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Vertical Alignment CE 2710 Spring 2014 Lecture 18 Originally Created by Chris McCahill

Vertical Alignment CE 2710 Spring 2014 Lecture 18 Originally Created by Chris McCahill. Components of The Alignment. Horizontal Alignment. Vertical Alignment. Cross-section. Vertical Alignment & Topography. Texas DOT. Today ’ s Class. Maximum/minimum grade

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Vertical Alignment CE 2710 Spring 2014 Lecture 18 Originally Created by Chris McCahill

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  1. Vertical AlignmentCE 2710Spring 2014Lecture 18Originally Created by Chris McCahill

  2. Components of The Alignment Horizontal Alignment Vertical Alignment Cross-section

  3. Vertical Alignment & Topography Texas DOT

  4. Today’s Class • Maximum/minimum grade • Properties of vertical curves (parabolic) • Technical design of vertical curves

  5. Vertical Alignment Tangents and Curves Like the horizontal alignment, the vertical alignment is made up of tangent and curves In this case the curve is a parabolic curve rather than a circular or spiral curve Crest Curve G2 G3 G1 Sag Curve

  6. Maximum Grade www.geograph.org.uk Harlech, Gwynedd, UK (G = 34%)

  7. Maximum Grade www.nebraskaweatherphotos.org

  8. Maximum Grade Dee747 at picasaweb.google.com

  9. Maximum and Minimum Grade • One important design consideration is the determination of the maximum and minimum grade that can be allowed on the tangent section • The minimum grade used is typically 0.5% • The maximum grade is generally a function of the • Design Speed • Terrain (Level, Rolling, Mountainous) • On high speed facilities such as freeways the maximum grade is generally kept to 5% where the terrain allows (3% is desirable since anything larger starts to affect the operations of trucks) • At 30 mph design speed the acceptable maximum is in the range of 7 to 12 %

  10. Properties of Vertical Curves BVC G1 G2 EVC PI L/2 L/2 L Change in grade: A = G2 - G1 where G is expressed as % (positive /, negative \) For a crest curve, A is negative For a sag curve, A is positive

  11. Properties of Vertical Curves BVC G1 G2 EVC PI L/2 L/2 L Rate of change of curvature: K = L / |A| Which is a gentler curve - small K or large K?

  12. Properties of Vertical Curves BVC G1 G2 EVC PI L/2 L/2 L • Rate of change of grade: r = (g2 - g1) / L • where, • g is expressed as a ratio (positive /, negative \) • L is expressed in feet or meters • Note – K and r are both measuring the same characteristic of the curve • but in different ways

  13. Properties of Vertical Curves BVC G1 Elevation = y G2 EVC PI L Equation for determining the elevation at any point on the curve y = y0 + g1x + 1/2 rx2 where, y0 = elevation at the BVC g = grade expressed as a ratio x = horizontal distance from BVC r = rate of change of grade expressed as ratio

  14. Properties of Vertical Curves • Distance BVC to the turning point (high/low point on curve) • xt = -(g1/r) • This can be derived as follows • y = y0 + g1x + 1/2 rx2 • dy/dx = g1 + rx • At the turning point, dy/dx = 0 • 0 = g1 + rxt • Therefore, • xt = -(g1/r) xt Low Point

  15. Properties of Vertical Curves BVC G1 G2 EVC PI Example: G1 = -1% G2 = +2% Elevation of PI = 125.00 m Station of EVC = 25+00 Station of PI = 24+00 Length of curve? L/2 = Sta. EVC – Sta. PI L/2 = 2500 m - 2400 m = 100 m L = 200 m

  16. Properties of Vertical Curves BVC G1 G2 EVC PI Example: G1 = -1% G2 = +2% Elevation of PI = 125.00 m Station of EVC = 25+00 Station of PI = 24+00 r - value? r = (g2 - g1)/L r = (0.02 - [-0.01])/200 m r = 0.00015 / meter

  17. Properties of Vertical Curves BVC G1 G2 EVC PI Example: G1 = -1% G2 = +2% Elevation of PI = 125.00 m Station of EVC = 25+00 Station of PI = 24+00 Station of low point? x = -(g1/r) x = -([-0.01] / [0.00015/m]) x = 66.67 m Station = [23+00] + 67.67 m Station 23+67

  18. Properties of Vertical Curves BVC G1 G2 EVC PI Example: G1 = -1% G2 = +2% Elevation of PI = 125.00 m Station of EVC = 25+00 Station of PI = 24+00 Elevation at low point? y = y0 + g1x + 1/2 rx2 y0 = Elev. BVC Elev. BVC = Elev. PI - g1L/2 Elev. BVC = 125 m - [-0.01][100 m] Elev. BVC = 126 m

  19. Properties of Vertical Curves BVC G1 G2 EVC PI Example: G1 = -1% G2 = +2% Elevation of PI = 125.00 m Station of EVC = 25+00 Station of PI = 24+00 Elevation at low point? y = y0 + g1x + 1/2 rx2 y = 126 m + [-0.01][66.67 m] + 1/2 [0.00015/m][66.67 m]2 y = 125.67 m

  20. Properties of Vertical Curves BVC G1 G2 EVC PI Elevation at station 23+50? y = 126 m + [-0.01][50 m] + 1/2 [0.00015/m][50 m]2 y = 125.69 m Elevation at station 24+50? y = 126 m + [-0.01][150 m] + 1/2 [0.00015/m][150 m]2 y = 126.19 m Example: G1 = -1% G2 = +2% Elevation of PI = 125.00 m Station of EVC = 25+00 Station of PI = 24+00

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