Highway design raport(final) (group 2)

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Highway design raport(final) (group 2)

  1. 1. Highway Design Senior Project 2010Part- ISection-1: IntroductionThere is a growing universal demand for well prepared professionals in all disciplines. Inaddition, increased pressure has consequently been placed in educational institution to preparethe required number of qualified professional to fulfill society’s need. It is imperative that thereis a large need in the industry for engineers with training and experience, and the academicshould move successfully to fill the need. This is especially true for in the situation of Ethiopiawhere there is a lack of well trained and experienced urban engineer’s.Therefore, the integration of academic program and exposing students to more practical projectresults in well-seasoned and, well-educated professionals.Thus, this high way design project is intended to equip the students with practical designreinforcing what they have attained theoretically in the class.It is already known that, for rapid economic, industrial and cultural growth of any country, agood system of transportation is very essential. One of the transportation systems that areeconomical for developing countries like Ethiopia is road. A well – designed road network playsan important role in transporting people and other industrial products to any direction with inshort time. Roads, to satisfy their intended purpose, must be constructed to be safe, easy,economical, environmentally friend and must full fill the needs of inhabitants. Being safe, thenumber of accidents that can occur will be minimized. Easiness decreases operation cost,pollution and even time cost. Economical roads assure their feasibility according to their plansand initiate further construction of roads. Schemes that do not satisfy the needs of localities maynot get the maximum utilization of the surplus man power that is really to exist in the ruralcommunity and also its economical value may also decrease. Therefore, from this project it isexpected to understand and to get acquainted with the above facts by going through on thefollowing design aspects.1.1 General Background 1ECSC, IUDS, Urban Engineering Department (UE)
  2. 2. Highway Design Senior Project 2010This high way design project is taken from the Hargele - Afder – Bare - Yet road project, whichis located in the Eastern part of the country in Somali National Regional State, AfderAdministrative Zone, Afder and Bare Woredas. The project is intended to facilitate the existingand for the expected traffic load in the future, because the town is developing.From this road we have given a stretch of 3 km emanating from station 12+500 to 15+500 forthis project to do geometric and pavement design in general.1.2 ObjectivesThis final year design project on high-way has the following major objectives:-  To expose the prospective graduates to a detail and organized design on road projects;  To implement the knowledge that the prospective graduates have learned theoretically in classes;  To ensure a good carrier development;1.3 Brief Description of The Project AreaThe Hargele - Afder – Bare - Yet road project, is located in the Eastern part of the country inSomali National Regional State, Afder Administrative Zone, Afder and Bare Woredas. Theproject starts at Hargele (5º13’N and 42º 11’E) and pass through Hargele, Afder, Bare, town andends at Yet. The project length is estimated to be 142.4km. The Location map together with thetopographic map of the project area is shown below. Fig. 1.1 Project Location Map 2ECSC, IUDS, Urban Engineering Department (UE)
  3. 3. Highway Design Senior Project 2010 Location of the Project Road Fig. 1.3.3 Digitized Proposed Project Alternative Alignments 3ECSC, IUDS, Urban Engineering Department (UE)
  4. 4. Highway Design Senior Project 2010Climate:One of the environmental factors that affect performance of pavements structures is climate.Hence, climate data of the project area mainly rainfall intensity, in terms of mean monthly andmean annual and, temperature are required. According to the map shown on National Atlas ofEthiopian Atlas, the project area is located in the region of the lowest annual rainfall. The meanannual rainfall in this region is 300mm per year. The rainfall of the project area is characterizedby the following rainfall distribution:  April, May and October  The wettest Months  And in the remaining months  The driest months.Topography:The terrain of the project area through which the road alignment traverses is rolling in substantialsection of the project which is intercepted by mountainous terrain in some sections.Potential of the area:In the project area limited crop production, livestock and livestock products are available in thearea of influence of the road project even though the area is under attention to reverse fooddeficit. There is an initiative to change the area that the potential resources of oil mining and saltproduction may attract private investors and governmental agencies.1.4 Scope of the projectThe scope of the project goes as far as designing the geometry and pavement of a given roadsection, with its appropriate drainage structures. 4ECSC, IUDS, Urban Engineering Department (UE)
  5. 5. Highway Design Senior Project 2010Section-2: Geometric design2.1 Geometric design Control and Criteria2.1.1 Terrain classification2.1.1.1 Contour generationThe surveying data x, Y and Z coordinate taken from the road corridor using Hand Held GPS areconverted to a contour using GIS software.2.1.1.2 Selection of center lineThe center line of the road is delineated on the given road corridor using the contour elevationsby considering to have minimum earth work along the corridor.2.1.1.3 Transverse terrain propertyIn order to know the type of the terrain along the selected center line or corridor, we tookhorizontal distance perpendicular to the center line and vertical elevation measurements acrossthe road. Each measurement is taken longitudinally along the rod at 20m interval to get betterterrain classification. The values obtained are summarized in index table 2-1.Slop= (vertical elevation / horizontal elevation)*100Therefore, we generalize the following terrains classification along the road corridor: STATION From To TERRAIN AVG. SLOPE CLASSIFICATION (%) 12+ 500 12+ 760 Rolling 23.14 12 + 760 13+ 080 Mountainous 26.63 13 + 080 13+ 520 Rolling 18.75 13 + 520 13+ 820 Mountainous 32.234 13 + 820 15 +500 Rolling 16.87 Table 2-2 Terrain Classification 5ECSC, IUDS, Urban Engineering Department (UE)
  6. 6. Highway Design Senior Project 2010Fig 2-1 Generated contour.2.1.2 Design traffic volume 6ECSC, IUDS, Urban Engineering Department (UE)
  7. 7. Highway Design Senior Project 20102.1.2.1 Traffic data analysisIn order to design the road, traffic data analysis is very important. Therefore, the secondary dataof traffic analysis we get from the project site comprises traffic volume before design, duringimplementation and up to the design life time of the road. As the secondary data shows theproject life is 15 year. The traffic volume data and the design life time are expressed in thefollowing table. T& Year Car 4 WD S/ Bus L/ Bus S/ Truck M/ Truck L/ Truck TOTAL T 2008 0 4 6 2 12 4 2 14 44 2009 0 5 7 2 13 5 3 16 51 2010 0 5 7 2 14 5 3 16 52 2011 0 6 8 3 14 5 3 17 56 2012 0 6 8 3 15 5 3 18 58 2013 0 15 16 6 31 20 28 34 149 2014 0 16 17 7 34 21 30 37 160 2015 0 19 19 8 36 22 32 39 174 2016 0 19 21 8 38 25 35 41 184 2017 0 19 21 9 40 26 36 44 193 2018 0 20 22 9 43 28 38 46 205 2019 0 21 25 11 44 31 42 49 221 2020 0 22 26 11 47 32 44 52 232 2021 0 22 26 12 49 34 46 53 241 2022 0 22 29 12 52 35 48 56 253 2023 0 25 30 13 55 36 51 59 267 2024 0 25 32 13 57 39 54 60 279 2025 0 26 33 14 60 40 57 64 292 2026 0 27 34 14 62 43 60 67 307 2027 0 28 37 16 66 44 63 70 323Table 2-3 Traffic data analysisFrom the above data, o Traffic volume when the road open =149 veh/day o Traffic volume at the end of the project life =323 veh/day 7ECSC, IUDS, Urban Engineering Department (UE)
