Class_19_Vertical_Curvature_I

Class_19_Vertical_Curvature_I - The Pennsylvania State...

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The Pennsylvania State University Department of Civil and Environmental Engineering CE 321: Highway Engineering Class 19 – Vertical Curvature (Part 1) Spring 2008
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Vertical Curves
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Crest Curve Design
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Sag Curve Design
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Types of Vertical Curves
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Vertical Curve Fundamentals y = ax 2 + bx + c assuming a constant rate of change of slope and equal tangent lengths. When x = 0, y = C = elevation on curve dy / dx = 2ax + b = slope (rise / run) at x = 0 => slope is dy/dx = b = G 1 (the initial slope)
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Vertical Curve Fundamentals d 2 y/dx 2 = 2a = rate of change of slope 2a = (G 2 - G 1 ) / L a = (G 2 - G 1 ) / 2L therefore: y x = a x 2 + G 1 x + Elevation at PVC Where: L (sta.), x (sta.)
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High / Low Point on Curve When dy/dx = 0 then a = (G 2 - G 1 ) / 2L therefore, dy/dx = 2ax + b = 0 2ax + G 1 = 0 2ax = - G 1 2(G 2 - G 1 )(x)/2L = - G 1 then: x = - G 1 L / (G 2 - G 1 )
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Vertical Curve Design Vertical Curves maximum grades depend on: Design Speed Type of Terrain Type of Highway Length of Grade
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Vertical Curve Design Grades also affect: fuel consumption speed safety or crashes (speed differential)
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Vertical Curve Design Design Speed Flat / Rolling Mountainous 50 mi/hr 5% 7% 60 mi/hr 4% 6% 70 mi/hr 3% 5% (+ 2 % for secondary / local roads)
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Example Problem 1 A 600 ft equal tangent sag vertical curve has the PVC at station 170+00 and elevation 1000 ft. The initial grade is -3.5 % and the final grade is 0.5 %. Determine the elevation and stationing of the PVI, PVT, and the lowest point on the curve.
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Example 1 (con’t) Solution: Station PVI = 170+00 + 300 = 173+00 Station PVT = 170+00 +
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