Vertical Curves (Part 2)

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Vertical Curves (Part 2)

  1. 1. 20/01/2013 1
  2. 2. General Equation of Parabola Equal Tangent Vertical Curves y  a x2  b x  c y  a x 2  g1 x  (elevation of BVC ) A 2a  r L r 2 y x  g1 x  (elevation of BVC ) 2 r is the rate of change of grade per station.20/01/2013 2
  3. 3. Design of a Vertical Curve A vertical curve design convex/concave means determining; • Radius of curvature R, • Central angle subtended by circular curve , which is the intersection angle of the two lines of alignments 1 and 2. • Length of the circular curve L. Curves Radii R (m) Design Speed Road Km/hr Classification Concave Convex 120 Highway 5,000 15,000 100 1st Order 3,000 10,000 80 2nd Order 2,000 5,000 60 3rd Order 1,500 3,000 40 4th Order 1,200 2,00020/01/2013 3
  4. 4. Design of a Vertical Curve Problem: Design a concave vertical curve appropriate for connecting two lines of alignment 1 and 2 of 1st order road, their grades are g1 = - 2% and g2 = +3%. The elevation of intersection point V is 390.55 m.1 R 2  V Solution: As the road is of 1st order, the design speed = 100 km/hr match R = 3,000.00 m. See previous Table. As the grades of the two lines of alignment 1 and 2 are very small,  = (g2 – g1)% = (+ 3 – (- 2))% = 0.05 rad. The curve length, L = R  = 150.00 m.20/01/2013 4
  5. 5. Passing a Vertical Curve Through a Fixed Point Problem: Using the grades, intersection station and elevation given in Figure below, it is required to the vertical curve to pass through station 18 + 25 at an elevation of 881.20 ft. What should be the new length of the curve? 18 + 00 V g1 = + 1.25% 886.10 ft g2 = - 2.75% 15 + 00 BVC 21 + 00 300 ft EVC 600 ft20/01/2013 5
  6. 6. Passing a Vertical Curve Through a Fixed Point Solution: The curve must be lengthened. The new value of L must be computed and a new value for r results. r 2 y x  g1 x  (elevation of BVC ) 2 r A g  g1  2.75  1.25   2  2 2L 2L 2L L x   0.25 2 gL elevation of BVC  886.10  1 2 r 2 y x  g1 x  (elevation of BVC ) 2  2.75  1.25 L L 1.25L 881.21  [ (  0.25) 2 ]  [1.25(  0.25)]  [886.10  ] 2L 2 2 2 L2  9.425L  0.25  020/01/2013 6
  7. 7. Passing a Vertical Curve Through a Fixed Point Solution Continued: L2 – 9.425L + 0.25 = 0 Solving for L gives: L = 9.3984 stations and L/2 = 4.6992 stations. The elevation of BVC is: 886.10 – 1.25  4.6992 = 880.23 ft. The station of BVC is: 18 – 4.6992 = 13.3008 stations. The value x is: 18 – 13.3008 = 4.9492 r/2 = (- 2.75 – 1.25)/(2  9.3984) = - 0.2128%/station. Then: y = - 0.2128  4.94922 + 1.25  4.9492 + 880.23 = 881.20 ft.20/01/2013 7
  8. 8. Unequal Tangent Vertical Curve In Figure below; l1 l2 a line v1v2 is drawn parallel with AB Av1 = v1V and Vv2 = v2B The curve from A to K is L consisted from two equal parabolic vertical curves. l1 l2 The curve from K to B is EVC consisted from two equal C parabolic vertical curves. BVC (B) K g v2 (A) v1 V20/01/2013 8
  9. 9. Unequal Tangent Vertical Curve In Figure below; The elevation of v1 is the average of that of A and V. The elevation of v2 is the average of that of V and B. L CK = KV l1 l2 CV = l1l2/L(g2 – g1) CK = l1l2/2L(g2 – g1) EVC C l1, l2 and L are distances in BVC (B) K stations and g2 and g1 are g v2 percent grades. (A) v1 V The grade g of v1v2 is the same as AB; g = (elev. B – elev. A)/L20/01/2013 9
  10. 10. Unequal Tangent Vertical Curve Problem: In Figure below, g1 = - 4%, g2 = + 3.5%, l1 = 250 ft, l2 = 400 ft, the elevation of V is 450.00 ft. Point V is at station 55 + 00. Compute the elevations of each half station between A and B. Solution: L The elevation of A is: 450.00 + 2.50  4 = 460.00 ft. The elevation of B is: l1 l2 450.00 + 4.00  3.5 = 464.00 ft. g = (464.00 – 460.00)/6.50 = + 0.615% EVC C Elevation of K: BVC (B) 460.00 - 1.25  4 + 1.25  0.615 = 455.77 ft. K g v2 (A) v1 Elevation of B: V 455.77+2.00  0.615+2.00  3.5 = 464.00 ft. (check).20/01/2013 10