  8. 8. Highway Design Senior Project 20102.1.3 Road functional classificationSome of the factors which affect road design control and criteria are functional classification ofthe road. In Ethiopian case, we have five functional classes based on AADT and importance ofthe road.Since, AADT of the project lies between 200-1000, and the road expected to serve centers ofprovisional importance, the road could be main access road (class II).2.2 Geometric Design StandardBased on the traffic data obtained from the above table we decide the project design standard tobe (DS4).Because:- a) Even if the AADT at the opening of the road (2013) is 149 veh/day it will be greater than 200 veh/ day after five year and it is 323 veh/day at the end of design life (15 years). So it fulfills the requirements of DS4. Since the recommended traffic volume for DS4 is 200- 1000 veh/day.(ERA) b) The second reason is that since the area is an oil mining area, we expect the road will accommodate the expected traffic volume during the design life time. c) Based on the above reason, we decide the road to be DS4, to get full knowledge from the whole project since the project is for academic purpose.Therefore, we took the entire design element based on DS4. Refer the above information fromERA manual Table 2.1.From Design Standards vs. Road Classification and AADT table of ERA for DS4,AADT=200 – 1000 vehicle/daySurface type = pavedCarriageway = 6.7mShoulder width =1.5m for rolling = 0.5m for mountainous 8ECSC, IUDS, Urban Engineering Department (UE)
  9. 9. Highway Design Senior Project 2010Design speed = 70km/hr for rolling = 60km/hr =for mountainous2.2.1 Horizontal AlignmentBased on our proposal of the center line of the road, we have tangents and curves. The curves arecurve1, curve2, curve3, curve4, curve5, and curve6.Based on our terrain classification, the curves fall in to different terrain classification that leadsus to determine the radius and different elements of each curve. Curve Terrain type Curve 1 Rolling Curve 2 Rolling Curve 3 Rolling Curve 4 Error! Not a valid link. Curve 5 Rolling Curve 6 RollingTable 2-4 Horizontal curves and their terrain classificationSince our road is DS4, the minimum radius of each curve based on the terrain is:- Minimum horizontal radius = 175m for rolling = 125m for mountainousRefer the following table for the rest of the design elements of DS4 (ERA standards) Design Element Unit Flat Rolling Mountainous Escarpment Urban/Peri- Urban Design Speed km/h 85 70 60 50 50Min. Stopping Sight Distance m 155 110 85 55 55Min. Passing Sight Distance m 340 275 225 175 175 % Passing Opportunity % 25 25 15 0 20Min. Horizontal Curve Radius m 270 175 125 85 85 9ECSC, IUDS, Urban Engineering Department (UE)
  10. 10. Highway Design Senior Project 2010 Transition Curves Required Yes Yes No No No Max. Gradient (desirable) % 4 5 7 7 7 Max. Gradient (absolute) % 6 7 9 9 9 Minimum Gradient % 0.5 0.5 0.5 0.5 0.5 Maximum Super elevation % 8 8 8 8 4 Crest Vertical Curve k 60 31 18 10 10 Sag Vertical Curve k 36 25 18 12 12 Normal Cross fall % 2.5 2.5 2.5 2.5 2.5 Shoulder Cross fall % 4 4 4 4 4 Right of Way m 50 50 50 50 50Table 2-5: Table 2-6 of ERA Geometric Design Parameters for Design Standard DS4 (Paved)2.2.1.1 Horizontal curve elementsCurve-1 Design computationa) Terrain type = Rollingb) Deflection angle Δ = 390 (by measurement)c) Point of intersection P.I=12+717.4md) Calculation of radius of the curve Vd 2 Rmin = 127(e + f )Where, Rmin=minimum radius Vd=70km/hr…………….ERA, table 2.6 ed= 8% (max design super elevation rate, ERA, table 2.6) f=0.14 (ERA. Table 8.1 for ed=8%) 70 2Then, Rmin = =175.3m 127(0.08 + 0.14) 10ECSC, IUDS, Urban Engineering Department (UE)
  11. 11. Highway Design Senior Project 2010The calculated Rmin has no significant change from the recommended in ERA manual standard(i.e., 175m), in addition to this, in order to minimize cut and fill, we use R min=175m from thestandard.Therefore, radius of curve=Rc=175me) Tangent (T1)  ∆ T1 = R * tan  2  39  T1 = 175 * tan   = 61.97 m  2 f) Point of curvature (PC) P.C1= P.I1 - T1 =12+717.4 – 0+061.97 =12+655.43mg) Length of the curve (L)  2Π  L1 = ∆ * R *    360   2Π  L1 = 390 *175 *   = 119.12m  360 h) Point of tangency (P.T) P.T1= P.C1+L1 =12+655.43+119.12 =12+774.55mi) External distance (E) 11ECSC, IUDS, Urban Engineering Department (UE)
  12. 12. Highway Design Senior Project 2010  ∆  E1 = R * sec  − 1  2    39   E1 = 175 * sec  −1 = 10.65m   2  j) Middle ordinate (M)   ∆  M 1 = R * 1 − cos    2    39  M 1 = 175 * 1 − cos  = 10.04m   2 k) Chord (Chord from P.C to P.T) ∆ C1 = 2 R sin   2  39 C1 = 2 *175 * sin   = 116.83m  2 Fig.2.2 elements 0f curve-1 12ECSC, IUDS, Urban Engineering Department (UE)
  13. 13. Highway Design Senior Project 2010Curve-2 Design computationa) Terrain type = Rollingb) Deflection angle Δ = 330 (by measurement)c) Point of intersection P.I=13+150.43md) Calculation of radius of the curve Vd 2 Rmin = 127(e + f )Where, Rmin=minimum radius Vd=70km/hr…………….ERA, table 2.6 ed= 8% (max design super elevation rate, ERA, table 2.6) f=0.14 (ERA. Table 8.1 for ed=8%) 70 2Then, Rmin = =175.3m 127(0.08 + 0.14)The calculated Rmin has no significant change from the recommended in ERA manual standard(i.e., 175m), in addition to this to prevent overlaps with curve 3, we use Rmin=175m from thestandard.Therefore, radius of curve=Rc=175me) Tangent (T1) Rmin = 175m  ∆T2 = R * tan   2  33 T2 = 175 * tan   = 51.84m  2 f) Point of curvature (PC) 13ECSC, IUDS, Urban Engineering Department (UE)
  14. 14. Highway Design Senior Project 2010 P.C2= P.I2 – T2 =13+150.43– 0+051.84 =13+098.59mg) Length of the curve (L)  2Π  L2 = ∆ * R *    360   2Π  L2 = 330 *175 *   = 100.79m  360 h) Point of tangency (P.T) P.T2= P.C2+L2 =13+98.59+100.79 =13+199.38mi) External distance (E)  ∆  E2 = R * sec  − 1  2    33   E2 = 175 * sec  −1 = 7.52m   2  j) Middle ordinate (M)   ∆  M 2 = R * 1 − cos    2    33  M 2 = 175 * 1 − cos  = 7.21m   2 k) Chord (Chord from P.C to P.T) 14ECSC, IUDS, Urban Engineering Department (UE)
  15. 15. Highway Design Senior Project 2010  ∆ C2 = 2 R sin   2  33  C2 = 2 *175 * sin   = 99.41m  2 Fig 2.3 elements of curve-2Curve-3 Design computationa) Terrain type = Rollingb) Deflection angle Δ = 59.620 (by measurement)c) Point of intersection P.I=13+363.64md) Calculation of radius of the curve Vd 2 Rmin = 127(e + f )Where, Rmin=minimum radius Vd=70km/hr…………….ERA, table 2.6 ed= 8% (max design super elevation rate, ERA, table 2.6) f=0.14 (ERA. Table 8.1 for ed=8%) 15ECSC, IUDS, Urban Engineering Department (UE)
  16. 16. Highway Design Senior Project 2010 70 2Then, Rmin = =175.3m 127(0.08 + 0.14)The calculated Rmin has no significant change from the recommended in ERA manual standard(i.e., 175m), in addition to this to prevent overlaps with curve 2, we use R min=175m from thestandard.Therefore, radius of curve=Rc=175me) Tangent (T3) Rmin = 175m ∆ T3 = R * tan   2  59.62  T3 = 175 * tan   = 100.26m  2 f) Point of curvature (PC) P.C3= P.I3 - T3 =13+363.64– 0+100.26 =13+263.38mg) Length of the curve (L)  2Π  L3 = ∆ * R *    360   2Π  L3 = 59.62 0 *175 *   = 182m  360 h) Point of tangency (P.T) 16ECSC, IUDS, Urban Engineering Department (UE)
  17. 17. Highway Design Senior Project 2010 P.T3= P.C3+L3 =13+263.38+182m =13+445.38mi) External distance (E)  ∆  E3 = R * sec  − 1  2    59.62   E3 = 175 * sec  −1 = 26.69m   2  j) Middle ordinate (M)   ∆  M 3 = R * 1 − cos    2    59.62  M 3 = 175 * 1 − cos  = 23.17 m   2 k) Chord (Chord from P.C to P.T) ∆ C3 = 2 R sin   2  59.62  C3 = 2 *175 * sin   = 173.99m  2 Curve-4 Design computationa) Terrain type = Rollingb) Deflection angle Δ = 90.810 (by measurement)c) Point of intersection P.I=14+045.5md) Calculation of radius of the curve 17ECSC, IUDS, Urban Engineering Department (UE)
