Two tangents of a reverse curve have a back tangent of Due East and a forward tangent of S50°E. The radius of the curve at PC is 220 m and that of the other curve is 450 m. A line traversing the back tangent originating from PT has a length of 325 m and a bearing of S22°E.Find the following:Central angle of the first curveCentral angle of the second curveStation of PTDraw and plot the curve. Compute all of the necessary elements of the curve. Include the proper units/dimensions and round-off the answers to 3 decimal places. PCT1V1Le1T1R2IzR1R2T2R1LczI1PRCV2T2Common tangentPTReversed CurvesPCV1back tangentR2forward tangentPRCPTI2V2R1R1Reversed Curves for Nonparallel TangentsPCV1back tangentR2R2PRCforward tangentV2PTR1IR1Reversed Curves for Parallel TangentsElements of Reversed Curve• PC = point of curvature%3D• PT = point of tangency• PRC = point of reversed curvatureT1 = length of tangent of the first curveT2 = length of tangent of the second curve%3DV1= vertex of the first curveV2 = vertex of the second curve• I1 = central angle of the first curveI2 = central angle of the second curveLc1 = length of first curveLc2 = length of second curveL1 = length of first chord%DL2 = length of second chordT1 + T2 = length of common tangentmeasured from Vị to V2%3DFinding the stationing of PTGiven the stationing of PCSta PT = Sta PC + Le1 + L2|Given the stationing of V1Sta PT = Sta Vị – Tị + Lcl + Le2-

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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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Two tangents of a reverse curve have a back tangent of Due East and a forward tangent of S50°E. The radius of the curve at PC is 220 m and that of the other curve is 450 m. A line traversing the back tangent originating from PT has a length of 325 m and a bearing of S22°E.

Find the following:

Central angle of the first curve
Central angle of the second curve
Station of PT

Draw and plot the curve. Compute all of the necessary elements of the curve. Include the proper units/dimensions and round-off the answers to 3 decimal places.

PC
T1
V1
Le1
T1
R2
Iz
R1
R2
T2
R1
Lcz
I1
PRC
V2
T2
Common tangent
PT
Reversed Curves
PC
V1
back tangent
R2
forward tangent
PRC
PT
I2
V2
R1
R1
Reversed Curves for Nonparallel Tangents
PC
V1
back tangent
R2
R2
PRC
forward tangent
V2
PT
R1
I
R1
Reversed Curves for Parallel Tangents
Elements of Reversed Curve
• PC = point of curvature
%3D
• PT = point of tangency
• PRC = point of reversed curvature
T1 = length of tangent of the first curve
T2 = length of tangent of the second curve
%3D
V1
= vertex of the first curve
V2 = vertex of the second curve
• I1 = central angle of the first curve
I2 = central angle of the second curve
Lc1 = length of first curve
Lc2 = length of second curve
L1 = length of first chord
%D
L2 = length of second chord
T1 + T2 = length of common tangent
measured from Vị to V2
%3D
Finding the stationing of PT
Given the stationing of PC
Sta PT = Sta PC + Le1 + L2
|
Given the stationing of V1
Sta PT = Sta Vị – Tị + Lcl + Le2
-

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