A rod 150 cm long and of diameter 2.0 cm is subjected to an axial pull of 20kN. If the modulus of elasticity of the material of the rod is 2 x 10^5 N/Sq(mm) ; determine : the elongation in rod?

Detailed Solution:

elongation of rod = original length x strain, we know that original length is equal to 150cm strain = stress/modulus of elasticity = force/ Area x modulus of elasticity = 20000/(0.785 x 20 x 20 x 200000) = 0.000318 by putting value of strain in elongation formula we get elongation of rod = strain x original length = 0.000318 x 150 = 0.0477 cm