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A beam of length L, simply supported at ends, carries a load W at a distance of a from one end and at a distance of b from other end such that a + b = L. If El is flexural rigidity of the beam, how much strain energy is absorbed in beam?
- a)W2ab3/3EI
- b)W2a2b3/6EIL
- c)Wa3b/6EIL
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
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A beam of length L, simply supported at ends, carries a load W at a di...
Problem Statement:
A beam of length L, simply supported at ends, carries a load W at a distance of a from one end and at a distance of b from the other end such that a+b=L. If E is the flexural rigidity of the beam, how much strain energy is absorbed in the beam?
Solution:
The strain energy stored in a beam is given by the expression:
U = (1/2) ∫M²/EI dx
where M is the bending moment, E is the modulus of elasticity, I is the moment of inertia, and x is the distance along the beam.
In this problem, the beam is simply supported at its ends and carries a load W at a distance of a from one end and at a distance of b from the other end such that a+b=L. The bending moment in the beam varies along its length and is given by:
M(x) = W(a-x) for 0 ≤ x ≤ a
M(x) = W(x-b) for a ≤ x ≤ L
Substituting these values of M into the expression for U, we get:
U = (1/2) ∫0^a (W(a-x))²/EI dx + (1/2) ∫a^L (W(x-b))²/EI dx
Solving the integrals, we get:
U = W²a³/6EI + W²b³/6EI
Substituting a+b=L, we get:
U = W²a²b³/6EIL
Therefore, the strain energy absorbed in the beam is given by option B, i.e., W²a²b³/6EIL.
A beam of length L, simply supported at ends, carries a load W at a distance of a from one end and at a distance of b from the other end such that a+b=L. If E is the flexural rigidity of the beam, how much strain energy is absorbed in the beam?
Solution:
The strain energy stored in a beam is given by the expression:
U = (1/2) ∫M²/EI dx
where M is the bending moment, E is the modulus of elasticity, I is the moment of inertia, and x is the distance along the beam.
In this problem, the beam is simply supported at its ends and carries a load W at a distance of a from one end and at a distance of b from the other end such that a+b=L. The bending moment in the beam varies along its length and is given by:
M(x) = W(a-x) for 0 ≤ x ≤ a
M(x) = W(x-b) for a ≤ x ≤ L
Substituting these values of M into the expression for U, we get:
U = (1/2) ∫0^a (W(a-x))²/EI dx + (1/2) ∫a^L (W(x-b))²/EI dx
Solving the integrals, we get:
U = W²a³/6EI + W²b³/6EI
Substituting a+b=L, we get:
U = W²a²b³/6EIL
Therefore, the strain energy absorbed in the beam is given by option B, i.e., W²a²b³/6EIL.
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A beam of length L, simply supported at ends, carries a load W at a distance of a from one end and at a distance of b from other end such that a + b = L. If El is flexural rigidity of the beam, how much strain energy is absorbed in beam?a)W2ab3/3EIb)W2a2b3/6EILc)Wa3b/6EILd)None of theseCorrect answer is option 'B'. Can you explain this answer?
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