  18. 18. Highway Design Senior Project 2010 Vd 2 Rmin = 127(e + f )Where, Rmin=minimum radius Vd=70km/hr…………….ERA, table 2.6 ed= 8% (max design super elevation rate, ERA, table 2.6) f=0.14 (ERA. Table 8.1 for ed=8%) 70 2Then, Rmin = =175.37 m 127(0.08 + 0.14)The calculated Rmin has no significant change from the recommended in ERA manual standard(i.e., 175m), so we use Rmin=175m from the standard.But to make the curve smooth, we took R=236m, I.e. =RC=236me) Tangent (T4) R = 236m ∆ T4 = R * tan  2  90.81  T4 = 236 * tan   = 239m  2 f) Point of curvature (PC) P.C4= P.I4 – T4 =14+045.5– 0+239 =13+806.5mg) Length of the curve (L)  2Π  L4 = ∆ * R *    360  18ECSC, IUDS, Urban Engineering Department (UE)
  19. 19. Highway Design Senior Project 2010  2Π  L4 = 90.810 * 236 *   = 374.m  360 h) Point of tangency (P.T) P.T4= P.C4+L4 =13+806.5+374m =14+180.5mi) External distance (E)  ∆  E4 = R * sec  − 1  2    90.81   E4 = 236 * sec  −1 = 100.12m   2  j) Middle ordinate (M)   ∆  M 4 = R * 1 − cos    2    90.810  M 4 = 236 * 1 − cos  2  = 70.31m    k) Chord (Chord from P.C to P.T)  ∆ C4 = 2 R sin   2  90.810  C 4 = 2 * 236 * sin   2  = 336.10 m   Curve-5 Design computationa) Terrain type = Rolling 19ECSC, IUDS, Urban Engineering Department (UE)
  20. 20. Highway Design Senior Project 2010b) Deflection angle Δ = 44.150 (by measurement)c) Point of intersection P.I=14+756.69md) Calculation of radius of the curve Vd 2 Rmin = 127(e + f )Where, Rmin=minimum radius Vd=70km/hr…………….ERA, table 2.6 ed= 8% (max design super elevation rate, ERA, table 2.6) f=0.14 (ERA. Table 8.1 for ed=8%) 70 2Then, Rmin = =175.4m 127(0.08 + 0.14)The calculated Rmin has no significant change from the recommended in ERA manual standard(i.e., 175m), in addition to this, in order to minimize cut and fill, we use R min=175m from thestandard.Therefore, radius of curve=Rc=175me) Tangent (T5) Rmin = 175m ∆ T5 = R * tan  2  44.15  T5 = 175 * tan   = 70.97 m  2 f) Point of curvature (PC) P.C5= P.I5 – T5 =14+756.69– 0+70.97m 20ECSC, IUDS, Urban Engineering Department (UE)
  21. 21. Highway Design Senior Project 2010 =14+685.72mg) Length of the curve (L)  2Π  L5 = ∆ * R *    360   2Π  L5 = 44.150 *175 *   = 134.85m  360 h) Point of tangency (P.T) P.T5= P.C5+L5 =14+685.72+134.85m =14+820.57mi) External distance (E)  ∆  E5 = R * sec  − 1  2    44.150   E5 = 175 * sec   −1 = 13.84m    2  j) Middle ordinate (M)   ∆  M 5 = R * 1 − cos    2    44.15  M 5 = 175 * 1 − cos  = 12.83m   2 k) Chord (Chord from P.C to P.T) ∆ C5 = 2 R sin   2 21ECSC, IUDS, Urban Engineering Department (UE)
  22. 22. Highway Design Senior Project 2010  44.150  C5 = 2 *175 * sin   2  = 131.54m   Curve-6 Design computationa) Terrain type = Rollingb) Deflection angle Δ = 32.480 (by measurement)c) Point of intersection P.I=15+226.73md) Calculation of radius of the curve Vd 2 Rmin = 127(e + f )Where, Rmin=minimum radius Vd=70km/hr…………….ERA, table 2.6 ed= 8% (max design super elevation rate, ERA, table 2.6) f=0.14 (ERA. Table 8.1 for ed=8%) 70 2Then, Rmin = =175.4m 127(0.08 + 0.14)The calculated Rmin has no significant change from the recommended in ERA manual standard(i.e., 175m), in addition to this, in order to minimize cut and fill, we use R min=175m from thestandard.Therefore, radius of curve=Rc=175me) Tangent (T6) Rmin = 175m  ∆ T6 = R * tan   2 22ECSC, IUDS, Urban Engineering Department (UE)
  23. 23. Highway Design Senior Project 2010  32.48  T6 = 175 * tan   = 50.97 m  2 f) Point of curvature (PC) P.C6= P.I6 – T6 =15+226.73m – 0+050.97 =15+175.76mg) Length of the curve (L)  2Π  L6 = ∆ * R *    360   2Π  L6 = 32.48 0 *175 *   = 99.20m  360 h) Point of tangency (P.T) P.T6= P.C6+L6 =15+175.76m +99.20m =15+274.96mi) External distance (E)  ∆  E6 = R * sec  − 1  2    32.480   E6 = 175 * sec   −1 = 7.27 m    2  j) Middle ordinate (M)   ∆  M 6 = R * 1 − cos    2  23ECSC, IUDS, Urban Engineering Department (UE)
  24. 24. Highway Design Senior Project 2010   32.480  M 6 = 175 * 1 − cos  2  = 6.98m    k) Chord (Chord from P.C to P.T) ∆ C6 = 2 R sin   2  32.48 0  C6 = 2 *175 * sin   2  = 97.88m   2.2.1.2 Transition curveWhen a vehicle traveling on a straight course enters a curve of finite radius, and suddenlysubjected to the centrifugal force which shock and sway. In order to avoid this it is customary toprovide a transition curve at the beginning of the circular curve having a radius equal to infinityat the end of the straight and gradually reducing the radius to the radius of the circular curvewhere the curve begins.Mostly transition curves are introduced between:- A/ between tangents and curves B/ between two curvesVarious forms of transition curves are suitable for high way transition, but the one most popularand recommended for use is spiral.Design of transition curveEven if there are places to design transition curve, ERA design manual standard recommendswhere and how to design this horizontal alignment design elements. Especially for Ethiopianroad, transition curves are a requirement for trunk and link road segments having a speed equalto or greater than 80km/hr. (ERA)But the characteristics of our project road segment is;- Speed=60km/hr (for mountainous terrain) 24ECSC, IUDS, Urban Engineering Department (UE)
  25. 25. Highway Design Senior Project 2010 Speed=70km/hr (for rolling terrain) Terrain= mostly rolling and mountainous Functional classification=Main access road.Therefore, based on the ERA standard all curves in the project will not have transition curve. So,it will be a simple curve with out transition curve.2.2.1.3 Super elevationCurve-1When a vehicles moves in a circular path, it is forced radially by centrifugal force. Thecentrifugal force is counter balanced by super elevation of the road way and/or the side frictiondeveloped between the tire and the road surface. The centrifugal force is the result of designspeed, weight of car, friction, and gravitational acceleration having the following relation ship. Wv 2 Fc = gRWhere, Fc= centrifugal force W=weight of the car V=design speed g= acceleration due to gravity R= radius of the curveSo, super elevation rate is changing the road cross section from the normal road to elevatetowards the center of the curve. I.e., it counteracts a part of the centrifugal force, the remainingpart being resisted by the lateral friction.Terms in super elevation:  Tangent run out(Lt)  Super elevation runoff(Lr) 25ECSC, IUDS, Urban Engineering Department (UE)
  26. 26. Highway Design Senior Project 2010Tangent run out (Lt)It is the longitudinal length along the road designed to remove the adverse crown to a zero slope.i.e., the outer edge of the road is raised from a normal cross slope to a zero slope which equal tothe grade level of the road (the level of the center line of the road).Super elevation runoff length (Lr)Super elevation run-off is a length of the road section from the point of removal of adversecrown of the road to the full super elevated point on the curve.Super elevation is equal to the length of transition curve when there is a transition curve. Whenthere is no transition curve i.e., when it is a simple curve,1/3 rd of the length is placed on the curveand 2/3rd of the length is placed on the tangent part(ERA). Therefore, we follow the secondstandard to design our super elevation since all the curves do not have transition curve.Design computationA/ computation of super elevation run-offSuper elevation runoff length can be obtained from table 8.5 (ERA) using radius (Rc) and superelevation rate (e), or it can be computed from the following formula. (AASHTO)Lr = ( wn1 ) ed ( b ) w GWhere, Lr=minimum super elevation run-off (m) G=maximum relative gradient (percent) n1=number of lanes rotated Bw=adjustment factor for number of lane rotated w=width of one traffic lane (in our case, w/2) ed=design super elevation rate, percent 26ECSC, IUDS, Urban Engineering Department (UE)
