Given broken line AB = 57.6 m., BC = 91,5 m. and CD = 91.5 m. arranged as shown. A reverse curve is to connect these three lines thus forming the center line of a new road. %3D a) Find the length of the common radius of the reverse curve, b) Find the total length of the reverse curve, CIf the P.C. is at Sta. 10+000, what are the stationing of P.R.C, and P.T. Solu tion: a) T, =Rtan 11 %3D T = R tan 32 T+T = 91.5 91.5 m R (tan 11 + tan 32 ) = 91.5 PT R = 111.688 m. 82 RI IR 7,5 7.5 64 56.7 G'16
Given broken line AB = 57.6 m., BC = 91,5 m. and CD = 91.5 m. arranged as shown. A reverse curve is to connect these three lines thus forming the center line of a new road. %3D a) Find the length of the common radius of the reverse curve, b) Find the total length of the reverse curve, CIf the P.C. is at Sta. 10+000, what are the stationing of P.R.C, and P.T. Solu tion: a) T, =Rtan 11 %3D T = R tan 32 T+T = 91.5 91.5 m R (tan 11 + tan 32 ) = 91.5 PT R = 111.688 m. 82 RI IR 7,5 7.5 64 56.7 G'16
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Transcribed Image Text:Given broken line AB = 57.6 m., BC = 91,5 m. and CD = 91.5 m. arranged
as shown. A reverse curve is to connect these three lines thus forming the center
line of a new road.
%3D
a) Find the length of the common radius of the reverse curve,
b) Find the total length of the reverse curve,
CIf the P.C. is at Sta. 10+000, what are the stationing of P.R.C, and P.T.
Solu tion:
a) T, =Rtan 11
%3D
T = R tan 32
T+T = 91.5
91.5 m
R (tan 11 + tan 32 ) = 91.5
PT
R = 111.688 m.
82
RI
IR
7,5
7.5
64
56.7
G'16
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