Railway Engineering: Geometric Design Prof. Dr. Padma Bahadur Shahi
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Geometric Design of Railway Track Geometric design should be such as to provide maxim ximum efficienc ency in the traffic oper eraation with maximum safety at reasonable cost. Gradient
ny epar ure o rac rom e eve s nown as grade or gradient. Purpose of providing gradient: •
To provi provide de unifo uniform rm ra rate te of rise rise or fa fall ll,,
•
To re redu duce ce cost cost of eart earth h work work..
•
To re reach ach diff differ erent ent stat statio ions ns at diff differ erent ent level level
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Types of gradient •
Ruling Ruling gradie gradient: nt: The The stee steepe pest st gra gradie dient allo allowe wed d on the tra track sec section. on. It determines the max load that the locomotive can haul that section. The steep gradient needs more powerful locomotives, smaller train loads, lower speed, resulting in costly hauling. – In plains: – In hilly regions:
• •
•
1 in 150 to 1 in 200 1 in 100 to 1 in 150
Momentum Gradient: The gradient on a section which are steeper than the ruling gradient acquire sufficient momentum to negotiate them are known as momentum gradient. Pusher gradient: As stated above a ruling gradient limits the maximum weight of a train which can be hauled over the section by a locomotive. If the ruling gradient is so severe on a section that it needs the help of extra engine to pull the same load than this gradient is known as pusher of helper gradient. In Darjeeling Railways Railways 1 in 37 pusher gradient is used on Western Ghat BG Track. Gradient at stations: at stations gradient are provided sufficient low due to following reason: – To prevent prevent movem movement ent of standi standing ng vehicle vehicle – To prevent prevent addition additional al resistance resistance due to to grade. grade.
On Indian railways, maximum gradient permitted is 1 in 400 in station yards. 3
Grade compensation on curves •
If a curve is provided on a track with ruling gradient, the resistance of the track will be increased this curve. In order to avoid resistance beyond the allowable limits, the gradients are known 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 4
Degree of curve: • A curve is defined by its degree or radius. The degree of a curve is the angle subtended at the center by a chord of 100 feet or 30.48m. • R is is the the radius radius of curve; curve; • Circumferenc Circumferencee of the curve= 2 ∏ R • Angle Angle subtend subtended ed at the center center by the the circle circle = 360 degree • Angle subtended subtended by the the arc arc of of 30.48m 30.48m = 360 2 ∏ R
X 30.48 =
1747.26
R
≈
1750
R
• Thus, a 1 degree degree curve has a radius of 1750 1750 m.
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Grade compensation • Numerical example: The ruling gradient on a BG track section has been fixed as 1 in 200. What should be the compensated gradient when a 4 degree horizontal curve is to be provided on this ruling gradient? • Solution: As per IS recommendation the grade com ens ensatio ation n on on BG BG tr track is 0. 0.04 % er de de ree of the the curve. • Then compensation for 4 degree curve = 0.04 X 4= 0.16% • Ruling Ruling gradient gradient is 1 in 200 = 0.5 0.5 % • Allowab Allowable le gradien gradientt to be prov provide ided d = 0.5 – 0.16 0.16 = 0.34 = 1 in 249 6
Elements of Circular Curve
Refer Transportation I for .
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Superelevation on Curves (Cant) •
Cant: It is defined as the difference in height between the inner and outer rails on the curve. It is provided by gradually raising the outer rail above the inner rail level. The inner rail is considered as the reference rail and normally is maintained at its original level. The .
•
Function of superelevation: –
Neutra Neutralize lizess the effect effect of lateral lateral force force
–
It provid provides es better better load load distrib distributi ution on on the two rails. rails.
–
It reduce reducess wear wear and tear tear of rails rails and rolling rolling stock. stock.
