Question

An aluminum alloy sample has a yield stress of 250 MPa and a modulus of elasticity of 70 GPa. What is the strain at yield point?

• A. 0.00357
• B. 0.0357
• C. 0.000357
• D. 0.357

Hint:

Elastic Strain. Within the elastic limit, Strain = Stress / Modulus of Elasticity (\(\epsilon = \sigma / E\)). Ensure consistent units for stress (Pa) and modulus (Pa). MPa = 10\(^6\) Pa, GPa = 10\(^9\) Pa.

Answer 1

The Correct Option is A

Within the elastic region, stress (\(\sigma\)) and strain (\(\epsilon\)) are related by Hooke's Law: $$ \sigma = E \epsilon $$ where E is the Modulus of Elasticity (Young's Modulus).
The yield point marks the end of the elastic region.
The strain at the yield point (\(\epsilon_y\)) corresponds to the yield stress (\(\sigma_y\)).
$$ \sigma_y = E \epsilon_y $$ Rearranging to find the yield strain: $$ \epsilon_y = \frac{\sigma_y}{E} $$ Given: Yield Stress \(\sigma_y = 250\) MPa = \(250 \times 10^6\) Pa.
Modulus of Elasticity \(E = 70\) GPa = \(70 \times 10^9\) Pa.
Substitute the values: $$ \epsilon_y = \frac{250 \times 10^6 \, \text{Pa}}{70 \times 10^9 \, \text{Pa}} = \frac{250}{70 \times 10^3} = \frac{25}{7 \times 10^3} $$ $$ \epsilon_y = \frac{25}{7000} \approx 0.
003571(4).
.
$$ The strain at the yield point is approximately 0.
00357.
Strain is dimensionless.