Problem 1.An equal-tangent vertical curve connects a +1.5% initial grade and a -3% final grade. The PVT is at station 120+52 and elevation197.665 ft. If the design speed is 50 mi/h, what is the elevation of the curve at station 118+00?MetricDesignspeed(km/h)889882888882030405060708090100110120130Stoppingsightdistance2035506585105130160185220250285Rate of verticalcurvature, KªCalculated Design120.61.922943.86.411.016.825.738.952.073.695.0123.44711172639527495124StoppingsightDesignspeed distance(mph)15 20 25 30 35 40 45 50 55 60 65 70 75 80US Customary88888888888880115155200250305360425495570645730820910Rate of verticalcurvature, KªCalculated Design33.06.111.118.529.043.160.183.7113.5150.6192.822486NNWRate of vertical curvature, K, is the length of curve per percent algebraic difference inintersecting grades (A). K = L/A1219294461114151193246.9 247311.6 312383.7 384Exhibit 3-76. Design Controls for Stopping Sight Distance and for Crest and Sag VerticalCurves***

Following are the given information.Initial grade, Final grade, Station of PVT Elevation of PVT…

An equal-tangent vertical curve connects a +1.5% initial grade and a -3% final grade. The PVT is at station 120+52 and elevation 197.665 ft. If the design speed is 50 mi/h, what is the elevation of the curve at station 118+00?

An equal-tangent vertical curve connects a +1.5% initial grade and a -3% final grade. The PVT is at station 120+52 and elevation 197.665 ft. If the design speed is 50 mi/h, what is the elevation of the curve at station 118+00?

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

See similar textbooks

Related questions

Q: An old equal-tangent crest vertical curve was designed with a PVI at station 117+45 and elevation of…

A: An equal tangent crest vertical curve is as below The point PVC is called as the initial tangent…

Q: 1. Consider a vertical curve with a +3/2% initial grade and -1.1% final grade. The point of…

Q: answer question 18 only

Q: the AASHTO method (“K” factors) (1/8 pts.) The station of the PVC/BVC (1/8 pts.) The station of the…

Q: A 1.5% grade passing at station 46+20.00 at an elevation of 369.93 ft meets a -2.5% grade passing at…

Q: 2300 ft equal tangent crest vertical curve is currently designed for 60 mph.this curve connects an…

Q: A 400-ft equal-tangent sag vertical curve has its PVC at station 78+00 and elevation 800 ft. The…

A: Length of curve (L) = 400 ftStation of PVC = 78+00Elevation of PVC = 800 ft

Q: What is the length in stations for a symmetrical vertical curve with an entering grade of +5% which…

Q: 600-ft crest vertical curve is to be developed from a +1.80% grade intersecting with a -3.4% grade…

A: Given Initial grade = 1.8% Final grade = -3.4% Station PVI= 15+50= 1550ft Elevation of PVI= 315.76ft…

Q: 5. A vertical sag curve consists of an initial grade of -3.5% with a final grade of +0.5%. The…

A: Given:- An initial grade of the sag curve = -3.5% A final grade of the sag curve = +0.5% The…

Q: Determine the elevation of the middle point of the curve.

Q: A 600 ft equal tangent vertical The initial grade is -3.5% and the final grade is +0.5%. Determine…

A: L=600ft

Q: A crest vertical curve (equal tangent) has PVI at station 210+00 and elevation 420.75 ft. The…

A: SPVI = Stationing of PVI = 210+00EPVI = Elevation of PVI = 420.75 ftg1 = Initial grade = 2.5% =…

Q: you have determined the curve to be 1100 ft long. Now, you need to when the BVC is located at…

Q: An equal-tangent vertical curve connects an initial grade of -4% and a final grade of +1% and is…

A: The objective of the problem is to determine the station and elevation of the lowest point on an…

Q: An SVC connects a -4% grade to a +4% grade. The curve goes under a pedestrian bridge at PVI. The…

A: Given: G1=-4%G2=+4% Vertical clearanceC=15 ft Sight distanceS=984 ft Height of drivers eyeh1=6 ft…

Q: An equal-tangent crest vertical curve is designed for 70 mi/h. The high point is at elevation 1011.4…

A: V=70mph To find elevation of PVT

Q: You are designing a vertical sag curve joining a -3% grade to a +2% grade. The design speed is 60…

A: Given: g1=-3%g2=+2% Design speedV=60 mph.

