Monday, July 13, 2015

WBUT B.Tech Civil Engineering Narula transport solve Question paper 2012

University:West Bengal University of Technology
Course:  B.Tech Civil Engineering
Subject :  Narula transport solve
Year of  Question Paper : 2012




Narula Transport 7th semester solve
Group B
 Draw a typical cross section of a permanent way. Discuss in brief the basic functions of various components of a railway track.
Ans:
Cross section of a permanent way:


Basic functions of various components of a railway track:





Ballast:
To provide firm and level bed for the sleepers to rest on
To allow for maintaining correct track level without disturbing the rail road bed
To drain off the water quickly and to keep the sleepers in dry conditions
To discourage the growth of vegetation
To protect the surface of formation and to form an elastic bed
To hold the sleepers in position during the passage of trains
To transmit and distribute the loads from the sleepers to the formation
To provide lateral stability to the track as a whole

Sleeper:
The primary function of a sleeper is to grip the rail to gauge and to distribute the rail loads to ballast with acceptable induced pressure.
The side functions of a sleeper include the avoidance of both longitudinal and lateral track movement.
It also helps to enhance correct line and level of the rails.
Rail:
Rails provide a continuous and level surface for the movement of the trains with minimum friction with steel wheels of the rolling stock.
Rails provide strength, durability and lateral guidance to the track.
Rails transmit the axle load to sleepers, which transfer the same load to the underlying ballast and formation.
Rails bear the stresses developed due to heavy vertical loads, breaking forces and temperature variance.

Bearing plates:
The bearing plates are required not only to distribute the load but also to reduce the rubbing action of the rail seat on the sleeper.
These also limit the compressive stresses on the sleeper.
The friction between the bearing plate and the sleeper reduces the lateral forces on the spike and thus these help to improve the gauge holding capacity of the sleeper.

A locomotive on M.G track has three pairs of driving wheels each carrying 20 tonnes. What minimum load can it pull on level track with curvature of 2degree at 50km/h ?
Ans:


Using a sleeper density of M+5 find out the number of sleepers required for constructing a railway track of 640 meter long B.G track.
Ans:






Difference:
Airport & airfield: Airport is the landing and taking-off area for civil aircraft, usually with surfaced runways and aircraft maintenance and passenger facilities but airfield is a landing and taking-off area usually with permanent buildings.
Dry dock & jetty: Dry dock is a large dock from which water can be pumped out; used for building ships or for repairing a ship below its waterline but jetty is a structure, such as a pier, that projects into a body of water to influence the current or tide or to protect a harbor or shoreline from storms or erosion.
Taxi way & terminal area: A taxiway is a path on an airport connecting runways with  ramps, hangars, terminals and other facilities but terminal area is a control area or portion thereof normally situated at the confluence of air traffic service routes in the vicinity of one or more major airfields.
Low tide & high tide: Low tide is when the moon pulls the sea out further creating more sand space but high tide is when the moon pulls the water up further onto the beach creating less sand space.

Draw a sketch showing coning of wheels on level track. What are the advantages of coning of wheels?
Ans:
 
Advantages of coning of wheels:
Smooth riding
Help vehicle to negotiate curves smoothly
Reduces wear and tear of wheel flanges

Group C

What is the ballast in permanent way? Mention the functions of ballast and state requirement of a good ballast material.
Ans:
Ballast:
The material used as an elastic cushion between the sleeper and the top of the formation, is called ‘Ballast’.
Functions:
To provide firm and level bed for the sleepers to rest on
To allow for maintaining correct track level without disturbing the rail road bed
To drain off the water quickly and to keep the sleepers in dry conditions
To discourage the growth of vegetation
To protect the surface of formation and to form an elastic bed
To hold the sleepers in position during the passage of trains
To transmit and distribute the loads from the sleepers to the formation
To provide lateral stability to the track as a whole

Requirements:
It should resist crushing under dynamic loads.
The designed depth of the ballast should be able to distribute the weight of passing trains on the formation underneath uniformly.
It should not make the track dusty due to powder formation under dynamic wheel loads.
It should be reasonably elastic.
It should have resistance to abrasion and weathering
It should be non-porous to provide durability to the ballast.
It should hold the sleepers laterally and longitudinally under all conditions traffic, especially on the curves.
It should be able to facilitate easy drainage to rain water.

Define: sub grade, embankment, cutting, formation rock
Sub grade: The level layer of rock or earth upon which the foundation of a road or railway is laid.
Embankment: A man-made ridge of earth or stone that carries a road or railway or confines a waterway.
Cutting: The formation at a level below the natural ground is called cutting.
Formation work: The formation is an important constituent of the track as it supports the entire structure.




If the ruling gradient is in 1 in 150 on a particular section of B.G and at the same time a curve of 4 degree is situated on this ruling gradient, what should be the allowable ruling gradient?
Ans:


Disadvantages of concrete sleepers:

They are not economical because of high cost of construction.
In case of derailments, heavy damage is caused.
High standard of maintenance of track is required.
The design and construction are both complicated.
They are more rigid.
They do not have any scrap value.


Short notes

Advantages of electric traction:
Lower running cost of locomotives and multiple units.
Higher power-to-weight ratio, resulting in
          Fewer locomotives
          Faster acceleration
          Higher practical limit of power
          Higher limit of speed
 Less noise pollution (quieter operation)
 Lower power loss at higher altitudes
 Lack of dependence on crude oil as fuel







Wind resistance:

When a vehicle moves with speed a certain resistance develops as t vehicle has to move forward against the wind. Wind resistance consists of side resistance, head resistance and tail resistance but its exact magnitude depends upon the size and shape of the vehicle, its speed and the wind direction as well as velocity.

Rw = 0.000017AV2

Where,

A= Exposed area of the train in m2
V= speed of the wind

Vcos θ


θ


Direction of wind and velocity off wind [V]





Vsin θ
θ




Resistance due to starting and accelerating:
Trains face these resistances at stations when they start, accelerate and decelerate. The values of these resistances are-
                                         Resistance on starting= RS = 0.15 W1 + 0.005 W2
                                         Resistance due to acceleration = Ra = .028 aW
Where,
W1 is the weight of the locomotive in tonnes
W2 is the weight of the vehicles in tonnes
W is the total weight of the locomotive and vehicle in tonnes
a is the acceleration
W= W1+ W2
a= (V2-V1)/t
V2 is the final velocity and V1 is the initial velocity and t is the time taken.






Hauling capacity of locomotive:
It is defined as the product of the co-efficient of friction and weight on the driving wheels. The co-efficient of friction between driving wheels and the rails depends largely upon the following two factors.
Condition of rail surface
Speed of the locomotive

Hauling capacity indicates power of locomotive
Hauling capacity = f × w × n
f =  co-efficient of friction
w = Load on each axle
n = No. of pair of driving wheels
  W = n x w
W = Total load on driving wheels
Case 1. Straight track on level
Hauling capacity = Train resistance
Case 2. Curve track on level
Hauling capacity = Train resistance + Curve resistance
Case 3. Curve track on grade
Hauling capacity = Train resistance + Curve resistance + Grade resistance








Calculate the maximum permissible speed on a curve of high speed B.G track with –
Degree of curve = 1 degree
Super elevation = 8 cm
Transition curve length = 130m
Maximum allowable speed= 153km/hr

Ans:
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