The stress in a steel rod for an extension of 0.2% of its length is —

Take Young’s modulus E = 2 × 10⁵ N/mm²

Question

Question

This question was previously asked in

UKPSC JE Civil 8 May 2022 Official Paper-I

Answer 1

Concept:

Stress, σ = \(\frac {F}{A}\)

Strain, ϵ = \(\frac {\Delta L}{L}\)

Young's Modulus, E = \(\frac {σ }{ϵ}\)

A = Area of cross-section

Calculation:

Given:

Strain = 0.2 % of length = 0.002, E = 2 × 10⁵ MPa

Using, E = stress/strain

\(2 \times 10^5 = \frac {\sigma }{0.002} = 400 N/mm^2\)