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 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 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, R875= m Length of the curve, L200= 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 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 trainh, kmp R:Radius of horizontal curve, m For broad gauge track,G 1.676= m 1.676VVV222 e === m 0.01319m 1.319cm 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

V4.3567max =− R =−=4.353006766.39km/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 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 ==40.040.16% 10.16 11 Allowable gradient =− = 150100 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 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 1.676V 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 === 0.722272.22cm 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. L LL 2 y(2).2 R−=−= yRYy 2 2 24 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 =− 1000.16 = 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 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 127127R 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. A4 B2 C3 D1 d. A1 B3 C2 D4

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 16 25. If the ruling gradient is 1 in 150 on a particular section of a broad gauge track, the allowable ruling gradient on a 40 curve in the track will be IES2011 a. 0.51% b. 0.53% c. 0.61% d. 0.67% Ruling gradient on BG track = 1 in 150 Degree of curve, D = 4° Allowable ruling gradient = ? Grade compensation for BG track on curve = 4 × 0.04 = 0.16% 1 Allowable ruling gradient = 100 − 0.16 = 0.51% 150

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 17 32. What is the hauling capacity of a railway locomotive with four axles, carrying an axle load of 24 tones each? (Assume 0.166 as the coefficient of friction) IES 2008 a. tonnes b. 16.0 tonnes c. tonnes d. tones Hauling capacity of railway locomotive, HC = ? Number of axles, n = 4

Load on each axle, Wd = 24 t Coefficient of friction,  = 0.166

HCnW====. .0.16642415.94t16td

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 18 35. What is the steepest gradient permissible on a 20Curve for BG line having ruling gradient of 1 in 200? IES 2007 a. 1 in 250 b. 1 in 238 c. 1 in 209 d. 1 in 198 Degree of curve, on track D = 2° Ruling gradient = 1 in 200 Permissible gradient = ? Grade compensation = 2 × 0.04 = 0.08% 10.081 Permissible gradient =−= 200100238

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 19 38. The ruling gradient on BG section of railway is 1 in 150 and a 40 curve is also there on it. What is the allowable ruling gradient? IES 2006 a. 1 in 146 b. 1 in 154 c. 1 in 196 d. 1 in 232 Ruling gradient on BG track = 1 in 150 Degree of curve, D = 4° Grade compensation for BG track on curve = 4 × 0.04 = 0.16% 1 0.16 1 Allowable ruling radiant = − = 150 100 197

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 20 39. What is the value of the steepest gradient to be provided on a 20 curve for BG line having ruling gradient of 1 in 200? IES 2006 a. 1 in 238 b. 1 in 227 c. 1 in 202 d. 1 in 198 Degree of curve for BG track = 2 ° As per Indian Railways, the maximum gradient permitted for all tracks = 1 in 200 (steepest) Grade compensation for BG track on curve = 2 × 0.04 = 0.08% 10.081 Permissible steepest gradient =−= 200100238

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 21 41. What is curve resistance for a 50 tones train on a track on a 4° curve? IES 2005 a. 0.05 tonne b. 0.06 tonne c. 0.08 tonne d. 0.10 tonne Curve resistance, R = ? Weight of train, W = 50t Degree of curve on BG track, D = 4° 40.04 R= Grade compensation × weight of train == 500.08t 100

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 22 42. Which one of the following is the correct expression for the versine (h) of a curve? (l = length of rail on the curve and radius of curvature) IES 2005 l 2 l 2 l 2 l 2 a. h = b. h = c. h = d. h = r 2r 8r 4r

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 23 48. Match List-I( Track parameter) with List-II (Equipments used) and select the correct answer using the codes given below the lists: IES 2003 List-I List-II A. Unevenness 1. Track recording car B. Gauge 2. Amsler car C. Superelevation 3. Feeler and spring D. Alignment 4. Gyroscopic pendulum

a. A2 B3 C1 D4 b. A3 B2 C1 D4 c. A2 B3 C4 D1 d. A3 B2 C4 D1

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 24 50. Grade compensation on cures in Indian Railways for BG is IES 2003 a. 0.40% per degree of curve b. 0.06% per degree of curve c. 0.04% per degree of curve d. 0.02% per degree of curve

