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A two hinged semi-circular arch of radius R carries a concentrated load W at the crown. The horizontal thrust is
- a)W2/π
- b)2W/3π
- c)W/π
- d)4W/3π
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
A two hinged semi-circular arch of radius R carries a concentrated loa...
H = W/π sin2α
If α = 90∘, H = W/π
Most Upvoted Answer
A two hinged semi-circular arch of radius R carries a concentrated loa...
2Rb)Wc)2W/Rd)W/R2
The correct answer is b) W.
Explanation:
For a two-hinged semi-circular arch, the horizontal thrust at the supports is equal to the load W. This can be derived using the principle of virtual work or the equilibrium equations.
Using the principle of virtual work, we can consider a small displacement of the arch due to the load W. The work done by the load is equal to the work done by the horizontal thrust at the supports. Since the arch is symmetric, the horizontal displacement at the supports is zero. Therefore, the work done by the horizontal thrust is also zero. This implies that the horizontal thrust is equal to the load W.
Using the equilibrium equations, we can consider the moments about one of the supports. The weight of the arch is balanced by the horizontal thrust and the reaction at the support. The moment due to the weight is zero since the weight acts through the center of curvature. The moment due to the horizontal thrust is W*R, where R is the radius of curvature. Therefore, the reaction at the support is also W.
Hence, the horizontal thrust at the supports of a two-hinged semi-circular arch carrying a concentrated load at the crown is equal to the load itself, which is W.
The correct answer is b) W.
Explanation:
For a two-hinged semi-circular arch, the horizontal thrust at the supports is equal to the load W. This can be derived using the principle of virtual work or the equilibrium equations.
Using the principle of virtual work, we can consider a small displacement of the arch due to the load W. The work done by the load is equal to the work done by the horizontal thrust at the supports. Since the arch is symmetric, the horizontal displacement at the supports is zero. Therefore, the work done by the horizontal thrust is also zero. This implies that the horizontal thrust is equal to the load W.
Using the equilibrium equations, we can consider the moments about one of the supports. The weight of the arch is balanced by the horizontal thrust and the reaction at the support. The moment due to the weight is zero since the weight acts through the center of curvature. The moment due to the horizontal thrust is W*R, where R is the radius of curvature. Therefore, the reaction at the support is also W.
Hence, the horizontal thrust at the supports of a two-hinged semi-circular arch carrying a concentrated load at the crown is equal to the load itself, which is W.
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A two hinged semi-circular arch of radius R carries a concentrated load W at the crown. The horizontal thrust isa)W2/πb)2W/3πc)W/πd)4W/3πCorrect answer is option 'C'. Can you explain this answer?
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