Railway Engineering:Geometric Design

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Prof. Dr. Padma Bahadur Shahi

Geometric Design of Railway Track

Geometric design should be such as to providemaximum efficiency in the traffic operationwith maximumsafety at reasonable cost.

GradientAny departureof track from the level is known as

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Any departureof track from the level is known asgrade or gradient. Purpose of providinggradient:

• To provide uniformrate of rise or fall,

• To reduce cost of earth work.

• To reach different stations at different level

Types of gradient• Ruling gradient: The steepest gradient allowed on the track section. It

determines the max load that the locomotive can haul that section. The steepgradient needs more powerful locomotives, smaller train loads, lower speed,resulting in costly hauling.– In plains: 1 in 150 to 1 in 200– In hilly regions: 1 in 100 to 1 in 150

• Momentum Gradient: The gradient on a section which are steeper than theruling gradient acquire sufficient momentum to negotiate them are known asmomentum gradient.

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• Pusher gradient: As stated above a ruling gradient limits the maximum weightof a train which can be hauled over the section by a locomotive. If the rulinggradient is so severe on a section that it needs the help of extra engine to pullthe same load than this gradient is known as pusher of helper gradient. InDarjeeling Railways 1 in 37 pusher gradient is used on Western Ghat BG Track.

• Gradient at stations: at stations gradient are provided sufficient low due tofollowing reason:– To prevent movement of standing vehicle– To prevent additional resistance due to grade.

On Indian railways, maximum gradient permitted is 1 in 400 instation yards.

Grade compensation on curves

• If a curve is provided on a track with rulinggradient, the resistance of the track will beincreased this curve. In order to avoid resistancebeyond the allowable limits, the gradients arereducedon curves. The reductionin gradientis

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reducedon curves. The reductionin gradientisknown as grade compensation for curves.

• BG track: 0.04% per degree of curve

• MG track: 0.03 % per degree of curve

• NG track: 0.02 % per degree of curve

Degree of curve:• A curve is defined by its degree or radius. The

degree of a curve is the angle subtended at thecenter by a chord of 100 feet or 30.48m.

• R is the radius of curve;• Circumference of the curve=2∏ R• Angle subtendedat thecenterby thecircle = 360

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• Angle subtendedat thecenterby thecircle = 360degree

• Angle subtended by the arc of 30.48m=

• Thus, a 1 degree curve has a radius of 1750 m.

RRX

R

175026.174748.30

2

360 ≈=∏

Grade compensation

• Numerical example: The ruling gradient on a BGtracksection has been fixed as 1 in 200. What should be thecompensated gradient when a 4 degree horizontal curveis to be provided on this ruling gradient?

• Solution: As per IS recommendation the gradecompensationon BG track is 0.04 % per degreeof the

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compensationon BG track is 0.04 % per degreeof thecurve.

• Then compensation for 4 degree curve = 0.04 X 4=0.16%

• Ruling gradient is 1 in 200 = 0.5 %• Allowable gradient to be provided = 0.5 – 0.16 = 0.34 =

1 in 249

Elements of Circular Curve

Refer Transportation I for this topic.

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this topic.

Superelevation on Curves (Cant)

• Cant: It is defined as the difference in height betweenthe inner and outer rails on the curve. It is provided bygradually raising the outer rail above the inner raillevel. The inner rail is considered as the reference railand normally is maintained at its original level. Theinnerrail is knownasthegradientrail.

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innerrail is knownasthegradientrail.

• Function of superelevation:– Neutralizes the effect of lateral force

– It provides better load distribution on the two rails.

– It reduces wear and tear of rails and rolling stock.

– It provides smooth running of trains and comforts to thepassengers

Speeds

• Equilibrium speed: It is the speed at which the effect ofcentrifugal force is exactly balanced by the superelevationprovided. It can also be said that when the speed of a vehiclerunning on a curved track is such that the resultant weight of thevehicle and the effect of radical acceleration is perpendicular tothe plane of rails and the vehicle is not subjected to an unbalancedradical acceleration, is in equilibriumthen its particular speed iscalledequilibriumspeed.

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calledequilibriumspeed.• Maximum permissible speed: This is the highest speed which

may be allowed or permitted on a curved track taking intoconsideration of the radius of curvature, actual cant, cantdeficiency, cant excess and the length of the transition curve.When, the maximumpermissible speed on the curve is less thanthe maximumsanctioned speed of the section of a line, permanentspeed restriction become necessary on such curves.

