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
Percentage elongation is determined by the formula:
\[\text{Elongation} (\%) = \frac{\text{Final Length} - \text{Original Length}}{\text{Original Length}} \times 100\]
Given an original length of 2.000 in and a final length of 2.195 in, the elongation is:
\[\text{Elongation} (\%) = \frac{2.195 - 2.000}{2.000} \times 100 = 9.75\%\]
Step 2: Calculate Percentage Reduction in Area
The formula for percentage reduction in area is:
\[\text{Reduction in Area} (\%) = \frac{\text{Original Area} - \text{Final Area}}{\text{Original Area}} \times 100\]
With an original diameter of 0.500 in and a final diameter of 0.398 in, the respective areas are:
\[\text{Original Area} = \pi \left( \frac{0.500}{2} \right)^2 = 0.196 \text{ in}^2, \quad \text{Final Area} = \pi \left( \frac{0.398}{2} \right)^2 = 0.124 \text{ in}^2\]
The percentage reduction is then calculated as:
\[\text{Reduction} (\%) = \frac{0.196 - 0.124}{0.196} \times 100 = 37.9\%\]