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 25 52. If the curve lead of a broad gauge railway turnout is 16.76 meters, the angle of crossing of the turnout will be given by IES 2002 1 1 a. ta n 10− 1 b. ta n 5− 1 c. ta n − 1 d. ta n−1 10 5 Curve lead of a BG railway turnout, CL= 16.76 m Angle of crossing of turnout, = ? As per Cole’s method, Number of crossings, N =cot Gauge length, G = 1.676 m CL=2GN 16.762= 1.676 cot

1 −1 1 cot =5 Tan = = Tan  5 5

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 26 54. Squaring of sleepers through packing consists of IES 2001 a. adjusting the sleepers to be perpendicular to the rails b. adjusting the ballast under sleepers to space them parallel to each other c. cutting the edges of the sleepers to a square shape. d. adjusting the rails to be perpendicular to the sleepers.

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 27 59. A cross-over of in exists between two broad gauge parallel tracks with centres at 5 m apart, the length of the straight track is IES 2000 a. 16.4 m b. 18.4 m c. 19.2 m d. 19.92 m Cross over between two BG parallel tracks = 1 in 10 N = 10 Centre to center of rail, D= 5m Length of straight track, S = ? Gauge length, G = 1.676 m S:Intermediate sight distance =−−+()1DGNGN 2 =(5 − 1.676)10 − 1.676 1 + 102 = 16.39 m

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 28 61. If A is the angle formed by two gauge faces, the crossing number will be a. tan A b. cot A c.sec A d.A rad IES 1999

Let be the angle formed by two gauge faces 1 Tan AN= = cot A N PQ: Point rail PR: Another point rail or splice rail QR: Spread at leg of crossing

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 29 62. For a sleeper density of (n + 5), the number of sleepers required for constructing a broad gauge (BG) railway track of length 650 m is IES 1999 a. 975 b. 918 c. 900 d. 880 Sleeper density = n + 5 n: Length of rail Length of BG railway track, = 650 m Number of sleepers required, = ? As per Indian Railways, Standard length of rail, n = 13 m 650 Total number of rails required = = 50 13 Sleeper density = 13 + 5 = 18 N = 50 × 18 = 900

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 30 63. In a BG railway track, the specified ruling gradient is 1 in 250. The horizontal curve of 30 on a gradient of 1 in 250 will have the permissible compensated gradient of IES 1999 a. 1 in 257 b. 1 in 357 c. 1 in 457 d. 1 in 512 Ruling gradient in a BG track = 1 in 250 Degree of curve, on BGtrack D = 30° Permissible gradient = ? Grade compensation for BG track on curve = 3 × 0.04 = 0.12% 10.12 1 Permissible gradient =− = 250100 357

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 31 64. The load on each axle of a locomotive is 22 tonnes. If the coefficient of friction is 0.2 then the hauling capacity due to 3 pair of driving wheels will be IES 1999 a. 26.4 t b. 19.8 t c. 13.2 t d. 6.6 t

Load on each axle of a locomotive, Wd = 225 Coefficient of friction,  = 0.2 Hauling capacity, HC= ? Number of axles, n= 3

HCnW= .. d = 0.2 × 3 × 22 = 13.2 t

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 32 67. What will be the curve lead for a in 8.5 turnout taking off from a straight broad gauge track? IES 1997 a. 28.49 m b. 21.04 m c. 14.24 m d. 7.45 m Curve lead, CL= ? 1 Number of crossings, N = 8 (1 in N) 2 Gauge length of 1.676BGm,G = CL ===2.21.6768.528.492G N m Curve lead is the distance between the theoretical nose of crossing (TNC) and the tangent point T measured along the length of main track.

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 33 70. A train is hauled by 2-8-2 locomotive with 22.5 tones load on each driving axle. Assuming the coefficient of rail-wheel friction to be 0.25, what would be the hauling capacity of the locomotive? IES 1996 a. 15.0 tones b. 22.5 tones c. 45.0 tones d. 90.0 tones Locomotive type: 2-8-2

Load on each driving axle, Wt1 = 22.5 8 driving wheels connected with 4 axles.