  27. 27. Highway Design Senior Project 2010Then, n1=1, since the number of lane rotated is =1(AASHTO, exhibit 3-31) bw=1, for one lane rotated(AASHTO, exhibit 3-31) G=0.55%, for Vd=70km/hr (AASHTO, exhibit 3-31) Design speed(Km/h)(Vd) Maximum relative Equivalent maximum relative gradient(%)(G) slope (%) 20 0.80 1:125 30 0.75 1:133 40 0.70 1:143 50 0.65 1:150 60 0.60 1:167 70 0.55 1:182 80 0.50 1:200 90 0.45 1:213 100 0.40 1:227 110 0.35 1:244 120 0.30 1:263 130 0.25 1:286Table2-6 (Exhibit 3-27 Maximum relative gradients of AASHTO)  6 .7   *1 * 0.08Therefore,  2  Lr = (1) = 48.87m 0.55But ERA recommends Lr=52m for ed=8% and Rc=175m. Thus, take Lr=52mB/ computation of tangent run out (Lt)Tangent run-out can be computed using the following equation. (AASHTO) eNCLt = * ( Lr ) edWhere, 27ECSC, IUDS, Urban Engineering Department (UE)
  28. 28. Highway Design Senior Project 2010 Lt =minimum length of tangent run-out eNC=normal cross slope rate, percent ed =design super elevation, percent Lr=super elevation runoff length 0.025Then, Lt = * ( 52 ) = 16.25m 0.08C/ Location of super elevation run-off (Lr)Since there is no transition curve (spiral) between the tangent and the curve in the project, 2/3 rdof the super elevation length is placed on the tangent and 1/3rd of the length is placed on thecurve part. 1 i.e., * 52 =17.33m (on the curve) 3 2 * 52 = 34.67 m (On the tangent) 3Then,The beginning of the super elevation runoff length is:- =P.C-34.67m =12+655.43-0+034.67 =12+620.76mThe end of the super elevation runoff length is:- =P.C+17.33m =12+655.43+0+017.33m =12+672.76m 28ECSC, IUDS, Urban Engineering Department (UE)
  29. 29. Highway Design Senior Project 2010D/ location of tangent run-out length Beginning=beginning of Lr minus Lt =12+620.76-16.25m =12+604.51m End=12+620.76mE/ station where outer and inner edge of the road will have the same normal cross fall i.e., 2.5%It is a length(R) where total crown removal is attained. So, R=2*Lt =2*16.25 =32.50m,Then, the station is, Beginning= station of beginning of adverse crown removal =12+604.51m End=station of beginning of adverse crown removal plus +R =12+604.51+32.50m =12+637.01mOn the same process we can do the super elevation at the exit of the curve.We know that the length of curve 1=119.12mThen the part of the curve to be full super elevated is =119.12-2*(1/3*Lr) =119.12-2*(1/3*52) =84.46m 29ECSC, IUDS, Urban Engineering Department (UE)
  30. 30. Highway Design Senior Project 2010F/ Then, the station of end of full super elevation is =12+672.76+84.46m =12+757.22mG/ station of end of super elevation runoff is =12+757.22+52m =12+809.22mH/ station of recovering adverse crown is =12+809.22+16.25m =12+825.47Attainment of full super elevation:-From three methods attaining full super elevation we use the method in which rotating thesurface of the road about the center line of the carriageway, gradually lowering the inner edgeand raising the upper edge, keeping the center line constant.Illustration: 30ECSC, IUDS, Urban Engineering Department (UE)
  31. 31. Highway Design Senior Project 2010Fig.2-4 Attainment of super elevationBased on the above super elevation attainment, the results are shown on the following figure. 31ECSC, IUDS, Urban Engineering Department (UE)
  32. 32. Highway Design Senior Project 2010Fig.2-5 Super elevation at entrance and exit for curve 1Curve-2 Design computationA/ computation of super elevation run-offLr = ( wn1 ) ed ( b ) w G n1=1, since the number of lane rotated is =1(AASHTO, exhibit 3-31) bw=1, for one lane rotated(AASHTO, exhibit 3-31) G=0.55%, (AASHTO, exhibit 3-31)  6.7   *1 * 0.08Therefore,  2  Lr = * (1) = 48.78m 0.55But ERA recommends Lr=52m for ed=8% and Rc=175m. Thus, take Lr=52mB/ computation of tangent run out (Lt) 32ECSC, IUDS, Urban Engineering Department (UE)
  33. 33. Highway Design Senior Project 2010Tangent run-out can be computed using the following equation. (AASHTO) eNCLt = * ( Lr ) ed 0.025Then, Lt = * ( 52 ) = 16.25m 0.08C/ Location of super elevation run-off (Lr)Since there is no transition curve (spiral) between the tangent and the curve in the project, 2/3 rdof the super elevation length is placed on the tangent and 1/3rd of the length is placed on thecurve part. 1 i.e., * 52 =17.33m (on the curve) 3 2 * 52 = 34.67 m (On the tangent) 3Then,The beginning of the super elevation runoff length is:- =P.C-34.67m =13+98.59-0+034.67 =13+63.92mThe end of the super elevation runoff length is:- =P.C+17.33m =13+98.59+0+017.33m =13+115.92mD/ location of tangent run-out length 33ECSC, IUDS, Urban Engineering Department (UE)
  34. 34. Highway Design Senior Project 2010Beginning=beginning of Lr minus Lt =13+63.92 -16.25m =13+47.67mEnd=13+63.92mE/ station where outer and inner edge of the road will have the same normal cross fall i.e., 2.5%It is a length(R) where total crown removal is attained. So, R=2*Lt =2*16.25 =32.50m,Then, the station isBeginning= station of beginning of adverse crown removal =13+047.67mEnd=station of beginning of adverse crown removal plus +R =13+47.67m +32.50m =13+080.17mOn the same process we can do the super elevation at the exit of the curve.We know that the length of curve-2=100.79mThen the part of the curve to be full super elevated is =100.79-2*(1/3*Lr) =100.79-2*(1/3*52) =66.12mF/ Then, the station of end of full super elevation is 34ECSC, IUDS, Urban Engineering Department (UE)
  35. 35. Highway Design Senior Project 2010 =end of Lr+L =13+115.92 +66.12m =13+182.04mG/ station of end of super elevation runoff is =13+182.04 +52m =13+234.04mH/ station of recovering adverse crown are: =13+234.04+16.25m =13+250.29mCurve-3 Design computationA/ computation of super elevation run-offSuper elevation runoff length can be obtained from table 8.5 (ERA) using radius (Rc) and superelevation rate (e), or it can be computed from the following formula. (AASHTO)Lr = ( wn1 ) ed ( b ) w GThen, n1=1, since the number of lane rotated is =1(AASHTO, exhibit 3-31) bw=1, for one lane rotated(AASHTO, exhibit 3-31) G=0.55%, (AASHTO, exhibit 3-31)  6.7   *1 * 0.08Therefore,  2  Lr = * (1) = 48.78m 0.55But ERA recommends Lr=52m for ed=8% and Rc=175m. Thus, take Lr=52mB/ computation of tangent run out (Lt) 35ECSC, IUDS, Urban Engineering Department (UE)
  36. 36. Highway Design Senior Project 2010Tangent run-out can be computed using the following equation. (AASHTO) eNCLt = * ( Lr ) ed 0.025Lt = * ( 52 ) = 16.25m 0.08C/ Location of super elevation run-off (Lr)Since there is no transition curve (spiral) between the tangent and the curve in the project, 2/3 rdof the super elevation length is placed on the tangent and 1/3rd of the length is placed on thecurve part. 1 i.e., * 52 =17.33m (on the curve) 3 2 * 52 = 34.67 m (On the tangent) 3Then,The beginning of the super elevation runoff length is:- =P.C-34.67m =13+263.38 -0+034.67 m =13+228.71mThe end of the super elevation runoff length is:- =P.C+17.33m =13+263.38 +0+017.33m =13+280.71mD/ location of tangent run-out lengthBeginning=beginning of Lr minus Lt 36ECSC, IUDS, Urban Engineering Department (UE)
  37. 37. Highway Design Senior Project 2010 =13+228.71-16.25m =13+212.46mEnd=13+228.71mE/ station where outer and inner edge of the road will have the same normal cross fall i.e., 2.5%It is a length(R) where total crown removal is attained. So, R=2*Lt =2*16.25 =32.50m,Then, the station isBeginning= station of beginning of adverse crown removal =13+212.46mEnd=station of beginning of adverse crown removal plus +R =13+212.46+32.50m =13+244.96mOn the same process we can do the super elevation at the exit of the curve.We know that the length of curve 3=182mThen the part of the curve to be full super elevated is =182-2*(1/3*Lr) =182-2*(1/3*52) =147.33mF/ Then, the station of end of full super elevation is =13+280.71m +147.33m 37ECSC, IUDS, Urban Engineering Department (UE)
