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Introduction to Transportation Engineering

Introduction to Transportation Engineering. Instructor Dr. Norman Garrick Hamed Ahangari 17th April 2014. Horizontal Curve. Δ. Elements. Δ – Intersection angle. R - Radius . R. Δ. Δ. Δ. Δ /2. Δ = Deflection Angle R = Radius of Circular Curve

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Introduction to Transportation Engineering

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  1. Introduction to Transportation Engineering Instructor Dr. Norman Garrick Hamed Ahangari 17th April 2014

  2. Horizontal Curve

  3. Δ Elements • Δ– Intersection angle • R - Radius R Δ

  4. Δ Δ Δ/2 Δ = Deflection Angle R = Radius of Circular Curve L = Length of Curvature (L = EC – BC) BC = Beginning of Curve (PC) EC = End of Curve (PT) PI = Point of Intersection T = Tangent Length (T = PI – BC = EC - PI) C = Chord Length M = Middle Ordinate E = External Distance PI T E T L EC (PT) BC (PC) M C R R

  5. Δ Δ Basic Definitions • BC/PC: Point of Curvature • BC = PI – T • PI = Point of Intersection • T = Tangent • EC/PT: Point of Tangency • EC = BC + L • L = Length BC (PC) EC

  6. Degree of Curvature • D used to describe curves • D defines Radius • Arc Method: • D/ Δ = 100/L(1)   (360/D)=100/(2R)    R = 5730/D (2)

  7. Curve Calculations • LengthL = 100.Δ/D (3) • Tangent T = R.tan(Δ /2)(4) • Chord C = 2R.sin(Δ /2)(5) • External Distance E = = R sec(Δ/2) – R(6) • Mid Ordinate M = R-R.cos(Δ /2))(7)

  8. Example 1 • A horizontal curve is designed with a 2000 ft. radius. The tangent length is 500 ft. and the EC station is 30+00. What are the BC and PI stations?

  9. Solution • Since we know R and T we can use T = R.tan(Δ /2) so Δ= 28.07 degrees • D = 5730/R. Therefore D = 2.86 • L = 100(Δ)/D = 100(28.07)/2.86 = 980 ft. • BC = EC – L = 3000 – 980 =2020~20+20 • PI = BC +T = 2020 + 500 = 2520~25+20

  10. Example 2 A curve has external angle of 20.30’ degrees, a degree of curvature is 2°30’ and the PI is at 175+00. Calculate: • Radius • Length of Curve • BC and EC • Chord • External Distance • Mid Ordinate

  11. Solution • Given: D = 2°30’, Δ=22.30’ • Part i) Radius: • Part ii ) Length of Curve • Part iii ) BC and EC

  12. Part iv ) Chord • Part v ) External Distance • Part vi ) Mid Ordinate

  13. Example 3 The central angle for a curve is 30 degrees - the radius of the circular curve selected for the location is 2000 ft. If a spiral with k=3 degrees is selected for use, determine the i) length of each spiral leg, ii) total length of curve iii) Spiral Central Angle

  14. Spiral Curves • k = 100 D/ Ls Δ = Δc + 2 Δs • Δs = Ls D / 200 http://www.nh.gov/dot/cadd/msv8/spiral.htm

  15. Solution Given: R=2000 , Δ=30, k=3 • D = 5730/2000. Therefore D = 2.86 • L = 100(Δ)/D = 100(30)/2.86 = 1047 ft. Part i) Ls k = 100 D/ Ls,  3=100*2.86/Ls  Ls= 95 ft

  16. Part ii) Total Length Total Length of curve = length with no spiral + Ls • Total Length= 1047+95=1142 ft Part iii) Spiral Central Angle Δs = Ls D / 200  Δs = 95*2.86/200= 1.35  Δs =1.35  Δc=Δ- 2*Δs=30-2*1.35  Δc = 27.30 degree

