A rod 300 cm long and of diameter 4.0 cm is subjected to an axial pull of 20 KN. If the modulus of Elasticity of the material of the rod is 4.0 x 105 N/mm2; Determine:
(i) The Stress,
(ii) The Strain, and
(iii) The Elongation of the rod.
Sol:
Given, Length of the rod, L = 300 cm,
Diameter, D = 4.0 cm = 40 mm,
Area, A = (π/4) D2 = (π/4) 402 = 400 π mm2
Axial pull, P = 20 KN = 20,000 N
Modulus of Elasticity, E = 4.0 x 105 N/mm2
(i) The Stress (σ) is given by equation as
σ = P/A = 20,000/400 π = 15.92 N/mm2
(ii) Using the equation, the strain is obtained as
E = σ/e
Strain e = σ/E = 15.92 / 4.0 x 105 = 0.0000398
(iii) Elongation is obtained by using equation as
e = dL/L
Elongation, dL = e x L = 0.0000398 x 300 = 0.01194 cm
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