–
It provides smooth running ing of tra trains and comforts to the the passengers 8
Speeds • Equilib Equilibriu rium m speed: speed: It is the speed at which the effect of centrif trifu ugal force is exactly balanced by the superelevation ion provided. It can also be said that when the speed of a vehicle running on a curved track is such that the resultant weight of the vehicle and the effect of radical acceleration is perpendicular to the plane of rails and the vehicle is not subjected to an unbalanced radical acceleration, is in equilibrium then its particular speed is called equilibrium speed. • Maximum permissible speed: This is the highest speed which may be allowed or permitt itted on a curved track taking ing into consideration of the radius of curvature, actual cant, cant deficiency, cant excess and the length of the transition curve. When, the maximum permissible speed on the curve is less than the maximum sanctioned speed of the section of a line, permanent speed restriction become necessary on such curves. 9
Cant…… • Cant Cant de defic ficie ienc ncy: y: Cant deficiency is the difference between the equilibrium cant (theoritical) necessary for the maximum permissible speed on a curve and the actual cant provided there. As per Indian Railways, Cant deficiency is recommended as follow: – BG Track: – MG track: – tr trac :
75 mm 50 mm mm
Excess: When a train travels on a curved rack at a • Cant Excess: speed lower than the equilibrium speed, then the cant excess occurs. It is the difference between the actual cant provided and the theoretical cant required for such lower speeds. Maximum value for cant excess is BG track: MG Track:
75 mm 65 mm 10
Centrifugal Force: • When a body moves on a circular curve, it has a tendency to move in a straight direction tangential to the curve. This tendency of the body is due to the fact that the body is subjected to a constant
Radial acceleration =
v2 R
This radial acceleration produces a force known as centrifugal force whose value is given by the following f ollowing relation:
F =
Wv
2
gR 11
e G
=
P
e
W
G
=
e=
e=
P W
W
*G
GV
2
gR 12
Superelevation
e=
GV
2
127 R
• Where, e is super elevation in mm; G is the gauge in mm + width o t e ra ea n mm; spee o t e tra n n mp ; ra us o the curve in m. – Fo For BG track: – For MG track: – For NG track:
G = 1676 mm+ 74 mm=1.75 m G = 1.058 m G= 0.772 m
G for: BG track-1.676+0.074=1.750 track-1.676+0.074=1.750 m MG :1.058m NG: 0.772m
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Maximum value of superlevation: • the maximum value of superlelevation has been decided on the basis of experiments carried out by many researchers. The maximum value of superelevation generally on many railways of the the gauge. As per Indian railways: – BG Track: Track:
165mm (normal condition); 185 (special permission)
– MG track:
90 mm (normal (normal condition), condition), 100 (speci (special al permission) permission)
– NG track:
65 mm (normal condition), 75 (special permission)
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Maximum degree of Curve: • It depends depends on on various various factor factorss such as gauge gauge,, wheel wheel base of vehi vehiccle, le, maxim aximum um supe supere rele leva vati tion on and and othe otherr such such factors. As per Indian railways:
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 15
Safe speed on curves On the curves the safe speed can be calculated empirically by the following formula: a) For BG and MG on transition curve:
V = 4 . 4
R − 70
For non-transition curve (80% of the speed on the transition curve):
V = 0 . 8 * 4 . 4
R − 70
b) For High speed track:
V = 4 . 58 R R is the radius in m, V is speed in Kmph:
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Rational formula •
Rational formula considering superelevation: a) BG track: V = 0 . 27
( Ca + Cd ) R
Ca is actual cant provided in mm; Cd is the cant deficiency permitted in mm; R is radius in m V is maximum s eed in km h.
b) On MG track: V = 0 . 347
( Ca + Cd ) R
C) On NG track: V = 3 . 65
R−6
R is the radius in m, V is speed in Kmph:
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Maximum permissible speed on curves a) Maximum sanctioned speed of the section: this is the maximum permissible speed authorized by the commissioner of rail safety. b) Safe speed on curves: •
For For BG and and MG on tran transi sitio tion n curv curve: e:
V = 4 . 4 •
n
g spee
R − 70
r ac
V = 4 . 58 R • For non-transition curve (80% of the speed on the transition curve):
V = 0 . 8 * 4 . 4
R − 70 18
Maximum permissible speed…… c)
Maximum speed of section taking into account the super elevation and cant deficiency cant deficiency: (use of rational r ational formula) –
BG Track:
V = 0 . 27
( Ca + Cd ) R
MG Track:
V = 0 . 347
( Ca + Cd ) R
NG track:
V = 3 . 65
R−6 19
Maximum permissible speed…… •
Spee Speed d corr corres espo pond ndin ing g to to the the leng length th of tran transi siti tion on curve: CaV m for speed up to 100 kmph. L = 125
L is the desirable length of transition curve; Ca is actual cant in mm; Vm is the maximum permissible speed, in Kmph
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Numerical Examples a)
On a BG rout routee inv invol olvi ving ng high high spee speed, d, A 100 100 m tra trans nsit itio ion n cur curve ve has been een prov rovided ided and and a supe supere rele lev vatio ation n of 80mm has bee been managed. The degree of curve is 10 and the maximum sanctioned speed for the curved section is 170 kmph. Determine maximum permissible speed on the curve. (Hint: assume cant deficiency as 100 mm)
b)
Find Find out out the supe supere rele lev vatio ation n to be prov provid ided ed and and the maxi maxim mum permissible speed for 20 BG transition curve on a high speed route having a maximum sanctioned speed of the section as 100 kmph. For calculating the equilibrium superelevation the speed given as 75 kmph and the booked speed for goods traffic is 50 kmph.
