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When the depth of a simply supported beam carrying a concentrated load at the midspan is doubled, the deflection of the beam at midspan will change by a change of
- a)1/6
- b)1/4
- c)1/2
- d)1/8
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
When the depth of a simply supported beam carrying a concentrated load...
Deflection at the center of S. S is WL^3/48Ei if depth is double it mean L = 2L then the formula become W (2L) ^3/48EI. The divided first term which is WL^3/48EI to W (2L) ^3/48EI. It becomes 1/8.
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When the depth of a simply supported beam carrying a concentrated load...
Given:
Depth of beam, d = d
Deflection of beam, y = ?
Change in depth of beam, d' = 2d
Solution:
We know that the deflection of a simply supported beam with a concentrated load at midspan is given by,
y = (P x L^3) / (48 x E x I)
Where,
P = Load applied at midspan
L = Span of the beam
E = Modulus of elasticity of the material
I = Moment of inertia of the beam
Now, when the depth of the beam is doubled, the moment of inertia of the beam will increase by a factor of 2^4 = 16.
So, the new moment of inertia of the beam will be,
I' = 16 x ((d')^4 / 12)
= 16 x ((2d)^4 / 12)
= 16 x (16d^4 / 12)
= 64 x (d^4 / 3)
= (64/3) x (d^4)
Now, substituting this value of I' in the deflection equation, we get,
y' = (P x L^3) / (48 x E x (64/3) x (d^4))
= (P x L^3 x 3) / (48 x E x 64 x (d^4))
= (P x L^3) / (3072 x E x (d^4))
So, the new deflection of the beam will be,
y' = (1/8) x y
Therefore, the change in deflection of the beam will be,
Δy = y' - y
= (1/8) x y - y
= -(7/8) x y
= -0.875y
≈ -1/8y
Hence, the correct option is D.
Depth of beam, d = d
Deflection of beam, y = ?
Change in depth of beam, d' = 2d
Solution:
We know that the deflection of a simply supported beam with a concentrated load at midspan is given by,
y = (P x L^3) / (48 x E x I)
Where,
P = Load applied at midspan
L = Span of the beam
E = Modulus of elasticity of the material
I = Moment of inertia of the beam
Now, when the depth of the beam is doubled, the moment of inertia of the beam will increase by a factor of 2^4 = 16.
So, the new moment of inertia of the beam will be,
I' = 16 x ((d')^4 / 12)
= 16 x ((2d)^4 / 12)
= 16 x (16d^4 / 12)
= 64 x (d^4 / 3)
= (64/3) x (d^4)
Now, substituting this value of I' in the deflection equation, we get,
y' = (P x L^3) / (48 x E x (64/3) x (d^4))
= (P x L^3 x 3) / (48 x E x 64 x (d^4))
= (P x L^3) / (3072 x E x (d^4))
So, the new deflection of the beam will be,
y' = (1/8) x y
Therefore, the change in deflection of the beam will be,
Δy = y' - y
= (1/8) x y - y
= -(7/8) x y
= -0.875y
≈ -1/8y
Hence, the correct option is D.
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When the depth of a simply supported beam carrying a concentrated load at the midspan is doubled, the deflection of the beam at midspan will change by a change ofa)1/6b)1/4c)1/2d)1/8Correct answer is option 'D'. Can you explain this answer?
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