6. The two tangents of a horizontal curve intersect 3000 ft. beyond Station 10+00. The backtangent has a bearing of N 45°00'00" W and the forward tangent has a bearing of N15°00'00"E. The decision has been made to design a 2000 ft radius horizontal curve between the twotangents.Forward tangentN 15°00'00" EP.I.P.T.3000'P.C./ R=2000'10+00Back tangentN 45°00'00" WDetermine the following:a. What is the length of the curve?b. What is the degree of the curve?c. What is the station of the PT?d. What is the long chord distance?e. What is the middle ordinate?f. Deflection angle and chord length to the first even 100 ft. station

Answered: Image /qna-images/answer/29632d29-0bd6-4613-9ae1-08e82aa3244b.jpg

Answered: 6. The two tangents of a horizontal…

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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A compound curve connects three tangents having an azimuths of 254’, 270’ and 280’ respectively. The length of the chord is 320m long measured from the P.C. to the P.T. of the curve and is parallel to the common tangent having an azimuth of 270’. If the stationing of the P.T. is 6 + (551), A. Determine the total length of the curve B. Determine the stationing of the P.C.C. C. Determine the stationing of the P.C.
Problem No. 1: Two tangents intersect at STA 2+550. A compound curve laid on their tangents has the following data: I1=31deg.;I2=36deg.;D1=3deg.;D2=5deg.; (a) Compute the stationing at PC,PCC and PT
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
Two tangents intersect at station 2 + 550. A compound curve laid on their tangents has the following data. I1 = 310, I2 = 360, D1 = 30, D2 = 50 a) Compute the stationing of the P.C., P.C.C. of the curve. b) If the P.T. is moved 15 m out, compute the station of the P.T. with the P.C.C. on the same point.
A compound curve is laid out 480 m. from the P.C. to the P.C.C. having a radius of 763.94 m then from the P.C.C another curve was laid out to the P.T. 250 m long with a radius of 208.85 m. If the stationing of the point of intersection of the tangents is 10 + 432.25. a. Determine the stationing of the P.C b. Determine the length of the long chord from the P.C. to the P.T. c. Determine the angle that the long chord makes with the tangent
Problem No. 1: Two tangents intersect at STA 2+550. A compound curve laid on their tangents has the following data: I1=31deg.;I2=36deg.;D1=3deg.;D2=5deg.; (a) Compute the stationing at PC,PCC and PT Problem No. 2: A compound curve has a central angle of 40deg. and radius 280m. The second curve has a degree of curve of 5deg. and central angle of 28deg. (a) Compute the length of the common tangent.
Two parallel roads 300 m. apart were connected by two equal simple curves and form a reverse curve. If the length of the intermediate tangent located between the two simple curves is 360 m. and the two curves have equal radii of 1200 m., find the central angle common to the two simple curves. a. 20deg 34’09” b. 18deg 06’24” c. 23deg 76’52” d. 21deg 33’03”
Two tangents intersect at station 26+050. A compound curve laid on their tangents has tne following data: I1 = 31deg I2 = 36deg D1 = 3 deg D2 = 5 deg determine: a) The stationing of the PC (format: 00+000.00) b) The stationing of the PT (format: 00+000.00) c) Total length of the common tangent **2 DECIMAL PLACES ONLY**
A simple curve has a central angle of 80° and a radius of 100ft Point C is at Sta. 0+25.00. If station of PC is Sta. 0+00.00. a) Determine the distance of Point J from the P.C. b) Determine the length of Line JC. c) Determine the chord distance from the P.C. to point C. d) Determine the station of Point J.
Situation 3] The bearing of the first tangent of a simple curve is N70.8°E while the second tangent has a bearing of S82°31'E. The degree of curve is 4.5° and chainage of PC is 3+801. It is proposed to decrease the central angle by an angle of 7° 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: i. New angle of intersection, I'. ii. New radius of the curve, R'. iii. New chainage of station PC. 4:37 PM /
The tangents of a sag curve connects at the point of intersection having Station 18+427.14 and elevation 157.50 meters. Point PT at the end of the curve lies at Station 18+571.93 and elevation 172.42 meters. The grade of the back tangent is -7.60%. Determine the value of the maximum offset in meters
Curve connects three tangents having an azimuth of 254°,270°, and 270°, respectively. The length of chord is 310 m long measured from P.C. to P.T. of the curve and is parallel to the common tangent having an azimuth of 270° if the stationing of P.T., is 6+ 720. Determine the total length of curve, stations of Common Tangent and P. C.

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