Q: An equal tangent vertical curve connects a 1% and 3% initial and final grade. Design speed is…

Q: A horizontal curve is being designed for a new four-lane roadway with 11-ft lanes. The PT is located…

A: Given;Road has four lane of 11 ft.PT is located at station 105+20Design speed = 45 mph =66…

Q: Step by step solutionProblem 1. An existing equal-tangent sag vertical curve is designed for 60 mph.…

A: Step 1:Step 2: Step 3: Step 4:

Concept explainers

Intersection Design

An intersection is a location where two or more roads or railways meet or join. The traffic movement is controlled at the intersection that depends on its design and type. It governs the operation, efficiency, speed, capacity, and safety of the highways, airways, and railways. The movement of the pedestrian causes hazards and delays at the intersection.

Question

Problem 1.
An equal-tangent vertical curve connects a +1.5% initial grade and a -3% final grade. The PVT is at station 120+52 and elevation
197.665 ft. If the design speed is 50 mi/h, what is the elevation of the curve at station 118+00?
Metric
Design
speed
(km/h)
88988288888
20
30
40
50
60
70
80
90
100
110
120
130
Stopping
sight
distance
20
35
50
65
85
105
130
160
185
220
250
285
Rate of vertical
curvature, Kª
Calculated Design
1
2
0.6
1.9
2294
3.8
6.4
11.0
16.8
25.7
38.9
52.0
73.6
95.0
123.4
4
7
11
17
26
39
52
74
95
124
Stopping
sight
Design
speed distance
(mph)
15 20 25 30 35 40 45 50 55 60 65 70 75 80
US Customary
888888888888
80
115
155
200
250
305
360
425
495
570
645
730
820
910
Rate of vertical
curvature, Kª
Calculated Design
3
3.0
6.1
11.1
18.5
29.0
43.1
60.1
83.7
113.5
150.6
192.8
22486NNW
Rate of vertical curvature, K, is the length of curve per percent algebraic difference in
intersecting grades (A). K = L/A
12
19
29
44
61
114
151
193
246.9 247
311.6 312
383.7 384
Exhibit 3-76. Design Controls for Stopping Sight Distance and for Crest and Sag Vertical
Curves
***

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

SEE SOLUTION Check out a sample Q&A here

Step 1: Step 1 Introduction

VIEW

Step 2: Step 2 Calculation of length of curve and stationing of PVC

VIEW

Step 3: Step 3 Calculation of elevation of PVC and elevation at station 118+00

VIEW

Solution

VIEW

Step by step

Solved in 4 steps with 10 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, civil-engineering and related others by exploring similar questions and additional content below.

Similar questions

2. A vertical parbolic curve was design in order to have a clear sight distance of 120 m. The grade lines intersect at Sta 9+000 at elev 160.50 . The curve was design such that when the height of the drivers eye is 1.50 m above the payment it would just see an object whose height is 0.10m above the pavement. Determine the max speed that a car could travel grade of 5% and a downgrade of -3%.
A crest vertical curve joining a + 3 percent and – 4 percent grade is to be designed for 75 mph. If the tangent intersect at station (345 + 6000) at an elevation of 250 ft, determine the stations and elevations of the PVC and PVT.
An equal-tangent curve connects a +1.0% and a -0.5% grade. The PVC is at station 54+24 and the PVI is at station 56+92. Is this curve long enough to provide passing sight distance for a 60-mi/r design speed? Problem 2
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.
I need the answer as soon as possible
1) An equal-tangent crest vertical curve has a 50 mi/h design speed. The initial grade is +3%. The high point is at station 33+37.43 and the PVT is at station 37+18.26. What is the elevation difference between the high point and the PVT? 2) A 1400-ft-long sag vertical curve (equal tangent) has a PVC at stateion 115+00 and elevation 750ft. The intial grade is -3.5% and the final grade is +6.5%. Determine the elevation and stationing of the low point, PVI, and PVT. If you could show as many steps taken as possible that would be helpful in me actually understanding how to do the problem. Thanks
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.