A cylindrical specimen of steel having an original diameter of 11 mm is tensile tested to fracture and found to have engineering fracture strength of 400 MPa. If the cross sectional diameter of fracture is 10 mm, the true stress at fracture is

Question

Question

This question was previously asked in

ISRO VSSC Technical Assistant Mechanical 14 July 2021 Official Paper

Answer 1

Concept:

True stress at fracture,

True stress = engineering stress × (1+ engineering strain) i.e.

𝜎𝑡 = 𝜎0 (1 + 𝑒)

Or, \({\sigma _t} = {\sigma _f} \times \frac{{{A_i}}}{{{A_f}}}\)

Calculation:

Given: di = 11 mm, d_f = 10 mm, σf = 400 MPa

Now,

\({\sigma _t} = 400 \times {\left( {\frac{{11}}{{10}}} \right)^2}\)

∴ 𝜎t = 484 MPa.