  38. 38. Highway Design Senior Project 2010 =13+428.04mG/ station of end of super elevation runoff is: =13+428.04 +52m =13+480.04mH/ station of recovering adverse crown is: =13+480.04+16.25m =13+496.29mCurve-4 Design computationA/ computation of super elevation run-offSuper elevation runoff length can be obtained from table 8.5 (ERA) using radius (Rc) and superelevation rate (e), or it can be computed from the following formula. (AASHTO)Lr = ( wn1 ) ed ( b ) w GThen, n1=1, since the number of lane rotated is =1(AASHTO, exhibit 3-31) bw=1, for one lane rotated(AASHTO, exhibit 3-31) G=.55%, for Vd=70km/hr, (AASHTO, exhibit 3-31)  6.7   *1 * 0.08Therefore,  2  Lr = * (1) = 48.7m 0.55But from ERA for ed=8% and v=70m/sec, by interpolation Lr=49.12m for Rc=236m. Thus, takeLr=49.12mB/ computation of tangent run out (Lt)Tangent run-out can be computed using the following equation. (AASHTO) 38ECSC, IUDS, Urban Engineering Department (UE)
  39. 39. Highway Design Senior Project 2010 eNCLt = * ( Lr ) ed 0.025Lt = * ( 49.12 ) = 15.35m 0.08C/ Location of super elevation run-off (Lr)Since there is no transition curve (spiral) between the tangent and the curve in the project, 2/3 rdof the super elevation length is placed on the tangent and 1/3rd of the length is placed on thecurve part. 1 i.e., * 49.12 =16.37 m (on the curve) 3 2 * 49.12 = 32.75m (On the tangent) 3Then,The beginning of the super elevation runoff length is:- =P.C-32.75m =13+806.5-0+032.75 =13+773.75mThe end of the super elevation runoff length is:- =P.C+16.37 =13+806.5+0+016.37m =13+822.87mD/ location of tangent run-out lengthBeginning=beginning of Lr minus Lt 39ECSC, IUDS, Urban Engineering Department (UE)
  40. 40. Highway Design Senior Project 2010 =13+773.75 -15.35m =13+758.4mEnd=13+839.25mE/ station where outer and inner edge of the road will have the same normal cross fall i.e., 2.5%It is a length(R) where total crown removal is attained. So, R=2*Lt =2*15.35 =30.7mThen, the station isBeginning= station of beginning of adverse crown removal =13+823.39mEnd=station of beginning of adverse crown removal plus +R =13+823.39m +30.70m =13+854.10mOn the same process we can do the super elevation at the exit of the curve.We know that the length of curve 4=374mThen the part of the curve to be full super elevated is =374-2*(1/3*Lr) =374-2*(1/3*49.12) =341.25mF/ Then, the station of end of full super elevation is 40ECSC, IUDS, Urban Engineering Department (UE)
  41. 41. Highway Design Senior Project 2010 =13+822.87+341.25m m =14+164.12mG/ station of end of super elevation runoff is: =14+164.12m +49.12m =14+213.24mH/ station of recovering adverse crown is: =14+213.24 +15.35m =14+228.59mCurve-5 Design computationA/ computation of super elevation run-offSuper elevation runoff length can be obtained from table 8.5 (ERA) using radius (Rc) and superelevation rate (e), or it can be computed from the following formula. (AASHTO)Lr = ( wn1 ) ed ( b ) w GThen, n1=1, since the number of lane rotated is =1(AASHTO, exhibit 3-31) bw=1, for one lane rotated(AASHTO, exhibit 3-31) G=.55%, for Vd=70km/hr, (AASHTO, exhibit 3-31)  6.7   *1 * 0.08Therefore,  2  Lr = * (1) = 48.78m 0.55But ERA recommends Lr=48m for ed=8% and Rc=175m. Thus, take Lr=52mB/ computation of tangent run out (Lt)Tangent run-out can be computed using the following equation. (AASHTO) 41ECSC, IUDS, Urban Engineering Department (UE)
  42. 42. Highway Design Senior Project 2010 eNCLt = * ( Lr ) ed 0.025Lt = * ( 52 ) = 16.25m 0.08C/ Location of super elevation run-off (Lr)Since there is no transition curve (spiral) between the tangent and the curve in the project, 2/3 rdof the super elevation length is placed on the tangent and 1/3rd of the length is placed on thecurve part. 1 i.e., * 52 =17.33m (on the curve) 3 2 * 52 = 34.67 m (On the tangent) 3Then,The beginning of the super elevation runoff length is:- =P.C-34.67m =14+685.72m -0+034.67m =14+651.05mThe end of the super elevation runoff length is:- =P.C+17.33m =14+685.72+0+017.33m =14+703.05mD/ location of tangent run-out lengthBeginning=beginning of Lr minus Lt 42ECSC, IUDS, Urban Engineering Department (UE)
  43. 43. Highway Design Senior Project 2010 =14+651.05-16.25m =14+634.80mEnd=14+651.05mE/ Station where outer and inner edge of the road will have the same normal cross fall i.e., 2.5%It is a length(R) where total crown removal is attained. So, R=2*Lt =2*16.25 =32.50m,Then, the station is;Beginning=station of beginning of adverse crown removal =14+634.80mEnd=station of beginning of adverse crown removal plus +R =14+634.80m +32.50m =14+667.30mOn the same process we can do the super elevation at the exit of the curve.We know that the length of curve 5=134.35mThen the part of the curve to be full super elevated is =134.35-2*(1/3*Lr) =134.35-2*(1/3*52) =99.68mF/ Then, the station of end of full super elevation is =14+703.05m +99.68m 43ECSC, IUDS, Urban Engineering Department (UE)
  44. 44. Highway Design Senior Project 2010 =14+802.73mG/ station of end of super elevation runoff are: =14+802.73m +52m =14+854.73mH/ station of recovering adverse crown is: =14+854.73m +16.25m =14+870.98mCurve-6 Design computationA/ computation of super elevation run-off:Super elevation runoff length can be obtained from table 8.5 (ERA) using radius (Rc) and superelevation rate (e), or it can be computed from the following formula. (AASHTO)Lr = ( wn1 ) ed ( b ) w GThen, n1=1, since the number of lane rotated is =1(AASHTO, exhibit 3-31) bw=1, for one lane rotated(AASHTO, exhibit 3-31) G=.55%, for Vd=60km/hr, (AASHTO, exhibit 3-31)  6.7   *1 * 0.08Therefore,  2  Lr = * (1) = 48.78m 0.55But ERA recommends Lr=48m for ed=8% and Rc=175m. Thus, take Lr=52m 44ECSC, IUDS, Urban Engineering Department (UE)
  45. 45. Highway Design Senior Project 2010B/ computation of tangent run out (Lt)Tangent run-out can be computed using the following equation. (AASHTO) eNCLt = * ( Lr ) ed 0.025Lt = * ( 52 ) = 16.25m 0.08C/ Location of super elevation run-off (Lr)Since there is no transition curve (spiral) between the tangent and the curve in the project, 2/3 rdof the super elevation length is placed on the tangent and 1/3rd of the length is placed on thecurve part. 1 i.e., * 52 =17.33m (on the curve) 3 2 * 52 = 34.67 m (On the tangent) 3Then,The beginning of the super elevation runoff length is:- =P.C-34.67m =15+175.76m -0+034.67m =15+141.10mThe end of the super elevation runoff length is:- =P.C+17.33m =15+175.76m +0+017.33m =15+193.10m 45ECSC, IUDS, Urban Engineering Department (UE)
  46. 46. Highway Design Senior Project 2010D/ location of tangent run-out lengthBeginning=beginning of Lr minus Lt =15+141.10m -16.25m =15+123.85mEnd=15+123.85mE/ station where outer and inner edge of the road will have the same normal cross fall i.e., 2.5%It is a length(R) where total crown removal is attained. So, R=2*Lt =2*16.25 =32.50m,Then, the station isBeginning= station of beginning of adverse crown removal =15+123.85mEnd=station of beginning of adverse crown removal plus + R =15+123.85m +32.50m =15+156.35mOn the same process we can do the super elevation at the exit of the curve.We know that the length of curve 6=99.20mThen the part of the curve to be full super elevated is =99.20-2*(1/3*Lr) =99.20-2*(1/3*52) =64.53m 46ECSC, IUDS, Urban Engineering Department (UE)
  47. 47. Highway Design Senior Project 2010F/ Then, the station of end of full super elevation is =15+193.10+64.53m =15+257.63mG/ station of end of super elevation runoff is: =15+257.63m +52m =15+309.63mH/ station of recovering adverse crown is: =15+309.63m +16.25m =15+325.88mSuper elevation overlaps:The end of tangent run out (super elevation runoff length) for curve 2 and the beginning oftangent run out (super elevation runoff length) of curve 3 overlaps with an amount of:Over lap= (13+250.29)-(13+212.46) =42.83mTherefore, this overlap length has to distribute on the curve part of each curve according to thefollowing.Half of the overlap distance has to be added to the part of the curve. I.e. if the overlap length is d,the part of super elevation on the curve will be =1/3rd (Lr) +d/2 =17.33+42.83/2m =38.475mBut this length has to be 40% of length of the corresponding curve. 47ECSC, IUDS, Urban Engineering Department (UE)
