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An equal-tangent crest vertical curve is being designed for a speed of 55 mph. The curve connects grades of 2% and -3.5% in the direction of interest. The curve high point is at station 110+00 and has an elevation of 500 ft. What is the station and elevation of the PVC and the PVI? Example prob below:

An equal-tangent crest vertical curve is being designed for a speed of 55 mph. The curve connects grades of 2% and -3.5% in the direction of interest. The curve high point is at station 110+00 and has an elevation of 500 ft. What is the station and elevation of the PVC and the PVI? Example prob below:

Structural Analysis
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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An equal-tangent crest vertical curve is being designed for a speed of 55 mph. The curve connects grades of 2% and -3.5% in the direction of interest. The curve high point is at station 110+00 and has an elevation of 500 ft. What is the station and elevation of the PVC and the PVI? Example prob below: An equal-tangent sag vertical curve is designed for a speed of 60 mph. The curve low point is at an elevation of 750.5 ft. If the initial and final grades are -2% and 2.5%, respectively, what is the elevation of the PVI?
Transcribed Image Text:An equal-tangent crest vertical curve is being designed for a speed of 55 mph. The curve connects grades of 2% and -3.5% in the direction of interest. The curve high point is at station 110+00 and has an elevation of 500 ft. What is the station and elevation of the PVC and the PVI? Example prob below: An equal-tangent sag vertical curve is designed for a speed of 60 mph. The curve low point is at an elevation of 750.5 ft. If the initial and final grades are -2% and 2.5%, respectively, what is the elevation of the PVI?
Given: V = 60 mph Ynt = 750.5 ft G₁ = -2%; G₂ = 2.5% Find: Elevation of PVT Curve equation: -2% PVC a= Solution: Assume S <L From Table 3.3 Kpesign = 136 and S=570 ft L = A x Koesign = 1-2-2.51 x 136 = 612' → Assumption ok XhiK|G₁|= 136 x |-2| = 272' Low point: Y-75058 G₂-G₂ 0.025-(-0.02) 2x612 2L b=G₁ = -0.02 c = elevation of PVC y = 0.0000367x² -0.02x + elev of PVC Knowing x and y of high point, plug into equation above to solve for c: 0.0000367 750.5 = 0.0000367(272)² -0.02(272)+c c = 753.2' Elevation of PVI: G₂L elevation of PVI = elevation of PVC + = 753.2 + 2 PVI -0.02 x 612 2 = 747.1' 2.5%
Transcribed Image Text:Given: V = 60 mph Ynt = 750.5 ft G₁ = -2%; G₂ = 2.5% Find: Elevation of PVT Curve equation: -2% PVC a= Solution: Assume S <L From Table 3.3 Kpesign = 136 and S=570 ft L = A x Koesign = 1-2-2.51 x 136 = 612' → Assumption ok XhiK|G₁|= 136 x |-2| = 272' Low point: Y-75058 G₂-G₂ 0.025-(-0.02) 2x612 2L b=G₁ = -0.02 c = elevation of PVC y = 0.0000367x² -0.02x + elev of PVC Knowing x and y of high point, plug into equation above to solve for c: 0.0000367 750.5 = 0.0000367(272)² -0.02(272)+c c = 753.2' Elevation of PVI: G₂L elevation of PVI = elevation of PVC + = 753.2 + 2 PVI -0.02 x 612 2 = 747.1' 2.5%
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