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 = 20.828KNBy = 20.828 KNFB DAn ↑Avс>H20.4321 KN =Max TENSION =20.4321 KNL=150.9876mEMc = 0(Av x 150.9876_) - (AH xh) - (275.890 x 150.9876 x 150.98724(20.828x 150.98762=(Anh) - 786, 193AhMax Tension = √√ (Av ²+ An ²)√√(20.828 2 +20.4321 2(20.8282 +2(786.193). = 16.335h2h² = 194.5223929=37838.961 m376) -(AHxh)-(275.890x2786.193h(786-193)2)(786.193)2)(N)2150.98762x2² ) =150.98764O=0 An4. A small suspension bridge for a walkway spans 150.9876 m between supports on the same level. Each of the twocables carries a uniform load of 28.1234 Kg/meter of horizontal walkway. If the maximum tension on the cable isKN at mid-span, the sag hfound to be 20.4321 KN, compute the tension To == 37838.961 meters and the angle =16.335degrees made by the cable with the horizontalat the supports.↑Av28.1234 kg /mL = 150.9876mсh≤FY = 0Av + Bv = 275.890 x 150.9876= 41.656 KNAv + By↑ByBhUniform Load = 28.1234 kg/m28.1234 x 9.81 = 275.890 N/mEMA = O(By x150.9876) = (275.890 x 150.9876x 150.9876-By = 20.828 KNsub By to E9 1Av + 20.828 = 41.65 6Av = 20.828 KNContinuation of SOLUTION on(соппon)OTHER PAPER

To Identify the difference in the solutions

Answered: I tried to solve the problem using your…

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

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Question

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

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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