Šituation: Simple curve connects two tangents XY and YZ The survey data are the following: Bearing of XY = N 85' 30' E Bearing of YZS 68' 30' E Station of Vertex / Intersection = 4 + 360.2 Station of Point of Curvature= 4 + 288.4 Given Angle of Intersection (1) = 26° Tangent Distance, T= 71.8m Radius of the Curve, R= 311m Degree of Curve, deg. (arc basis) = 3.68 Degree of Curve, deg. (chord basis) =3.69 External Distance, E= 8.18m Middle Ordinate, M = 7.97m Stations Interval - 20 m Chord/Sub-Chord Arc/Sub-arc distance First Station Next Station Deflection Angle Distance PC (Sta. 4 + 288.4) 4+300 4+320 4+340 PT (Sta. 4 + 360.2) 4+300 4+320 4+340 Complete the table and find a) Chord distance (from PC to PT) b) Length of Curve (from PC to PT) c) Station point of tangency
Šituation: Simple curve connects two tangents XY and YZ The survey data are the following: Bearing of XY = N 85' 30' E Bearing of YZS 68' 30' E Station of Vertex / Intersection = 4 + 360.2 Station of Point of Curvature= 4 + 288.4 Given Angle of Intersection (1) = 26° Tangent Distance, T= 71.8m Radius of the Curve, R= 311m Degree of Curve, deg. (arc basis) = 3.68 Degree of Curve, deg. (chord basis) =3.69 External Distance, E= 8.18m Middle Ordinate, M = 7.97m Stations Interval - 20 m Chord/Sub-Chord Arc/Sub-arc distance First Station Next Station Deflection Angle Distance PC (Sta. 4 + 288.4) 4+300 4+320 4+340 PT (Sta. 4 + 360.2) 4+300 4+320 4+340 Complete the table and find a) Chord distance (from PC to PT) b) Length of Curve (from PC to PT) c) Station point of tangency
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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horizontal circular curve. provide complete solution. some elements are solved, please help us to the rest. thank you.
Transcribed Image Text:Situation:
Simple curve connects two tangents XY and YZ
The survey data are the following:
Bearing of XY = N 85' 30' E
Bearing of YZS 68' 30' E
Station of Vertex / Intersection = 4 + 360.2
Station of Point of Curvature= 4 + 288.4
Given
Angle of Intersection (I) = 26*
Tangent Distance, T= 71.8m
Radius of the Curve, R = 311m
Degree of Curve, deg. (arc basis) = 3.68"
Degree of Curve, deg. (chord basis) =3.69°
External Distance, E = 8.18m
Middle Ordinate, M = 7.97m
Stations Interval = 20 m
Chord/Sub-Chord Arc/Sub-arc
distance
First Station
Next Station
Deflection Angle
Distance
PC (Sta. 4 + 288.4) 4+300
4+300
4+320
4+320
4+340
4+340
PT (Sta, 4 + 360.2)
Complete the table and find
a) Chord distance (from PC to PT)
b) Length of Curve (from PC to PT)
c) Station point of tangency
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