Question

Aluminium alloy after failure has a final length of 2.195 in and a final diameter of 0.398 in at the fractured surface. The percentage of elongation and reduction in area is:

• A. Elongation = 9.75%, Reduction in area = 37.9%
• B. Elongation = 19.75%, Reduction in area = 37.9%
• C. Elongation = 9.75%, Reduction in area = 47.9%
• D. Elongation = 19.75%, Reduction in area = 47.9%

Hint:

When calculating elongation or reduction in area, ensure consistent units. For elongation, always compare the final length to the original length. For area reduction, use cross-sectional diameters to determine the change in area.

Answer 1

The Correct Option is A

Step 1: Calculate Percentage Elongation}
Elongation is given by:
\[\text{Elongation} (\%) = \frac{\text{Final Length} - \text{Original Length}}{\text{Original Length}} \times 100\]
Given: Original length = 2.000 in, Final length = 2.195 in:
\[\text{Elongation} (\%) = \frac{2.195 - 2.000}{2.000} \times 100 = 9.75\%\]
{Step 2: Calculate Reduction in Area}
Reduction in area is given by:
\[\text{Reduction in Area} (\%) = \frac{\text{Original Area} - \text{Final Area}}{\text{Original Area}} \times 100\]
Original diameter = 0.500 in, Final diameter = 0.398 in:
\[\text{Original Area} = \pi \left( \frac{0.500}{2} \right)^2 = 0.196 \text{ in}^2, \quad \text{Final Area} = \pi \left( \frac{0.398}{2} \right)^2 = 0.124 \text{ in}^2\]
\[\text{Reduction} (\%) = \frac{0.196 - 0.124}{0.196} \times 100 = 37.9\%\]