You are required to set up a vertical curve for a highway alignment. Gradient of the backward and forward tangents are +5% and -3%, respectively. The stations of the beginning of the vertical curve (BVC) and the end of the vertical curve (EVC) are 100+00 and 112+00 (ft), respectively. Elevation of EVC is 260.00 ft. Please calculate 1. Station of Point of vertical intersection (PVI). 2. Elevations of the curve at 110+00. 3. Station and elevation of the point in the curve that has the maximum elevation. 4. Tangent offset (distance from the tangent to the curve) at the point calculated in (b).

You are required to set up a vertical curve for a highway alignment. Gradient of the backward and forward tangents are +5% and -3%, respectively. The stations of the beginning of the vertical curve (BVC) and the end of the vertical curve (EVC) are 100+00 and 112+00 (ft), respectively. Elevation of EVC is 260.00 ft. Please calculate 1. Station of Point of vertical intersection (PVI). 2. Elevations of the curve at 110+00. 3. Station and elevation of the point in the curve that has the maximum elevation. 4. Tangent offset (distance from the tangent to the curve) at the point calculated in (b).

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

See similar textbooks

Related questions

Q: An equal tangent vertical curve is to be constructed between grades of -6.75% (initial) and 4.75%…

Q: 19. What is the stationing (ft) of the EVC?

A: Length of curve, L = 1100ft ⇒ 11 stations. Stationing of BVC = 12+15 ft.

Q: 3.48 A horizontal curve is being designed for a new two-lane highway (12-ft lanes). The PI is at…

A: Given that :- PI is at station = 250+50 = 25050 ft Design speed = v = 65 mi/h maximum super…

Q: Determine variables requested below for a horizontal curve with the following characteristics.…

A: given- length of curve= 1255' radius of curve=1200' sta of PC 12+00 ROADWAY IS THREE LANE. design…

Q: Q2) A proposed highway has two tangents of bearings N 45°54'36" E and N 1°22'30" W. The highway…

A: “Since you have posted a question with multiple sub-parts, we will solve the first three sub-parts…

Q: conditions require a highway curve to pass through a fixed point. Compute a suitable equal-tangent…

Q: An equal-tangent sag curve of length 920ft needs to be designed for 55 mph to connect an initial…

A: Given:- The length of the sag curve = 920 ft The design speed = 55 mph The initial downgrade = 5%…

Q: 3.3 Vertical Alignment 53 ELEMENTS OF VERTICAL CURVE DESIGN An equal-tangent vertical curve is to be…

A: I have tried to solve this problem and also write the given data in diagram plz have gone through it

Q: An equal-tangent crest vertical curve is designed for 45 mi/h. The high point is at elevation 380.10…

A: Solution Given That G1=+1.90%G2=-3.10%

Q: A horizontal curve on a two-lane highway has its PC at station 123+70 and its PI at station 130+90.…

Q: The initail and final grades of a vertical curve having 2000 ft length are 2 % (plus) and 4 %…

Q: An equal-tangent crest vertical curve is being designed for a speed of 45 mph. The curve connects…

A: We have to find station and elevation of the PVC and the PVI according to the given data like speed…

Q: A car drives through portion AB of the S-curve at constant speed, decelerates in BC, and accelerates…

A: In physics, the acceleration vector is a vector that points in the direction of the net force acting…

Q: 3.35 A horizontal curve on a two-lane highway (10-ft lanes) is designed for 50 mi/h with a 6%…

A: Two lane highway -10 ft lanes Designed for 50 mi/h with a 6% superelevationCentral angle = 350PI…

Q: A horizontal curve on Texas Highway 83 (Tx83) comprises of a two-lane rural highway with a lane…

A: Stopping Sight Distance: It is the minimum sight distance available at any spot having sufficient…

Q: An equal-tangent vertical curve connects a +1.5% initial grade and a -3% final grade. The PVT is at…

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

Q: you have determined the curve to be 1100 ft long. Now, you need to when the BVC is located at…

Q: A simple circular curve is specified with a 12° degree (arc basis) horizontal curve and e=0.08. A…

A: Degree of horizontal curve = 120 Super elevation (e) = 0.08 Radius of curve (R) = 500 ft Coefficient…

Q: A 750-ft equal-tangent crest vertical curve connects tangents that intersect at elevation 572 ft.…

Q: Determine variables requested below for a horizontal curve with the following characteristics.…

Q: A simple horizont tangents that intersect at an angle of 66°30'. Compute the parts of the curve,…