Cant……

• Cant deficiency: Cant deficiency is the differencebetween the equilibriumcant (theoritical) necessary forthe maximumpermissible speed on a curve and theactual cant provided there. As per Indian Railways, Cantdeficiency is recommended as follow:

– BG Track: 75 mm– MG track: 50 mm– NG track: 40mm

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– NG track: 40mm

• Cant Excess: When a train travels on a curved rack at aspeed lower than the equilibriumspeed, then the cantexcess occurs. It is the difference between the actual cantprovided and the theoretical cant required for such lowerspeeds. Maximumvalue for cant excess is

BG track: 75 mmMG Track: 65 mm

Centrifugal Force:

• When a body moves on a circular curve, it has atendency to move in a straight direction tangentialto the curve. This tendency of the body is due tothe fact that the body is subjected to a constantradialacceleration.

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radialacceleration.

This radial acceleration produces a force known as centrifugal force whose value is given by the following relation:

gR

WvF

2

=

R

vonacceleratiRadial

2

=

W

P

G

e =PW

P

G

e =

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WGG

W

Pe *=

gR

GVe

2

=

Superelevation

• Where,e is super elevation in mm; G is the gauge in mm+ widthof the rail headin mm; V speedof the train in kmph; R radiusof

R

GVe

127

2

=

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of the rail headin mm; V speedof the train in kmph; R radiusofthe curve in m.– For BG track: G = 1676 mm+ 74 mm=1.75 m– For MG track: G = 1.058 m– For NG track: G= 0.772 m

G for: BG track-1.676+0.074=1.750 mMG :1.058mNG: 0.772m

Maximum value of superlevation:

• the maximumvalue of superlelevation has beendecided on the basis of experiments carried outby many researchers. The maximumvalue ofsuperelevation generally on many railways of theworld hasbeenadoptedabout1/10th to 1/12th of

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world hasbeenadoptedabout1/10th to 1/12th ofthe gauge. As per Indian railways:– BGTrack: 165mm (normal condition); 185 (special permission)

– MG track: 90 mm (normal condition), 100 (special permission)

– NG track: 65 mm (normal condition), 75 (special permission)

Maximum degree of Curve:

• It depends on various factors such as gauge, wheel baseof vehicle, maximumsuperelevation and other suchfactors. As per Indian railways:

Gauge On plain track On turn out

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Gauge On plain track On turn out

Max degree Min radius, m Max degree Min radius, m

BG 10 175 8 218

MG 16 109 15 116

NG 40 44 17 103

Safe speed on curves

On the curves the safe speed can be calculatedempirically by the following formula:

a) For BG and MG on transition curve:

704.4 −= RV

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704.4 −= RVFor non-transition curve (80% of the speed on the transition curve):

704.4*8.0 −= RV

RV 58.4=

b) For High speed track:

R is the radius in m, V is speed in Kmph:

Rational formula

• Rational formula considering superelevation:a) BG track:

RCdCaV )(27.0 +=

Ca is actual cant provided in mm;Cd is the cant deficiency permitted in mm;R is radiusin m; V is maximumspeedin kmph.

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R is radiusin m; V is maximumspeedin kmph.

RCdCaV )(347.0 +=b) On MG track:

665.3 −= RV

C) On NG track:

R is the radius in m, V is speed in Kmph:

Maximum permissible speed on curves

a) Maximumsanctioned speed of the section: this is the maximumpermissible speed authorized by the commissioner of rail safety.

b) Safe speed on curves:• For BG and MG on transition curve:

704.4 −= RV• On high speed track

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RV 58.4=

• On high speed track

• For non-transition curve (80% of the speed on the transition curve):

704.4*8.0 −= RV

Maximum permissible speed……

c) Maximum speed of section taking into account the super elevation and cantdeficiency cant deficiency: (use of rational formula)

– BG Track:

RCdCaV )(27.0 +=

MG Track:

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RCdCaV )(347.0 +=

NG track:

665.3 −= RV

Maximum permissible speed……

• Speed corresponding to the length of transitioncurve:

for speed up to 100 kmph.125

mCaVL =

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L is the desirable length of transition curve; Ca is actual cant in mm;Vm is the maximum permissible speed, in Kmph

Numerical Examples

a) On a BGroute involving high speed, A 100 m transition curvehas been provided and a superelevation of 80mmhas beenmanaged. The degree of curve is 10 and the maximumsanctioned speed for the curved section is 170 kmph. Determinemaximumpermissible speed on the curve.

(Hint: assume cant deficiency as 100 mm)

b) Find out the superelevationto be provided and the maximum

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b) Find out the superelevationto be provided and the maximumpermissible speed for 20 BGtransition curve on a high speedroute having a maximumsanctioned speed of the section as 100kmph. For calculating the equilibriumsuperelevation the speedgiven as 75 kmph and the booked speed for goods traffic is 50

kmph.

