Thank you for answering :) PC1PI10+000LetPT1R111STATION NORTHINGSEASTINGS0+00011475.558927.24PC2P1111478.809291.91PI211179.379546.66R2V1PI310804.769282.44Lc2PI410512.379433.481312V111245.489490.41PCCPI2V211074.509472.699016.99END10203.85Lc3R3V2LINEAZIMUTHDISTANCE0-PI1PI1-V1V1-V2PT2V2-PI3PI3-P14PI4-ENDPC3R4CURVERT.LC1+140PI31stCurve2nd Curve3rd Curve4th Curve14Lc45th CurvePRCSTATIONING15PCPTLe5PC2PI4PCCR5PT2PC3PRCPT3PT3ENDEND OF ROAD Shown on the drawing are the application of the different horizontal curves for aroadway project.GIVEN:Coordinates of Intersecting PointsLength of Chord of the First Simple is 132.24 mDegree of Curve of the first curve of the compound Curve is 8 degreesThe deflection angle from PC3 to Sta. 1+140 is 15°28'REQUIRED:1. The azimuth and distance of the tangent lines.2. The Intersecting Angle (1), radius (R), tangent distance (T), Length of Curve (Lc) ofall the curves3. The Stationing of PC, PT, PC2, PCC, PT2, PC3, PRC, PT3 and End of ProjectFill up your answer on the given table.Show solutions.IFICEVESITY

Length of a chord is 132.24 mDegree of curve = 8 degreesDeflection angle = 15028'To DetermineTable…

STATION NORTHINGS 0+000 11475.55 PI1 11478.80 11179.37 P12 P13 10804.76 P14 10512.37 V1 11245.48 V2 11074.50 END 10203.85 LINE 0-P11 PI1-V1 V1-V2 V2-P13 P13-P14 P14-END CURVE 1stCurve 2nd Curve 3rd Curve 4th Curve 5th Curve PC PT PC2 PCC PT2 PC3 PRC PT3 AZIMUTH 1 STATIONING EASTINGS 8927.24 9291.91 9546.66 9282.44 9433.48 9490.41 9472.69 9016.99 DISTANCE R T LC 13 R5 R1 11 R3 P130 PC3 PT3 PT1 R2 1+140 LC4 LC5 12 PRC LC2 Lc3 PT2 PC2 R4 14 P14 V1 PCC V2

STATION NORTHINGS 0+000 11475.55 PI1 11478.80 11179.37 P12 P13 10804.76 P14 10512.37 V1 11245.48 V2 11074.50 END 10203.85 LINE 0-P11 PI1-V1 V1-V2 V2-P13 P13-P14 P14-END CURVE 1stCurve 2nd Curve 3rd Curve 4th Curve 5th Curve PC PT PC2 PCC PT2 PC3 PRC PT3 AZIMUTH 1 STATIONING EASTINGS 8927.24 9291.91 9546.66 9282.44 9433.48 9490.41 9472.69 9016.99 DISTANCE R T LC 13 R5 R1 11 R3 P130 PC3 PT3 PT1 R2 1+140 LC4 LC5 12 PRC LC2 Lc3 PT2 PC2 R4 14 P14 V1 PCC V2

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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Concept explainers

Design of structures using FEA

An arrangement, in which various members are interconnected with their constrained relative motion, the whole arrangement is then known as a structure. The degree of freedom of such a structure is zero. In such a structure, the load can act throughout the length of the structure or at the joints.

Question

Thank you for answering :)

PC1
PI1
0+000
Let
PT1
R1
11
STATION NORTHINGS
EASTINGS
0+000
11475.55
8927.24
PC2
P11
11478.80
9291.91
PI2
11179.37
9546.66
R2
V1
PI3
10804.76
9282.44
Lc2
PI4
10512.37
9433.48
13
12
V1
11245.48
9490.41
PCC
PI2
V2
11074.50
9472.69
9016.99
END
10203.85
Lc3
R3
V2
LINE
AZIMUTH
DISTANCE
0-PI1
PI1-V1
V1-V2
PT2
V2-PI3
PI3-P14
PI4-END
PC3
R4
CURVE
R
T.
LC
1+140
PI3
1stCurve
2nd Curve
3rd Curve
4th Curve
14
Lc4
5th Curve
PRC
STATIONING
15
PC
PT
Le5
PC2
PI4
PCC
R5
PT2
PC3
PRC
PT3
PT3
END
END OF ROAD

Shown on the drawing are the application of the different horizontal curves for a
roadway project.
GIVEN:
Coordinates of Intersecting Points
Length of Chord of the First Simple is 132.24 m
Degree of Curve of the first curve of the compound Curve is 8 degrees
The deflection angle from PC3 to Sta. 1+140 is 15°28'
REQUIRED:
1. The azimuth and distance of the tangent lines.
2. The Intersecting Angle (1), radius (R), tangent distance (T), Length of Curve (Lc) of
all the curves
3. The Stationing of PC, PT, PC2, PCC, PT2, PC3, PRC, PT3 and End of Project
Fill up your answer on the given table.
Show solutions.
IFICE
VESITY

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Step 4: Calculation of curve

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