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A cantilever beam of rectangular cross-section is subjected to a load W at its free end. If the depth of the beam is doubled and the load is halved, the deflection of the free end as compared to original deflection will be:
- a)Half
- b)One-eighth
- c)One-sixteenth
- d)Double
Correct answer is option 'C'. Can you explain this answer?
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A cantilever beam of rectangular cross-section is subjected to a load ...
Deflectionin cantilever 
If h is doubled, and W is halved, New deflection 
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A cantilever beam of rectangular cross-section is subjected to a load ...
Explanation:
The formula for deflection of a cantilever beam is given by:
δ = (Wl^3) / (3EI)
where,
δ = deflection
W = load
l = length of beam
E = modulus of elasticity
I = moment of inertia
When the depth (height) of the beam is doubled, the moment of inertia (I) also doubles. This is because the moment of inertia is proportional to the cube of the height of the beam.
When the load is halved, the deflection becomes one-fourth of the original deflection. This is because the deflection is proportional to the load and inversely proportional to the cube of the length of the beam.
Therefore, the new deflection can be calculated as:
δ_new = (W/2) * (2l)^3 / (3E * 2b * (2h)^3)
δ_new = (Wl^3 / 8EI)
Comparing with the original deflection formula, we can see that the new deflection is one-sixteenth of the original deflection.
Therefore, the correct answer is option C, i.e., the deflection of the free end will be one-sixteenth of the original deflection.
The formula for deflection of a cantilever beam is given by:
δ = (Wl^3) / (3EI)
where,
δ = deflection
W = load
l = length of beam
E = modulus of elasticity
I = moment of inertia
When the depth (height) of the beam is doubled, the moment of inertia (I) also doubles. This is because the moment of inertia is proportional to the cube of the height of the beam.
When the load is halved, the deflection becomes one-fourth of the original deflection. This is because the deflection is proportional to the load and inversely proportional to the cube of the length of the beam.
Therefore, the new deflection can be calculated as:
δ_new = (W/2) * (2l)^3 / (3E * 2b * (2h)^3)
δ_new = (Wl^3 / 8EI)
Comparing with the original deflection formula, we can see that the new deflection is one-sixteenth of the original deflection.
Therefore, the correct answer is option C, i.e., the deflection of the free end will be one-sixteenth of the original deflection.
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A cantilever beam of rectangular cross-section is subjected to a load W at its free end. If the depth of the beam is doubled and the load is halved, the deflection of the free end as compared to original deflection will be:a)Halfb)One-eighthc)One-sixteenthd)DoubleCorrect answer is option 'C'. Can you explain this answer?
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