  17. Vertical Curve

  18. Horizontal Alignment Vertical Alignment Crest Curve G2 G3 G1 Sag Curve

  19. Design of Vertical Curves

  20. Parabolic Curve BVC G1 G2 EVC PI L/2 L/2 L Change in grade: A = G2 - G1G is expressed as % Rate of change of curvature: K = L / |A| Rate of change of grade: r = (g2 - g1) / L Equation for determining the elevation at any point on the curve y = y0 + g1x + 1/2 rx2 where, y0 = elevation at the BVCg = grade expressed as a ratio x= horizontal distance from BVC r = rate of change of grade (ratio)

  21. Example 1 • The length of a tangent vertical curve equal to 500 m. The initial and final grades are 2.5% and -1.5% respectively. The grades intersect at the station 10+400 and at an elevation of 210.00 m • Determine: a)the station and the elevation of the BVC and EVC points b) the elevation of the point on the curve 100 and 300 meters from the BVC point c) the station and the elevation of the highest point on the curve

  22. PI EVC BVC 2.5% -1.5%

  23. Solution • Part a) the station and the elevation of the BVC and EVC points • Station-BVC= 10400-250=10150~10+150 • Station-EVC=10400+250=10650~10+650 • Elevation-BVC= 210-0.025*250= 203.75 m • Elevation-EVC= 210-0.015*250= 206.25 m

  24. Part b) the elevation of the point on the curve 100 and 300 meters from the BVC point. y = y0 + g1.x + 1/2 .r.x^2, y0= 203.75,g1= 0.025 • r=(g2-g1)/L • r=(-0.015-(0.025))/500=-0.04/500r=-0.00008 y = 203.75 + 0.025.x - 0.00004.x^2,  y(100)=203.75+0.025*100-0.00004*100^2  y(100) = 205.85  y(300)= 207.65

  25. Part c) the station and the elevation of the highest point on the curve • The highest point can be estimated by setting the first derivative of the parabola as zero. • Set dy/dx=0, • dy/dx= 0.025-0.0008*x=0 X=0.025/0.00008= 312.5 • y(312.5) = 203.75+0.025*31.25-0.00004*(31.25)^2  y(312.5)= 207.65

  26. Example 2 • For a vertical curve we know that G1=-4%, G2=-1%, PI: Station 20+00, Elevation: 200’, L=300’ • Determine: i)K and r ii) station of BVC and EVC iii) elevation of point at a distance, L/4, from BVC iv) station of turning point v) elevation of turning point vi) elevation of mid-point of each curve vii) grade at the mid-point of each curve

  27. Solution -4% -1% • i) K and r • K = L / |A| • A=G2-G1, A=-4-(-1)=-3 K=300/3=100 • r=(g2-g1)/L • r=(-0.01-(-0.04))/300=0.03/300 r=0.0001

  28. ii) station of BVC and EVC • L=300’, L/2=150’ • BVC= 2000-150=1850’, ~ 18+50 • EVC= 2000+150=2150’, ~ 21+50 • iii) elevation of point at a distance, L/4, from BVC • y = y0 + g1.x + 1/2 r.x^2, • r= -0.0001, g1=0.04 • y0=200+150*0.04=206 y=206-0.04x+0.00005.x^2 L/4~ x=75  Y(75)=203.28’

  29. iv) station of turning point • at turning point: dy/dx=0 • dy=dx=- 0.04+0.0001x=0 • x(turn)=400’ • v)elevation of turning point • x=400’ • Y (400) = 206-0.04*(150)+0.00005.(150)^2 • Y(400)=198’ This point is after vertical curve (Turning point is not in the curve)

  30. vi) elevation of mid-point • x=150’ • Y(150)= 206-0.04*(150)+0.00005.(150)^2 y (150)=201.12’ • vii) grade at the mid-point of each curve • Grade at every points: dy/dx= =- 0.04+0.0001x • if x= 150  Grade (150)=-0.04+0.0001*150  grade(150)= -0.025

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