Question

A 5 mm diameter aluminium alloy test bar is subjected to a load of 500 N. If the diameter of the bar at this load is 4 mm, the true strain is
1. 0.56
2. 0.22
3. 0.25
4. 0.45

Answer 1

Correct Answer - Option 4 : 0.45

Concept:

Engineering strain: It is the ratio of change in dimension to the original dimension.

ϵe = \(\frac{{{\rm{\Delta }}L}}{L_0}\)

True strain: It is the ratio of instantaneous elongation to the instantaneous dimension.

\({\epsilon_t} = \ln \left( {\frac{{{A_0}}}{{{A_f}}}} \right)\)

Calculation:

Given:

d_o = 5 mm,d_f = 4 mm

⇒ \({\epsilon_t} = \ln \left( {\frac{{{A_0}}}{{{A_f}}}} \right) = 2\ln \left( {\frac{{{d_0}}}{{{d_t}}}} \right) = 2ln\left( {\frac{5}{4}} \right) = 0.446\)