Q6. To help improve the safety of at-grade intersection of two roads, the intersection is being redesigned so that the E-W road will pass underneath the N-S road. Currently the vertical alignment of the E-W road consists of a crest vertical curve joining a 4% upgrade to a 3% downgrade. The existing vertical curve is 138 m long, the PVC of this curve is at station 14+47, and the elevation of the PVC is 477 m. The centerline of the N-S road is at station 15+45. Your job is to find the shortest vertical curve that provides 6.0 m of clearance between the new E-W road and the N-S road.

Q6. To help improve the safety of at-grade intersection of two roads, the intersection is being redesigned so that the E-W road will pass underneath the N-S road. Currently the vertical alignment of the E-W road consists of a crest vertical curve joining a 4% upgrade to a 3% downgrade. The existing vertical curve is 138 m long, the PVC of this curve is at station 14+47, and the elevation of the PVC is 477 m. The centerline of the N-S road is at station 15+45. Your job is to find the shortest vertical curve that provides 6.0 m of clearance between the new E-W road and the N-S road.

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

5. A 600 ft equal tangent vertical curve has the PVC at station 170+00 and elevation 1000 ft. The initial grade is -3.5% and the final grade is +0.5%. Determine the stationing and elevation of the PVI, the PVT, and the lowest point on the curve.
A crest vertical curve (equal tangent) has PVI at station 210+00 and elevation 420.75 ft. The initial grade is 2.5% and the final grade is -4%. The length of the curve is 950 ft. Determine the following, 1. Stationing of the high point, PVC, and PVT. 2. Elevation at station 212+00, PVC, high point, and PVT.
An SVC connects a -4% grade to a +4% grade. The curve goes under a pedestrian bridge at PVI. The vertical clearance under the bridge should be 15 ft and the required sight distance is 984 ft. The height of the driver's eye is 6 ft, and the assumed hazardous object is 1.5 ft tall. What should be the length of the curve? Assume S > L.
An equal-tangent crest vertical curve is designed for 70 mi/h. The high point is at elevation 1011.4 ft. The initial grade is +2% and the final grade is -1%. What is the elevation of the PVT?
A 500-m long equal tangent vertical curve has a point of vertical curvature at station 2 + 600 and elevation 450 m. The initial grade is -5% c) determine the desing speed for this vertical curve( in 10 km/h increments )
A crest and sag curve connect a 0% road segment with a +2% road segment. The two curves connect with each other (PVI=PVC) and share a 3% common grade. If the design speed of the curves is 45 mi/h, what is the elevation difference between the two road segments?
Questions 5. a) A vertical crest curve has an approach grade +3% and a departure grade -2%. The Intersection Point (IP) level is 100m Australian Height Datum (AHD), on Chainage 1000 m. It is proposed to use a curve 550m long. Calculate: i) Chainage and Level at TP1 ii) Chainage and Level at TP2. iii) Chainage and level at the apex point.
A horizontal curve is being designed for a new four-lane roadway with 11-ft lanes. The PT is located at station 1050+20, the design speed is 45 mph and maximum superelevation of 4%. If the central angle of the curve is 30 degrees, what is the radius of the curve and location of the PC and PI?
West and east highway segments are separated by 1000 ft horizontally. The west segment has a 0% constant grade, and the east segment has a –1% grade. The east segment has a higher elevation than the west segment, and the two segments are connected by a joining sag and crest curve combination (so PVTs = PVCc). If the road is designed for 60 mi/h, what is the elevation difference between the west and east highway segments?
Q 1: Determine the minimum length of a highway sag curve (based on SSD Criterion) that is designed to join a - 0.035 grade to a + O.025 grade, if the design speed is 90 km/h. Assume a = 3.4 m/s2, t = 2.5 s.
Step by step solutionProblem 1. An existing equal-tangent sag vertical curve is designed for 60 mph. The initial grade is -3% and the elevation of the PVT is 750 ft. The PVC of the curve is at station 134+00 and the PVI is at 137+00. An overpass is being constructed directly above the PVI. The overpass design assumes the driver’s eye height is 5 ft. Determine: a. the stopping sight distance b. the curve lengthc. the clearance heightd. the mid-curve offset e. the elevation of the PVIf. the elevation of the bottom of the overpass structureg. Is there a linear relationship between curve length and K-value in this problem? Note: No need to show calculations for case (g).
2. A vertical parbolic curve was design in order to have a clear sight distance of 120 m. The grade lines intersect at Sta 9+000 at elev 160.50 . The curve was design such that when the height of the drivers eye is 1.50 m above the payment it would just see an object whose height is 0.10m above the pavement. Determine the max speed that a car could travel grade of 5% and a downgrade of -3%.