Question

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.

Answer 1

Step-by-step explanation:

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Answer 2

(a) Elongation of rod = 0.19732 mm

(b) Change in diameter = 0.00651 mm

Step-by-step explanation:

Circular area at end of steel rod = pi * diameter^2 / 4

`Area = pi * (22 * 10^-^3)/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