Lecture 5 Final

CE-363

Lecture – 5: Stresses in

Components of Track

Dr. Ankit Gupta, Assistant Professor

Department of Civil Engineering

National Institute of Technology Hamirpur

Lecture Outline

Stresses in Rails

Stresses in Sleepers

Stresses in Ballast

Stresses in Formation

Coning of Wheels

Tilting of Rails / Adzing of Sleepers

Stresses in Rails

The number of sleepers per rail have little effect on the stresses in rails.

Eccentric vertical loads causes bending moment along with torsion of beam.

Lateral thrust at the rail head produces lateral deflection and twisting of the rail

Lateral deflection also causes lateral movement of sleepers on either side.

Contact shear stress under repeated load condition causes metals of the wheel tread and rail gets fatigued

Stresses in Rails

Very high stresses occur in lower fillets on

sharp curves.

Temperature causes alteration of tensile

and compressive stresses.

Tractive effort and braking force results in

flow of metal and deformation of rail head.

The effect is higher with small diameter

wheels.

Track irregularities causes heavy impact

and deflection

Stresses in Rails

Bending stress is rails (due to vertical

loads)

Assumption is that rail is a long bar

continuously supported by an elastic

foundation.

On account of vertical loads the rail is

subjected to bending and flexural

stresses.

Based on the elastic theory, the bending

moment is computed as:

Stresses in Rails


loads)

M = 0.25Ple-x/l {Sin (x/l) – Cos (x/l)}

where P = isolated vertical load

I = Characteristics length

= {4EI / µ}1\4

EI = Flexural stiffness of the rail

µ = Track Modulus

x = distance of the point under

consideration from the load

Stresses in Rails


loads)

Bending Stress = BM / Sectional Modulus of rail

BM is zero at points x = πl/4, 3 πl/4 and is

maximum at x = 0, πl/2, 3 πl/2, etc.

+20

-20

-40

-60

-80

-

100

Πl/4 Πl/2 3Πl/4 Πl 5Πl/4

Length of track

Max BM = Pl/4

Stresses in Rails

Bending Moment can also be computed

using:

Mmax = 0.318 W . x (in cm .tons)

Where W = Isolated wheel load in tons

x = Distance from the load to the

point of contra flexure in cm

= 42.3 [I/µ]1/4 (in cm)

µ = sub-grade modulus in kg/cm2

I = Moment of inertia of rail section

in cm4

Stresses in Rails

Bending stress then can be computed as:

f = Mmax/Z (in kg/cm2) OR

f = 13.8 W/Z . [I/µ]1/4 (in tons/cm2)

Where Mmax = maximum BM in cm . tons

f = stress due to BM immediately

under the load W in kg/cm2

(permissible value 22 to 25 kg/mm2)

Z = Modulus of rail section in cm3

Stresses in Rails

Deflection due to vertical load can be

computed as:

d = 9.25 W/[Iµ3]1/4 (in cm)

Where d = deflection of track in cm

W = Isolated wheel load in tonne

µ = Modulus of track in kg/cm2

I = Moment of inertia of rail section

in cm4

Stresses in Rails

Reduction

Factors


Various factors causing stresses in

sleepers are:

Dead and moving loads

Wheel load

Weight transfer from one wheel to another

wheel

Effect of speed on the rail and

Dynamic augment


Factors causing stresses in sleepers

Track components

Track modulus

Elasticity of rail and its stiffness

Design and strength of sleeper

Sleeper density

Type and efficiency of fastenings


Factors causing stresses in sleepers

Maintenance

Irregularities in rails

Their maintenance

Maintenance of wheels of coaches

Effectiveness of supervision for ensuring

good maintenance


Distribution of stresses in sleepers

Due to alteration in ballast reaction under

sleepers the pattern of distribution of

stresses gets modified.

The wave motion of rail causes movement

of load on the rail seat from one edge of

the sleeper to the other and makes its

distribution non-uniform.



End bound sleepers:

In case of newly constructed and

compacted sleepers, more deflection is

caused at the center as compared to the

ends. This is called end bound sleepers.

Load transferred

through rails

Sleepers



Centre bound sleepers:

With the repeated application of loads a

depression is caused at the end of the

sleepers. This results in higher depression

at the ends as

compared to the centre.

Load transferred

through rails

Sleepers



Load on Rail Seat:

Max. load on rail seat

= (P/Zµl) . µ . S

= P.S / Z . l

where P = wheel load

S = sleeper spacing

l = characteristics length

Z = Modulus of rail section



This is approximately 30 to 50% of the dynamic wheel load.

