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A simply supported beam of span ‘L’ carries a uniformly distributed load, 'w' per unit length over the entire span. The deflection at the center is 'y'. If the distributed load per unit length is doubled and also the depth of beam is doubled, the deflection at the center would be
  • a)
    2 y
  • b)
    4 y
  • c)
    0.5 y
  • d)
    0.25 y
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
A simply supported beam of span ‘L’ carries a uniformly distributed l...
δmax = y
w2 = 2w
d2 = 2d
y2 = y / 4
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Most Upvoted Answer
A simply supported beam of span ‘L’ carries a uniformly distributed l...
Given data:
Span of the beam = L
Uniformly distributed load = w per unit length
Deflection at the center = y

To find:
Deflection at the center when distributed load per unit length is doubled and the depth of the beam is also doubled.

Solution:
Let's consider the deflection formula for a simply supported beam with a uniformly distributed load:

y = (5*w*L^4)/(384*E*I)

where E is the modulus of elasticity and I is the moment of inertia of the cross-section.

When the distributed load per unit length is doubled, the new load w' = 2w.

When the depth of the beam is doubled, the new moment of inertia I' = 2^4*I = 16I.

Therefore, the new deflection y' can be calculated as:

y' = (5*w'*L^4)/(384*E*I')
= (5*2w*L^4)/(384*E*16I)
= (5/768)*(w*L^4)/(E*I)

Now, we can see that the new deflection y' is 1/16th of the original deflection y.

Therefore, the correct answer is option D, i.e., the deflection at the center would be 0.25 y.
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A simply supported beam of span ‘L’ carries a uniformly distributed load, 'w' per unit length over the entire span. The deflection at the center is 'y'. If the distributed load per unit length is doubled and also the depth of beam is doubled, the deflection at the center would bea) 2 yb) 4 yc) 0.5 yd) 0.25 yCorrect answer is option 'D'. Can you explain this answer?
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