In a standard tensile test a steel rod of 22-mm diameter is subjected to a tension force of 75kN. Knowing that v = 0.30 and E =
200 GPa, determine (a) the elongation of the rod in a 200-mm gage length, (b) the change in diameter of the rod.
2 answers:
Answer:
(a) Elongation of rod = 0.19732 mm
(b) Change in diameter = 0.00651 mm
Explanation:
Circular area at end of steel rod = pi * diameter^2 / 4
Area = 3.801 * 10^(-4) meter squared
Stress = force / area
Stress = 75000 / (3.801 * 10^-4)
Stress = 197316495 Pa OR 0.197 GPa
Modulus of elasticity = stress / strain
200 = 0.19732 / Strain
Strain = 0.0009866 (longitudinal)
(a) Strain = change in length / initial length
0.0009866 = Elongation of rod / 200
Elongation of rod = 0.19732 mm
(b) Poisson ratio = lateral strain / longitudinal strain
0.3 = lateral strain / 0.0009866
Lateral strain = 0.000296
Lateral strain = Change in diameter / original diameter
0.000296 = Change in diameter / 22
Change in diameter = 0.00651 mm
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