A rod of length 2 m is at a temperature of 20^@ C. find the free expansion of the rod, if the temperature is increased to 50^@C, then find stress produced when the rod is (i) fully prevented to expand, (ii) permitted to expand by 0.4mm. Y=2xx10^(11)N//m^(2), alpha=15xx10^(-6//^(@))C.
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Explanation:
Free expansion of the rod =αLΔθ
=15×10^−60C×2m×(50−20)°C
=9×10^-4m=0.9mm
If the expansion is fully prevented, then
Strain= {9×10^−4}/2
=4.5×10^−4
Temperature Stress = Strain ×Y
=4.5×10^−4 ×2×10^11
=9×10^7 N/m^2
If 0.4 mm expansion is allowed, then length restricted to expand
=0.9−0.4=0.5m
Strain= {5×10^−4}/2
=2.5×10^−4
Temperature stress = Strain×Y=2.5×10^−4×2×10^11
=5×10^7N/m^2
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