An equal-tangent crest vertical curve is designed for 45 mi/h. The high point is at elevation 380.10 ft. The initial grade is +1.90% and the final grade is -3.10%. What is the elevation of the PVT? You answer must be in ft and rounded to the second digit (e.g., 0.00), if you need to use values from tables, use "design" values (usually last column), not "calculated"

An equal-tangent crest vertical curve is designed for 45 mi/h. The high point is at elevation 380.10 ft. The initial grade is +1.90% and the final grade is -3.10%. What is the elevation of the PVT? You answer must be in ft and rounded to the second digit (e.g., 0.00), if you need to use values from tables, use "design" values (usually last column), not "calculated"

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

Home Work No. 2 An engineer has been decided to connect a tunnel with floor elevation (100.00m) and station (0.00) with bridge elevation (112.50 m) and station (0+450). He suggested vertical curves with equal tangents. Sketch the Figure. Is the crest vertical curve safe to provide sight distance for design speed (70 km/hr.)? Use t=2.5 sec. Bridge Elevation =+112.5m Bridge Station 0+450m Tunnel elevation=100m Tunnel station + 0.00
An equal tangent vertical curve is intended to connect grades of 2.6% and -0.8%. If this curve is to be a part of a highway designed with a PSD of 550ft, determine the minimum length (ft) of the vertical curve. Use exact solution (do not use the table). Round your answer to 3 decimal places.
Given: L = 25 stations, PT station = 50+00, G1=-3%, G2=+1%; What is the stationing at zero % grade on the curve? Given: L = 20 stations, G1=-2%,G2=+1%; The elevation difference from the PC to the low point of the curve is ? Given: G1=-1%, G2=+4%; What percent of L is the grade of +3% from the PC ? Given: L=1500 ft.; G1=+2%, G2=+4%; What is the change in grade from a station 3 stations from the PC to a station 5 stations from the PT?
A crest vertical curve has the following information: L = 160 m, Station PVI = 4 + 310, Elevation PVI = 411 m, Elevation PVC = 407.8 m, Elevation PVT = 409 m, Find (1) Initial and final grades ? (2) Elevation and stationing of the highest point?
Just do question 2
An equal-tangent equal tangent vertical curve connects a +2.10% initial grade and -2.70% final grade. The PVT is at station 127+18 and elevation 206.52 ft. If the design speed is 75 mi/h, what is the elevation of the curve at station 121+58? You answer must be in ft and rounded to the second digit, if you need to use values from tables, use "design" values (usually last column), not "calculated".
Please solve the problem in the picture, make sure you show all of the work and give the right answer.
A curve with a -6% downgrade intersects an upward slope of +2%. The elevation at the intersection of these grades at station 50+00 is 550.97 feet. The length of the curve is 800 feet. What would be the elevation of the catch basin?
Determine the minimum length (ft) for a vertical curve with 55mph design speed if the grades to be connected are -1.93% and 1.79%. Use the table provided by AASHTO. Round your answer to 3 decimal places.
Please show all of your work and formulas you use please.
Grades of +3.00% and -2.00%, sight distance 530 ft, h1 = 3.30 ft and h2 = 2.50 ft. What is the minimum length of a vertical curve to provide a required sight distance for the conditions given? Express your answer in feet to five significant figures.