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If the depth of a simply supported rectangular beam subjected to concentric load at the Centre is doubled, the deflection at the Centre will be reduced to
- a)25%
- b)12.5%
- c)37.5%
- d)40%
Correct answer is option 'B'. Can you explain this answer?
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If the depth of a simply supported rectangular beam subjected to conc...
Given: Simply supported rectangular beam subjected to concentric load at the Centre
To find: The reduction in deflection at the Centre when the depth of the beam is doubled
Solution:
Let us consider a simply supported rectangular beam of length L, breadth B and depth d subjected to a concentric load P at the Centre. The deflection at the Centre can be calculated using the formula:
δ = (P*L^3)/(48*E*I)
where E is the modulus of elasticity of the material and I is the moment of inertia of the cross-section of the beam.
Now, let us assume that the depth of the beam is doubled to 2d. The new moment of inertia I' can be calculated as:
I' = (B*(2d)^3)/12 = 8*I
Therefore, the new deflection δ' can be calculated as:
δ' = (P*L^3)/(48*E*I')
δ' = (P*L^3)/(48*E*8*I)
δ' = δ/8
Hence, the deflection at the Centre is reduced to 1/8th of its original value when the depth of the beam is doubled.
To calculate the percentage reduction in deflection, we can use the formula:
Percentage reduction = (Original value - New value)/Original value * 100
Percentage reduction = (δ - δ')/δ * 100
Percentage reduction = (δ - δ/8)/δ * 100
Percentage reduction = 87.5%
Therefore, the correct answer is option B (12.5%).
To find: The reduction in deflection at the Centre when the depth of the beam is doubled
Solution:
Let us consider a simply supported rectangular beam of length L, breadth B and depth d subjected to a concentric load P at the Centre. The deflection at the Centre can be calculated using the formula:
δ = (P*L^3)/(48*E*I)
where E is the modulus of elasticity of the material and I is the moment of inertia of the cross-section of the beam.
Now, let us assume that the depth of the beam is doubled to 2d. The new moment of inertia I' can be calculated as:
I' = (B*(2d)^3)/12 = 8*I
Therefore, the new deflection δ' can be calculated as:
δ' = (P*L^3)/(48*E*I')
δ' = (P*L^3)/(48*E*8*I)
δ' = δ/8
Hence, the deflection at the Centre is reduced to 1/8th of its original value when the depth of the beam is doubled.
To calculate the percentage reduction in deflection, we can use the formula:
Percentage reduction = (Original value - New value)/Original value * 100
Percentage reduction = (δ - δ')/δ * 100
Percentage reduction = (δ - δ/8)/δ * 100
Percentage reduction = 87.5%
Therefore, the correct answer is option B (12.5%).
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If the depth of a simply supported rectangular beam subjected to concentric load at the Centre is doubled, the deflection at the Centre will be reduced toa) 25%b) 12.5%c) 37.5%d) 40%Correct answer is option 'B'. Can you explain this answer?
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