An equal-tangent vertical curveconnects a +3.2 % and a -1.1 %grade. The PVI is at station100+20 and elevation 950 ftand the PVT is at station105+20. Answer the following5 questions Question 3 toQuestion 7.The station of the PVC is at:sta 93+20sta 95+20sta 97+20sta 99+20sta 105+20 For the curve above, (an equal-tangent vertical curve connectsa +3.2 % and a -1.1 % grade.The PVI is at station 100+20and elevation 950 ft and thePVT is at station 105+20.),what is the design speed inmph? Use K values, i.e., youcan assume G=0 for thestopping sight distancecalculations and assumeAASHTO standards apply.6065707580

As per Bartleby's guidelines we are supposed to answer only first question. Kindly repost other…

An equal-tangent vertical curve connects a +3.2 % and a -1.1 % grade. The PVI is at station 100+20 and elevation 950 ft and the PVT is at station 105+20. Answer the following 5 questions Question 3 to Question 7. The station of the PVC is at: sta 93+20 sta 95+20 sta 97+20 sta 99+20 O sta 105+20

An equal-tangent vertical curve connects a +3.2 % and a -1.1 % grade. The PVI is at station 100+20 and elevation 950 ft and the PVT is at station 105+20. Answer the following 5 questions Question 3 to Question 7. The station of the PVC is at: sta 93+20 sta 95+20 sta 97+20 sta 99+20 O sta 105+20

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 incoming grade of +1.800% meets an exiting grade of -2.500% at a PVI having station = 27+32.20…

A: Incoming grade = +1.800% Exiting grade = -2.500% PVI station = 27+32.20 PVI elevation = 231.87…

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: answer question 18 only

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

A: Given:Vertical clearanceSight distanceHeight of drivers eyeHeight of objectTo find: Length of curve.

Q: A curve with a -6% downgade intersects an upward slope of +2%. The elevation at the intersection of…

A: To Find :- station of EVC

Q: 19. What is the stationing (ft) of the EVC?

A: Length of curve, L = 1100ft ⇒ 11 stations. Stationing of BVC = 12+15 ft.

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

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: have designed a vertical crest curve which joins a +3% E mph and you have determined the curve to be…

Q: 3. What is AASHTO's recommended perception reaction time (s)?

A: Reference : AASHTO, American Association of State Highway and Transportation To determine : AASHTO…

Q: An equal-tangent sag curve of length 920ft needs to be designed for 55 mph to connect an initial…

A: Given:- The length of the sag curve = 920 ft The design speed = 55 mph The initial downgrade = 5%…

Q: highway engineer must stake a symmetrical vertical curve where an entering grade of +0.80% meets an…

Q: Given a vertical alignment connecting a -4% grade with a +2% grade with the PVC station at 34+64…

Q: nly answer question 6

A: Stopping Sight Distance is the minimum distance that must be available in front of driver so that he…

Q: You have designed a vertical crest curve which joins a +3% grade and a -4% grade. The design speed…

Q: Vertical highway curve: Given PVI at 128 + 00, L = 1,000 ft, g1 = -5%, g2 = +4%, and PVI elevation =…

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

Q: Design a vertical curve connecting PVI₁ (station = 4+980, Elevation = 198 m), PVI₂ (station = 5+990,…

A: PVI₁ (station = 4+980, Elevation = 198 m) PVI₂ (station = 5+990, Elevation = 163 m) and PVI₃…

Q: A +2% grade intersects with a -1% grade at station (535-24.25) at an elevation of 300 ft. If the…

A: Solution:-Given,Elevation=300ftDesign speed=60 mi/hg1=2%g2=-1%

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: 2. A vertical parbolic curve was design in order to have a clear sight distance of 120 m. The grade…

Concept explainers

Electronic Spreadsheet

Tabular listing has many advantages ranging from time management to efficient scheduling of tasks. Electronic spreadsheets allow the user to tabulate and analyze data, using its easy-to-use interface. An electronic spreadsheet is a staple application used among accountants, management staff, and IT professionals. It is used in various sectors worldwide for data management. An electronic spreadsheet is a system and computer software that could be used to do numerical computations automatically. Spreadsheet application programs are often organized as a table with rows and columns. Each row and column intersect to form a cell, which may be used to store data. These data can take the form of a written label, a number, or a formula that combines information from other cells.

Question

An equal-tangent vertical curve
connects a +3.2 % and a -1.1 %
grade. The PVI is at station
100+20 and elevation 950 ft
and the PVT is at station
105+20. Answer the following
5 questions Question 3 to
Question 7.
The station of the PVC is at:
sta 93+20
sta 95+20
sta 97+20
sta 99+20
sta 105+20

For the curve above, (an equal-
tangent vertical curve connects
a +3.2 % and a -1.1 % grade.
The PVI is at station 100+20
and elevation 950 ft and the
PVT is at station 105+20.),
what is the design speed in
mph? Use K values, i.e., you
can assume G=0 for the
stopping sight distance
calculations and assume
AASHTO standards apply.
60
65
70
75
80

Expert Solution

This question has been solved!

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

This is a popular solution!

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Step 3

VIEW

Trending now

This is a popular solution!

Step by step

Solved in 3 steps with 2 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

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.
answer question 22 only
answer question 17 only
A -1% grade meets a +3% grade at sta. 12+00 and elevation 520 ft. by a vertical curve of length 200 ft. Calculate the following: i. K and r ii. Station of PVC and PVT iii. Elevation of point at a distance, L/4, from PVC and PVT iv. station of turning point v. elevation of turning point vi. elevation of mid-point of 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.
4. Horizontal Curve: P.C. 12+00.00 P.I T 4= 30° P.T. This highway has a 60 mph speed limit requiring a minimum radius for any curve to be 1500'. If the curve is to begin at Station 12+00.00 and is to have a radius of 1610.10', how long will the curve be? At what station will the PI and PT go? How long will E be? (Answers should all be in hundredths.)
Vertical highway curve: Given PVI at 128 + 00, L = 1,000 ft, g1 = -5%, g2 = +4%, and PVI elevation= 457.25 ft, compute the elevations of the curve low point and the 100 ft stations (meansstations 124+00, 125+00, etc. until reaching the end of the curve) as well as for the beginningand end of the curve.
answer question 21 only