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With the same problem the answers should be: R = 510 m T = 306.46 m  Station point "x" = 10+329.8 Last subchord = 11.87m Sketch the curve  Additional Formula: L(Curve Length) = R x I(Angle of intersection) x pi /180

With the same problem the answers should be: R = 510 m T = 306.46 m  Station point "x" = 10+329.8 Last subchord = 11.87m Sketch the curve  Additional Formula: L(Curve Length) = R x I(Angle of intersection) x pi /180

Structural Analysis
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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With the same problem the answers should be:

R = 510 m
T = 306.46 m 
Station point "x" = 10+329.8

Last subchord = 11.87m

  • Sketch the curve 
  • Additional Formula: L(Curve Length) = R x I(Angle of intersection) x pi /180
Tangent Distance Middle Ordinate External Distance Long Chord T = R tan (2) M = R(1-cos( (→)) E = R(sec() - 1) Lc = 2R sin / 2
Transcribed Image Text:Tangent Distance Middle Ordinate External Distance Long Chord T = R tan (2) M = R(1-cos( (→)) E = R(sec() - 1) Lc = 2R sin / 2
Degree of Arc basis Chord Basis R R= 1145.916 D 10 sin
Transcribed Image Text:Degree of Arc basis Chord Basis R R= 1145.916 D 10 sin
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With the same problem, compute the length of the last subchord. ANS = 11.87m

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Follow-up Question

Where did θ = 14.04 and 2θ = 28.08 came from? And in Rcos(28.08) why is it multiples to cos? 

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Can you show an illustration on how the tan(θ) = 60/40 was determined? where did the 40 came from?

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