B-Unequal-Tangent Parabolic (Vertical) Curve: Example: A grade g1 of -2% intersects g2 of +1.6% at a vertex whose station and elevation are 87+00 and 743.24, respectively. A 400' vertical curve is to be extended back from the vertex, and a 600' vertical curve forward to closely fit ground conditions. Compute and tabulate the curve for stakeout at full stations. 83 +00 BVC (751.24) 84 +00 - 85 +00 A (747.24) ناوه شرق 40 1-86 +00 00+ 28- CVC (747.56) -2.00% Stations 00 +88 89 +00 +1.60% V (743.24) 00+ 06- 1-91 +00 B (748.04) Point A STA 85 +00: Elev. = 743.24+2 (2) = 747.24' Point B STA 90+00: Elev. = 743.24 +1.6 (3)=748.04' 00 + 26 وه ی 600 93 +00 EVC (752.84) ونا تسلیة شنانه و پوینتة نوت ده دوام وه Solution: The CVC is defined as a point of Compound Vertical Curvature. We can determine the station and elevation of points A and B by reducing this unequal tangent problem to two equal tangent problems. Point A is located 200' from the BVC and Point B is located 300' from the EVC. Knowing this we can compute the elevation of points A and B. Once A and B are known we can compute the grade from A to B thus allowing us to solve this problem as two equal tangent curves.

B-Unequal-Tangent Parabolic (Vertical) Curve: Example: A grade g1 of -2% intersects g2 of +1.6% at a vertex whose station and elevation are 87+00 and 743.24, respectively. A 400' vertical curve is to be extended back from the vertex, and a 600' vertical curve forward to closely fit ground conditions. Compute and tabulate the curve for stakeout at full stations. 83 +00 BVC (751.24) 84 +00 - 85 +00 A (747.24) ناوه شرق 40 1-86 +00 00+ 28- CVC (747.56) -2.00% Stations 00 +88 89 +00 +1.60% V (743.24) 00+ 06- 1-91 +00 B (748.04) Point A STA 85 +00: Elev. = 743.24+2 (2) = 747.24' Point B STA 90+00: Elev. = 743.24 +1.6 (3)=748.04' 00 + 26 وه ی 600 93 +00 EVC (752.84) ونا تسلیة شنانه و پوینتة نوت ده دوام وه Solution: The CVC is defined as a point of Compound Vertical Curvature. We can determine the station and elevation of points A and B by reducing this unequal tangent problem to two equal tangent problems. Point A is located 200' from the BVC and Point B is located 300' from the EVC. Knowing this we can compute the elevation of points A and B. Once A and B are known we can compute the grade from A to B thus allowing us to solve this problem as two equal tangent curves.

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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Curve connects three tangents having an azimuth of 254°, 270° and 270° respectively. The length of chord is 310 m long measured from PC to PT of the curve and is parallel to the common tangent having an azimuth of 270° if the stationing of PT is 6 + 720. Determine the total length of curve, stations of Common Tangent and PC.
Situation 3] The bearing of the first tangent of a simple curve is N70.7ºE while the second tangent has a bearing of S82037'E. The degree of curve is 4.50 and chainage of PC is 3+777. It is proposed to decrease the central angle by an angle of 70 by changing the direction of the second tangent, in such a way that the position of PT and the direction of the first tangent shall remain unchanged. Determine the following: 主 New angle of intersection, I New radius of the curve, R' New chainage of station PC
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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?
Sample Problem #1: The back tangent with grade of +3.4% and forward tangent with grade of -4.8% intersects at station 14 + 750 and elevation 76.30m. The two tangents are connected by a 320m parabolic curve. A. Find the location of the highest point from PC. B. Compute the stationing of the highest point of the curve. C. Determine the elevation of the highest point.
A symmetrical parabolic summit curve connects two grades of +6% and -4%. It is to pass though a point "P" the stationing of which is 35+280 and the elevation is 193.13 meters. If the elevation of the grade intersection is 200 m with stationing 35 + 300. 1. What is the length of the curve? a. 120 m b. 123 m c. 121 m d. 122 m
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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.
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Show complete solution and drawing. COMPOUND CURVE Problem: Determine the tangent distance through the PC if the tangent distance through the PT is 175.82m. Radii of the 1st and 2nd simple curves are 382m and 230m respectively. The tangents intersecting at an angle of 67˚. Determine also the angles of the intersection of the 1st and 2nd curve at the compound curve.
The reduced level at the intersection of a rising gradient of 1.7% and a falling gradient of 2.3% on a proposed road is 52.50m. Given that K value is 55. The throught chainage of the intersection point is 4020m and proposed vertical curve is to have equal tangent. As a civil engineering you are required to prepare the following vertical curve design calculations : A) The through chainage of the points of the vertical curve if the minimum required length is to be used (LV) and the vertical curve setting out data at 20 metres interval. b) The chainage and reduced level of the highest point on the curve. c) Through chainage at highest point was located exactly under a bridge, if the RL of soffit is 57.42m, calculate the headroom based on proposed vertical curve design.