It is required to layout a simple curve by deflection angles. The curve is to connect two tangents with an intersection angle of 32° and a radius of 240m. Compute the deflection angles to each full stations in the curve, if the transit is set-up at PC which is station 5+767.2. What is the stationing of the PT? Sta Occ Sta Obs Deflection Angle Remarks 5+767.2 5+780 1°31'30" 1 5+800 3°54'30" 2 5+820 6°17'30" 3 5+8400 4 5+860 5+880 5+900 7 5+901.47 16° PT 6.

It is required to layout a simple curve by deflection angles. The curve is to connect two tangents with an intersection angle of 32° and a radius of 240m. Compute the deflection angles to each full stations in the curve, if the transit is set-up at PC which is station 5+767.2. What is the stationing of the PT? Sta Occ Sta Obs Deflection Angle Remarks 5+767.2 5+780 1°31'30" 1 5+800 3°54'30" 2 5+820 6°17'30" 3 5+8400 4 5+860 5+880 5+900 7 5+901.47 16° PT 6.

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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Q.9 is what i need the answer to pls. 1. A simple horizontal curve of radius 750 ft connects two tangents that intersect at an angle of 66°30'. Compute the parts of the curve, including T, L, LC, E, and M. 9. For the simple curve in Problem 1, if the station of the PI is 22 +50, what are the stations of the PC and the PT?
Stations for each curve: STA PC¹ = 65+48.7 STA PT¹ = 68+11.6 STA PC² = 71+46.9 STA PT² = 72+79.9 STA PC³ = 77+62.7 STA PT³ = 78+73.5 Intersection angles for each curve: A¹ = 63°15' A² = 48°20' 4³ = 42°45' Disclaimer: the angles and stations given are not the real ones. Cant 2 3 Profe Determine the radius for each of the three curves shown. Also, determine the PI stations for each curve.
3 Horizontal curves You are asked to design a 5-degree horizontal curve with an intersection angle at the center equals 60°15'. The station of point of curvature is 155 + 26.5 ft. Calculate the following: A. Length of the curve B. Station of the point of tangency (PT) C. For first two whole stations next to PC, calculate the degree of deflection from the PC and chord length from the previous point along the horizontal curve.
A simple curve has a radius of 123.45 meters and radius of 40 degrees. The stationing of PI is 5+105. This simple curve will be layout with six stations of the curve: PC, 1, 2, 3,4 and PT using Offset Distance. 1. What is the offset distance of Point 3? 2. What is the offset distance of PC? 3. What is the central angle of the curve from Point 4 to PT? 4. What is the offset distance of Point 1? 5. What is the offset distance of PT?
10-1. A simple horizontal curve of radius 750' connects two tangents that intersect at an angle of 66° 30’ Compute T, L, LC, E, and M. 10-3. What is the degree of curve (using the arc definition) in problem #1? 10-5. A simple curve is to be laid out so that its middle ordinate is at least 75'. If the tangents intersect at an angle of 40°, what is the highest degree of curve that can be used? 10-7. The radius of a simple curve is twice its tangent distance. What is the angle of intersection? 10-9. For the simple curve in problem #1, if the station of the PI is 22+50, what are the stations of the PC and PT?
3. A -9% grade meets a +5% grade at station 2+123.56 of elevation 55.56 m. The lengths areunequal L1=100 m, L2=80 m. Calculate the elevation at every full station. Compute thestationing and elevation of the lowest point on the curve.
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need please
In a compound curve, the line connecting the P.I. at point V and the P.C.c. is an angle bisector. AV is 270 meters and BV is 90 m. The stationing of A is 6 + 421 and that of B is 6 + 721. Point A is along the tangent passing thru the P.C. while point B is along the tangent passing thru the P.T. The P.c.C. is along line AB. (a) Compute the radius of the first curve passing thru the P.C. in m. (b) Compute the radius of the second curve passing thru the P.T. in m. (c) Determine the length of the long chord from P.C. to P.T. in m.
Situation 2] A simple curve having a radius of 234.0 m has tangents of N28.5°E and N66.0°E. A line running parallel to the first tangent crosses the centerline of the curve at a distance 10 meters away from the first tangent. If chainage of station PC is 17+584, What is the deflection angle at the point of intersection of the line and the simple curve measured from PC? i. What is the chainage of the point of intersection? Compute the chord distance from PC to the point of intersection of the line and the ii. iii. curve.
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Q1: The figure below represents a circular curve having coordinates given in the Table. The chainage/station of PC is 270 m. Calculate: 1. The curve length (L), R, and Tangent (T). 2. Coordinates of the required station (Easting, Northing). E pc 258.1 N E PI N E PT N 220.1 572.1 534.1 886.1 220.1 The required station 538.8