ans No. 1 only. asap PROBLEM 1. The long chord of a compound curve is 120m long which makes an angle of 14° fromthe tangent of the first curve passing through the PC and 20° from the tangent of the second curvepassing through the PT. If the common tangent is parallel to the long chord.a.) Compute the length of chord from PC to PCb.) Compute the length of chord from PCC to PTc.) Compute the difference in radius of the first and second curve.PROBLEM 2. A compound curve passes thru a common tangent AB having a length of 300m. Theradius of the first curve is equal to 290m and a central angle of 42°. If the angle of intersection ofthe curve is 70°36'.a.) Compute for the Radius of the second curveb.) If the stationing of the PI is at (20+542.2), Compute for the Stationing of PT..PROBLEM 3. Two parallel tangents 20m apart are to be connected by a reversed curve with equalradius. The tangent of the 1st curve is 37.817m which is along the tangents of common direction.The stationing of the PC is 10+200.a.) Compute for the long chord.b.) Compute for the common radius.c.) Compute for the stationing of PT.PROBLEM 4. The tangents of a simple curve have bearings of N 20° E and N 80° E respectively.The radius of the curve is 200 m.Determine the following:a. The external distance of the curve.b. The middle ordinate of the curve.c. The stationing of point A on the curve having a deflection angle of 6 from the P.C. whichis at 1+200.00.PROBLEM 5. A simple curve have tangents XY and YZ intersecting at a common point Y. XY hasan azimuth of 180° and YZ has an azimuth of 230°. The stationing of the point of curvature at X is10+140.26. If the degree of curve of the simple curve is 4°.Determine the following:d. The length of the long chord from X..e. The tangent distance XY.f. The stationing at a point A on the curve on which a line passing through the center ofthe curve makes an angle of 58° with the line XY, intersects the curve at point A.

Given Data: The length of a long chord is C1=120 m. The angle from the tangent of the first curve is…

PROBLEM 1. The long chord of a compound curve is 120m long which makes an angle of 14° from the tangent of the first curve passing through the PC and 20° from the tangent of the second curve passing through the PT. If the common tangent is parallel to the long chord. a.) Compute the length of chord from PC to PCC b.) Compute the length of chord from PCC to PT c.) Compute the difference in radius of the first and second curve.

PROBLEM 1. The long chord of a compound curve is 120m long which makes an angle of 14° from the tangent of the first curve passing through the PC and 20° from the tangent of the second curve passing through the PT. If the common tangent is parallel to the long chord. a.) Compute the length of chord from PC to PCC b.) Compute the length of chord from PCC to PT c.) Compute the difference in radius of the first and second curve.

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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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.
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]
p1. handwritten
A reversed curve is to connect three lines thus forming the center line of a new road. The lines are AB = 60.2 m, BC = 100.4 m and CD = 100.4 m arranged as shown. Find the length of the common radius of the reverse curve. a.) Find the length of the common radius of the reverse curve. b.) If station of B is 4+450 m, what is the station of P.C.? c.) If station of B is 4+450 m, what is the station of P.T.?
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4. A curve formed by a Back Tangent with a bearing of N 30° W and a Forward Tangent with a bearing of N 85° E, has a degree of 15° defined by 100 feet of arc. The station of the point of intersection is 135+76. a. Calculate the station of PC. b. Calculate the station of PT. c. Calculate the longest distance between the long chord and the curve. d. Calculate shortest distance between the point of intersection and the curve.
D1=5°; 1=36° and D2=8°; 12=28°. Station of the tangent intersection is at 2+222. Determine the stations of the point of curvature, point of tangency and common point of tangency for: (a) Compound Curve (b) Reversed Curve
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Q3- Two straight AB and CD intersect at E. BD is common tangent of length=215 m. It is proposed to introduce a reverse curve consists of two curves, the radius of the second curve is greater than the first curve by 1.10, the deflection angles ABD and CDB are 28°30' and 23°30' respectively. calculate: 1- The elements of the both curves. 2- Stations of PC, Pre, and PT, if the St. of B=65+22.24
p2. handwritten
Problem 1 A simple curve connects two tangents AB and BC with bearings N 75°12' E and S 78°36' E, respectively. If the stationing of P.C. is 10+185.28. and its middle ordinate is 7.46 m, Determine the following: a. Radius of the curve. b. Tangent Distance C. External Distance Long Chord. d. e. Length of the curve. f. Stationing of the vertex g. Degree of the curve using Arc Basis