Q5/A reverse curve must be fit to the tangents shown below: PC1 P11 N 58-30′25 E S87 1045 E N47'5930 E P12 Point Station PC1 21+83.05 PII 27+35.80 P12 38+43.05 What are the radii of the two curves which fit this alignment? Compute to 0.01 ft.

Q5/A reverse curve must be fit to the tangents shown below: PC1 P11 N 58-30′25 E S87 1045 E N47'5930 E P12 Point Station PC1 21+83.05 PII 27+35.80 P12 38+43.05 What are the radii of the two curves which fit this alignment? Compute to 0.01 ft.

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: The tangents for a D = 4º30' circular curve (arc definition) meet at PI Sta 34+18.19 and the…

A: Degree of Curve = 4°30'PI of Station = 34+18.19Deflection angle (I) = 25°48'

Q: It is required to layout a simple curve by deflection angles. The curve is to connect two tangents…

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: Two tangents intersecting at V which is inaccessible has an angle of intersection of 30 degree . Two…

Q: The deflection angle of 3 curve is 50 25. If the PC is located at the station (331+38.75), determine…

A: The given data is: Intersection angle=50°25' The degree of curvature=3° Station of PC=(331+38.75)

Q: uation 1] A simple curve connects two tangent roads AB and BC with bearings N85048'E and 4040°E,…

A: "Since you have posted a question with multiple sub-parts, we will solve first three sub parts for…

Q: A compound curve has the following properties: R1 = use 480 ft; R2 is 70% of R1 1 = 24°, 2 = 34°…

A: As per the company policy we can only solve 3 sub-parts in one question. For other parts please…

Q: What is the length of the curve, G1, highest point of curve, and the elevation of the highest point…

A: To find : Length of the curve, G1, highest point of curve, and the elevation of the highest point on…

Q: 2. On a rail road line, two tangents that intersects at station 10 + 243 so as to form an angle of…

A: Roadway transportation is the method for detail conveyance between homes, shops, plants, and so on…

Q: Q6/A grade gl of+ 1.23% intersects grade g2 of -2.75% at a vertex whose station 18+00 and elevation…

A: Given:- The gradient of first tangent = g1 = +1.23% The gradient of second tangent = g2 = -2.75%…

Q: The bearing of the forward tangent of a compound curve is S82°15’E while the backward tangent has…

A: In the surveying of the transport road, there are two types of curves are used like, vertical curve…

Q: A circular curve having coordinates given in the late: e curve length (L), Chord (C), Tangent (T),…

A: given data, Chainage of PC = 3+40 m to find chord length, tangent length, radius, deflection, length…

Q: A vertical summit curve is having tangents which intersects at point Pl at Station 12+584.57 and…

A: Given Data:PVI=12+584.57Elv=139.32mgrade=-5.25%PVC=12+527.65Elv=122.32m

Q: Given a simple circular curve, in feet, with the following known information: = 25°34'00" o D₂ =…

A: I = 25°34'00"Da= 2°15'00"STA P.I. = 50+16.25

Q: SURVEYING 103 C2 Problem set: as An unsymmetrical parabolic curve has a forward tangent of-8% and…

Q: Determine variables requested below for a horizontal curve with the following characteristics.…

A: We have to determine the value of tangent of the given horizontal curve as per the given data in the…

Q: Compute the new tangent distance

A: Point of tangency (PT): - It is defined as the end point of the curve from where straight alignment…

Q: Situation 2] A simple curve having a radius of 231.0 m has tangents of N28.8°E and N66.0°E. A line…

Q: The angle of intersection of a circular curve is 45° 30' and its radius is 198.17 m. PC is at Sta. 0…

A: Given : Angle of intersection of circular curve = 45o30' Radius = 198.17 m PC at Sta. 0 + 700 To…

Q: Two tangents that cross at station 10 + 243 to produce an angle of 36°28' must be linked on a…

Q: 2. Two tangents intersecting at Station 8+095.21 have Azimuth of 260° and 284° respectively, and are…

A: Let New tangent distance be T new T new = T - Shifted distance T = Tangent length T = R tan ϕ2 R =…

Q: 23.1 A 1,500-ft vertical curve is to be used for joining a +3.0% grade line and a-4.0% grade line.…

A: The length of curve, L= 1500 ftThe grade of upgrade, g1 = +3%The grade of downgrade, g2 = -4%The…

Q: wo straight tangents of a two-line highway intersect at point B. The tangents have azmuths of 270…

A: azmuths- 2700 and 1100 Radius - 7300 ft To find - a)the tangent lengths b)the length of curve c) PI…

Q: Stations for each curve: STA PC¹ = 65+48.7 STA PT¹ = 68+11.6 STA PC² = 71+46.9 STA PT² = 72+79.9 STA…

Q: PI Elevation 81 82 PC Elevation What is the total length of the curve in meters consider the…

A: G1 = 6.82% G2 = -2.38%

Q: A compund curve is laid out 486.80 meters from the P.C. to P.CC. having a radius of 710.10 meters.…

Q: Choose the letter of the answer. A simple curve having a radium of 200m have tangents with bearing…

Q: An unsymmetrical parabol of 5% and a forward tanfent of -3%. The length of the cur on the left side…

Q: A highway curve was designed to connect intersecting tangents intersecting at an angle of 72°. The…

Q: Situation 1] A circular road is proposed to be laid out with tangent AB whose bearing is N87054'E…

Q: A circular highway curve has the following design elements: . I = 65°32'17" D = 3°52'00" • Pl…

A: Solution-The given values are as follows,I=65°32'17''D=3°52'00''PI sataion=36+4500PI…