Number of axles, n4= Coefficient of rail wheel friction,  = 0.25

Hauling capacity = ..nWd = 0.25 × 4 × 22.5 = 22.5 t Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 34 75. The grade compensation on a curve on a Broad Gauge railway track is a. 0.20% b. 0.16% c. 0.12% d. 0.08% IES 1995 Grade compensation on BG railway track = ? Degree fo curve,D = 40 Grade compensation standards is given below: For BG track: 0.04% per degree of curve For MG track: 0.03% per degree of curve For NG tack: 0.02% per degree of curve Grade compensation for BG track = 4 × 0.04 = 0.16%

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 35 Airport Planning Introduction: • Airport planning is a systematic process used to establish guidelines for the efficient development of airports that is consistent with local, state and national goals. • A key objective of airport planning is to assure the effective use of airport resources in order to satisfy aviation demand in a financially feasible manner. • Airport planning may be as broad based as the national system plan or more centrally focused as an airport master plan for a specific airport.

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 36 The primary types of airport planning may basically be classified as follows: • National System Planning • State Airport System Planning • Metropolitan Airport System Planning • Airport Master Planning Airports are classified by two organizations: • ICAO: International Civil Aviation Organization. • FAA: Federal Aviation Agency. ICAO classified airport into two categories- i. Based on basic runway length A → longest runway B → shortest runway ii. Equivalent Single Wheel Load of the aircraft.

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 37 A runway of at least 1,800 m in length is usually adequate for aircraft weights below approximately 91 t. For larger aircraft including widebodies will usually require at least 2,400 m at sea level and some what more at higher altitude airports.

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 38 Important Points related to Wind- Rose Diagrams: i. Length of runway requirements will be more if landing and take-off operation are performed along the wind direction. ii. Wind parameters (direction and intensity) are graphically represented by wind rose diagrams. iii. Wind parameters should be collected for a period of 3 years.

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 39 iv. Normal component of the wind is cross-wind component and it may interrupt safe landing and take-off of the aircraft. • Maximum Permissible limit of normal component of the wind for the smaller size of aircraft is 15 km/hour and that for bigger aircraft is 25 km/hour • Wind coverage is the percentage of time during which in a year, the cross wind component remains within the permissible limit. • Wind Rose Diagrams, showing wind direction, duration and intensity are an essential requirement for planning the best direction of main runway of the airport. • The orientation of runway is decided by • Maximum wind coverage and least cross winds • Scope for future expansion • Obstruction-free approaches

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 40 Basic Runway length requirements • Airport altitude is at MSL. • Temperature of airport is standardized at 15°c at MSL. • Runway is levelled in the longitudinal direction. • Aircraft is loaded through its full loading capacity. • Speed of wind should be zero on the runway. Correction for Elevation • ICAO recommends that the basic runway length should be increased by 7% per 300 m rise in elevation above MSL.

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 41 Correction for temperature Standard Atmospheric temperature at rest altitude is calculated as = 150C - [0.0065×change in Elevation above MSL] Airport Reference Temperature is the Local body temperature of a particular area and it is defined for hottest month of the year.

TTma− ARTT=+a  3 Tm = Monthly mean of maximum daily temperature. Ta = Monthly mean of average daily temperature. • ICAO recommends that the basic runway length after corrected for elevation should be increased by 1% for every 1°C rise of temperature above the standard atmospheric temperature at that elevation. • If the total correction for elevation and temperature is less than or equal to 35% of basic runway length, no corrections are made. If the correction is greater than 35%, scientific analysis must be performed at site conditions again.

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 42 Correction for gradient After corrected runway length for elevation and temperature, runway length should be increased by 20% for every 1% of effective gradient.

• In case of take-off, elevation, temperature and gradient corrections are necessary. • In case of landing, only elevation correction is necessary.

