A simple curve connects two tangents AB and BC with bearing N84°E and S65°E, respectively. If the stationing of the vertex is 4,362 and the stationing of PC is 4,285. Determine the 1. Radius of the curve. 2. The external distance. 3. The middle ordinate 4. Chord distance 5. Length of the curv

A simple curve connects two tangents AB and BC with bearing N84°E and S65°E, respectively. If the stationing of the vertex is 4,362 and the stationing of PC is 4,285. Determine the 1. Radius of the curve. 2. The external distance. 3. The middle ordinate 4. Chord distance 5. Length of the curv

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 simple curve connects two tangents AB and BC, with bearings N 85°30' E and S 68°30' E, respectively. If the stationing of the vertex (P.I.) is 4+360.2 and the stationing of P.C. is 4+288.4 a. Determine the radius= 311m Determine the external distance = 8.18m Determine the middle ordinate = 7.971m Determine the chord distance = 139.920m Determine the length of the curve = 141. 127m Determine the stationing at P.T. C. d. e. f.
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
A simple curve connects two tangents AB and BC with bearings N 36°12' E and N 58°53' E respectively. The PC is a A which is at Sta. 5+234.65 and the B is at Sta. 5+265.09. * Find the radius of the curve. a. 169. 09 m c. 201.12 m b. 137.85 m d. 151.76 m * Using chord basis, compute the degree of the curve. c. 6.78° a. 5.70° b. 7.56° d. 8.31° * Find the external distance. c. 3.36 m a. 2.75 m b. 3.02 m d. 4.01 m * Find the middle ordinate. c. 2.69 m a. 2.96 m b. 3.93 m d. 3.30
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)
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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.
1. Find the area of the regions shaded by the following curves: 2cose andr = r = 2sine