Question

A circular rod of 100 mm diameter and 500 m long is subjected to a tensile force of 1000KN. Determine the modulus of rigidity, bulk modulus and change in volume if poisons ratio = 0.3 and Young'* Modulus = 2 × 105 N/mm2.

Answer 1

Modulus of rigidity G = E/2.(1 + µ) = 2 × 105/ 2(1 + 0.3) = 0.769 × 105 N/mm2

Bulk modulus K = E/3(1 – 2µ) = 2 × 105/3(1 – 2 × 0.3) = 1.667 × 105 N/mm2

Normal stress σ = P/A = 1000 × 103 / π/4(100)2 = 127.388 N/mm2

Linear (Longitudinal ) strain = δL/L = Normal stress/ Young'* modulus

= 127.388/ 2 × 105 = 0.000637

Diametral (Lateral) strain

= δd/d = µ.δL/L = 0.3 × 0.000637 = 0.0001911

Now volume of a circular rod = V = π/4.d2.L

Upon differentiation

δV = π/4[ 2.d.δd.L + d2.δL]

Volumetric strain

δV/V = π/4[ 2.d.δd.L + d2.δL]/ π/4.d2.L = 2δd/d + δL/L

Substituting the value of δd/d and δL/L as calculated above, we have

δV/V = 2 (-0.0001911) + 0.000637 = 0.0002548

The -ive sign with δd/d stems from the fact that whereas the length increases with tensile force, there is decrease in diameter.

Change in volume

δV = 0.0002548 [π/4(100)2 × 500] = 1000.09 mm3.