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 43 For normal take off condition, clearway shall not be more than 1 (1.5 Take off distance –1.15Lift off diste anc ) 2

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 44 eg. Basic runway length = 1620 m. ART = 32.94°c Elevation of airport site = 270 m above MSL. Effective Gradient = 0.2% Corrected runway length=?. 7270 Ans. Correction for elevation == 102.06m 100300 Corrected runway length = 1620 + 102.06 = 1722.06 m. Change in standard temperature = 15° – (0.0065×change in elevation above MSL) = 15° – (0.0065 ×270)= 13.24°C

TTma− 0 ARTTC=+=a 32.94 3

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 45 =−=T0000 CCCC32.9413.2419.7 1 Correction for temperature1722.06== 19.7 339 .24m 100 Corrected runway length= 2061.3m 2061.3− 1620 Check100= 27.2% = 35% 1620 200.2% Correction for gradient2061.382.45== m 1001% Corrected runway length=+= 2061.3 82.45 2143.75m

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 46 Example:

Basic runway length = 1800 m. Elevation of airport site = 600 m, Ta = 15°C, Tm = 21.6°C, Effective gradient = 0.6% Ans: 7 600 Correction for elevation=  1800  = 252m 100 300 Corrected length= 1800 + 252 = 2052m Temperature correction:

21.6− 15 0 ART= 25 + = 17.2 C 3 Standard temperature= 1500 − (0.0065  600) = 11.1 C Rise in ART= 17.2 − 11.1 = 6.10 C

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 47 16.1 0 C Correction for temperature2052125.2m== 1001 0 C So Corrected length=+= 2052 125.2 2177.2m 2177.2− 1800 Check100== 20.954% 35% 1800 200.6% Correction for gradient2177.2261.3m== 1001% Final corrected length=+= 2177.2 261.3 2438.5m

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 48 Taxiway system: Factors are to be considered while designing the Taxiway system • Volume of air traffic • Runway configuration • Location of terminal building • Location of hangars Turning radius at taxiway is maximum of V 2 1. R = R : Radius of curve, m 125 f V: Speed, km/hr 0.388W 2 2.R = f : Friction coefficient = 0.13 T − s W : Wheel base, m 2 T : Width of Taxiway, m (22.2m) 3.R 180 m for supersonic 4.R>120 m for Subsonic jets Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 49 Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 50 eg. Evaluate taxiway radius for supersonic aircraft with W = 35 m Tread of landing gear = 7.5 m. Speed of aircraft = 50 km/hr V 2250 1. R153.8mm= == 125f 125 0.3 0.388W 2 2. R = T − S 2 TLG 7.5 Let T== 22.5m, + S66 = + 9.75m = 22 0.388 35 35 R ==316.86m 22.5 − 9.75 2 3. For supersonic, R 180 Hence,R = 316.86m

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 51 15.8 Airport Engineering (GATE)

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 52 01.The appropriate design length of a clearway is calculated on the basis of ‘Normal Take- off’ condition. Which one of the following options correctly depicts the length of the clearway? (Note: None of the options are drawn to scale) GATE CE 2020

a. b.

c. d.

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 53 01.c For normal take off condition, 1 Clearway  (1.5 Take off distance –1.15 lift off distacn e) 2 1  (1.151625 –1.15 875) 2  431.25

Clearway is less than 431m.

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 54 02.For the hottest month of the year at the proposed airport site, the monthly mean of the average daily temperature is 39°C. The monthly mean of the maximum daily temperature is 48°C for the same month of the year. From the given information, the calculated Airport Reference Temperature (in °C), is GATE CE 2020 a. 36 b. 39 c. 42 d. 48 02.c 0 Monthly mean of the average daily temperature, TCa = 39 0 Monthly mean of the maximum daily temperature, TCm = 48 TT− Airport reference temperature =+ T ma a 3 4839− =+=3942 0 C 3