  48. 48. Highway Design Senior Project 2010Check:Lc of curve 2=100.79mThen, 40%*100.79=40.32>38.745m…………….OK!Lc of curve 3=182m,Then, 0.4*182=72.8>38.475m………………………OK!Re-adjustment for super elevation stations.Curve-21. The beginning of the super elevation runoff length is:- =P.C-(34.67-21.415) m =13+98.59-(0+013.25) =13+085.34m2. The end of the super elevation runoff length is:- =P.C+17.33m =13+98.59+ (0+017.33+21.415) m =13+137.34m3. Location of tangent run-out lengthBeginning=beginning of Lr minus Lt =13+085.34m -16.25m =13+069.09mEnd=13+085.34m4. Station where outer and inner edge of the road will have the same normal cross fall i.e., 2.5%It is a length(R) where total crown removal is attained. 48ECSC, IUDS, Urban Engineering Department (UE)
  49. 49. Highway Design Senior Project 2010 So, R=2*Lt =2*16.25 =32.50m,Then, the station isBeginning= station of beginning of adverse crown removal =13+069.09mEnd=station of beginning of adverse crown removal plus +R =13+069.09m +32.50m =13+101.59mOn the same process we can do the super elevation at the exit of the curve.We know that the length of curve-2=100.79mThen the part of the curve to be full super elevated is =100.79-2*(1/3*Lr+21.415) =100.79-2*(1/3*52+21.415) =23.29m5. Then, the station of end of full super elevation is =end of Lr+23.29 =13+137.34m +23.29m =13+160.63m6. Station of end of super elevation runoff is =13+160.63+52m =13+212.63m 49ECSC, IUDS, Urban Engineering Department (UE)
  50. 50. Highway Design Senior Project 20107. Station of recovering adverse crown is: =13+212.63m +16.25m =13+228.88mCurve-31. The beginning of the super elevation runoff length is:- =P.C-(34.67-21.415) m =13+263.38 – (0+013.25) m =13+250.13m2. The end of the super elevation runoff length is:- =P.C+ (17.33+21.415) m =13+263.38 + (0+38.75) m =13+302.13m3. Location of tangent run-out lengthBeginning=beginning of Lr minus Lt =13+250.13m -16.25m =13+233.88mEnd=13+250.13m4. Station where outer and inner edge of the road will have the same normal cross fall i.e., 2.5%It is a length(R) where total crown removal is attained. So, R=2*Lt =2*16.25 =32.50m, 50ECSC, IUDS, Urban Engineering Department (UE)
  51. 51. Highway Design Senior Project 2010Then, the station isBeginning= station of beginning of adverse crown removal =13+250.13mEnd=station of beginning of adverse crown removal plus +R =13+250.13m +32.50m =13+282.63mOn the same process we can do the super elevation at the exit of the curve.We know that the length of curve 3=182mThen the part of the curve to be full super elevated is =182-2*(1/3*Lr+d/2) =182-2*((1/3*52) +42.83/2) =104.50m5. Then, the station of end of full super elevation is =13+302.13m +104.50 =13+406.63m6. Station of end of super elevation runoff is: =13+406.63m + 52m =13+458.63m7/ station of recovering adverse crown is:=13+458.63m +16.25m =13+474.88m 51ECSC, IUDS, Urban Engineering Department (UE)
  52. 52. Highway Design Senior Project 2010Fig 2-6 profile, section and station of super elevation, tangent run out for all curves 52ECSC, IUDS, Urban Engineering Department (UE)
  53. 53. Highway Design Senior Project 2010 STATIONS CURVE NUMBER A B C D E F G H Curve 1 12+604.51 12+620.76 12+637.01 12+672.76 12+757.22 12+792.97 12+809.22 12+825.47 Curve 2 13+069.09 13+085.34 13+101.59 13+137.34 13+160.63 13+196.38 13+212.63 13+228.88 Curve 3 13+233.88 13+250.13 13+282.63 13+302.13 13+406.63 13+442.38 13+458.63 13+474.88 Curve 4 13+756.4 13+773.75 13+789.10 13+822.87 14+164.12 14+197.89 14+213.24 14+228.59 Curve 5 14+634.80 14+651.05 14+667.30 14+703.05 14+802.73 14+838.48 14+854.73 14+870.98 Curve 6 15+123.85 15+141.10 15+156.35 15+193.10 15+257.63 15+293.38 15+309.63 15+325.88Table 2-7 stations of super elevation, tangent run out for all curves. 53ECSC, IUDS, Urban Engineering Department (UE)
  54. 54. Highway Design Senior Project 20102.2.1.4 Curve wideningWidening on a curve is giving extra width on a road curves. This is because:-  It has been found that the drivers on curves have difficulty in steering their vehicles to outer edge of road as they are able to on the straight because the rear wheels do not follow precisely the same path as the front wheels when the vehicles negotiates a horizontal curve or makes a turn.  Also there is psychological tendency to drive at greater clearance, when passing vehicle on curved than on straights. Hence, there is dire necessity for widening the carriage way on curves.  On curves the vehicles occupy a greater width because the rear wheels track inside the front wheels.Analysis of extra widening on horizontal curvesWhen vehicles negotiate a curve, the rear wheel generally do not follow the same track asthat of the front wheels. It has been observed that except at very high speed, the rear axleof a motor vehicles remains in line with the radius of the curve. Since the body of thevehicle is rigid, therefore, the front wheel will twist themselves at one angle to their axle,such that vertical plane passing through each wheel is perpendicular to the radius of thecurve in order to trace the path on the curve. This is known as ‘off tracking’.To determine width (W) it is necessary to select an appropriate design vehicle. Thedesign vehicle should usually be a truck because the off tracking is much greater fortrucks than for passenger car. (AASHTO) There fore, widening on horizontal curvesdepend on:  The length and width of the vehicle  Radius of curvature 54ECSC, IUDS, Urban Engineering Department (UE)
  55. 55. Highway Design Senior Project 2010 Fig 2-7 widening of pavements on horizontal curvesLet;L= length of wheel base of vehicle in m.b=width of the road in m,w=extra width in m,R1=radius of the outer rear wheel in m,R2= radius of the outer front wheel in m,n=number of lanesRc= radius of curvatureThe formula obtained from the above geometries for extra widening for more than onelane (mechanical widening) is:- n * L2 mechanical..widening = wm = 2 * RcThe extra widening needed for psychological reasons mentioned above is assumed as:- 55ECSC, IUDS, Urban Engineering Department (UE)
  56. 56. Highway Design Senior Project 2010 v psycho log icalwidening = w p = 10 RcThere fore, total widening w will be:- n * L2 v w= + 2 * Rc 10 RcWidening attainment on curvesThe following rules apply for attaining widening on both ends of the curve. (AASHTO)A. widening should be done gradually and has to be realized on the inside edge of un-spiraled curve (on simple curve) pavements.B. In the case of a circular curve with transition curves, widening may be applied on theinside edge or divide equally on either side of the center line.C. On highway curves without transition curves widening should preferably be attainedalong the length of super elevation runoff. A smooth fitting alignment would result fromattaining widening on-one half to two-third along the tangent and the remaining along thecurve.D. Widening is not necessary for large radius greater than 250m.Curve-1, 2, 3, 5, and 6 Design computationsDesign data: Rc = 175m, n=2L= take 6m (for the design vehicle usually a truck, corresponding to AASHTO, Singleunit (SU))V=70m/sec n * L2 vw= + 2 * Rc 10 Rc 2 * 62 70w= + = 0.73m 2 *175 10 175 56ECSC, IUDS, Urban Engineering Department (UE)