Numerical Examples

c) Calculate the superelevation and maximum permissible speedfor a 30 curve on a high speed BG track with the followingdata:

– maximum sanctioned speed = 130 kmph– equilibrium speed = 85 kmph– booked speed for goods train = 50 kmphSolution hints:

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Solution hints:– Calculate R– Equilibrium superelevation (with V = 85kmph):– Equilibrium superelevation for sanctioned speed(130kmph):– Cant deficiency: which is more than permissible value of 100mm– Actual cant then to be provided = should be maximum value of

165mm.– Equilibrium superelevation for goods train for 50kmph (e=59mm)– Cant access= 165-59=106mm but (limited to the 75mm.)

) Calculate the maximum permissible on a curve of high speedBG route with the following data:

– Degree of curve = 10– Superelevation = 85 mm– Length of transition curve = 125m– Sanctioned speed of the section = 170 kmph

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Transition Curves

Objectives of providing transition curves:

– To attain gradual rise of the outer rail;

– To decrease the radius of the curve gradually frominfinite at the straight end of the rack to that of thecircular curve at the junction with the circular curve

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of the selected radius.

– To provide smooth running of vehicles and providecomfort to the passengers

– To reduce chance of derailment.

Types of Transition Curve

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This is adopted in Indian railways. Rate of decrease of the radius of curvature increases rapidlyEquation of the cubic parabola is

RL

XY

6

3

=

Length of transition curve:

• The distance along the center line of the track between thestarting point on the straight portion of the track and themeeting point on the junction with the circular curve iscalled the transition curve length.

• For Indian railways it can be calculated with the help offollowing equation and the greatest of all should be

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following equation and the greatest of all should beadopted.– Based on the arbitrary gradient (1 in 720): L = 7.2 x e– Based on the rate of change of cant deficiency: L = 0.073D x Vmax– Based on the change of superelevation: L = 0.073 x e x Vmax

• L is the length of transition curve in m; e is the actual cantin cm; D is the cant deficiency in cm; V is the maximumspeed in kmph.

Length of transition curve

• Length of the transition curve should be taken asthe maximum value given by the followingformula:– Railway code: L = 4.4√R; where R is the radius of

thecurve(L andR in meter)

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thecurve(L andR in meter)

– At the rate of change of superelevation of 1 in 360;i.e, 1 cmfor every 3.6 m length of track.

• Based on the rate of change of radial acceleration:

• Based on the maximumpermissible speed:R

VL

2066.0=

134mCaV

L =

Numerical example:

Find out the length of transition curve for a fourdegree BGcircular curved track having a cant of15 cm. the maximumpermissible speed on thecurve is 90 kmph. Find out the shift and offset atevery 15 m interval of the curve. Draw the

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every 15 m interval of the curve. Draw thetransition curve also. Assume maximumpermissible cant deficiency is 75 mm.

• Solution hints:– Length of transition curve:

– L = 7.2 x e

=7.2 x 15 = 108 m

– L= 0.073D x Vmax.

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– L= 0.073D x Vmax.=0.073 x 7.5 x 90 = 49.28

– L = 0.073 x e Vmax

= 0.073 x 15 x 90 = 98.55

Hence the length of Curve will be the greatest of three values i.e 108m.

• Shift = L2/24R = (108 x 108)/(24 x 1750/4) =11664/10500 = 1.11m

• Offset at every 15 m interval is calculated by the cubicparabola equation as follows

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• At 15 m;

• At 30 m; Y2 = 9.52cm

• At 45m; Y3 = 32.2 cm

• At 105m Y7 = 408 cm

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Vertical curves

• Types of vertical curves:– Summit curve– Valley curve

• As per existing provision the vertical curves are provided onlyat the junctions of the grades where algebraic differencebetweenthe gradesis equalto or morethan4 mm per meteror

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betweenthe gradesis equalto or morethan4 mm per meteror0.4% theminimum radius of the vertical curve should be asfollows:

BG Track: MG track:

A: 4000m; B: 3000m; C, D & E:2500m

minimum 2500m

Length of vertical curve

• Length of vertical curve: L=Rθ– L is the length of vertical curve

– R is the radius of vertical curve as per given table

– Θ is the difference in percentage of gradients(expressedin radian)

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(expressedin radian)

Numerical Example

• A rising gradient of 1 in 150 m meets a fallinggradient of 1 in 250 on agroup A route. Theintersection point has a chainage of 1000m and itsRL is 100 m Calculate following:– i) Lengthof theverticalcurve

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– i) Lengthof theverticalcurve

– ii) RL of vertical length and chainage of variouspoints

Solution Hints:

• First gradient = +0.67 %

• Second gradient = -0.4 %

• Difference in gradient = +0.67-(-0.4) = 1.07 %

• Radius of the curve as per given value R = 4000m

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– ThenL = Rxθ = 4000x(1.07/100)= 42.8 m.

– Chainage of point A = 1000-21.4 = 978.6m

– Chainage of point B 1000+21.4 = 1021.4 m

– RL of point A = 100-(21.4/150) = 99.860m

– RL of point B = 100 – (21.4/250) = 99.914m