A steel bar (area A_s, modulus E_s, coefficient α_s) and a copper bar (area A_c, modulus E_c, coefficient α_c) of equal initial length L are rigidly connected in parallel and subjected to a uniform temperature rise of t°. Find the common extension δ at temperature t°.
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Aδ = L t * (α_s + α_c) / 2
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Bδ = L t * (α_s E_s + α_c E_c) / (E_s + E_c)
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Cδ = L t * (α_s E_s A_s + α_c E_c A_c) / (E_s A_s + E_c A_c)
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Dδ = L t * (α_s A_s + α_c A_c) / (A_s + A_c)
Answer
Correct Answer: δ = L t * (α_s E_s A_s + α_c E_c A_c) / (E_s A_s + E_c A_c)
Explanation
Introduction / Context:Compound (composite) bars of dissimilar materials bonded in parallel develop internal self-equilibrating forces under temperature change because each material tends to expand by a different free amount. Compatibility enforces a common extension, while equilibrium requires net axial force to vanish when there is no external axial load.
Given Data / Assumptions:
- Bars in parallel: same initial length L, rigidly connected at ends.
- Material data: A_s, E_s, α_s for steel; A_c, E_c, α_c for copper.
- Uniform temperature rise t°, no external axial load.
- Elastic, small-strain behavior.
Concept / Approach:Let ε be the common strain, δ = ε*L. Free thermal strains would be α_s t and α_c t. Actual stresses: σ_s = E_s(ε − α_s t), σ_c = E_c(ε − α_c t). Force equilibrium (no external load) gives σ_s A_s + σ_c A_c = 0. Solve for ε, then δ.
Step-by-Step Solution:1) Write stresses: σ_s = E_s(ε − α_s t), σ_c = E_c(ε − α_c t).2) Equilibrium: E_s A_s (ε − α_s t) + E_c A_c (ε − α_c t) = 0.3) Factor ε: ε (E_s A_s + E_c A_c) = t (E_s A_s α_s + E_c A_c α_c).4) Solve: ε = t (E_s A_s α_s + E_c A_c α_c) / (E_s A_s + E_c A_c).5) Extension: δ = ε * L = L t * (α_s E_s A_s + α_c E_c A_c) / (E_s A_s + E_c A_c).
Verification / Alternative check:Special case A_s E_s ≫ A_c E_c → δ ≈ L t α_s (steel dominates); conversely if A_c E_c ≫ A_s E_s → δ ≈ L t α_c (copper dominates). This limiting behavior validates the formula.
Why Other Options Are Wrong:
- Arithmetic averages (A, D) ignore stiffness weighting.
- Option B ignores areas; stiffness depends on E*A, not just E.
Common Pitfalls:
- Forgetting that the net axial force must be zero (no external axial load).
- Using α-weighted averages without E*A weighting.
Final Answer:δ = L t * (α_s E_s A_s + α_c E_c A_c) / (E_s A_s + E_c A_c).