  57. 57. Highway Design Senior Project 2010For all curves having a radius between 120 to 250m ERA recommends a minimum ofwidening width equal to 0.6m. But we recommend the calculated value 0.73m. So, all thecurves will have the corresponding value unless they are no less than the recommendedvalue by ERA. Therefore, this widening will be introduced at the inner edge of thecurves. Because all the curves are un spiraled curves.Fig2-8.widening of pavement on curvesWIDENING STARTING STARTING LAST PT OF END POINT REMARKWIDTH(M) POINT OF POINT OF FULL OF WIDENING FULL WIDENING WIDENING WIDENING0.73 12+620.76 12+672.76 12+757.22 12+809.22 12+620.76Table 2-8 widening stations for curve 1Curve-4 Design computationDesign data: Rc=236m, N=2, L= take 6m, V=70m/se 57ECSC, IUDS, Urban Engineering Department (UE)
  58. 58. Highway Design Senior Project 2010 n * L2 vw= + 2 * Rc 10 Rc 2 * 62 70w= + = 0.61m 2 * 236 10 236CUR WIDENI STARTING STARTING LAST PT OF END POINT VE NG POINT OF POINT OF FULL OFNO. WIDTH( WIDENING FULL WIDENING WIDENING M) WIDENING C1 0.73 12+620.76 12+672.76 12+757.22 12+809.22 C2 0.73 13+085.34 13+137.34 13+160.63 13+212.63 C2 0.73 13+250.13 13+302.13 13+406.63 13+458.63 C3 0.73 13+839.25 13+822.87 14+164.12 14+213.24 C4 0.61 14+651.05 14+703.05 14+802.73 14+854.73 C5 0.73 15+141.10 15+193.10 15+257.63 15+309.63 C6 0.73 12+620.76 12+672.76 12+757.22 12+809.22Table2-9 Widening length and stations for all curves.2.2.1.4 Site distanceAnother element of horizontal alignment is the site distance across the inside of thecurves. Sight distance is the distance visible to the driver of a passenger car or theroadway ahead that is visible to the driver. For highway safety, the designer must providesight distances of sufficient length that drivers can control the operation of their vehicles.They must be able to avoid striking an unexpected object on the traveled way.Where there are site obstruction( such as walls, cut slops, buildings and longitudinalbarriers) on the inside of curves or the in side of the median lane on divided highways, adesign may need adjustment in the normal high way cross section or change in thealignment if removal of the obstruction is impractical to provide adequate site distance.Because of the many variables in alignment, in cross section and in the number, type and 58ECSC, IUDS, Urban Engineering Department (UE)
  59. 59. Highway Design Senior Project 2010location of potential obstructions, specific study is usually need for each individual curve.With site distance for the design speed as a control, the designer should check the actualconditions on each curve and make the appropriate adjustment to provide adequatedistance.Two-lane rural highways should generally provide such passing sight distance at frequentintervals and for substantial portions of their length.Stopping site distanceStopping sight distance is the distance required by a driver of a vehicle traveling at agiven speed to bring his vehicle to a stop after an object on the road way becomes visible.The minimum stopping sight distance is determined from the following formula, whichtakes into account both the driver reaction time and the distance required to stop thevehicle. The formula is:d= (0.278) (t) (v) +v2/ 254fWhere:d = distance (meter)t = driver reaction time, generally taken to be 2.5 secondsV = initial speed (km/h)F = coefficient of friction between tires and roadway (see Table 7-1)OR the stopping site distance is given in ERA manual in the following table.Design Speed Coefficient Stopping Sight Passing Sight Reduced Passing Sight Distance(km/h) of Friction (f) Distance (m) Distance (m) for design (m) from formulae 20 0.42 20 160 50 30 0.40 30 217 75 59ECSC, IUDS, Urban Engineering Department (UE)
  60. 60. Highway Design Senior Project 2010 40 0.38 45 285 125 50 0.35 55 345 175 60 0.33 85 407 225 70 0.31 110 482 275 85 0.30 155 573 340 100 0.29 205 670 375 120 0.28 285 792 425Table 2-10: Sight DistancesThe coefficient of friction values shown in Table 2-10 have been determined from testusing the lowest results of the friction tests. The values shown in the third column of theabove table for minimum stopping sight distance are rounded from the above formula.For the general use in the design of horizontal curve, the sight line is a chord of the curve,and the stopping site distance is measured along the center line of the inside lane aroundthe curve.The horizontal site line offset needed for clear site areas that satisfy stopping site distancecan be derived from the geometry for the several dimension explained in the followingfigure. 60ECSC, IUDS, Urban Engineering Department (UE)
  61. 61. Highway Design Senior Project 2010Fig 2-9 Site distance for horizontal curvesRelevant formulae are as follows: ∆Siteline( S ) = 2 R sin 2  ∆Middle..ordinate(d ) = R1 − cos   2Where ∆ = Deflection angle R=radius (from the center line of the inner lane)Design computationUsing the above formulas the stopping site distance(d), the line of site(S) and middleordinate(M) of each horizontal curves can be calculated from the data’s of each curveorganized in the following table below. driver deflection Radius speed(V) reaction Coefficient ofcurve no time angle(D) (R),m km/hr friction(f) (t) in sec.Curve 1. 39 173.325 70 2.5 0.31Curve 2. 33 173.325 70 2.5 0.31Curve 3. 59.62 173.325 70 2.5 0.31Curve 4. 90.81 234.325 70 2.5 0.31Curve 5. 44.15 173.325 70 2.5 0.31Curve 6. 32.48 173.325 70 2.5 0.31Table 2-11 different data about each curve ∆Siteline( S ) = 2 R sin 2 61ECSC, IUDS, Urban Engineering Department (UE)
  62. 62. Highway Design Senior Project 2010  ∆Middle..ordinate( d ) = R1 − cos   2 v2Stoppingsitedist..(d ) = 0.278vt + 254 f Curve Site line (S) Middle Stopping site distance(m) in m. ordinate (M) in m. Calculated Recommended by distance in m ERAcurve 1 115.714 9.94 510.55 110curve 2 98.454 7.14 510.55 110curve 3 172.329 22.93 510.55 110curve 4 333.72 69.81 510.55 110curve 5 130.278 12.76 510.55 110curve 6 96.945 6.92 510.55 110Table2-12 Site distance elements 62ECSC, IUDS, Urban Engineering Department (UE)
  63. 63. Highway Design Senior Project 2010Fig 2-10 stopping site distance of curve 1Passing site distancePassing sight distance is the minimum sight distance on two-way single roadway roadsthat must be available to enable the driver of one vehicle to pass another vehicle safelywithout interfering with the speed of an oncoming vehicle traveling at the design speed.Within the sight area the terrain should be the same level or a level lower than theroadway. Otherwise, for horizontal curves, it may be necessary to remove obstructionsand widen cuttings on the insides of curves to obtain the required sight distance. Thepassing sight distance is generally determined by a formula with four components, asfollows:d1 = initial maneuver distance, including a time for perception and reactiond2 = distance during which passing vehicle is in the opposing laned3 = clearance distance between vehicles at the end of the maneuverd4 = distance traversed by the opposing vehicleThe formulae for these components are as indicated below: 63ECSC, IUDS, Urban Engineering Department (UE)
  64. 64. Highway Design Senior Project 2010d1 = 0.278 t1 (v – m + at1/2)Where,t1 = time of initial maneuver, sa = average acceleration, km/h/sv = average speed of passing vehicle, km/hm = difference in speed of passed vehicle and passing vehicle, km/hd2 = 0.278 vt2Where,t2 = time passing vehicle occupies left lane, sec.v = average speed of passing vehicle, km/hd3 = safe clearance distance between vehicles at the end of the maneuver, is dependent onambient speeds as per Table 7-2 of ERA standard:Table 7-2: Clearance Distance (d3) vs. Ambient SpeedsSpeed Group (km/h) Speed group(km/hr) 50-65 66-80 81-100 101-120 D3(m) 30 55 80 100d4 = distance traversed by the opposing vehicle, which is approximately equal to 2/3 rd ofd2 whereby the passing vehicle is entering the left lane, estimated at:d4 = 2d2/3The minimum Passing Sight Distance (PSD) for design is therefore:PSD = d1+ d2 + d3 + d4 64ECSC, IUDS, Urban Engineering Department (UE)
