ce-461 coning of wheels train resistances hauling capacity

8/12/2019 CE-461 Coning of Wheels Train Resistances Hauling Capacity

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Coning of wheels, train resistances,hauling capacity

Dr. Indrajit Ghosh, Assistant Professor

Department of Civil Engineering

Indian Institute of Technology Roorkee


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Flat Surface Coned Surface


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Problems with flat wheel Lateral Sway on straight track

Wearing of flanges and side of rail head

Unequal movement on curved rails Longer distance to be moved on outer

curved rail as compared to inner curvedrail

Flexibility not available due to rigidity ofvehicle base.


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Coning causes

On a straight track Bringing back wheel to average diameter

by slipping the wheel On a curved track

Shifting the outer wheel outwards (due tocentrifugal force) thus causing an increase

in diameter that helps it in moving longerdistance on outer curve as compared toinner wheel for which the diameter reducesthus making it to traverse shorter distance.


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Coning helps in

Controlling differential movement of frontand rear axles caused due to rigidity of

frame and axle, thus acting as a balancingfactor.

On curves the rear axle has a tendency to

move towards inner rail. Reducing wear and tear of wheel flanges

Smooth riding.


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Problem with Coning of Wheels

Wear and tear due to slipping action

Slip of wheel = [2

/360] x G = angle made by rigid wheel base atcenter of the curve

BG slip = 0.029 m per degree curve


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Pressure on outer rail is more

Results in wear of outer rail

Horizontal component of centrifugalforce tend to turn rail out

Gauge has widening tendency

Sleepers under the edge of rail gotdamaged

If no base plate is used


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Rails are tilted at an angle of 1 in 20

Controls

Eliminate or minimize demerits Adzing of sleepers or use of canted

bearing plates


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Train consists of two units Locomotives

Provides power

Trailing unit

Passenger compartment/goods wagon

Traction The source through which the locomotive derives power

Sources: Steam

Diesel fuel

Electric supply (AC/DC)


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Source has an important bearing upon

Load carrying capacity

Speed

Economy Efficiency of service


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Various forces offer resistance to the movementof a train on track

As well as speed of train

Resistances may be the result of Movement of various parts

Friction between various parts

Irregularities in track profile

Atmospheric resistances to a train

When moving


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Tractive forces employed by locomotivesshould be adequate enough to overcome theseresistances

Haul train at a specified speed


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Resistance due to friction

Resistance due to wave action, trackirregularities, and speed

Resistance due to wind or atmosphericresistance

Resistance due to gradient

Curve resistance Resistance due to starting

Resistance due to acceleration


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Resistance offered by friction between internalparts of locomotives and wagons As well as between metal surfaces of rail and wheel

To a train moving at a constant speed

Independent of speed Can be broken into different categories

a. Journal friction

Friction of locomotive, wagons and compartment itself

Depends on

Bearing type

Lubricant used

Temp and Condition of bearing

For Roller Bearing, it varies between 0.5-1.0 kg per tonne


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b. Internal resistances

Consequential to movement of various parts oflocomotives and wagons

c. Rolling resistances

Due to rail-wheel interaction

On account of movement of steel wheels on steel rails

Total frictional resistance

R1= 0.0016W


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b. Internal resistances

Consequential to movement of various parts oflocomotives and wagons

c. Rolling resistances

Due to rail-wheel interaction

On account of movement of steel wheels on steel rails

Total frictional resistance

R1= 0.0016W


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Caused due to

Wave action in rail

When trains move with speed

Track irregularities Resistance due to improper maintenance of tracks

Longitudinal unevenness and differences in cross levels


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Total train resistances due to wave action,track irregularities, and speed

R2= 0.00008 WV

whereW is weight of train in tonnes and

V is speed of train in km ph.