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Numerical Examples c) Calculate the superelevation and maximum permissible speed for a 30 curve on a high speed BG track with the following data: –
maximum sanctioned speed =
130 kmph
–
equilibrium speed =
8 5 k m ph
–
booked speed for goods train =
50 kmph
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.)
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)
Calculate the maximum permissible on a curve of high speed BG route with the following data: –
Degree of curve =
10
–
Superelevation =
85 mm
–
Length of transition curve =
125m
–
Sanctioned spe peeed of th thee section tion =
170 kmph
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Transition Curves Objectives of providing transition curves: –
To attain attain gradu gradual al rise rise of the outer outer ra rail; il;
–
To dec decre reas asee the the radi radius us of of the the curv curvee grad gradua uall lly y from from infinite at the straight end of the rack to that of the circular curve at the junction with the circular curve of the selected radius.
–
To provid providee smo smooth oth runni running ng of vehicl vehicles es and and pro provi vide de comfort to the passengers
–
To re reduc ducee chanc chancee of derai derailm lmen ent. t.
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Types of Transition Curve
This is adopted in Indian railways. Rate of decrease of the radius of curvature increases rapidly 3 X Equation of the cubic parabola is Y =
6 RL
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Length of transition curve: • The distan distance ce along along the the center center line line of the the track track betwee between n the starting point on the straight portion of the track and the meeting point on the junction with the circular curve is called the transition curve length. • For For Indi Indian an railw railway ayss it can be calcu calcula late ted d with with the help help of followin e uation and the reatest of all should be adopted. – Base Based d on the the arbi arbitr trar ary y grad gradie ient nt (1 in 720): 720): L = 7.2 7.2 x e – Based Based on the rate rate of change change of cant cant defici deficiency: ency: L = 0.073D x Vmax Vmax – Based Based on the change change of superele superelevatio vation: n: L = 0.073 0.073 x e x Vmax
• L is the length length of of transit transition ion curve curve in m; m; e is the actua actuall cant in cm; D is the cant deficiency in cm; V is the maximum speed in kmph. 26
Length of transition curve • Length Length of the trans transiti ition on curv curvee shou should ld be taken taken as the maximum value given by the following formula: – Railw ailway ay code code:: L = 4.4 4.4 √R; where R is the radius of the curve curve L and R in meter meter – At the rate of change of superelevation of 1 in 360; i.e, 1 cm for every 3.6 m length of track. • Based on the rate of change of radial acceleration: acceleration: L =
• Based on the maximum maximum permissible speed: L =
0 . 066 V
2
R
CaV m 134
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Numerical example: Find out the length of transition curve for a four degree BG circular curved track having a cant of 15 cm. the maximum permissible speed on the curve is 90 kmph. Find out the shift and offset at transition curve also. Assume maximum permissible cant deficiency is 75 mm.
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• Solution hints: – Length of transition curve: – L = 7.2 x e =7.2 x 15 = 108 m –
.
.
=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 108 m. 29
• Shift = L2/24R = (108 x 108)/(24 x 1750/4) = 11664/10500 = 1.11m • Offset Offset at at every every 15 m interv interval al is calc calculat ulated ed by the the cubic cubic parabola equation as follows
• At 15 m;
30
•
At 30 m; Y2 = 9.52cm
•
At 45m; Y3 = 32.2 cm
•
At 105m Y7 = 408 cm
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Vertical curves •
Type Typess of ver erti tica call curv curvees: – –
•
Summit curve Valley curve
As per per exi exist stin ing g pro prov visio ision n the the vert vertic ical al curv curves es are prov provid ided ed only only at the jun junction ions of the grades where algebraic difference 0.4% the minimum radius of the vertical curve should be as follows: BG Track:
MG track:
A: 4000m; B: 3000m; C, D & E: minimum 2500m 2500m
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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
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Numerical Example • A rising gradient of 1 in 150 m meets a falling gradient of 1 in 250 on a group A route. The intersection point has a chainage of 1000m and its RL is 100 m Calculate following: – – ii) RL of vertical length and chainage of various points
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Solution Hints: • First First gradien gradientt = +0.67 +0.67 % • Second Second gradie gradient nt = -0.4 -0.4 % • Difference Difference in gradient gradient = +0.67-(-0.4) +0.67-(-0.4) = 1.07 % • Radius of the the curve as per given value R = 4000m 4000m – Then L = 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 35
