Simple Curves: A highway curve was designed to connect intersecting tangents at Pl. Back tangent = N 30° 25' E Forward tangent = S 69°35' E Degree of curve = 6⁰ Station PI = 2+222.22 8. Determine the length of tangent distance (PI to PT). a) 145.66m b) 123.56m 9. Compute the distance from the PI to the midpoint of the curve. a) 58.33m b) 35.67m c) 37.78m c) 160.26m 10. Determine the length of curve (PC to PT, arc length). a) 256.78m b) 234.78m c) 266.54m 11. Determine the stationing of PT. a) 2 + 244.56 b) 2 + 328.50 c) 2 + 289.89 d) 142.33m d) 25.56m d) 276.45m d) 2 + 301.34

Simple Curves: A highway curve was designed to connect intersecting tangents at Pl. Back tangent = N 30° 25' E Forward tangent = S 69°35' E Degree of curve = 6⁰ Station PI = 2+222.22 8. Determine the length of tangent distance (PI to PT). a) 145.66m b) 123.56m 9. Compute the distance from the PI to the midpoint of the curve. a) 58.33m b) 35.67m c) 37.78m c) 160.26m 10. Determine the length of curve (PC to PT, arc length). a) 256.78m b) 234.78m c) 266.54m 11. Determine the stationing of PT. a) 2 + 244.56 b) 2 + 328.50 c) 2 + 289.89 d) 142.33m d) 25.56m d) 276.45m d) 2 + 301.34

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 is composed of three simple curves (CURVE1, CURVE2 & CURVE3).Compute the stationing of PT if the stationing of PC is 6 + 623.56.CURVE1 Radius = 1,157.40 metersCURVE1 Long Chord = 470.80 metersCURVE2 Radius = 550 metersCURVE2 Long Chord = 185.50 metersCURVE3 Radius = 467.87 metersCURVE3 Long Chord = 153.75 meters
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A simple curve connects two tangents AB and BC with bearings N 85°30'00" E and S 68°30'00"E, respectively. If the stationing of B is 4+360.2, and the stationing of PC is 4+288.4, determine the following: a. Radius Answer: R= b. External Distance Answer: E= c. Middle Ordinate Answer: M= d. Chord Distance Answer: Lc= e. Length of Curve Answer: L= m m m m E
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QUESTION 2 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= b. Compute the long chord of the curve. Answer: Lc = m 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 = Note: Unit has been provided after the blank space. No need to indicate units on your answer.
SITUATION 12: AB and BC are tangents of a simple curve. AB is due north and BC is N50°E. Degree of simple curve is 4° and PC is at station 20 + 130. 33. Compute the length of the long chord. a. 242.14 m c. 386.12 m d. 284.51 m b. 184.52 m 34. Compute the distance from point D on the simple curve to point of intersection of tangents that makes an angle of 54° with the tangent passing through the PC. а. 30.2 m b. 26.8 m С. 23.4 m d. 31.5 m 35. Compute the stationing of point D. a. 20 + 639.3 b. 20 + 739.3 c. 20 + 439.3 d. 20 + 249.3
TOPIC: Surveying (Horizontal Curves)
4. A parallel reversed curve has an equal radius of 360m and a central angle of 8º. a. Compute the distance between parallel tangents b. Compute the length of the Chord from PC to PT If Sta PC = 3 +960.40, what is Sta PT? C.
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SITUATION 10: Two parallel tangents 12 m apart are connected by a reverse curve having equal radii. The long chord connecting the P.C to P.T is 140 m. 28. Determine the radius of the reverse curve. c. 200.41 m d. 512.41 m a. 408.33 m b. 813.64 m 29. Find the length of the curve. a. 175.75 m b. 131.29 m c. 279.30 m d. 140.17 m
In a compound curve, the line connecting the P.l. at point V and the PC is an angle bisector. VA is 270 meters and BV =90 m. The stationing of A is 6+421 and that of B is 6+721. Point A is along the tangent passing through PC while point B is along the tangent passing through PT. The PC is also along AB. a. Compute the radius of the first curve b. Compute the radius of the second curve c. Determine the length of the line connecting PC and PT.
QUESTION 7 Determine the common tangent of a reversed curve given the following properties: 11 = 38.90° D₁ = 5.75° 12 = 37.95⁰ D2 = 3.81° NOTE: Solution is not required.