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's 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's 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.