Answer fast A tangent with a bearing of N 56° 48' 20" E meets another tangent with a bearing N 40° 10' 20" E at PI STA 6+ 26.57. A horizontal curve with radius 1000' will be used to connect the two tangents. Compute the following:a) degree of curvatureb) tangent distancec) length of curved) chord distancee) middle ordinatef) externalg) PC and PT Stations.For the toolbar, press ALT+F10 (PC) or ALT+FN+F10 (Mac).

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Answered: A tangent with a bearing of N 56° 48'…

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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Given: Stationing B= 18 + 685 Stationing C= 19 + 125 Tangent Lines: AB: Azimuth from the south = 224° 24' BC: Azimuth from the south = 264° 29' CD: Azimuth from the south = 315° 28' A proposed highway curve is to connect these three tangents. Find: a. Radius of simple curve that connects these tangents. b. Sta PC. c. Length of the curve from PC to PT.
For your Assignment 5, provide what is required by the problem. PI Elev 108.00m 1.50m -3% +6% 8.00m A Sta 1+060 PT PC Sta 1+000 Elev 100.00m Answer the following: • What is the total length of the curve in meters? • What is the length of the first curve in meters? • What is the length of the second curve in meters? • What is the elevation of point A? • What is the distance of the highest point from PC in meters? • What is the elevation of the highest point?
Question t Pow t Edge Calculate the length and radius for the following curve. The PC is at Sta 8+10. The PT is at Sta 10+50. The delta is 17d 32m 00s. t Excel t Word t Word t Pow t Pow t Wore t Pow
Complete the table below. Unit Simple Curve Intersection angle Degree of Curve Radius 1 2 22 25 Tangent Distance m |Long Chord m Length of Curve m Compound Curve Elements ICommon Tangent m Length of Curve Angle of Intersection Sta of Tangent intersection 4+200 m IStation of the start of the curve USEFUL EQUATIONS 1145.916 D° = C = 2Rsin R [D° sin 10m 2.TR LC 360° 0° 2 R = T = Rtan 20 LC E = R[sec(0/2)-1] 27- M = R[1- cos(0 /2)]] 10m sin(D° /2) LC 360°
Show your COMPLETE solution in 3 decimal places. Provide illustration and label accordingly. 2.The tangent throughthe PC has a direction due north and the tangentthroughthe PT has a bearing of N50 ̊E. It has a radius of 200m. Station of the PC is 12+060. (a) Compute the tangent distance of the curve, (b) Compute the long chord of the curve, and(c) If the line making an angle of 62 ̊with the tangent throughthe PC intersects the curve at point B,what is the stationing of B if this line passes through the center of the curve.
Highway curve has R = 1000 ft, I = 25° 00', and Pl station = 26 + 40.00 ft. Part A Determine degree of curvature Da of the curve. Express your answer in degrees to five decimal places. Da = Submit Part B Submit V— ΑΣΦ L, T = 436.33,221.69 ft Request Answer Determine curve length L and tangent distance T of the curve. Express your answers in feet to two decimal places separated by a comma. Part C Correct Previous Answers Submit vec VGI ΑΣΦ M, E= 23.7,20.40 Determine middle ordinate M and external distance E of the curve. Express your answers in feet to two decimal places separated by a comma. ? vec Previous Answers Request Answer ? ft
SHOW THE COMPLETE SOLUTION IN NEAT MANNER. SHOW THE ILLUSTRATIONS. Two tangents that intersect at an angle of 44°36` are to be connected by a compound curve. The tangent at the beginning of the curve at P.C. is 125.70m long. The degree of curve of the first curve on P.C. is 4°. Using arc basis. Compute the radius of the second curve. Compute the central angle of the second curve. Compute the central angle of the first curve.
Given an equal tangent vertical curve, in feet, with the following known information: • STA A (Beginning of Curve) = 13+00 o Elevation A = 220.00 feet • L= 900.0 feet G1 = -4.25% and G2 = +3.25% What is elevation of the low point on the curve, in feet? rrect Answer O 209.16 u Answered O 220.00 O 209.76 O 215.51
1. Two tangents AB & BC are intersected by a line having angles 40 o & 35 o respectively. The radius of 1 st arc is 1000m and of 2 nd arc is 800m. Carry out all the necessary calculation to find the chainage of tangent points points and the point of compound curvature given that the chainage of intersection point is 5000 m. (CLO2; C3; PLO1)
Given: D₁ 10°30'; D₂ = 10°; 1₁ = 15°; 1₂ = 16°; STA V = 00 + 876 STATION CENTRAL ANGLE PC-1 PC-PCC PCC-2 PCC-3 PC-CPT N/A DEFLECTION ANGLE REMARKS PC 1 PCC 2 3 PT Show computations on R, t, T, Lc, c1, c2, d1, and d2 of the first and second curve and Sta PC, Sta PCC, Sta PT, central angle, and deflection angle.
The azimuth from north of the forward and backward tangents of a 4 degree curve is 260 degree and 280 degree, respectively. At the commencing of the construction, it was found out that the forward tangent is shifted inward and parallel to itself at a distance of 12m without changing the direction of the tangents with P.T. moved along the radius. If the P.T. is at 10 + 525.25, compute the: a. New tangent distance b. Radius pf the new curve c. Stationing of new P.I. and P.T.
Given G₁ & in decimals and a & L in feet, which of the following equations for the elevation of the vertical curve would you use? (only one of the following equations will give the correct answer) a) Elx = ELBvc+x+4x² 100 BVC b) ELx = ELBvc + G₁x + x² BVC 2L -x² c) ELx = ELBvc +x+ BVC 100 d) ELx = ELBvc + G₁x + BVC 200L A 200L

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