  65. 65. Highway Design Senior Project 2010Even if it is calculated using the above formula ERA recommends passing site distance,so we use the value given by ERA design manual.Sample calculationCurve 1Data:Design speed=70km/hr=v of passing vehicleAssume the following valuesT1=3.5 sec, T2=3sec, a=1.0m/sec2V of passing vehicle=70km/hrV of passed vehicle=65km/hri.e., m=70-65=5km/hrThen,d1= 0.278 t1 (v – m + at1/2)d1 = 0.278 *3.5* (70 – 5 + (1*3)/2) =64.71md2= 0.278 vt2= 0.278 *70*3 =58.38md3=55m, for design speed group=66km/hr-80km/hrd4= 2d2/3 = (2*58.38)/3 =38.92mTherefore, total passing site distance is,PSD=d1+d2+d3+d4 = Error! Not a valid link.Error! Not a valid link.Error! Not a validlink.Error! Not a valid link. =218.95m 65ECSC, IUDS, Urban Engineering Department (UE)
  66. 66. Highway Design Senior Project 2010 Fig 2-11 Components of passing maneuver used in passing site distance.2.2.2 Design of vertical alignmentThe two major aspects of vertical alignment are vertical curvature, which is governed bysight distance criteria, and gradient, which is related to vehicle performance and level ofservice. The purpose of vertical alignment design is to determine the elevation of selectedpoints along the roadway, to ensure proper drainage, safety, and ride comfort. So it isimportant to use different series of grades and to create a smooth transition between thesegrades parabolic curves are used. The vertical alignment includes:  Joining the grades with smooth curve.  Location of appropriate gradients.2.2.2.1 Design consideration2.2.2.1.1 Gradient and grade controlsChanges of grade from plus to minus should be placed in cuts, and changes from a minusgrade to a plus grade should be placed in fills.Highway should be designed to encourageuniform operation throughout the stretch.In the analysis of grades and grade control, oneof the most important considerations is the effect of grades on the operating of the motor 66ECSC, IUDS, Urban Engineering Department (UE)
  67. 67. Highway Design Senior Project 2010vehicle.Determination of grades for vertical alignment the following are taken in toconsideration for;1. The maximum limit of grades.  Visibility related to sight distance.  Stopping sight distance.  Passing sight distance.  Rider and passenger comfort.  Cost of vehicle operation  General appearance  Cut and fill (earth work)2. The minimum limit of grades.  Drainage purposeIn this project the two cases are taken in to account as recommended by ERA 2001.2.2.2.1.2 Vertical curvesA vertical curve provides a smooth transition between two tangent grades. There are twotypes of vertical curves. Crest vertical curves and sag vertical curves.  When a vertical curve connects a positive grade with a negative grade, it is referred to as a crest curve.  When a vertical curve connects a negative grade with a positive grade, it is termed as a sag curve.In this project crest and sage curves are applied to create a smooth transition betweenthese grades.Length of vertical curvesCrest curves: 67ECSC, IUDS, Urban Engineering Department (UE)
  68. 68. Highway Design Senior Project 2010For crest curves, the most important consideration in determining the length of the curveis the sight distance requirement.  Sight distance — stopping and — passing sight distanceSag curves:For sag curves, the criteria for determining the length of the curve are:  vehicle headlight distance,  rider comfort,  drainage control and  General appearance.When the computed curve length for the above requirements is less than the minimumcurve length recommended by ERA2001, this recommended value is taken as curvelength.Error! Not a valid link.Site distance (Both stopping and passing)For Crest Vertical CurveThe stopping sight distance is the controlling factor in determining the length of a crestvertical curve.Minimum Length required for safe stopping calculated (from AASHTO) When Sd ≥ Lvcmin When Sd ≤ Lvcmin The 100 in the above equations are to convert A from % into decimals. 68ECSC, IUDS, Urban Engineering Department (UE)
  69. 69. Highway Design Senior Project 2010Where Lvc min = Minimum length of vertical curve compute Sd = Min. Stopping Sight Distance = 85 m for mountainous terrain. Psd = Min. Passing Sight Distance = 225 m for mountainous terrain.Sight distances should be checked during design, and adjustments made to meet theminimum requirements. The following values should be used for the determination ofsight lines. Shown in the figures below:Fig 2-12 Site distance for crust curveERA Manual recommends that: h1= Drivers eye height = 1.07 meters h2 = Object height for stopping sight distance = 0.15 meters = Object height for passing sight distance: = 1.30 metersFor sag Vertical Curve 69ECSC, IUDS, Urban Engineering Department (UE)
  70. 70. Highway Design Senior Project 2010Figure below shows the driver’s sight limitation when approaching a sag vertical curve.The problem is more obvious during the night time when the sight of the driver isrestricted by the area projected by the headlight beams of vehicle. Hence, the angle of thebeam from the horizontal plane is also important. This design control criteria is known asheadlight sight distance. The headlight height of h = 0.6 m and upward angle for theheadlight projection cone of β =1° is normally assumed. The governing equations are(from AASHTO) When Sd ≥ Lvcmin When Sd ≤ LvcminFig 2-13 Site distance for sag curveA driver may experience discomfort when passing a vertical curve. The effect ofdiscomfort is more obvious on a sag vertical curve than a crest vertical curve with thesame radius, because the gravitational and centripetal forces are in the oppositedirections. Some of the ride discomfort may be compensated by combination of vehicleweight, suspension system and tire flexibility. The following equation has been 70ECSC, IUDS, Urban Engineering Department (UE)
  71. 71. Highway Design Senior Project 2010recommended by AASHTO as the minimum length of a vertical curve that will providesatisfactory level of ride comfort.Design standards from ERA manual: Urban/Peri- UrbanDesign Element Unit Flat Mountainous Escarpment RollingDesign Speed km/h 85 70 60 50 50Min. Stopping Sight Distance m 155 110 85 55 55Min. Passing Sight Distance m 340 275 225 175 175% Passing Opportunity % 25 25 15 0 20Max. Gradient (desirable) % 4 5 7 7 7Max. Gradient (absolute) % 6 7 9 9 9Minimum Gradient % 0.5 0.5 0.5 0.5 0.5Crest Vertical Curve k 60 31 18 10 10Sag Vertical Curve k 36 25 18 12 12Table 2-13 Design Parameters for Design Standard DS4 (Paved)Phasing: Even if we face phasing problem on vertical curve 1 with horizontal curve 3and vertical curve 3 with horizontal curve 5, we took a corrective action by separatingthem again vertical curve 2 and horizontal curve 4 corrected by making the ends of thecurves to end at a common station in the design process according to ERA.2.2.2.2. Computation of gradients1. Gradient of the first alignment (g1) 71ECSC, IUDS, Urban Engineering Department (UE)
  72. 72. Highway Design Senior Project 2010 To calculate the first gradient; Elevation of the first point = 1386 m Elevation of the second point = 1395.4 m Elevation difference = 1395.4-1386 = 9.4 m Horizontal distance b/n the two points = (13+572)-(12+500) = 1072 m Gradient (Slope) = elevation difference/horizontal distance = (9.4/1072) = 0.0088 Gradient (Slope) g1 = 0.88 %2. Gradient of the second alignment (g2)To calculate the second gradient; Elevation of the first point = 1395.4 m Elevation of the second point = 1375 m Elevation difference = 1375-1395.4 = -20.4 m Horizontal distance b/n the two points = (14+000)-(13+572) = 428 m Slope (gradient) = elevation difference/ horizontal distance = -20.4/430 = -0.0477 Gradient (Slope) g2 = -4.77 %3. Gradient of the third alignment (g3)To calculate the third gradient Elevation of the first point = 1375 m Elevation of the second point = 1377 m Elevation difference = 1377-1375 = 2m Horizontal distance b/n the two points = (14+480)-(14+000) = 480m 72ECSC, IUDS, Urban Engineering Department (UE)

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