The central angles of a compound curve are 30° and 40° respectively. In addition, the radii of the curves are 400 m and 240 m respectively. PC is at 0+999. a. What is the length of the common tangent? b. What is the station of PCC? c. What is the 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.

Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
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The central angles of a compound curve are 30° and 40° respectively. In addition, the radii of the curves are 400 m and 240 m respectively. PC is at 0+999.

a. What is the length of the common tangent?

b. What is the station of PCC?

c. What is the 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.

ヘー
COMPOUND CURVE
COMMON
TANGENT
PCC
PT
PC
LONG
CHORD (LC)
PI
LC
PC
POINT OF TANGENCY
KNOWN STA. PC
Sta PT- Sta PC+Le9tLez
KNOWN STA PI
Sta PT - Sta PI - x- TtLe+Lez
PI
I = I1 + I2
y
Common tangent
V1 I
1
РСС
Iz
V2
La
T1
T2
L2
L1
MATH=I0mp.com
Halin-
PT
L
PC
Iz
R2
R1
I1
Compound Curves
Elements of compound curve
• PC = point of curvature
%3D
• PT = point of tangency
%3D
• PI = point of intersection
• PCC = point of compound curve
%3D
T1 = length of tangent of the first curve
T2 = length of tangent of the second curve
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
• I = angle of intersection = I1 + I2
%3D
• L¢1 = length of first curve
Lc2 = length of second curve
L1 = length of first chord
%D
L2 = length of second chord
• L = length of long chord from PC to PT
%3D
T1 + T2 = length of common tangent
measured from Vị to V2
• 0 = 180° - I
x and y can be found from triangle V1-V2-
PI.
• L can be found from triangle PC-PCC-PT
Finding the stationing of PT
Given the stationing of PC
Sta PT = Sta PC + Le1 + Le2
Given the stationing of PI
Sta PT = Sta PI – x – T1 + Lc1 + Lc2
-
T2 Y
Transcribed Image Text:ヘー COMPOUND CURVE COMMON TANGENT PCC PT PC LONG CHORD (LC) PI LC PC POINT OF TANGENCY KNOWN STA. PC Sta PT- Sta PC+Le9tLez KNOWN STA PI Sta PT - Sta PI - x- TtLe+Lez PI I = I1 + I2 y Common tangent V1 I 1 РСС Iz V2 La T1 T2 L2 L1 MATH=I0mp.com Halin- PT L PC Iz R2 R1 I1 Compound Curves Elements of compound curve • PC = point of curvature %3D • PT = point of tangency %3D • PI = point of intersection • PCC = point of compound curve %3D T1 = length of tangent of the first curve T2 = length of tangent of the second curve 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 • I = angle of intersection = I1 + I2 %3D • L¢1 = length of first curve Lc2 = length of second curve L1 = length of first chord %D L2 = length of second chord • L = length of long chord from PC to PT %3D T1 + T2 = length of common tangent measured from Vị to V2 • 0 = 180° - I x and y can be found from triangle V1-V2- PI. • L can be found from triangle PC-PCC-PT Finding the stationing of PT Given the stationing of PC Sta PT = Sta PC + Le1 + Le2 Given the stationing of PI Sta PT = Sta PI – x – T1 + Lc1 + Lc2 - T2 Y
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