A compound curve is having a common tangent of 84.50 meters long. The central anglesmade by the common tangent for the first and second simple curve is 16°and20° respectively. The length of the tangent of the first curve is 38.60 meters and Sta. PI is10+180.Determine:Station of PCStation of PCCStation of PT

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Answered: A compound curve is having a common…

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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The bearings of forward and backward tangent are N 65°30' E and N 75°40' W, respectively. If the curve is 2° for a full station of 20m and the stationing of the vertex is 12+ 582.50, determine the: Elements of the curve Stationing of P.C and P.T Lengths of sub-chords and its corresponding sub-angles
Two tangent intersects at station 0+550. Acompound curve laid on their tangents has the following data: I1= 30° D1 = 4° I2 =35° D2 = 8° a.) Compute the station of the curve PC, PCC and PT of the curve b.) Determine the first and Second radius of the curve WITH COMPLETE SOLUTION AND FIGURE. THANKS
A two vertically parallel tangents, 20 meters apart, were connected by a reverse curve. The length of the common tangent is 180 meters and the station of the first vertex is 18+100. Determine: Length of curve from PC to PT Station of PC Station of PCC Station of PT
A compound curve is having a common tangent of 104.78 meters long. The central angles made by the common tangent for the first and second simple curve is 18° and 29° respectively. The length of the tangent of the first curve is 30.21 meters and Sta. Pl is 10+180. determine : show solution - station of pcc - station of pt
A horizontal curve is designed with a ‘500’ m radius. The curve has a tangent length of 120 m and the Pl is at station 3 + 140. Determine the stationing of the PT. a. Station of PT= 3 + 255.53 b. Station of PT= 3 + 257.09 C. Station of PT= 3 + 256.99 d. Station of PT= 3 + 256.74 e. Station of PT= 3 + 257.80
Two tangents intersect at station 5+800. A compound curve laid on their tangents has the following data. = 32° A2 = 38° D1 = 4° D2 = 7° а. Determine the stationing of BC. b. Determine the stationing of PCC. Determine the stationing of EC. c.
A simple curve of the proposed extension of Mantabahadra Highway have a direction of tangent AB which is due north and tangent BC bearing N45°E. Point A is at the P.C. who stationing is 20 + 130.46. The degree of the curve is 4º. 1.1. Compute the long chord of the curve 1.2. Compute the stationing at point D on the curve along a line joining the center of the curve which makes an angle of 54° with the tangent line passing thru the P.C. 1.3. What is the length of the line from D to the intersection of the tangent AB.
A compound curve is having a common tangent of 104.78 meters long. The central angles made by the common tangent for the first and second simple curve is 18° and 29° respectively. The length of the tangent of the first curve is 30.21 meters and Sta. Pl is 10+180. determine : show solution - length of long chord - length of curve (Lc1) - length of curve (Lc2)
The tangent thru the PC has a direction due to North; and the tangent through the PT has a bearing of N 50° E. It has a radius of 200 m. Stationing of PC is 12+060. % a. Compute the tangent distance of the curve. Answer: T= [tangent] m b. Compute the long chord of the curve. Answer: Lc = [Lchord] m c. A line is drawn from the center of the curve (at the central angle) and intersects the curve at point B. This line, when extended, makes an angle of 62° with the tangent thru PC. What is the stationing of point B? Answer: Sta. B = [StaB]
2. The tangent lines of a simple curve have azimuths 300° and bearing N 04 E, respectively. A third tangent line AB intersects the two tangent lines at bearing S 34° E. Stationing of the PI of the curve is 16 + 464.35 and the distance from point B to the PI of the curve is 277.6 m (ie BV = 277.6 m). Determine the following: a. Radius of the simple curve that shall be tangent to these three lines. b. Stationing of the PC c. Stationing of the PT
Given for a Simple Curve, a. R=750m b. I-34 deg c. STA. V = 20+050 Detemine the Elements of the given Simple Curve. (Use 100m-Stations) i. Degree of the Curve ii. Tangent Distance iii. Length of the Chord iv. Length of the Curve v. Extemal Distance vi. Middle of Ordinate vii. Sta. PC and Sta. PT
The bearings of forward and backward tangent are N 65°30' E and N 75°40' W, respectively. If the degree of curve is 2° for a full station of 20m and the stationing of the vertex is 12 + 582.50, determine the: a. Elements of the curve b. Stationing of P.C. and P.T. c. Lengths of sub-chords and its corresponding sub-angles

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