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I tried to solve the problem using your solution and got a different answer. Also where is the angle in degrees made by cable with the horizontal at the supports?

I tried to solve the problem using your solution and got a different answer. Also where is the angle in degrees made by cable with the horizontal at the supports?

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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I tried to solve the problem using your solution and got a different answer. Also where is the angle in degrees made by cable with the horizontal at the supports?

4) Av = 20.828KN By = 20.828 KN FB D An ↑ Av с >H 20.4321 KN = Max TENSION =20.4321 KN L=150.9876m EMc = 0 (Av x 150.9876_) - (AH xh) - (275.890 x 150.9876 x 150.987 2 4 (20.828x 150.9876 2 = (Anh) - 786, 193 Ah Max Tension = √√ (Av ²+ An ²) √√(20.828 2 + 20.4321 2 (20.8282 + 2 (786.193). = 16.335 h 2 h² = 194.5223929 = 37838.961 m 376) -(AHxh)-(275.890x 2 786.193 h (786-193)2) (786.193)2) (N) 2 150.9876 2 x 2² ) = 150.9876 4 O =0
Transcribed Image Text:4) Av = 20.828KN By = 20.828 KN FB D An ↑ Av с >H 20.4321 KN = Max TENSION =20.4321 KN L=150.9876m EMc = 0 (Av x 150.9876_) - (AH xh) - (275.890 x 150.9876 x 150.987 2 4 (20.828x 150.9876 2 = (Anh) - 786, 193 Ah Max Tension = √√ (Av ²+ An ²) √√(20.828 2 + 20.4321 2 (20.8282 + 2 (786.193). = 16.335 h 2 h² = 194.5223929 = 37838.961 m 376) -(AHxh)-(275.890x 2 786.193 h (786-193)2) (786.193)2) (N) 2 150.9876 2 x 2² ) = 150.9876 4 O =0
An 4. A small suspension bridge for a walkway spans 150.9876 m between supports on the same level. Each of the two cables carries a uniform load of 28.1234 Kg/meter of horizontal walkway. If the maximum tension on the cable is KN at mid-span, the sag h found to be 20.4321 KN, compute the tension To = = 37838.961 meters and the angle = 16.335 degrees made by the cable with the horizontal at the supports. ↑ Av 28.1234 kg /m L = 150.9876m с h ≤FY = 0 Av + Bv = 275.890 x 150.9876 = 41.656 KN Av + By ↑ By Bh Uniform Load = 28.1234 kg/m 28.1234 x 9.81 = 275.890 N/m EMA = O (By x150.9876) = (275.890 x 150.9876x 150.9876- By = 20.828 KN sub By to E9 1 Av + 20.828 = 41.65 6 Av = 20.828 KN Continuation of SOLUTION on (сопп on) OTHER PAPER
Transcribed Image Text:An 4. A small suspension bridge for a walkway spans 150.9876 m between supports on the same level. Each of the two cables carries a uniform load of 28.1234 Kg/meter of horizontal walkway. If the maximum tension on the cable is KN at mid-span, the sag h found to be 20.4321 KN, compute the tension To = = 37838.961 meters and the angle = 16.335 degrees made by the cable with the horizontal at the supports. ↑ Av 28.1234 kg /m L = 150.9876m с h ≤FY = 0 Av + Bv = 275.890 x 150.9876 = 41.656 KN Av + By ↑ By Bh Uniform Load = 28.1234 kg/m 28.1234 x 9.81 = 275.890 N/m EMA = O (By x150.9876) = (275.890 x 150.9876x 150.9876- By = 20.828 KN sub By to E9 1 Av + 20.828 = 41.65 6 Av = 20.828 KN Continuation of SOLUTION on (сопп on) OTHER PAPER
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