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 55 03. An aircraft approaches the threshold of a runway strip at a speed of 200 km/h. The pilot decelerates the aircraft at a rate of 1.697 m/s2 and takes 18 s to exit the runway strip. If the deceleration after existing the runway is 1 m/s2, then the distance (in m, up to one decimal place) of the gate position for the location of exit on the runway is ….. 03. 312.85 CE1 2018 5 Speed of aircraft = 200 km/h = 200 = 55.56 m/s 18 Deceleration of aircraft, a = 1.697 m/s2. Time taken to exit the runway strip = 18 s Speed of the aircraft at the exit of the runway, v = u + at = 55.56 – 1.697×18 = 25.014 m/s After existing the runway, deceleration = 1 m/s2. Distance of the gate position from the location of exit on the runway, s=? v22−= u2 as 0− (25.014)2 = − 2  1 SS  = 312.85m Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 56 04. A runway is being constructed in a new airport as per the International Civil Aviation Organization (ICAO) recommendations. The elevation and the airport reference temperature of this airport are 535 m above the mean sea level and 22.65°C, respectively. Consider the effective gradient of runway as 1%. The length of runway required for a design-aircraft under the standard conditions is 2000 m. Within the framework of applying sequential corrections as per the ICAO recommendations, the length of runway corrected for the temperature is ……. CE1 2017 a.2223 m b. 2250 m c.2500 m d.2750 m 04. c Elevation of airport above MSL = 535 m Airport reference temperature = 22.65°C Effective gradient of Runway = 1% Length of runway required for a design aircraft under standard conditions =2000 m

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 57 Correction for elevation = 7% increase per 300 m height above MSL 535 Correction =0.072000 = 249.7 m 300 Corrected length = 2000 + 249.7 = 2249.7 m Correction for temperature: Standard atmospheric temperature at MSL = 15°C Reduction of temperature at 0.0065° C per m above MSL Standard atmospheric temperature at Airport = 15-0.0065×535 =11.52°C Change in temperature, = 22.65 – 11.52= 11.13°C 11.13 Correction =2249.7 = 250.32 m 100 The corrected length of runway for temperature = 2249.7 + 250.32= 2500.02 m Total correction of runway should not be more than 35% 2500.02− 2000 Total correction = = 100 25% < 35% ok 2000 Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 58 15.8 Airport Engineering (IES)

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 59 01.Which one of the following instance of performance of aircraft is not considered for determining basic runway length? IES 2019 a. Normal landing case b. Normal take-off case c. Engine failure case d. Emergency landing case

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 60 02.Conisder the following data for designing a taxiway for operating Boeing 707-320 aeroplane : IES 2019 Wheel base =17.70m Tread of main loading gear=6.62 m Turning speed =40 km/h Coefficient of friction between tyres and pavement surface =0.13 The turning radius of the taxiway will be a. 98.5m b.94.5m c.89.5m d.86.5m 02.a Wheel base, b=17.7 m Tread of main loading gear =6.62 m Turning speed, V = 40 km/h Coefficient of friction between tyres and pavement surface, f = 0.13 Turning radius of taxi way, R = ? V 2240 R = = = 98.5m 125f 125 0.13 Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 61 03. What shall be the radius of an exit taxiway with design exit speed of 90 kmph and coefficient of friction 0.13? IES 2015 a. 550 m b. 500 m c. 475 m d. 449 m Radius of exit taxiway, R = ? Design exit speed, V = 90 kmph Coefficient of friction, f = 0.13

V 2290 fR=== 0.13498.5m 125125RR

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 62 04. The magnetic azimuth of one end of a runway is 80° measured clockwise from the magnetic north. The other end of the runway will be numbered as IES 2015 a. 16 b. 24 c. 26 d. 8 Magnetic azimuth of one end of runway = 80°

Magnetic azimuth of other end of runway = 80° + 180° = 260° Therefore, the other end of runway will be numbered as 26.

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 63 05. The runway length for an airport located at 460 m above MSL, corrected for elevation, is 3670 m. The monthly means of maximum and mean daily temperatures for the hottest month of the year are 27°C and 18°C, respectively. What will be the final corrected length of the runway with correction incorporated also due to temperature effects? IES 2015 a. 4500 m b. 4000 m c. 3750 m d. 3400 m Location of runway above MSL = 460 m Runway length corrected for elevation, L = 3670 m

Monthly mean of maximum daily temperature, Tm = 27°C Monthly mean of average daily temperature, Ta = 18°C 1 1 Airport reference temperature, TTTT=+− ( ) =+−18(2718) = 210 C ama 3 3

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 64 Correction for temperature: 1% for every 1°C rise of airport reference temperature from standard atmospheric temperature. Standard temperature at runway = 15°-0.0065×460 = 12.01°C Correction for temperature = 0.01×3670 (21-12.01) = 329.9 m Corrected length of runway = 3670 + 329.9 = 3999.9 = 4000 m

