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EХAMPLE 1: PI Given the following data of highway: D,=4° b-31. D,=30 1,= 28° Sta. Pl= 30+125.5 m Determine: Radius of the first Curve Radius of the second curve 1. 5. Long Chord 1 2. 6. Long Chord 2 РСС Sta. PCC Sta. PT 3. 7. Length of the first curve Length of the second curve 4. 8. T2 Lcz 5. Long chord 1, L4 L1 PC Using cosine law: PT L = R; + R – 2(R4)(R)cos l4 = 286.4789 m R, L = 184.8147 m = 381,9719 m 31° 28

EХAMPLE 1: PI Given the following data of highway: D,=4° b-31. D,=30 1,= 28° Sta. Pl= 30+125.5 m Determine: Radius of the first Curve Radius of the second curve 1. 5. Long Chord 1 2. 6. Long Chord 2 РСС Sta. PCC Sta. PT 3. 7. Length of the first curve Length of the second curve 4. 8. T2 Lcz 5. Long chord 1, L4 L1 PC Using cosine law: PT L = R; + R – 2(R4)(R)cos l4 = 286.4789 m R, L = 184.8147 m = 381,9719 m 31° 28

Structural Analysis
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
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EХAMPLE 1: PI Given the following data of highway: D,=4° k=31° D,=30 1,= 28° Sta. Pl= 30+125.5 m Determine: 1. Radius of the first Curve 5. Long Chord 1 2. Radius of the second curve 6. Long Chord 2 РСС Sta. PCC Sta. PT 3. 7. Length of the first curve Length of the second curve 4. 8. T2 Lcz 5. Long chord 1, L L1 PC Using cosine law: PT L = R; + R; – 2(R4)(R1)cos I4 = 286. 4789 m R, L = 184.8147 m a = 381. 9719 m 319 28
Transcribed Image Text:EХAMPLE 1: PI Given the following data of highway: D,=4° k=31° D,=30 1,= 28° Sta. Pl= 30+125.5 m Determine: 1. Radius of the first Curve 5. Long Chord 1 2. Radius of the second curve 6. Long Chord 2 РСС Sta. PCC Sta. PT 3. 7. Length of the first curve Length of the second curve 4. 8. T2 Lcz 5. Long chord 1, L L1 PC Using cosine law: PT L = R; + R; – 2(R4)(R1)cos I4 = 286. 4789 m R, L = 184.8147 m a = 381. 9719 m 319 28
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