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Gate – Civil Engineering GATE – CIVIL ENGINEERING TRANSPORTATION ENGINEERING Online Lecture: 12 (22.06.2020) Prof.B.Jayarami Reddy Professor and Head Department of Civil Engineering Y.S.R. Engineering College of Yogi Vemana University, Proddatur, Y.S.R.(Dt.), A.P-516360. E.mail : [email protected] Prof. B. Jayarami Reddy 22-06-2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR Length of transition curve: a. L = maximum of (i), (ii), (iii) and (iv) (i) As per railway code, LR = 4.4 where L and R are in m (ii) At the change of superelevation of 1 in 360. (iii) Rate of change of cant deficiency. Say 2.5 cm is not exceeded (iv) Based on rate of change of radial acceleration with radial acceleration of 0.3048 m/s² 3.28V3 L = R Where V is in m/s b. Length of transition curve, L = 0.073 × e × Vmax Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 2 RAILWAY ENGINEERING Previous GATE Questions Prof. B. Jayarami Reddy 3 22-06-2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 01.A broad gauge railway line passes through a horizontal curved section (Rradius=875 m) of length 200 m. The allowable speed on this portion is 100 km/h. For calculating the cant, consider the gauge as centre-to-centre distance between the rail heads, equal to 1750 mm. The maximum permissible cant (in mm, round off to 1 decimal place) with respect to the centre-to-centre distance between the rail heads is …… CE2 2019 01. 157.5 Radius of the curve,R = 875 m Length of the curve,L = 200 m Allowable speed,V = 100 km/h Gauge, G = 1750 mm Maximum permissible cant, e = ? GV 221750(100) e == = 157.5 mm 127R 127 875 Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 4 02. For a broad gauge railway track on a horizontal curve of radius (in m), the equilibrium cant e required for a train moving at a speed of (in km per hour) is V 2 V 2 V 2 V 2 a. e =1.676 b. e =1.315 c. e = 0.80 d. e = 0.60 CE2 2017 R R R R GV. 2 Equilibrium cant, e = 127R V :speed of the train, kmp h R:Radius of horizontal curve, m For broad gauge track,G = 1.676 m 1.676VVV2 2 2 e =m = 0.01319 m = 1.319 cm 127 RRR Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 5 06. Which one of the following types of steel is used in the manufacturing of metro and mono rails? IES 2018 a. Mild steel b. Cast steel c. Manganese steel d. Bessemer steel Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 6 07. A transition curve is to be provided for a circular railway curve of 300 m radius, the gauge being 1.5 m with the maximum superelevation restricted to 15 cm. What is the length of the transition curve for balancing the centrifugal force? a. 72.3 m b. 78.1 m c. 84.2 m d. 88.3 m IES 2018 Ans. (i) Length of transition curve,L = 0.073 × e × Vmax Vmax = 4.35 R − 67 =4.35 300 − 67 = 66.39km/hr L = 0.073 × 15× 66.39= 72.3m Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 7 09. The gradient for a B.G. railway line such that the grade resistance together with curve resistance due to a 4° curve which will be equivalent to a simple ruling gradient of 1 in 150 is IES 2016 a. 1:180 b. 1:200 c. 1:300 d. 1:400 Degree of curve for BG track, D = 4° Allowable gradient = ? Ruling gradient = 1 in 150 Grade compensation for BG track on curve =4 0.04 = 0.16% 1 0.16 11 Allowable gradient =− = 150 100 197.4 200 Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 8 10. The maximum speed of a train on B.G. track having a curvature of 3° and a cant of 10 cm with allowable cant deficiency of 76 mm, for conditions obtaining in India, is IES 2016 a. 87.6 km/h b. 99.6 km/h c. 76.6 km/h d. 65.6 km/h Maximum speed of train on track, = ? Degree of curve, D = 3° Cant = 10 cm Allowable