 = 2

Now, the hauling capacity is given as:

H = 0.3 × 23 × 2 = 13.8 tonnes.

Explanation:

On a horizontal curve, the pavement is widened slightly more than the normal width. The prime objective of widening the pavement is that when a vehicle takes a turn to negotiate a horizontal curve, the rear wheels do not follow the same path as that of the front wheels. This phenomenon is known as off tracking.

Off tracking can be calculated as:

\({\rm{Off\;tracking}} = \frac{{{{\rm{l}}^2}}}{{2{\rm{R}}}}\)

Where,

Comparison of the allowable grade of railway and highway:

1. Maximum allowable graded are lower for railways than for highways due to steel wheel on steel rails have lower frictional co-efficient than rubber tyres on pavements.

2. Maximum allowable grades are lower for railways than for highways because the construction cost becomes prohibitive for the railways at high grades.

3. Higher grade in railways is avoided due to the high traction effort required and you might have seen two engines pulling the whole train on high grades.

Concept:

Note:

(i) The maximum gradient permitted in railway track is 1 in 400. The gradient is limited to prevent the downward movement of standing vehicles on the track due to effect of gravity.

(ii) In addition to above, a minimum gradient of 1 in 1000 must be provided in station yard in order to ensure efficient drainage.

Important Points

Order of different gradients in railways are as follows:

(Gradients in station yards) < (Ruling Gradient) < (Momentum Gradient) < (Pusher or helper Gradient)

Explanation:

In order to minimize the disadvantages due to coning of wheels, canting of rails is done which means that rails are not laid flat but are tilted inwards.

This reduces wear on the rails as well as on the tread of the wheel.

The slope of the base plate is 1 in 20 which is the slope of the wheel flange.

Tilting of rails can be achieved by:

• Adzing of sleepers
• Use of canted base plates

Adzing: 

To accommodate the coned wheel of wagons, the rails too should be inclined inward in 1:20 slope, which is done by raising the sleepers along ends

Concept:

Hauling capacity = f × n × w

Where, 

f = Coefficient of friction, n = Number of driving axles and w = Total load on each axle

Calculation:

Given,

Coefficient of friction = 0.4

Total load on each axle = 22.5 tonnes

2-8-2 locomotive means 2 boggie carriers, 8 central or driving or coupled wheels and 2 rear or trailing wheels.

Thus,

Number of driving axles (n) = 8/2 = 4

Hauling capacity = f × n × w = 0.4 × 4 × 22.5 = 36 tonnes