23.1 A 1,500-ft vertical curve is to be used for joining a +3.0% grade line and a-4.0% grade line. The vertex or intersection of the grade lines is at station 57 + 00 and the elevation there is 583.25 ft. Compute the elevations at 100-ft stations throughout the curve. (Ans.: Curve elevation at stations 53, 55, and 59 = 568.39 ft, 570.19 ft, 568.19ft)

23.1 A 1,500-ft vertical curve is to be used for joining a +3.0% grade line and a-4.0% grade line. The vertex or intersection of the grade lines is at station 57 + 00 and the elevation there is 583.25 ft. Compute the elevations at 100-ft stations throughout the curve. (Ans.: Curve elevation at stations 53, 55, and 59 = 568.39 ft, 570.19 ft, 568.19ft)

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

3. A grade g1 of +3.25% intersects grade g2 of -2.75% at a vertex whose station and elevation are STA 257+17.65 and 3,820.14 ft, respectively. An equal-tangent parabolic curve 900 ft long has been selected to join the two tangents. a. Compute the station and elevation of PVC. b. Compute the station and elevation of the first full station inside the curve. c. Identify the station and elevation of the highest point on the curve.
3.9 An equal-tangent crest curve connects a +2% initial grade with a -1% final grade, and is designed for 55 mi/h. The station of the PVI is 233 + 40 with elevation 1203 ft. What is the elevation of the curve at station 234 +00?
3. A -9% grade meets a +5% grade at station 2+123.56 of elevation 55.56 m. The lengths areunequal L1=100 m, L2=80 m. Calculate the elevation at every full station. Compute thestationing and elevation of the lowest point on the curve.
Note : If you don't know the solution please leave it but don't give me wrong solution. A vertical curve joining entering grade of -4% with an exiting grade of - 7% begins at station 2+607 and is being designed with K value of 72. What is the station of the highest point on the curve . Code the station as the total distance from the origin, i.e., if the station is 1+100 code it as 1100.
A vertical curve has its highest point at a distance 48 meters from point PT. The curve is having two tangents having a grade of +6% and -4% which intersects at point Pl. The curve passes thru point A at station 25+140. The elevation of point Pl is 100 meters above sea level at 25+160 station. a. Determine the height of the maximum Offset (H) b. The Length of the Curve (L)
A -4% grade line meets another line of +3.2% grade at station 11+ 4O. If a 225-ft parabolic curve connects these grade lines, at what station is the lowest point along the curve located? Your answer
A symmetrical parabolic summit curve connects two grades of +6% and -4%. It is to pass through a point, p, the stationing of which is 35 + 280 and the elevation is 198.13 m. If the elevation of the grade intersection is 200 m. with stationing 35 + 300, 1. Determine the length of the curve. a. 120 m b. 167 m c. 234 m d. 436 m 2. Determine stationing of P.C. a. 35 + 260 b. 35 + 240 c. 35 + 160 d. 35 + 120 3. Determine elevation of P.C. a. 324.30 m b. 167.40 m c. 523.40 m d. 196.40 m 4. determine the location of the highest point on the curve from P.C. a. 84 m b. 57 m c. 72 m d. 66 m 5. determine elevation of station 35 + 260 on the curve.. a. 132.43 m b. 157.85 m c. 197.43 m d. 162.46 m
2. An unsymmetrical parabolic curve has a forward tangent of -8% and back tangent of +5%. The length of curve on the left side is 40 m, while that of the right side is 60 m long. PC is at Sta 6+780 and at elevation 110.1 m. Determine the following: a) Value of the common grade b) Elevation at Sta 6+820 c) Station and elevation at the highest/lowest point
3. A grade of +5% grade intersects a grade of -3.0% P.C.has a of 10 + 110.00 which has an elevation of 30MASL.These two center gradelines are to be connected by a240 meter symetrical vertical parabolic curve.What is the station of the highest point of the vertical curve.What will be the elevation of the curve at station 10+210What will be the elevation of the curve at station 10+250
A vertical parabolic curve has a back tangent of -5% and a forward tangent of +3% intersecting at station 1 + 240 at an elevation of 100.00 m. If the stationing of P.C. is at 1 + 120. Evaluate the stationing of the lowest point in the curve. Evaluate the elevation of the lowest point in the curve. Evaluate the rate of change of grade of the sag curve per 20 m length.
A vertical curve is to be design to joint (+3) percent grade with a (-3.5) percent grade, sight distance (SSD=120.00 m). The minimum length of the curve equal to: 162.204 m. 150.00 ft/sec.. O 142.422 m. O 467.1982 m. O 154.402 ft.
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.