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Trains move with speed Certain resistance develops as they have to move

against wind

Wind resistance

Side resistance

Head resistance

Tail resistance

Exact magnitude depends on Size

Shape

Speed

Wind direction and speed


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Wind resistance depends on

Exposed area of vehicle

Velocity and direction of wind

Horizontal component opposes movement of trainR3= 0.000017 AVwind

2

R3= 0.0000006 WVtrain2


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When a train has to move along a risinggradient

W = weight of train acting at CG

N = Normal pressure on rails

Computed as -

R = (W x % slope) / 100

NW

Rg

G


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Resistance due to curves Factors controlling are:

Rigidity of wheel base

Wear on inner side of outer rail due to flange of leadingaxle and inner side of inner rail due to flange of trailingaxle, causing mount on rail

In other two cases it tends to derail


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Resistance due to curves Factors controlling are:

Slippage of wheel (longitudinal and transverse) Longitudinal: Opposes forward movement

Transverse: Friction between wheel flanges and rails,Increases curve resistance

Insufficient super-elevation: increased pressure oninner rail

Extra super-elevation: greater pressure on outer rail

Poor maintenance of track and components Improper gauge, poor alignment, worn out rails, high or

low joint, etc


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Curve resistance increases with increase inspeed

G

R

D2

D1

D


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Resistance due to curves (Rc)

Therefore, Resistance gets affected by Force ofsliding friction (F), Gauge of track (G) and degreeof curvature (R)

Recommended values of curve resistances:

Broad gauge Rc= 0.0004w.D

Meter gauge Rc= 0.0003w.D

Narrow gauge Rc= 0.0002w.D


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Get induced due to- Starting operation

Acceleration given to a locomotive


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Resistance due to starting (RS)

Varies according to the type of an object

For locomotives RS= 0.15 W1

For vehicles RS= 0.005 W2where

W1is weight of locomotive in tonnes

W2is weight of a vehicle in tonnes


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Resistance due to acceleration

Caused due to change in speed with respect to time

RA= 0.028 W (V2V1) / t

WhereV1= velocity at the beginning (kmph)

V2= velocity at the end (kmph)

t = Time taken in seconds for achieving the speed from V2to V1

W = total weight of train in tonnes


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Defined as the load that can be handled by thelocomotive. It is an indicative of poweravailable to a locomotive.

It can be computed as a product of coefficientof friction and weight on the driving wheels.

At the minimum level it should be equal toTraction Resistances.


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The factors controlling the capacity are: Weight coming on the driving wheels, and

Coefficient of friction It largely depends up on:

Condition of rail surface, and

Speed of the locomotive


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Coefficient of friction - value

Condition of rail surface:

Very wet / very dry 0.25

Greasy 0.03Average dampness 0.166

In tunnels / frosty condition 0.125

With respect to speed it varies between 0.1 athigh speeds to 0.2 at low speeds


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Hauling Capacity = .w.n = .W

Where = coefficient of friction

w = weight on driving axle

n = number of pairs of driving wheelsW = Total load on driving wheels

Maximum axle load in India

BG = 28.56 tons

MG = 17.34 tons

NG = 13.26 tons


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It is usually equal to or little in excess ofhauling capacity.

Computed by equating work done by tractiveeffort to the total power developed by thelocomotive.


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For steam locomotive:

It depends up on

Difference in pressure on two sides of the

cylinder (p) Length of stroke (L)

Area of piston (a)

Diameter of piston (d)

Diameter of wheel (D)


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If Teis the mean tractive effort then,

Work done by a two cylinder engine

= 2.p.a.(2L) = .p.L.d2

Work done in one revolution of driving

wheel = . D. Te

Therefore, equating the work done

Te= p.d2.L / D

Hence, a small diameter wheel will increase thetractive effort, but it will reduce the speed ofmovement.


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For diesel locomotive

Te= 308 x HPr/ V

Where

HPris rated horse power of the engine,V = Velocity in km ph

ce-461 coning of wheels train resistances hauling capacity

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