Practice portal › Mechanical Properties of Solids › Stress, Strain and Hooke's Law
Asked in NEET 2024 · Thermal stress
Given: Y=0.5×10¹¹ N m⁻², α=10⁻⁵ °C⁻¹, L=1 m, A=10⁻³ m², Δ T=100 °C.
Idea: the clamps hold the length fixed, so the expansion the heating would have produced is undone by an elastic compression of exactly the same size.
Free expansion that is prevented: Δ L=Lα Δ T, so the strain forced on the bar is (Δ L)/L=α Δ T.
α Δ T=10⁻⁵×100=10⁻³
Stress: σ=Yα Δ T=0.5×10¹¹×10⁻³=0.5×10⁸ N m⁻²
Force: F=σ A=0.5×10⁸×10⁻³=0.5×10⁵ N
F=50×10³ N
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