A vertical summit parabolic curve has a vertical offset of 0.375 m. from the curve to the grade tangent at STA. 10+050. The curve hasa slope of + 4% and- 2% grades intersecting at the PI. The offset distance of the curve at PI is equal to 1.5 m. If the stationing of thePC is at 10 + 000a. Compute the required length of curve.b. Compute the horizontal distance of the vertical curve turning point from the point of intersection of the gradesc. Compute the elevation of the vertical curve turning point if the elevation of PT is 86.45m

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Answered: A vertical summit parabolic curve has a…

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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23.17 A -5% slope meets a +3% grade at elevation 548.97 ft at station 43 +00. Assuming a 1000-ft vertical curve, determine the elevation on the curve at each 100- ft station and find the station and elevation at the low point on the curve. (Ans.: Low point at 44 +25, elevation 558.34 ft)
The symmetrical parabolic curve passes through point A whose elevation is 23.23m at a distance of 54m from the PC. The elevation of the PC at station 4+100 is 22.56m. The grade of the back tangent is 2% and the length of the curve is 160m. Compute for the stationing of the highest point of the curve.
Vertical Curve
A compound curve connects three tangents having an azimuth of 240°, 270° and 280° respectively. The length of the chord is 382.3 m long measured from the PC to the PT of the curve and is parallel to the common tangent having an azimuth of 270°. If the stationing of the PT is 6 + 382.3 . a. Determine the total length of the compound curve. b. Determine the stationing of the PCC. c. Determine the stationing of the PC.
A symmetrical parabolic summit curve connects two grades of +6% and -4%. It is to pass though a point "P" the stationing of which is 35+280 and the elevation is 193.13 meters. If the elevation of the grade intersection is 200 m with stationing 35 + 300. 1. What is the length of the curve? a. 120 m b. 123 m c. 121 m d. 122 m
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3.9 An equal-tangent crest curve connects a +2% initial grade with a -1% final grade, and is designed for 55 mi/h. The station of the PVI is 233 + 40 with elevation 1203 ft. What is the elevation of the curve at station 234 +00?
The length of a summit curve is 188 meters. Compute the radius of the curve if the tangentgrades are +3% and -5%.
A vertical curve has its highest point at a distance 48 meters from point PT. The curve is having two tangents having a grade of +6% and -4% which intersects at point Pl. The curve passes thru point A at station 25+140. The elevation of point Pl is 100 meters above sea level at 25+160 station. a. Determine the height of the maximum Offset (H) b. The Length of the Curve (L)
An unsymmetrical parabolic summit curve AB has a slope of +7% on the back tangent and -4% on the forward tangent. The length of the curve on the side of the back tangent is 60 m and 40 m on the side of the forward tangent. What is the vertical offset at the intersection of the tangents from the curve? How far is the HP of the curve from A?
A reversed curve is connecting the two tangent lines AB and CD having directions of due East and D 60° E, respectively. The radius of the first curve at A (PC) is 200 m and that of the second curve at D (PT) is 400 m. Stationing of A (PC) is at 10 + 120.50. If ED is 300 m long and has a bearing of S 20° E. 1. Find the central angle of first curve 2. Find the central angle of the second curve 3. Find the stationing of PT
The length of a summit curve is 188 meters. Compute the radius of the curve if the tangentgrades are +3% and -5%.

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