CE-461 Coning of Wheels Train Resistances Hauling Capacity
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Transcript of 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