A uniform metal rod fixed at its ends of 2 mm2 cross-section is cooled from 40°C to 20°C. The coefficient of the linear expansion of the rod is 12 × 10–6 per degree & it’s young modulus of elasticity is 1011 N/m2. The energy stored per unit volume of the rod is :

Engineering

Physics

Stress Strain and Hookes Law New

Question

A uniform metal rod fixed at its ends of 2 mm² cross-section is cooled from 40°C to 20°C. The coefficient of the linear expansion of the rod is 12 × 10^–6 per degree & it’s young modulus of elasticity is 10¹¹ N/m². The energy stored per unit volume of the rod is :

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When a constrained rod cools, it cannot contract freely, so thermal stress develops. The energy stored per unit volume (u) is given by u = (1/2) × Stress × Strain. Using Hooke's law, Stress = Y × Strain, where Y is Young's modulus. The thermal strain is αΔT, where α is the coefficient of linear expansion and ΔT is the temperature change. Therefore, the stress is YαΔT, and the strain is αΔT. Substituting into the energy formula:

$u = \frac{1}{2} \times (Y α ΔT) \times (α ΔT) = \frac{1}{2} Y α^{2} {(ΔT)}^{2}$

Given: Y = 10¹¹ N/m², α = 12×10^–6 /°C, ΔT = 20°C – 40°C = –20°C (we use the magnitude, 20°C).

$u = \frac{1}{2} \times 10^{11} \times {(12 \times 10^{- 6})}^{2} \times {(20)}^{2} = \frac{1}{2} \times 10^{11} \times 144 \times 10^{- 12} \times 400 = 2880 J/m³$

Final Answer: 2880 J/m³