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 4mm, the true strain is
- A. 0.56
- B. 0.22
- C. 0.25
- D. 0.45
Answer: D. 0.45
Explanation: True strain = 2 ln(d0/d) = 2 ln(5/4) = 2 * 0.223 = 0.446.