The length of the sleeper and the rail seat is so selected that under normal loading, the middle part of the sleeper is stress less

Provision of bearing plates also reduces the maximum stress in sleepers by about 32% on BG track.

Stresses in Ballast

Depends up on Live load and dead load of superstructure

and trains

Elastic properties of the sleepers

Section and length of sleepers

Spacing of sleepers

Degree of compaction of ballast

Nature of ballast bed

Size, gradation, depth and compaction of ballast

Type of sub-grade

Pattern of distribution of load / stresses

Stresses in Ballast

It is observed that

Pressure on the sleeper is maximum at

the centre of its width and this pressure

decreases from centre towards the ends.

The vertical pressure under the sleeper is

uniform at a depth which is approximately

equal to the spacing of the sleepers.

Stresses in Ballast

Distribution of stresses / pressure in ballast

section

40 Percentage

of average

pressure

50

100

30

10

1

Depth of

ballast

Spacing of

sleeper

Line of uniform pressure

Stresses in Ballast


section

Depth of

ballast (z)

Spacing of

sleeper

Bottom of

Tie

Pressure at the bottom of tie = Pa

Pressure at depth ‘z’ = Pz

Pressure at distance ‘x’ for depth ‘z’ =

Px

Pz

Px

Pa

P

Width of sleeper

(w)

Length of

sleeper

(L)

Stresses in Ballast


section

Pressure at the bottom of the tie (sleeper) is

given by:

Pa = P/ (w . L/2) in kg/sq.m

Where, P = isolated wheel load, in kg

w = width of the tie or sleeper, in m

L = length of the sleeper, in m

Stresses in Ballast

Distribution of stresses / pressure in

ballast section

Pressure at depth ‘z’ (in cm) below centre of

tie width is computed as:

Pz = 5.24 Pa / z1.25 (in kg/sq.m)

Pressure at distance ‘x’ (in cm) at depth ‘z’ is

given by:

Px = 0.48 (Pa / z) . 10-2.06 (x / z) (x / z)

(in kg/sq.m)

Stresses in Sub-grade

Bearing pressure of sub-grade soil

Alluvial soil below 0.70 kg/cm2

Soft clay 1.12 – 1.41kg/cm2

Wet or loose sand 1.12 – 1.41 kg/cm2

Dry clay or firm sand 1.48 – 2.11 kg/cm2

Compacted soil 2.88 & over

Stresses in Sub-grade

Maximum formation pressure permitted

on Indian Railways

For motive power 3.5 kg/cm2 for BG

2.5 kg/cm2 for MG

For goods wagons 3.0 kg/cm2 for BG

2.3 kg/cm2 for MG

Permissible Values

Bending stress in rails 36.0 kg/mm2

Contact stress between

rail and wheel 21.6 kg/mm2

Dynamic over loads at rail joints due to unsuspended masses

27 tons for locos

19 tons for wagons

Formation pressure 3.5 kg/cm2 for locos

3.0 kg/cm2 for wagons

Permissible Values

Fish plate stresses 30 kg/mm2

Bolt hole stresses 27 kg/mm2

Minimum ultimate tensile strength

72 kg/mm2

Assumed yield point 42.5 kg/mm2

Coning of Wheels

Flat Surface Coned Surface

Coning of Wheels

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 curved rail

Flexibility not available due to rigidity of vehicle base.

Coning of Wheels

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 to centrifugal force) thus causing an increase in diameter that helps it in moving longer distance on outer curve as compared to inner wheel for which the diameter reduces thus making it to traverse shorter distance.

Coning of Wheels

Coning of Wheels

Coning helps in –

Controlling differential movement of

front and rear axles caused due to

rigidity of frame and axle, thus acting

as a balancing factor.

On curves the rear axle has a tendency

to move towards inner rail.

Reducing wear and tear of wheel

flanges

Smooth riding.

Coning of Wheels

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 at

center of the curve

BG slip = 0.029 m per degree curve

Eccentric loading on rails

Coning of Wheels

Problem with Coning of Wheels

Coning of Wheels

RAIL


Rails are tilted at an angle of 1 in 20

Controls –

Lateral bending stresses due to

eccentric loading

Reduces wear and tear at inner edge of

rail and on tread of wheel

Adzing of sleepers or use of canted

bearing plates


Lecture 5 Final

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