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

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Presentation on theme: "Introduction to Transportation Engineering Instructor Dr. Norman Garrick Hamed Ahangari 17th April 2014 1."— Presentation transcript:

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

2 Horizontal Curve 2

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

4 4 Δ 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 R R BC (PC) EC (PT) PI T T L M E C Δ Δ /2

5 Basic Definitions 5 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 ΔΔ

6 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 7  LengthL = 100.Δ/D (3)  TangentT = 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 What are the BC and PI stations? 8

9 Since we know R and T we can use T = R.tan(Δ /2) so Δ = 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 = = 2520~ Solution

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

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

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

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 13

14 Spiral Curves 14 Δ s = L s D / 200 Δ = Δ c + 2 Δ s k = 100 D/ Ls

15 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 15 Solution

16 Part ii) Total Length Total Length of curve = length with no spiral + Ls  Total Length= =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 = degree 16

17 Vertical Curve 17

18 18 Horizontal Alignment Vertical Alignment Crest Curve Sag Curve G1G1 G2G2 G3G3

19 Design of Vertical Curves

20 Parabolic Curve 20 BVC EVC L G2G2 G1G1 L/2 PI Change in grade: A = G 2 - G 1 G is expressed as % Rate of change of curvature: K = L / |A| Rate of change of grade: r = (g 2 - g 1 ) / L Equation for determining the elevation at any point on the curve y = y 0 + g 1 x + 1/2 rx 2 where, y 0 = elevation at the BVCg = grade expressed as a ratio x = horizontal distance from BVCr = 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 and at an elevation of 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 21

22 22 BVC EVC PI 2.5% -1.5%

23 Part a) the station and the elevation of the BVC and EVC points – Station-BVC= =10150~ – Station-EVC= =10650~ – Elevation-BVC= *250= m – Elevation-EVC= *250= m 23 Solution

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= ,g1= – r=(g2-g1)/L – r=( (0.025))/500=-0.04/500 r= y = x x^2,  y(100)= * *100^2  y(100) =  y(300)=

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= *x=0  X=0.025/ =  y(312.5) = * *(31.25)^2  y(312.5)=

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 26

27 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= Solution -4% -1%

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

29 iv) station of turning point – at turning point: dy/dx=0 – dy=dx= x=0  x(turn)=400’ v) elevation of turning point – x=400’ – Y (400) = *(150) (150)^2  Y(400)=198’  This point is after vertical curve (Turning point is not in the curve) 29

30 vi) elevation of mid-point – x=150’ – Y(150)= *(150) (150)^2  y (150)=201.12’ vii) grade at the mid-point of each curve – Grade at every points: dy/dx= = x – if x= 150  Grade (150)= *150  grade(150)=


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