Correct Answer - Option 4 : 1.75
Concept:
Stress in axial loading is given by;
\(σ=\frac{F}{A}\)
where σ = Normal stress, F = Load/Weight, A = Area of cross-section
\(A=\frac{\pi d^2}{4}\)
where d = Diameter
Calculation:
Given:
F = 2200 N, d = 4 cm = 40 mm
\(A=\frac{\pi\times (40)^2}{4}=1256.63\ mm^2\)
\(σ=\frac{2200}{1256.63}\)
σ = 1.75 N/mm2
Hence the stress induced in the metal rod will be 1.75 N/mm2.