Q: 1. Two tangents intersecting at V with bearings of N 75°12'E and S 78°36' E at station 2+092.021 are…

A: Given , Two tangents bearings are N75°12'E and S78°36'E Simple curve =4° Shorten the curve =100 m

Q: Determine variables requested below for a horizontal curve with the following characteristics.…

A: The given data is shown below: Curve length=1255'Radius=1200'PC ta=12+00Lane width=14…

Q: 2. A vertical parbolic curve was design in order to have a clear sight distance of 120 m. The grade…

Concept explainers

Curve setting

A curve is a turn that is provided horizontally or vertically in the design of highways, railways, and runways. A curve connects two straight lines that are separated by the deflection angle. It helps to change the direction and alignment of the highways, railway tracks, and runways. Therefore, the gradual change in their alignment is made by using a curve to provide comfort to the passengers and this also provides an easy turn to change the direction of the moving vehicle.

Question

You are required to set up a vertical curve for a highway alignment. Gradient of the backward and
forward tangents are +5% and -3%, respectively. The stations of the beginning of the vertical curve
(BVC) and the end of the vertical curve (EVC) are 100+00 and 112+00 (ft), respectively. Elevation of
EVC is 260.00 ft. Please calculate
1. Station of Point of vertical intersection (PVI).
2. Elevations of the curve at 110+00.
3. Station and elevation of the point in the curve that has the maximum elevation.
4. Tangent offset (distance from the tangent to the curve) at the point calculated in (b).

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

SEE SOLUTION Check out a sample Q&A here

Step 1: Introduction of the problem and calculate length of the curve

VIEW

Step 2: Determine station of point of vertical intersection and elevation of the curve 110+00

VIEW

Step 3: Determine station and elevation of the point in the curve that has maximum elevation

VIEW

Solution

VIEW

Step by step

Solved in 4 steps with 4 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, civil-engineering and related others by exploring similar questions and additional content below.

Similar questions

A horizontal curve on Texas Highway 83 (Tx83) comprises of a two-lane rural highway with a lane width of 12 ft. and superelevation of 8%. The posted speed limit is 50 mph. along a 0.5-mile section of highway, both a horizontal and vertical curve exists. The vertical curve has an initial grade of -2.00% and a final grade of +4.00%. The PVI is at station 156+40. A driver traveling eastbound strikes a stationary roadway object. The daytime crash results in a fatality and is being investigated for safe design speed. Evaluate and comment on the roadway design.
A crest vertical curve joining a + 3 percent and – 4 percent grade is to be designed for 75 mph. If the tangent intersect at station (345 + 6000) at an elevation of 250 ft, determine the stations and elevations of the PVC and PVT.
answer question 17 only
Two tangents in profile meet at a vertical point of intersection (VPI). The entrance tangent grade is +5%. The exit tangent grade is -3%. Point on Tangent 1 (POT1) is on the entrance grade. Point on Tangent 2 (POT2) is on the exit grade. An equal tangent vertical curve of 1000.00 ft has been designed to provide a smooth transition between them. Point Station Elevation ft POT1 61+00.00 4900.00 POT 2 242+00.00 4500.00 Elevation of the VPI = O 1) 4773.23 2) 5089.47 3) 4998.45 4) 4941.33 5) 4989.38 6) 5080.00 7) 5118.80 8) none 9) 5087.28 10) 5117.75
A horizontal curve is designed with a 2000-ft radius. The curve has a tangent length of 400 ft and the PI is at station 103+00. Determine the stationing of the PC, PT, deflection angle (A/2), and chord length for each station between the PC and PT.
Consider the same vertical curve from Problem 1 that has a +3.2% initial grade and -1.1% final grade. The point of vertical intersection is at station 98+20. The highest point of the curve is at station 100+79.35 to meet drainage design. What is the station of the point of vertical curvature (Also known as beginning of vertical curve). 92+88.95 92+90.81 92+88.05 (INCORRECT) 92+91.23 This is Question 1: Consider a vertical curve with a +3/2% initial grade and -1.1% final grade. The point of vertical intersection is at station 98+20. The highest point of the curve is at station 100+79.35 to meet drainage design. What is the design speed, in mph? Correct answer here is 70 MPH
Question 2 Field conditions require a highway curve to pass through a fixed point. Compute a suitable equal-tangent vertical curve and full-station elevations: (Compute L, and the elevations at the full stations). Grades of g₁ = -2.50% and 82 = +1.50%, VPI elevation 2560.00 ft at station 315 +00. Fixed elevation 2567.00 ft at station 314 + 00.