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# Vertical Curves (Part 1)

## by haroldtaylor1113 on Jan 19, 2013

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## Vertical Curves (Part 1)Presentation Transcript

• 1Vertical Curves 20/01/2013
• Vertical curves are the curves used in a vertical plane to provide a smooth transition between the grade lines of highways and railroads. PVI 2 g1 BVC g2 EVC L0 + 00 Forward Stations 20/01/2013
• In previous Figure;• g1 is the slope of the lower chainage grade line.• g2 is the slope of the higher chainage grade line.• BVC is the beginning of the vertical curve.• EVC is the end of the vertical curve. 3• PVI is the point of intersection of the two adjacent gradelines.• L is the length of vertical curve.• A = g2 – g1, is the algebraic change in slope direction. 20/01/2013
• General Equation of ParabolaThe vertical axis parabola is used in vertical alignment design. Theparabola has two desirable characteristics such as;• A constant rate of change of slope. 4• Ease of computation of vertical offsets. y  a x2  b x  c The slope : dy  2a x  b dx The rate of change of slope : d2y  2a (const.) dx A  2a L 20/01/2013
• General Equations of Parabola Y Y PVI EVC EVCX X BVC X BVC X PVI 5 b) Crest curvea) Sag curve Y YIf the origin of the axes is placed at BVC, the general equation becomes; y  a x2  b x y  a x 2  g1 xThe slope at the origin is g1 : dy  slope  2 a x  g1 dx 20/01/2013
• L L/2 L/2 E V x ax2 g1x 6 m BVC C E EVC Crest CurveGeometric Properties of the Parabola 20/01/2013
• Geometric Properties of the ParabolaFrom previous Figure;• g1x is the difference in elevation between the BVC and a point on the g1gradeline at a distance x. 7• ax2 is the tangent offset between the gradeline and the curve.• BVC + g1x – ax2 = curve elevation at distance x from the BVC.• BVC to V = L/2 = V to EVC• Offsets from the two gradelines are symmetrical with respect to the PVI (V).• Cm = mV 20/01/2013
• Computation of the High/Low Point on a Vertical Curve Low Point EVC BVC 8 Tangent Through Low Point PVI Slope = 2 ax + g1 = 0The tangent drawn through the low point is horizontal with a slope of zero;2 ax + g1 = 0x = - g1 (L/A)Where x is the distance from the BVC to the high or low point. 20/01/2013
• Computing a Vertical CurveProcedures for computing a vertical curve:1. Compute the algebraic difference in grades A = g1 – g2.2. Compute the stationing of BVC and EVC. Subtract/add L/2 to the PVI.3. Compute the distance from the BVC to the high or low point; x = - g1(L/A). Determine the station of the high or low point. 94. Compute the tangent gradeline elevation of the BVC and the EVC.5. Compute the tangent gradeline elevation for each required station.6. Compute the midpoint of the chord elevation: [(elevation of BVC + elevation of EVC)/2].7. Compute the tangent offset d at PVI, Vm, d = (difference in elevation of PVI and C)/2).8. Compute the tangent offset for each individual station (see line ax2), tangent offset = d (x)2/(L/2)2 = (4d) x2/L2 where x is the distance from the BVC or EVC (whichever is closer) to the required station.9. Compute the elevation on the curve at each required station by combining the tangent offsets with the appropriate tangent gradeline elevations. Add for sag curves and subtract for crest curves. 20/01/2013
• Problem: You are given the following information: L = 300 ft, g1 = - 3.2%, g2 = + 1.8%,PVI at 30 + 30 with elevation = 465.92. Determine the location of the low point and theelevations on the curve at even stations, as well as the low point. L = 300 ft L/2 = 150 ft 1.875 469.67 10 BVC 467.795 EVC 28 + 80.00 31 + 80.00 470.72 468.62 d = 1.875 PVI 30 + 30.00 Low Point 465.92Solution:1. A = g2 – g1 = 1.8 – (- 3.2) = 5.02. Station BVC = PVI – L/2 = (30 + 30.00) – (1 + 50) = 28 + 80.00 Station EVC = PVI + L/2 = (30 + 30.00) + (1 + 50) = 31 + 80.0Ukkk Check EVC – BVC = L; (31 + 80.00) – (28 + 80.00) = 300 20/01/2013
• Solution:3. Location of the low point; x = - g1L/A = (3.2  300)/5 = 192.00 ft from BVC.4. Elevation BVC = Elevation PVI + (150 ft at 3.2%) = 465.92 + 4.80 = 470.72 Elevation EVC = Elevation PVI + (150 ft at 1.8%) = 465.92 + 2.70 = 468.625. Tangent gradeline elevations are computed and entered in the table below: e.g. Elevation at 29 + 00 = 470.72 – (0.032  20) = 470.72 – 0.64 = 470.08 11 Station Tangent Elevations + Tangent Offset (x/L/2)2 .d = Curve Elevation BVC 28 + 80 470.72 (0/150)2  1.875 = 0 470.72 29 + 00 470.08 (20/150)2  1.875 = 0.03 470.11 30 + 00 466.88 (120/150)2  1.875 = 1.20 468.08 PVI 30 + 30 465.92 (150/150)2  1.875 = 1.875 467.80 Low 30 + 72 466.68 (108/150)2  1.875 = 0.97 467.65 31 + 00 467.18 (80/150)2  1.875 = 0.53 467.71 EVC 31 + 80 468.62 (0/150)2  1.875 = 0 468.62 30 + 62 466.50 (118/150)2  1.875 = 1.16 467.66 30 + 72 466.68 (108/150)2  1.875 = 0.97 467.65 30 + 82 466.86 (98/150)2  1.875 = 0.80 467.66 20/01/2013
• Solution:6. Mid-chord elevation; [470.72 (BVC) + 468.62 (EVC)]/2 = 469.67 ft.7. Tangent offset at PVI (d): d = (difference in elevation of PVI and mid-chord)/2 d = (469.67 – 465.92)/2 = 1.875 ft.8. For other stations, tangent offsets are computed by multiplying the distance ratio squared (x/L/2)2 by the maximum tangent offset (d). Refer to previous Table.9. 12 The computed tangent offsets are added to the tangent elevation to determine the curve elevation. 20/01/2013
• Computing Vertical Curve Directly From the General Equationy = ax2 + bx + cWhere;a = (g2 – g1)/2L 13L, horizontal length of vertical curve.c, elevation at BVC.x, horizontal distance from BVCy, elevation on the curve at distance x from the BVC.Using previous given data, compute the vertical curve elevationsdirectly from parabolic general equation; y = ax2 + bx + c. 20/01/2013
• Computing Vertical Curve Directly From the General Equation y = ax2 + bx + c Station Distance from ax2 bx c y (Elevation on the BVC (x) CurveBVC 28 + 80 0 0 0 0 470.72 14 29 + 00 20 0.03 - 0.64 470.72 470.11 30 + 00 120 1.20 - 3.84 470.72 468.08PVI 30 + 30 150 1.88 - 4.80 470.72 467.80Low 30 + 72 192 3.07 - 6.14 470.72 467.65 31 + 00 220 4.03 - 7.04 470.72 467.71EVC 31 + 80 300 7.50 - 9.60 470.72 468.62 20/01/2013