PRIN.OF HIGHWAY ENGINEERING&TRAFFIC ANA.
7th Edition
ISBN: 9781119610526
Author: Mannering
Publisher: WILEY
Question
100%
Chapter 3, Problem 1P
To determine

The elevation and station of the high point, PVC, and PVT.

The elevation of PVC is 1,324.90ft ,and station of PVC is 33,740ft, elevation of PVT is 1,325.06ft, and StationofPVT is 342+60Stations, and the station of high point is 340+60stations, the elevation of high point is 1,331ft.

Explanation of Solution

Given:

The distance of vertical curve is 520ft.

The vertical curve intersects at station 340+00 and the elevation 1325ft.

Formula used:

The formula to calculate the distance when the curve is an equal-tangent vertical curve is given by

D1=L2 ...... (I)

Here, D1 is thedistance between the point of vertical intersection PVI and initial point of vertical curve PVC, L is the distance of the vertical curve.

The formula to calculate the station of the initial point of vertical curve is given by

StationofPVC=StationofPVID1100ft ...... (II)

Here, stationofPVC is the station of the initial point of vertical curve and StationofPVI is the station of the initial point of vertical intersection of the vertical tangents.

The formula to calculate the station of the final point of vertical curve is given by

StationofPVT=StationofPVI+D1100ft ...... (III)

Here, StationofPVT is the station point of the vertical tangent.

The formula to calculate the elevation of the PVC is given by

ElevationofPVC=ElevationofPVI(G1×N1) ...... (IV)

Here, G1 is the initial grade and N1 is the number of stations between PVC, and PVI.

The formula to calculate the distance when the curve is an equal-tangent vertical curve is given by

D2=L2 ...... (V)

Here, D2 is thedistance between the final point of vertical tangent PVT and initial point of vertical intersection PVI.

The formula to calculate the elevation of the PVT is given by

ElevationofPVT=ElevationofPVI(G2×N2) ...... (VI)

Here, G2 is the final grade and N2 is the number of stations between PVT, and PVI.

The equation of the vertical curve is given by

y=ax2+bx+c ...... (VII)

Here, y is the roadway elevation at distance x from the beginning of the vertical curve PVC in stations, x is the distance from the beginning of the vertical curve station, aandb are the coefficients and c is the elevation of the PVC.

Differentiate equation (V) with respect to x ,to find the station of the high point on the vertical curve.

dydx=d( a x 2 +bx+c)dxdydx=2ax+b ...... (VIII)

The relation between b and G1 is given by

b=G1

The formula to calculate the coefficient a by using the above relation is given by

a=G2G12L ...... (IX)

The formula to calculate the station of the high point from the PVC ,

StationofHighPoint=StationofPVC+x ...... (X)

Calculation:

Substitute 520ft for L in equation (I).

D1=520ft2=260ft

The above calculation represents that the point PVT, final point of the vertical curve is exactly 260ft away from the intersection point PVI .It can be represent as (2stations+60ft).

The point PVT, final point of the vertical curve is exactly 520ft away from the initial point of the curve PVC .It can be represent as (5stations+20ft).

Substitute 340 for StationofPVI and 260ft for D1 in equation (II).

StationofPVC=340260ft100ft=340×100ft260ft=33,740ft

Divide the length of above calculation in ft by 100 and put the remainder in section as follows:

StationofPVC=337+40Stations.

Hence, the stationing of PVC is at 337+40.

Substitute 340 for StationofPVI and 260ft for D1 in equation (III).

StationofPVT=340+260ft100ft=340×100ft+260ft=34,260ft

Divide the length of above calculation in ft by 100 and put the remainder in section as follows:

StationofPVT=342+60Stations.

Hence, the stationing of PVT is at 342+60.

The number of stations between PVC and PVI is 2.6stations or (2stations+60ft).

Substitute 1,325 for ElevationofPVI, +4.0%ft/station for G1 and 2.6stations for N1 in equation (IV).

ElevationofPVC=1,325ft(+4.0% ft station×2.6stations)(1 100)=1,324.90ft

Substitute 520ft for L in equation (V).

D2=520ft2=260ft

The number of stations between PVT and PVI is 2.6stations or (2stations+60ft).

Substitute 1,325 for ElevationofPVI, 2.5%ft/station for G2 and 2.6stations for N2 in equation (VI).

ElevationofPVT=1,325ft(2.5% ft station×2.6stations)(1 100)=1,325.06ft

The L is 5.20 or (5stations+20ft).

Substitute +4.0 for G1 and 2.5 for G2 in equation (IX).

a=( 2.5)( +4.0)2( 5.20)=0.625

Substitute 0.625 for a and 4.0 for b in equation (VIII).

dydx=2(0.625)x+(4.0)=1.25x+4.0

Substitute 0 for dydx in above calculation to find x.

0=1.25x+4.01.25x=4.0x=4.01.25x=3.2stations

Substitute 337,40ft for StationofPVC and 3.2×100ft for x in equation (X).

StationofHighPoint=337,40ft+3.2×100ft=34,060ft=340+60stations

Substitute 3.2stations for x, 0.625 for a and 4.0 for b, and 1,324.90ft for c in equation (VII).

y=0.625(3.2)2+(4×3.2)+1,324.90ft=1,331ft

Conclusion:

Thus, the elevation of PVC is 1,324.90ft and station of PVC is 33,740ft, elevation of PVT is 1,325.06ft and StationofPVT is 342+60Stations and the station of high point is 340+60stations, the elevation of high point is 1,331ft.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
14:26
Students have asked these similar questions
A -6% grade and a +2% intersect at Sta 12 + 200 whose elevation is at 14.375 m. The two grades are to be connected by a parabolic curve, 160 m long. Find the elevation of the first quarter point on the curve.
A descending grade of 6% and an ascending grade of 2% intersect at Sta 12 + 200 km whose elevation is at 14.375 m. The two grades are to be connected by a parabolic curve, 160 m long. Find the elevation of the third quarter point on the curve.
A 400-ft equal tangent sag vertical to curve has its PC at station 100+00 and elevation 500ft. The initial grade is -4.0% and the final grade is +2.5%. Determine the elevation of the lowest point of curvature
Knowledge Booster
Recommended textbooks for you
Structural Analysis
Civil Engineering
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:Cengage,
Structural Analysis (10th Edition)
Civil Engineering
ISBN:9780134610672
Author:Russell C. Hibbeler
Publisher:PEARSON
Principles of Foundation Engineering (MindTap Cou...
Civil Engineering
ISBN:9781337705028
Author:Braja M. Das, Nagaratnam Sivakugan
Publisher:Cengage Learning
Fundamentals of Structural Analysis
Civil Engineering
ISBN:9780073398006
Author:Kenneth M. Leet Emeritus, Chia-Ming Uang, Joel Lanning
Publisher:McGraw-Hill Education
Sustainable Energy
Civil Engineering
ISBN:9781337551663
Author:DUNLAP, Richard A.
Publisher:Cengage,
Traffic and Highway Engineering
Civil Engineering
ISBN:9781305156241
Author:Garber, Nicholas J.
Publisher:Cengage Learning
• Copyright, Community Guidelines, DSA & other Legal Resources: Learneo Legal Center
• bartleby, a Learneo, Inc. business