Structural Analysis
6th Edition
ISBN: 9781337630931
Author: KASSIMALI, Aslam.
Publisher: Cengage,
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Two tangents converge at an angle of 30° and is connected by a reverse curve of equal radii. The direction of the second tangent is due east. The perpendicular distance from PI1 to PI2 is 159.5 m. The central angle of the second curve is 57°. Determine the radius of the curves.
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- A compound curve has the following data. I1 = 38 D1 = 4 12 = 22 D2 = 5 1.) Determine the length of the first curve 2.) Determine the length of the common tangent.arrow_forwardThree simple curves are connected to each other such that the fırst and the second form a compound curve. While the second and the third form a reversed curve. The distance between the point of curvature and the point of tangency of the compound curve (which is also the point of reversed curvature of the reversed curve is 500 m. If the angle of convergence between the second and the third tangents is 16°, I, = 47°. R1 =200 meters, I2 = 65°, R3 =160 m, and stationing at PC is 5+600. Hint: sine law. determine the stationing at PT at the end of the long chord. Answer for part A should be in this Format: x+xx.xXxx (FOUR DECIMAL PLACES i.e. 0+123.4567) determine the angle between the long chord of the compound curve and the first tangent. Answer in four decimal places (i.e. 12.3456)arrow_forwardTwo tangents AB and VB are connected by a compound curve at points A (P.C.). and B (P.T.). Point V is the point of intersection of the tangents (P.I.). Angle VAB =30° and angle VBA =40°. Distance AB is 200m. and the radius of the second curve R2=100m. Determine the central angle of the first curve b. Determine the central angle of the 2nd curve C. Determine the radius of the first curvearrow_forward
- the two-centered compound curve has centers at 01, 02 with central angle equal to. A1, A2. Given R1 = 220 m and R2 = 290 m and A1 = 152°13' , A2 =47°12' and Pl station 1+000 Draw a neat drawing for compound curve Find Ta and Tb and station value BC1, EC1, BC2, and EC2 by using Traverse method and repeat the same calculation by using vertex methodsarrow_forwardQUESTION 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.arrow_forward1. From a known point V (vertex), the tangent lines of a compound curve are drawn - having azimuths 300 (forward tangent) and bearing N 04 E (back tangent), respectively. A common tangent line CD intersects the two tangent lines at bearing S 34° E. Stationing of the vertex of the compound curve is 16+ 464.35 and the distance from point D to the vertex of the compound curve is 137.6 m. If the external distance of the curve passing through the PC is 1.42 times the external distance of the simple curve that passes through the PT, solve for the following. a. Radius of the curve that passes through the PC b. Radius of the curve that passes through the PT c. Stationing of the PTarrow_forward
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