na certain road construction, it was decided to connect a forward tangent of 3% and a back tangent of -5% by a 200m symmetrical parabolic curve. It was discovered that the grade intersection at station 10+100 whose elevation is 100m falls on a rocky section with an outcrop of 2.67m directly above the grade ntersection. To avoid rock excavation, the project engineer decided to adjust the vertical parabolic curve in such a way that the curve will just clear the rock without altering the position of PVC and the grade of the tangents.

na certain road construction, it was decided to connect a forward tangent of 3% and a back tangent of -5% by a 200m symmetrical parabolic curve. It was discovered that the grade intersection at station 10+100 whose elevation is 100m falls on a rocky section with an outcrop of 2.67m directly above the grade ntersection. To avoid rock excavation, the project engineer decided to adjust the vertical parabolic curve in such a way that the curve will just clear the rock without altering the position of PVC and the grade of the tangents.

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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For a road construction, it is decided to connect a forward tangent of 3% and a back tangent of -5% by a 200m symmetrical parabolic curve. It was discovered that the grade intersection at station 10+100 whose elevation is 100m falls on a rocky section with an outcrop of 2.67m. directly above the grade intersection. To avoid rock excavation, it was agreed to adjust the vertical parabolic curve in such a way that the curve will just clear the rock without altering the position of PC and the grade of the tangents. Determine the total length of the new parabolic curve.
IN A CERTAIN ROAD CONSTRUCTION IS UNDERTAKEN DPWH, IT WAS DECIDED TO CONNECT A FORWARD TANGENT OF 3% AND A BACK TANGENT OF -15% BY A 300m SYMMETRICAL PARABOLIC CURVE. IT WAS DISCOVERED THAT THE GRADE INTERSECTION AT STA 10+100 WHOSE ELEVATION IS 150m FALLS ON ROCKY SECTION WITH THE EXPOSED BOULDER AT ELEVATION 1102.67 TO AVOID ROCK EXCAVATION, THE ENGINEER DECIDED TO ADJUST THE VERTICAL PARABOLIC CURVE IN SUCH AWAY THAT THE CURVE WILL JUST CLEAR THE ROCK WITHOUT ALTERING THE POSITION OF PC AND THE GRADE OF TANGENTS. DETERMINE THE ELEVATION AND STA OF THE NEW PT.
Help pls In a certain road construction is undertaken DPWH, It was decided to connect a forward tangent of 3% and a back tangent of -5% by a 300 m symmetrical parabolic curve. It was discovered that the grade intersection at sta 10+100 whose elevation is 150 m falls on rocky section. With the exposed boulder at elevation 102.67 to avoid rock excavation , the engineer decided to adjust the vertical parabolic curve in such away that the curve will just clear the rock without altering the position of pc and the grade of tangents. Determine the elevation and sta of the new PT.
CONNECT A FORWARD TANGENT OF 3% AND A BACK TANGENT OF-15% BY A 300 M SYMMETRICAL PARABOLIC CURVE. IT WAS DISCOVERED THAT THE GRADE INTERSECTION AT STA 10+100 with ELEVATION 50 M FALLS ON ROCKY SECTION WITH THE EXPOSED BOULDER AT ELEVATION 102.67. TO AVOID ROCK EXCAVATION, IT WAS DECIDED TO ADJUST THE VERTICAL PARABOLIC CURVE THAT THE CURVE WILL JUST CLEAR THE ROCK WITHOUT ALTERING THE POSITION OF PC AND THE GRADE OF EACH TANGENTS. DETERMINE THE ELEVATION AND STA OF THE NEW PT.
. A grade line AB having a slope of +5% intersect another grade line BC having a slope of –3% atB. The elevations of pointsA,BandCare 95 m, 100 m and 97 m respectively. Determine the elevation of the summit of the 100 m parabolic vertical curve to connect the grade lines. Resolve using equations for a parabola. (y=Ax^2+Bx+C)
UNSYMMETRICAL PARABOLIC CURVE1. An unsymmetrical parabolic curve has a forward tangent of -8% and a back tangent of +5%. The length of curve on the left side of the curve is 40m. long while that of the right side is 60m long. The PVC is at station 6+780 and has an elevation of 110m. An outcrop is found at station 6+800 and has an elevation of 108.40ma. Compute the height of fill needed to cover the outcrop.b. Compute the elevation of curve at station 6+820.c. Compute the elevation of the highest point of the curve.
Fundamentals of Surveying: Please show your solution to every question. 1.) The length of the unsymmetrical vertical curve is 609. The gradient of PVC is +4% while that of PVT is -6%. If the vertical offset was found to be 6.5 feet, find the horizontal distance of PVI from PVC. 2.) A +3 % grade @ PC intersects with a -4.2 % grade at PI forming a symmetrical vertical curve with length L = 259. Determine the horizontal distance from the summit to PC. 3.) A simple curve has 269m radius and 64° deflection angle. What is its middle ordinate? 4.) Using arc basis, what is the degree of curve of a simple curve whose long chord is 69 m and middle ordinate equal to 11.5 m? 5.) A compound curve laid on their tangents have the following data: 11 = 34', 12 = 39", and D1 = 80. If the length of the common tangent passing thru the P.C.C. is 109 m, find the degree of curve (D2) of the second curve. 6.) Compute the distance (in meter, along the curve) from P.C. to P.T. of the compound curve…
An unsymmetrical parabolic curve has a forward tangent of -8% and a back tangent of +5%. The length of the curve on the left side of the curve is 40 m while that of the right is 60 m. The PC is at station 6+780 and has an elevation of 110 m. An outcrop is found at station 6+800 has an elevation of 108.40 m. a. Compute the height of fill needed to cover the outcrop. b. Compute the elevation of the curve at station 6+820. c. Compute the elevation of the highest point of the curve.
3. A grade descending at the rate of -4% intersects another grade ascending at the rate of +8% at station 2+000, elevation 100m. A vertical curve is to connect the two such that the curve will clear a boulder located at station 1+980, elevation 101.34. Determine the necessary length of the curve. Sketch the curve.
A vertical parabolic sag curve has tangent grades of -1.428% and +0.707%. If the grade changes uniformly at 0.16% per 20 m, find the length of the curve
A vertical sag curve has tangent grades of -3.6% and +4.8% meeting at point S whose elevation is 446.5 m. If the length of the curve is 420 m, find the elevation of the PC.
A -6% grade and a +2% grade intersect at STA 12+200 whose elevation is at 25.632m. Two grades are to be connected by a parabolic curve, 160m long. A. Find the grade of the first quarter point on the curve. In decimal form B. Find the stationing of the lowest point of the curve.