Q: The common tangent CD of a reversed curve is 280.5m and has an azimuth from the south (s)of 312°…

Q: Situation 2] A simple curve having a radius of 237.0 m has tangents of N28.7°E and N66.0°E. A line…

Q: Given an equal tangent vertical curve, in feet, with the following known information: STAA…

Q: A proposed highway has two tangents of bearings N 45°54'36" E and N 1º22'30" W. The highway design…

A: The bearings of tangents are N45∘54'36''E, and N1∘22'30''W. The external ordinates is E=13 m. Draw…

Q: QUESTION 6 For a horizontal curve with the following pargmeters what is the station of the PT? Give…

A: Horizontal Curve: The horizontal curve occurs at the intersection of two highways. The curve which…

Q: #9dmul jabutz STAPI=135+ 76.45 4. There is a curve with a Back Tangent with a bearing of S 30° E and…

A: Bearing of back tangent = S30oE Bearing of forward tangent = S85oE Degree of curve, D = 10o

Q: Situation 3] The bearing of the first tangent of a simple curve is N70.8°E while the second tangent…

Q: The common tangent BC of a reversed curve is 280.5m and has a bearing of S47°31’E. AB is the tangent…

A: To Find the stationing of the PT.

Q: My REG NO= 5202 Question-1-B A compound curve is to consist of an arc of (Reg. No) m radius followed…

A: Radius of first arc = 5202 mRadius of second arc = 5202+400 = 5602 mIntersection angle of two…

Q: A 5° curve intersects s property line CD at point D. The back Tangent intersects the property line…

A: Given - A 5° curve intersects s property line CD at point D. The back Tangent intersects the…

Q: the intersection point is 6427.464 m and pegs are required on the centre line of the curve at exact…

Concept explainers

Construction Surveys

A construction survey is a survey done on a construction site to mark construction activities and for laying out positions for construction equipment. It includes evaluating and estimating construction activities before the beginning of the project.

Question

Q5/ A reverse curve must be fit to the tangents
shown below:
PC1
N 58 30′25 E
ETABBoard
P11
S87 1045 E
N47'5930 E
P12
Point Station
PC1 21+83.05
PII 27+35.80
P12 38+43.05
What are the radii of the two curves which fit
this alignment? Compute to 0.01 ft.

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: Given data:

VIEW

Step 2: Length of the tangent:

VIEW

Step 3: Radii of two curves:

VIEW

Solution

VIEW

Step by step

Solved in 4 steps with 8 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

SHOW THE COMPLETE SOLUTION AND ILLUSTRATION. In a compound curve, the line connecting the P.I. at point V and the P.C.C. is an angle bisector. AV is 270m and BV is 90m. The stationing of A is 6+421 and that of B is 6+721. Point A is a along the tangent passing thru P.C. while point B is passing thru P.T. The P.C.C. is along line AB.a. Compute the radius of the first curve passing thru P.C.b. Compute the radius of the second curve passing thru P.T.c. Determine the length of chord from P.C. to P.t.
Q5. Grades of gl=-2.50% and g2=+1.50%, the intersection of grade lines at station 315+00 is at elevation 2560.00 ft. Fixed elevation 2567.00 ft at station 314+00. Detremine curve length also stations and elevations of BVC and EVC 0001
The symmetrical parabolic curve passes through point A whose elevation is 23.23m at a distance of 54m from the PC. The elevation of the PC at station 4+100 is 22.56m. The grade of the back tangent is 2% and the length of the curve is 160m. Compute for the stationing of the highest point of the curve. O 4+163.88 O 4+171.11 O 4+170.59 O 4+194.12
ROUTE SURVEYING
Skip if not too familiar thanks
Given below are the details of a compound curve. • Pl is located at 55+69.23 • Deflection angle between tangents is 85°00'00"R. • The compound curve begins 463.64 ft before the PI • The compound curve ends 405.60 ft after Pl • Central angles of the two curves are 30°00'00" and 55°00'00", respectively. Determine the radius of each curve. Draw
fe p5 a b and c
[Situation 1] A simple curve connects two tangent roads AB and BC with bearings N88°40'E and S61°40'E, respectively. If the chainage (or stationing) of the point of intersection of the two tangents is 4+331.42 and the radius of the curve is 300 m, determine the following: i. Tangent distance, T ii. External distance, E ii. Middle Ordinate, M iv. Versine, V v. Long Chord, C vi. Length of the curve, LC
A compound curve with intersection angle 49, PC = 13+25 , the first curve (from PC to PCC) has a radius of 152m , the second curve (from PCC to PT) has a radius of 140m. find station PT assuming the length of the second curve is minimum. a) 1448.73 b) 1450.19 c) 1445.52 d) 1451.48 e) 1446.08
The common tangent BC of a reversed curve is 280.50m and has a bearing of S 47deg 31min E. AB is the tangent of the first curve whose bearing is N 72deg 45min E. CD is a tangent of the second curve whose bearing is N 38deg 13min E. A is at the PC while D is at the PT . The radius of the first curve is 180m . The PI1 is at Sta 12 + 523.37. Find the stationing of the PT12 + 878.1612 + 883.6512 + 893.2412 + 889.54 Please base your answer on the given choices thank you for the help
Stations for each curve: STA PC¹ = 65+48.7 STA PT¹ = 68+11.6 STA PC² = 71+46.9 STA PT² = 72+79.9 STA PC³ = 77+62.7 STA PT³ = 78+73.5 Intersection angles for each curve: A¹ = 63°15' A² = 48°20' 4³ = 42°45' Disclaimer: the angles and stations given are not the real ones. Cant 2 3 Profe Determine the radius for each of the three curves shown. Also, determine the PI stations for each curve.
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