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 65 06. The lowest height above the runway where the pilots make the decision to continue the landing manoeuvre or to cut it short is called the IES 2015 a. Runway height b. Decision height c. Threshold height d. Runway visual range

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 66 09. For safe landing and take-off the following factors need to be carefully considered: 1. Cross-wind 2. Runway grade 3. Runway width and side clearance 4. Obstructions a. 1,2 and 3 only b. 1,2,3 and 4 c. 1,3 and 4 only d. 2,3 and 4 only

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 67 10. Wind-rose diagram is useful in deciding on the orientation of IES2011 a. Taxiway b. Hanger c. Apron d. Runway

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 68 16. The orientation of runway is decided by which factors? IES 2008 1. Maximum wind coverage and least cross winds 2. Landing characteristics of aircrafts 3. Scope for future expansion 4. Obstruction-free approaches Select the correct answer using the codes given below: a. 1 and 2 only b. 1, 2 and 3 c. 1, 3 and 4 d. 2, 3 and 4

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 69 17. As the elevation increases, the runway length has to be changed at what rate? a. Decreased @ 7% per 300 m rise in elevation above MSL b. Increased @ 7% per 300 m rise in elevation above MSL c. Decreased @ 5% per 300 m rise in elevation above MSL d. Increased @ 5% per 300 m rise in elevation above MSL IES 2007

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 70 18. The total correction percentage for altitude and temperature, in calculating the runway length from basic runway length, normally does not exceed. IES 2006 a. 7 b. 14 c. 28 d. 35

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 71 th 19. If Mi = Proportion of i type of aircraft in the mix, IES 2006 th Ti = “Gate occupancy time” of the i type of aircraft, G = Number of gates (considering that all the available gates can be used by all aircrafts) Which one of the following equations gives the ultimate gate capacity C ? n GM  i G 60 n M G a. C = i=1 b. C = c. i d.C = n n CT=i n T i=1 G  i  MTii  MTii i=1 i=1 i=1

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 72 20. For design of a runway length, match List-I (Factor) with List-II (Correction) and select the correct answer using the codes given below the list: IES 2006 List-I List-II A. Standard basic length 1. 7% for every 300 metres B. Elevation 2. 1% for every 1°C rise C. Temperature 3. 20% for every 1% of effective gradient D. Effective gradient 4. Depends upon aircraft and obtained from Standard Tables a. A4 B1 C2 D3 b. A2 B3 C4 D1 c. A4 B3 C2 D1 d. A2 B1 C4 d3

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 73 21. A runway is located 450 m above the mean sea level. If the aeroplane reference field is 1800 m, what is the approximate corrected runway length for elevation? a. 1849 m b. 1889 m c. 1987 m d. 2013 m IES 2006 Location of airport above MSL = 450 m Aeroplane reference field, L = 1800 m

Corrected runway length for elevation, L1 = ? Correction for elevation = 7% of basic runway length per 300 m rise in elevation 450 =0.071800 =189m 300

L1 = 1800 + 189 = 1989 m

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 74 22. At a certain station, the mean of the average temperature is 250 C and mean of the maximum daily temperature is 400 C. What is the airport reference temperature (ART)? IES 2005 a. 46.60 C b. 450 C c. 350 C d. 300 C

Mean of the daily average temperature, Ta = 25°C Mean of the daily maximum temperature, Tm = 40°C Airport reference temperature, T = ?

1 1 TTTT= +( − ) =25 + (40 − 25) = 300 C a3 m a 3

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 75 23. Match List-I (Unit) with List-II (Purpose) and select the correct answer using the code given below the lists: IES 2005 List-I List-II A. Width and length of safety area of airport 1. Basic runway length B. Engine failure case 2. Runway geometric design C. Location of exit taxiways 3. Airport drainage D. Gradient of airport size 4. Runway capacity a. A2 B4 C1 D3 b. A3 B1 C4 D2 c. A2 B1 C4 D3 d. A3 B4 C1 D2

