Math, asked by renu7154, 4 months ago

Example 41 : A copper bar 16 mm Ø is heated through 50°C. If 50% of its elongation is prevented, find themagnitude and the nature of stress induced. Take E = 100 kN/mm², Qc = 18 x10C/C, L = 1200 mm.(V.V. Imp./4 M)

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Answered by Anonymous

Given data; L = = 16 mm d = 15mm ∆T = 50 C αC = 18.5 × 10–6 /°C EC = 1.25 × 105N/mm2 (1) Expansion when free to expand δL = α.∆T.L = 18.5 × 10–6 × 50 × 800 = 0.74mm (2) When the ends do not yield Thermal stress = σ = αE∆T = 18.5 × 10–6 × 1.25 × 105 × 50 σ = 115.63 N/mm2 P = σA = 115.63 × π/4(15)2 = 20.43KN (Compressive) (Since gripping is provided so the force is compressive) (3) The ends yield by 0.5mm σ = (α.L.∆T – ∆L).E/L = {(18.5 × 10–6 × 800 × 50 – 0.5)/800 } × 1.25 × 105 = 37.5 N/mm2 P = σA = 37.5 × π/4(15)2 = 6.63KN (Compressive)Read more on Sarthaks.com - https://www.sarthaks.com/514081/copper-rod-15-mm-diameter-long-is-heated-through-50c-what-is-its-expansion-when-free-expand?show=514099#a514099

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Example 41 : A copper bar 16 mm Ø is heated through 50°C. If 50% of its elongation is prevented, find themagnitude and the nature of stress induced. Take E = 100 kN/mm², Qc = 18 x10C/C, L = 1200 mm.(V.V. Imp./4 M)​

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Example 41 : A copper bar 16 mm Ø is heated through 50°C. If 50% of its elongation is prevented, find themagnitude and the nature of stress induced. Take E = 100 kN/mm², Qc = 18 x10C/C, L = 1200 mm.(V.V. Imp./4 M)