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:
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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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?
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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