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 76 24. Which one of the following gives the number of gate positions in an airport? Capacityof runway a. ×averagegateoccupancy time 60×2 Capacityof apron b. ×number of aircraft movements 60×2 Capacityof taxiway c. ×average gate occupancy time 60×2 Capacityof holding apron d. ×average gate occupancy time IES 2005 60×2

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 77 25. Consider the following aircraft operations: IES 2004 1. Normal landing 2. Normal take-off with all engines 3. Engine failure at take-off 4. Emergency landing with all engines shut 5. Landing with maximum payload with the help of ILS Which of the above aircraft operations are taken into consideration in deciding the basic runway length required for an aircraft? a. 1, 2 and 3 b. 2, 3 and 4 c. 3, 4 and 5 d. 1 and 5

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 78 26. The orientation of preferential runway in an airport is influenced by IES 2004 a. direction of prevailing wind, adequate length, obstruction-free landing and take-off zones. b. adequate waiting and service facilities c. convenience of terminal and control facilities d. stable ground and adequate turning space.

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 79 28. Which one of the following imaginary surface in airport is circular in plan with centre located at an elevation of 150 m above the airport reference point? a. Conical surface b. Transitional surface c. Inner horizontal surface d. Outer horizontal surface IES 2003

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 80 29. The monthly mean of maximum daily temperature and monthly mean of average daily temperature of the hottest month of the year are 490 C and 400 C respectively. Then airport reference temperature is IES 2003 a. 430 C b. 69.60 C c. 370 C d. 520 C

Monthly mean of maximum daily temperature, Tm = 49°C Monthly mean of average daily temperature, Ta = 40°C Airport reference temperature, T = ? 1 1 0 TTTT=a +( m − a ) =40 + (49 − 40) = 43 C 3 3

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 81 33. Which of the following factors are to be taken into consideration while designing the taxiway system? IES 2001 1. Volume of air traffic 2. Runway configuration 3. Location of terminal building 4. Location of hangars Select the correct answer using the codes given below: a. 1, 2 and 3 b. 2 and 3 c. 1 and 4 d. 1, 2, 3 and 4

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 82 35. For a runway at an elevation of 1000 m above MSL and airport reference temperature of 160 C , the rise in temperature to be taken into account as per ICAO is IES 2000 a. 24.50 C b. 150 C c. 7.50 C d. 60 C Elevation of runway above MSL = 1000 m Airport reference temperature, T = 16°C Temperature at a given elevation = 15° - 0.0065×1000 = 8.5°C Correction for temperature: Rise of temperature = 16-8.5 = 7.5°C

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 83 37. The runway length after correcting for elevation and temperature is 2845 m. If the effective gradient on runway is 0.5 per cent, the revised runway length will be a. 2845 m b. 2910 m c. 3030 m d. 3130 m IES 1999 Runway length after correcting for elevation and temperature, L = 2845 m Effective gradient on runway = 0.5% Revised runway length = ? For every 1% of gradient, increase in length of runway is at the rate of 20% Correction for gradient = 0.2×2845×0.5 = 284.5 m Revised length of runway = 2845 + 2845 = 3129.5 m

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 84 40. In an airport, If 4 groups of 5 gates each located well-separated are considered for traffic and the future-to present traffic ratio is 3, then the total requirement of future gates will be IES 1998 a. 32 b. 36 c. 44 d. 68

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 85 41. Which of the following factors are taken into account for estimating the runway length required for aircraft landing? IES 1997 1. Normal maximum temperature 2. Airport elevation 3. Maximum landing weight 4. Effective runway gradient. Select the correct answer using the codes given below:- a. 1, 2, 3 and 4 b. 1, 3 and 4 c. 2 and 4 d. 1, 2 and 4

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 86 42. Consider the following statements: Wind rose diagram is used for the purpose(s) of IES 1997 1. runway orientation 2. estimating the runway capacity 3. geometric design of holding apron Which of these statements is/are correct? a. 1 and 2 b. 2 and 3 c. 1 and 3 d. 1 alone

Prof. B. Jayarami Reddy 6/22/2020 Y.S.R. ENGINEERING COLLEGE OF YOGI VEMANA UNIVERSITY, PRODDATUR 87