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What will be the value of Ue if material is linear elastic? Axial force is increased from 0 to P gradually.
- a)1⁄4 P Δ
- b)1⁄3P Δ
- c)1⁄2P Δ
- d)P Δ
Correct answer is option 'C'. Can you explain this answer?
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What will be the value of Ue if material is linear elastic? Axial forc...
Answer: c
Explanation: Due to linear elasticity, we can substitute force in terms of P and Δ and then integrate wrt x to get the final answer.
Explanation: Due to linear elasticity, we can substitute force in terms of P and Δ and then integrate wrt x to get the final answer.
Most Upvoted Answer
What will be the value of Ue if material is linear elastic? Axial forc...
If the material is linear elastic and the axial force is increased gradually from 0 to P, the value of Ue will depend on the material's stiffness or Young's modulus (E), the initial length of the material (L0), and the change in axial force (ΔP).
The formula for strain (ε) in a linear elastic material is given by:
ε = ΔL / L0
Where ΔL is the change in length and L0 is the initial length.
The formula for stress (σ) in a linear elastic material is given by:
σ = P / A
Where P is the axial force and A is the cross-sectional area.
Since the material is linear elastic, stress is directly proportional to strain:
σ = E * ε
Combining the above equations, we get:
P / A = E * ΔL / L0
Solving for ΔL, we get:
ΔL = (P * L0) / (E * A)
The value of Ue is equal to the change in potential energy, which is given by:
Ue = (1/2) * P * ΔL
Substituting the value of ΔL, we get:
Ue = (1/2) * P * (P * L0) / (E * A)
Therefore, the value of Ue when the axial force is increased gradually from 0 to P in a linear elastic material is:
Ue = (1/2) * (P^2 * L0) / (E * A)
The formula for strain (ε) in a linear elastic material is given by:
ε = ΔL / L0
Where ΔL is the change in length and L0 is the initial length.
The formula for stress (σ) in a linear elastic material is given by:
σ = P / A
Where P is the axial force and A is the cross-sectional area.
Since the material is linear elastic, stress is directly proportional to strain:
σ = E * ε
Combining the above equations, we get:
P / A = E * ΔL / L0
Solving for ΔL, we get:
ΔL = (P * L0) / (E * A)
The value of Ue is equal to the change in potential energy, which is given by:
Ue = (1/2) * P * ΔL
Substituting the value of ΔL, we get:
Ue = (1/2) * P * (P * L0) / (E * A)
Therefore, the value of Ue when the axial force is increased gradually from 0 to P in a linear elastic material is:
Ue = (1/2) * (P^2 * L0) / (E * A)
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