Design the curve, of An equal tangent vertical curve is to be constructed between grades of -2.0% (initial) and 1.0% (final). The PVI is at station 11 + 000.000 and at elevation 420 m. Due to a street crossing the roadway, the elevation of the roadway at station 11 + 071.000 must be at 421.5 m.

Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
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Design the curve, of An equal tangent vertical curve is to be constructed between grades of -2.0% (initial) and 1.0% (final). The PVI is at station 11 + 000.000 and at elevation 420 m. Due to a street crossing the roadway, the elevation of the roadway at station 11 + 071.000 must be at 421.5 m. 

GIVEN DIMENSIONS ON THE ATTACHED PICTURE

From the given values
PVC station = 11 +000.000- /
PVC elevation = PVI elevation - G₁ X = 420+ 0.01L
x = (11 +071.000) - (11 +000.000)- /
x = 71 +0.5L
a=
b=G₁ -0.02
=
C = 420 + 0.01L
G₂-G 0.01-(-0.02)
21.
21.
=
0.015
L
y = ax²+bx+c
421.5 = 0.015 (71+0.5L)² + (−0.02) (71 +0.5L) +420 + 0.01L
L
simply the equation
1.855 =
75.613
+0.00375L
L
0.00375L² - 1.855L + 75.615 = 0
L = 449.842 m
Length of curve =449.842 m
Station of PVC = (10+775.079)
Elevation of PVC = 424.49842 m
Station of PVT =(11+224.921)
Elevation of PVT = 422.24981 m
Transcribed Image Text:From the given values PVC station = 11 +000.000- / PVC elevation = PVI elevation - G₁ X = 420+ 0.01L x = (11 +071.000) - (11 +000.000)- / x = 71 +0.5L a= b=G₁ -0.02 = C = 420 + 0.01L G₂-G 0.01-(-0.02) 21. 21. = 0.015 L y = ax²+bx+c 421.5 = 0.015 (71+0.5L)² + (−0.02) (71 +0.5L) +420 + 0.01L L simply the equation 1.855 = 75.613 +0.00375L L 0.00375L² - 1.855L + 75.615 = 0 L = 449.842 m Length of curve =449.842 m Station of PVC = (10+775.079) Elevation of PVC = 424.49842 m Station of PVT =(11+224.921) Elevation of PVT = 422.24981 m
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