A structural iron rod has a radius of 10 mm and a length of 1.0 m. A 1000 kN force stretches it along its length. Calculate 

(a) stress, 

(b) elongation, and 

(c) strain on the rod. Young’s modulus, of iron is 1.9 × 1011 N m–2.

Let us assume that the rod is fixed at one end and the force is applied on the other end parallel to the rod.

Stress on the rod:

Stress = ^F/_A = ^F/_{π r²}

= ^{100 × 103 N}/_{3.14 × 10^-2 × 10^-6 m²}

= 3.18 × 10⁸ N m^-2

Now, we know that Y = ^{F L}/_{A ΔL}

∴ ΔL = ^{F L}/_{A Y}

Therefore, the elongation is:

ΔL = ^{3.18 × 108 N m-2 × 1 m}/_{1.9 × 10¹⁰ N m^-2}

∴ ΔL = 16 mm

Therefore, the strain in the rod is:

Strain = ^ΔL/_L = ^{1.67 × 10-2 m}/_{1 m}

= 1.67%