Problem 1.An equal-tangent vertical curve connects a +1.5% initial grade and a -3% final grade. The PVT is at station 120+52 and elevation197.665 ft. If the design speed is 50 mi/h, what is the elevation of the curve at station 118+00?MetricDesignspeed(km/h)889882888882030405060708090100110120130Stoppingsightdistance2035506585105130160185220250285Rate of verticalcurvature, KªCalculated Design120.61.922943.86.411.016.825.738.952.073.695.0123.44711172639527495124StoppingsightDesignspeed distance(mph)15 20 25 30 35 40 45 50 55 60 65 70 75 80US Customary88888888888880115155200250305360425495570645730820910Rate of verticalcurvature, KªCalculated Design33.06.111.118.529.043.160.183.7113.5150.6192.822486NNWRate of vertical curvature, K, is the length of curve per percent algebraic difference inintersecting grades (A). K = L/A1219294461114151193246.9 247311.6 312383.7 384Exhibit 3-76. Design Controls for Stopping Sight Distance and for Crest and Sag VerticalCurves***

Following are the given information.Initial grade, Final grade, Station of PVT Elevation of PVT…

An equal-tangent vertical curve connects a +1.5% initial grade and a -3% final grade. The PVT is at station 120+52 and elevation 197.665 ft. If the design speed is 50 mi/h, what is the elevation of the curve at station 118+00?

An equal-tangent vertical curve connects a +1.5% initial grade and a -3% final grade. The PVT is at station 120+52 and elevation 197.665 ft. If the design speed is 50 mi/h, what is the elevation of the curve at station 118+00?

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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