cant deficiency = 76 mm Superelevation, e = 100 + 76 = 176 mm 1718.9 1718.9 D =3 =R = 573.0 m R R GV 2 1.676V 2 e =0.176 =V = 87.4 kmph 127R 127 573 Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 9 11. What will be the optimum depth of ballast cushion required for a BG railway track below the sleepers with sleeper density of (M+5) and bottom width of 22.22 cm? IES 2015 a. 25 cm b. 21 cm c. 28 cm d. 30 cm Optimum depth of ballast cushion required for BG = D = ? As per Indian railways, Length of each rail on BG track = 12.8 m Sleeper density = M+5 = 13 + 5 = 18.0 Bottom width = 22.22 cm 13.0 Spacing = =0.7222 = 72.22cm 18.0 SW−−72.22 22.22 D = = = 25cm 22 Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 10 12. In the layout of an MG track, the versine of a horizontal circular curve is measured over a 11.8 m chord length. What would be the radius of the curve if the value of the versine was 2 cm? IES 2015 a. 900 m b. 800 m c. 870 m d. 850 m Chord length of circular curve, L = 11.8 m Radius of the curve, R = ? Versive of the curve = 2cm From the property of circular curve, CD. (2 OD-DC) = AC.BC y is small; y2 is neglected. LLL2 y(2 R− y ) = . 2 RY − y2 = 2 2 4 LL2 211.8 2 2RY= R = = = 870.2m 4 8y 8 0.02 Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 11 13. What would be the admissible gradient for a BG track when the grade resistance coupled with a 40 curve resistance shall equal the resistance due to a ruling gradient of 1 in 200? IES 2015 a. 0.30% b. 0.40% c. 0.24% d. 0.34% Admissible gradient = ? Degree of curve for BG track, D = 4° Ruling gradient = 1 in 200 Grade compensation for BG track on curve = 4 × 0.04 = 0.16% 1 Admissible gradient = 100 − 0.16 = 0.34% 200 Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 12 14. The steepest gradient on a 20 curve on a Broad Gauge line with a stipulated ruling gradient of 1 in 200, given that grade compensation is per degree of curve, is IES 2014 a. 1 in 200 b. 1 in 150 c. 1 in 238 d. 1 in 283 Degree of curve for BG track, D = 2° Ruling gradient = 1 in 200 Steepest permissible gradient = ? Grade compensation for BG track on curve = 2 × 0.04 = 0.08% 1 0.08 1 Steepest permissible gradient =−= 200 100 238 Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 13 15. Two parallel railway tracks are to be connected by a reverse curve, both segments having the same radius. If the centre lines of the tracks are 8 m apart and the maximum adaptable distance between the tangent points is 32 m, the allowable radius for the curve is IES 2014 a. 4 m b. 8 m c. 32 m d. 64 m Centre line distance of tracks = 8m Maximum adoptable distance between the tangent points = 32 m Allowable radius of the curve, R = ? v 8 tan= = = 28.070 2h 32 h=2 R sin 32 = 2 R sin 28.07 R = 34m Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 14 19. An electric locomotive running at 60 kmph on a curved track of 1.68 m gauge laid at 800 m radius should be provided with superelevation of the rail by an amount of IES 2014 a. 50.5 mm b. 55.5 mm c. 59.5 mm d. 65.5 mm Speed of locomotive, V = 60 kmph Gauge length, G = 1.68 m Radius of curve, R = 800 m Superelevation of rail, e = ? GV 221.68 60 e = = =0.0595m = 59.5 mm 127R 127 800 Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 15 24. Match List-I With List- II and select the correct answer using the code given below the lists. List – I List - II A. Rails 1. Connect one section of rail to next B. Sleepers 2. Convert line load into uniformly distributed load C. Ballast 3. Convert point load into uniformly distributed load D. Fish Plates 4. Convert rolling loads into point load (s) a. A4 B3 C2 D1 b. A1 B2 C3 D4 c.
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