Question:

A cylindrical specimen is subjected to plastic deformation in tension up to a uniform elongation of 10%. The final cross-sectional area of the gage section is found to be 20 mm\(^2\). The initial cross-sectional area of the gage section is _________ mm\(^2\) (rounded off to an integer).

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When dealing with plastic deformation, the relationship between the initial and final areas can be calculated using the strain (which is the change in length or area divided by the original length or area). Make sure to account for the reduction in area due to the strain.
Updated On: Apr 29, 2025
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Solution and Explanation

The elongation (or strain) is given as 10%, which means the final area is 90% of the original area (because 10% elongation corresponds to 10% reduction in cross-sectional area). Let \( A_0 \) be the initial area and \( A_f \) be the final area. We have: \[ A_f = A_0 \times (1 - {strain}) \] Substituting the values: \[ 20 = A_0 \times (1 - 0.10) \] \[ 20 = A_0 \times 0.90 \] Solving for \( A_0 \): \[ A_0 = \frac{20}{0.90} = 22.22 \, {mm}^2 \] Thus, the initial cross-sectional area of the gage section is 22 mm\(^2\). 
Answer: 22 mm\(^2\).

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