Question

A bar of 1.2 cm diameter gets stretched by 0.3 cm under a steady load of 8KN. What stress would be produced in the bar by a weight of 0.8KN. Which falls through 8 cm before commencing the stretching of the rod, which is initially unstressed. Take E = 200GN/m2.

Answer 1

Cross-sectional area of the bar = A = π/4.d2 = π/4(1.2/1000)2 = 0.0001131 m2

Steady load = 8 kN

Elongation under steady load, δL = 0.3 cm = 0.003 m

Falling load = 0.8 kN

Distance through which the weight falls, h = 8 cm = 0.08 m

Modulus of elasticity, E = 200 GN/m2

Instantaneous stress produced due to the falling load, σi = ?

In order to first find length of the bar, using the following relation, we have

Now to calculate instantaneous stress si due to falling load 0.8 kN using the following relation, we have

= 7073386.3 (1 + 23.119)

σi = 170.6 × 106 N/m2 or 170.6 MN/m2 .