  11. 11. Unequal Tangent Vertical Curve Solution Continued: For the vertical curve from A to K: r = (0.615 – (- 4.00))/2.50 = 1.846%/station Elevation of BVC at A = 460.00 ft. The equation of the curve is: y = 0.923 x2 – 4x + 460.00 For the vertical curve from K to B: r = (3.5 – 0.615)/4.00 = + 0.721%/station Elevation of BVC at K = 455.77 ft. The equation of the curve is: y = 0.361 x2 + 0.615x + 455.77 The computations for the two curves are shown in Table below: * The low point on this curve is at station 54 + 66.68 at which the elevation is 455.67.20/01/2013 11
  12. 12. Unequal Tangent Vertical Curve Solution Continued: Station x x2 (r/2)x2 g1x Elev. BVC Elev. Curve 52 + 50 0 0 0 0 460.00 460.00 53 + 00 0.5 0.25 + 0.23 - 2.00 460.00 458.23 53 + 50 1 1 + 0.92 - 4.00 460.00 456.92 54 + 00 1.5 2.25 + 2.08 - 6.00 460.00 456.08 54 + 50 2 4 + 3.69 - 8.00 460.00 455.69 55 + 00 2.5 6.25 + 5.77 - 10.00 460.00 455.77 55 + 50 0.5 0.25 + 0.09 + 0.31 455.77 456.17 56 + 00 1 1 + 0.36 + 0.62 455.77 456.75 56 + 50 1.5 2.25 + 0.81 + 0.92 455.77 457.50 57 + 00 2 4 + 1.44 + 1.23 455.77 458.44 57 + 50 2.5 6.25 + 2.26 + 1.54 455.77 459.57 58 + 00 3 9 + 3.25 + 1.85 455.77 460.87 58 + 50 3.5 12.25 + 4.42 + 2.15 455.77 462.34 59 + 00 4 16 + 5.78 + 2.46 455.77 464.0020/01/2013 12
  13. 13. Minimum Length of Vertical Curve The length of a vertical curve on a highway should be ample to provide a clear sight that is sufficiently long to prevent accidents. The American Association of State Highway and Transportation Officials, AASHTO has developed criteria for the, safe passing sight distance, Ssp and safe stopping sight distance, Snp Assumption: • The eyes of the driver of a vehicle are about 3.75 ft above the pavement. • The top of an oncoming vehicle is about 4.50 ft above the pavement. • An obstruction ahead of the vehicle is 0.50 ft.20/01/2013 13
  14. 14. Minimum Length of Vertical Curve The values of length of Ssp and Snp are given in the following Table: AASHTO Sight-Distance Recommendations Design Speed (mph) Ssp (ft) Snp (ft) 30 1,100 200 40 1,500 275 50 1,800 350 60 2,100 475 70 2,500 600 80 2,700 75020/01/2013 14
  15. 15. Minimum Length of Vertical Curve The values of minimum length of a vertical curve are given here without proof: a ) If , S sp  L S sp ( g1  g 2 ) 2 L 33 b) If , S sp  L 33 L  2 S sp  g1  g 2 a ) If , S np  L S np ( g1  g 2 ) 2 L 14.0 b) If , S np  L 14.0 L  2 S np  g1  g 220/01/2013 15
  16. 16. Minimum Length of Vertical Curve Problem: The grades at a crest are g1 = + 2% and g2 = - 3% and the design speed is 60 mph. Compute the lengths of the vertical curves required for the safe passing sight distance, Ssp and the safe stopping sight distance, Snp recommended by the AASHTO. Solution: From previous Table, the safe passing sight distance, Ssp = 2100 ft and the safe stopping sight distance, Snp = 475 ft. If , S sp  L S sp ( g1  g 2 ) 2 2100 2  0.05 L   6682 ft 33 33 If , S np  L S np ( g1  g 2 ) 2 475 2  0.05 L   806 ft 14.0 14.020/01/2013 16
  17. 17. Minimum Length of Vertical Curve Continued Solution: Since the safe passing sight distance, Ssp = 2100 ft  L = 6682 ft and the safe stopping sight distance, Snp = 475 ft  L = 806 ft, it would not be necessary to apply the cases b) of Ssp and Snp longer than the curve.20/01/2013 17
  18. 18. Crowns & Superelevations x 0.015 – 0.020 ft/ft y1 y W/2 Pavement Cross-Section Showing W Pavement Crown Horizontal Curves, 8% superelevation W = 24 ft, e = 0.08 (24) = 1.92 m The curve to be banked20/01/2013 18
  19. 19. Crowns & Superelevations • Design Speed of the highway • Side Friction Climate and Area Classification (Urban/Rural) Snow (7% - 8%) 10% - 12% the highest for gravel roads with cross drainage Urban streets (4% - 6%) Runoff, is the transition from the normal crown cross section on a tangent to the fully superelevated cross section. The length of the runoff transition may range from (500 ft – 600 ft) or (100 ft– 250 ft).20/01/2013 19
  20. 20. Crowns & Superelevations 4 3 Circular curve 2 1 3 2 1 Normal 3 3 4 2 1 2 1 Convex Transition section Normal crowned section Section 1 - 1 Level Convex Maximum superelevation Transition section Section 2 - 220/01/2013 20

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