Question

find the total extension of a solid bar of
length 500mm and 50 mm Diameter with
Central anis tensile force of 400 KN - Assume
E = 200 GPa .​

Answer 1

The total extension of a solid bar is 2 × 10⁻⁶

Explanation:

On applying Hooke's law, the solid extension is given by the formula:

δ = (Pl)/(AE)

Where,

P = Force applied on the object = 400 KN = 4 × 10⁵ N (given)

l = Length of the solid bar = 500 mm = 0.5 m (given)

A = Area of the solid bar = 0.012 m² (given)

E = Young's modulus of the solid bar = 200 GPa = 2 × 10¹¹ (given)

On substituting the values, we get,

δ = (4 × 10⁵ × 0.5)/(0.012 × 2 × 10¹¹)

δ = (2 × 10⁵)/(1 × 10¹¹)

∴ δ = 2 × 10⁻⁶