An unsymmetrical parabolic curve has a forward tangent of -8% and a backward tangent of 5%. The length of the curve on the left side (PVC) is 60m. and the vertex is at station 0+500 is at elevation of 100m. The elevation of the road just below the vertex is 97.3m. a. Calculate the total length of the unsymmetrical curve. b. Find the stationing of the highest point. С. Find the elevation of station 0+550.

An unsymmetrical parabolic curve has a forward tangent of -8% and a backward tangent of 5%. The length of the curve on the left side (PVC) is 60m. and the vertex is at station 0+500 is at elevation of 100m. The elevation of the road just below the vertex is 97.3m. a. Calculate the total length of the unsymmetrical curve. b. Find the stationing of the highest point. С. Find the elevation of station 0+550.

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

With full solution pls
Three simple curves are connected to each other such that the fırst and the second form a compound curve. While the second and the third form a reversed curve. The distance between the point of curvature and the point of tangency of the compound curve (which is also the point of reversed curvature of the reversed curve is 500 m. If the angle of convergence between the second and the third tangents is 16°, I, = 47°. R1 =200 meters, I2 = 65°, R3 =160 m, and stationing at PC is 5+600. Hint: sine law. determine the stationing at PT at the end of the long chord. Answer for part A should be in this Format: x+xx.xXxx (FOUR DECIMAL PLACES i.e. 0+123.4567) determine the angle between the long chord of the compound curve and the first tangent. Answer in four decimal places (i.e. 12.3456)
Show solution w/fbd
TOPIC: COMPOUND CURVES please answer, i will upvote your solution
A compound curve has a common tangent parallel to the long chord from PC to PT which measures 348 meters in length. The angle it makes with the longer and shorter tangents are 17° and 22° respectively. a. Find the radius of the first curve. b. Find the radius of the second curve. c. If the stationing of PC is at 10+916.07, find the stationing of PT.
QUESTION 1 The long chord from the PC to the PT of a compound curve is 300 m. A common tangent, from point A to B, has an angle of 12° with AV and 15⁰ with BV. If the common tangent is parallel to the long chord, determine: a. the radius of the first curve Answer: R₁= b. the radius of the second curve Answer: R₂= c. the stationing of PT if PC is 10+204.30 Answer: Sta PT= m m Note: Unit has been provided after the blank space. No need to indicate units on your answer.
A curve having tangents of UV and VW intersects at point V. UV and VW has an azimuth (from south) of 165° and 215° respectively. U is at 10+340.26. The degree of the curve is 4°. a. What is the length of the long chord UW. b. What is the tangent distance UV. c. What is the external distance of the curve? d. What is the middle ordinate of the curve? e. A line passes through the center of the curve to the curve itself at point X, then to UV, making an angle of 60°. What is the station of point X. Draw and plot the curve. Compute all of the necessary elements of the curve. Include the proper units/dimensions and round-off the answers to 3 decimal places.
For a simple curve, the stationing of the vertex is 11+361.2 while the stationing of the PC is 11+309. If the degree of curve is 6⁰, find the deflection angle of point 3 on curve.
Assignment No. 1 1. A simple curve have the following elements: middle ordinate = 2 ma and length of long chord = 70 m. Find the degree of curve using chord basis. 2. The central angle of a circular curve is 36° 30'. Determine the radius if the external distance is 12.02 m. 3. A compound curve has R1=150m and R2-315m respectively, the backward tangent and the forward tangent are to form a right angle. Compute the values of the central angles and the value of the Long Chord
A simple curve has tangents AB and BC intersecting at a common point B. AB has a south azimuth of 180° and BC has a south azimuth of 230°. The stationing of the point of curvature at A is 10 + 140.26. If the degree of curve of the simple curve is 4°. a.) Compute the length of the long chord from A b.) Compute the tangent distance AB of the curve. c.) Compute the stationing of a point "x" on the curve on which a line passing through the center of the curve makes an angle of 58° with the line AB, intersects the curve at point "x."