Find out stress - strain and elongation of bar:

A circular rod having diameter 20 m and 500 m long is subjected to the tensile force of 45kN. The modulus of elasticity for steel can be taken as200 kN/m2. Find out stress, strain and elongation of bar because of applied load.            

Sol.: Given:

D = 20m

L = 500m

P = 45KN = 45000N

E = 200KN/m² = (200 × 1000 N/mm²  = 200000 N/m²

By using the relation; σ = P/A  = P/(Π /4 × D²)

σ = 45000/( Π /4 × 20²)

σ = 143.24 N/m²                                                                                    .......ANS

E = σ / e

200000 = 143.24 /e

e = 0.000716                                                               .......ANS

Now, e = dL_A/L

0.000716 = dL_A/500

dL_A = 0.36                                                                        .......ANS