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GEOMETRIC DESIGN: HORIZONTAL ALIGNMENT

GEOMETRIC DESIGN: HORIZONTAL ALIGNMENT. CE331 Transportation Engineering. Objectives. Describe components of horizontal alignment Determine design parameters for circular curve Understand the impact of superelevation and stopping sight distance on the design of horizontal curve.

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GEOMETRIC DESIGN: HORIZONTAL ALIGNMENT

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  1. GEOMETRIC DESIGN:HORIZONTAL ALIGNMENT CE331 Transportation Engineering

  2. Objectives • Describe components of horizontal alignment • Determine design parameters for circular curve • Understand the impact of superelevation and stopping sight distance on the design of horizontal curve

  3. General Concepts • Components • Tangents • Curves (circular) • Design values • Function of design speed and superelevation • Stopping sight distance at any point • Length measured along centerline of the curve

  4. πR Δ/180 Arc L = Δ Cord C = 2R sin(Δ/2) R tan(Δ/2) Tangent T = M Δ/2 C R R Δ Horizontal Curves PI T M = R [1 – cos (Δ/2)] PT PC Sta PC = Sta PI – T Sta PT = Sta PC +L

  5. 30o 1257.3 Example Given: R = 1800 ft, Δ = 30 o, Sta PI Find: L, Sta PC, Sta PT? Sta 12+57.3 T L Sta 17+17.5 Sta 7+75 Sta PC = Sta PI - T Sta PT = Sta PC + L PT PC L = π R Δ/180 = π(1800)(30)/180 = 942.5 ft T = R tan(Δ/2) = 1800 tan(30/2) = 482.3 ft

  6. Example A highway has a design speed of 70mph and a superelevation rate of 0.01. If fs = 0.15, What should be the radius of the curve? R = V2/[15(fs+e)] = 702/[15*(0.15+0.01)] = 2042 (ft)

  7. Example (cont’d) If the curve is fitted through two tangents with central angle Δ = 25°, How long should the curve be? L = πRΔ/180 = π*2042*25/180 = 891 (ft)

  8. SSD C C L L M’ R R’ Sight Distance on Horizontal Curves • Clearance from roadside obstruction • M’ measured from the CL of the inside lane • M’ = R’ {1 – cos[90SSD/(πR’)]}

  9. M’ C L 60 mph 0.00 11.2 2 1000-12/2 Example Min. distance to wall from centerline to ensure SSD? Given: R=1000 ft; tr= 2 sec; V=60mph; a=11.2ft/sec2; G=0%; lane width (W) = 12ft SSD = 1.47Vtr + V2/[30(a/32.2+G)] = 521.4 (ft) M’ = R’ {1 – cos[90SSD/(π R’)]} = 34.0 (ft)  Clearance from centerline of the road = 34.0 + 12/2 = 40 ft

  10. M’ C L Example (cont’d) What if the clearance from the centerline is 35 ft? Change V. M’ = 35 – 12/2 = 29 ft Find SSD associated with M’: M’ = R’ {1 – cos[90SSD/(π R’)]}  SSD = πR’ cos-1(1- M’/R’)/90 = 481.4 (ft) SSD = 1.47Vtr + V2/[30(a/32.2+G)] • V = 57.1 (mph) • Speed limit = 55 mph

  11. Deflection Angles • Used in laying out curves • Incremental central angle δi = 180 xi/(Rπ) • Subtended angles θi = δi/2=180 xi/(2Rπ) • Deflection angles • Cumulative of subtended angles • Chords for laying out curve • Ci = 2R sin(δi/2) = 2R sin(θi)

  12. 30o Example(1/4) R = 600 m Deflection angles, cords? Sta 1+556.91 Sta 1+396.14 Sta 1+710.30 PT PC

  13. 30o θ2 θ1 1+500.00 1+400.00 1+396.14 1+450.00 θ3 Laying Out a Curve(2/4) PC

  14. Example(3/4) • 1st Station • 1400-1396.14 = 3.86 m • θ1 = 180 (3.86)/[2 (600) π] = 0.18o • C1 = 2 (600)sin(0.18) = 3.86 m • 2nd Station • 1450-1400 = 50.00 m • θ2 = 180 (50)/[2 (600) π] = 2.39o • C2 = 2 (600)sin(2.39) = 49.99 m

  15. Curve Layout Data(4/4)

  16. Used to help surveyors stakeout curve Arc -> Roadways Chord -> Railroads

  17. Superelevation • Pavement cross slope to keep vehicles on road • e + fs = V2/(15R) • e = tan(α), superelevation rate • fs: coefficient of side friction • V: design speed, mph • R: radius ofthe curve, ft

  18. Superelevation Issues • emax – Lower in Maine than in Florida – Why? • fs is a function of driver comfort and safety • ranges from 0.17 for 20mph to 0.08 at 80mph • Max superelevation sets Min radius • Practice: AASHTO Green Book tables

  19. 0% 2% e% 0% Superelevation Transition To From • Tangent Runout • Superelevation Runoff • Pavement rotation rate 1:200 • L = 200 W e

  20. 0% 2% e% Superelevation Transition PC Example W = 3.30 m TR = 200(3.3)(0.02)=13.2 m SR = 200(3.3)(0.08)=52.8 m A-PC = SR(2/3) = 35.2 m A

  21. Where do we rotate the roadway? • Rotate pavement about the centerline • most common for undivided roadways • Others mainly for drainage or terrain • Rotate pavement about the inner edge • Rotate pavement about the outside edge • Rotate about the center of the median

  22. Super-elevation Transition

  23. Superelevation Road Section View Road Plan View CL 2% 2%

  24. Superelevation Road Section View Road Plan View CL 1.5% 2%

  25. Superelevation Road Section View Road Plan View CL 1% 2%

  26. Superelevation Road Section View Road Plan View CL 0.5% 2%

  27. Superelevation Road Section View Road Plan View CL -0.0% 2%

  28. Superelevation Road Section View Road Plan View CL -0.5% 2%

  29. Superelevation Road Section View Road Plan View CL -1% 2%

  30. Superelevation Road Section View Road Plan View CL -1.5% 2% 2%

  31. Superelevation Road Section View Road Plan View CL -2% 2%

  32. Superelevation Road Section View Road Plan View CL -3% 3%

  33. Superelevation Road Section View Road Plan View CL -4% 4%

  34. Superelevation Road Section View Road Plan View CL -3% 3%

  35. Superelevation Road Section View Road Plan View CL -2% 2%

  36. Superelevation Road Section View Road Plan View CL -1.5% 2%

  37. Superelevation Road Section View Road Plan View CL -1% 2%

  38. Superelevation Road Section View Road Plan View CL -0.5% 2%

  39. Superelevation Road Section View Road Plan View CL -0.0% 2%

  40. Superelevation Road Section View Road Plan View CL 0.5% 2%

  41. Superelevation Road Section View Road Plan View CL 1% 2%

  42. Superelevation Road Section View Road Plan View CL 1.5% 2%

  43. Superelevation Road Section View Road Plan View CL 2% 2%

  44. How do we transition into a super-elevated curve? • No Spiral • Spiral

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