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The document provides the solution to determining the reaction forces at supports A and C (Ay=15.46 kN, Cy=9.545 kN) and the tension in the cable (TA=4.319 kN) for a tied three-hinged arch subjected to 15 kN and 10 kN loads. The solution involves writing the equations of equilibrium for the entire arch and individual members AB and BC.

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MarielVillaluna
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226 views3 pages

First

The document provides the solution to determining the reaction forces at supports A and C (Ay=15.46 kN, Cy=9.545 kN) and the tension in the cable (TA=4.319 kN) for a tied three-hinged arch subjected to 15 kN and 10 kN loads. The solution involves writing the equations of equilibrium for the entire arch and individual members AB and BC.

Uploaded by

MarielVillaluna
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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Example 5-8

The tied three-hinged arch is subjected to the loading shown. Determine the
components of reaction at A and C and the tension in the cable.

15 kN B
10 kN
2m
A D

2m 2m
0.5 m 1m

39
SOLUTION
15 kN B
10 kN
2m
A
Ax D
0
Ay C

2m 2m Cy
0.5 m 1m

Entire arch :

+ ΣMA = 0: C y (5.5) − 10( 4.5) − 15(0.5) = 0

Cy = 9.545 kN

+ ΣFy = 0: Ay − 15 − 10 + 9.545 = 0

Ay = 15.46 kN
40
15 kN B B
Bx Bx
10 kN
2m By By
A
TA TD D

Ay = 15.46 kN C
Cy = 9.545 kN
2m 2m
0.5 m 1m

Member AB :

+ ΣMB = 0: 15( 2) − 15.455(2.5) + TA (2) = 0 TA = 4.319 kN

+ ΣFy = 0: 15.455 − 15 − B y = 0 By = 0.455 kN

+ ΣF = 0: 4.319 − Bx = 0
x Bx = 4.319 kN
Member AB :
+ ΣF = 0: 4.319 − TD = 0
x TD = 4.319 kN 41

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