A surveyor must stake a symmetrical vertical curve where anentering grade of+0.80% meets an exiting grade of -0.40% atstation 90+00.00, which has an elevation of 100.00 ft. Themaximum allowable change in grade per 100.00 ft station is-0.2%.What is the length, in station, of the vertical curvea)described?What is the elevation of 92+00.00 on the verticalb)curve?с)What is the highest point on the curve previouslydescribed?Surveying Solved Problems for the FS and PS Exams4th Edition by Jan Van Sickle.

An asymmetrical vertical curve is the curve with the horizontal distance from the PVI to the PVC is…

Answered: A surveyor must stake a symmetrical…

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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A grade of -3.0% is followed by a grade of +2.1% which intersects at station 10+250 and elevation of 110 m. If the maximum change in grade per 20 m station is 0.20%, determine the following A. Length of parabolic curve that shall connect these two gradelines. B. Elevation of PC. C. Elevation of a point on the curve @ Sta.10+250.
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Grades of g1= -2.00% and g2= +2.00%, VPI elevation 750.00 ft at station 30 + 00. Fixed elevation 753.00 ft at station 30 + 00.Field conditions require a highway curve to pass through a fixed point. Compute a suitable equal-tangent vertical curve and full-station elevations. Give the elevations in order of increasing X. Please give 6 elevations.
answer question 17 only
1. An upgradient of 2.5% meets a downgradient of 2% at station kilometer 10+834.26 and at elevation of 64.46 m. The length of vertical curve is 140 m. Calculate ; a) Elevations on the curve at even 25 m. St. Km.s b) Elevations on the curve using 25 m. from PVC and PVT c) St. Km. and elevation of the highest point on the curve.
3. A grade of-8% id followed by a grade of 3%, the grade intersecting at station 11+ 020 of elevation 423. 10 m. The change of grade is restricted to 0.5% in 25 meters. a. Compute the length of vertical parabolic curve b. Solve the lowest point of the curve from the P.T. c. Calculate the elevation at station 11+ 000.
5
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
A vertical summit curve is having tangents which intersects at point Pl at Station 12+618.37 and 133.75 meters elevation. The grade for the forward tangent is -513%. Station PC is at station 12+515.99 and having an elevation of 116 26 meters. What is the elevation of point A along the curve with Station 12+548.27? cs Seanned with CamScanner
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?
Find reduced levels of the various points on a parabolic vertical curve, which is to be set to connect two uniform grades 0.6% and - 0.7%. The chainage and required level of the point of intersection are 2525 m and 335.65 m respectively. A ssume the rate of change of grade to be 0.05 per 20 m chain.
A grade of -3.6% is followed by a grade of +2.1% which intersects at station 10+256 and elevation of 116 m. If the maximum change in grade per 20 m station is 0.20%, determine the following A. Length of parabolic curve that shall connect these two grade lines. B. Elevation of PC. C. Elevation of a point on the curve @ Sta.10+250.

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