Copper Elongation Calculation

This problem requires calculating the elongation of a copper piece under tensile stress, assuming the deformation remains within the elastic limit.

Problem Setup

• Original Length ($L_0$): 305 mm
• Applied Stress ($\sigma$): 276 MPa
• Young's Modulus for Copper ($E$): 110 GPa

Elastic Deformation Formula

For elastic deformation, the relationship between stress ($\sigma$), strain ($\epsilon$), and Young's modulus ($E$) is given by Hooke's Law: $ \sigma = E \epsilon $ Strain ($\epsilon$) is also defined as the change in length ($\Delta L$) divided by the original length ($L_0$): $ \epsilon = \frac{\Delta L}{L_0} $ Combining these equations, we get: $ \sigma = E \frac{\Delta L}{L_0} $ Rearranging the formula to solve for elongation ($\Delta L$): $ \Delta L = \frac{\sigma L_0}{E} $

Calculation Steps

1. Ensure Consistent Units: Convert Young's modulus to MPa to match the stress unit. Note that 1 GPa = 1000 MPa. $E = 110 \text{ GPa} = 110 \times 1000 \text{ MPa} = 110000 \text{ MPa}$
2. Substitute Values into the Formula: $ \Delta L = \frac{(276 \text{ MPa}) \times (305 \text{ mm})}{110000 \text{ MPa}} $
3. Compute the Elongation: $ \Delta L = \frac{84180}{110000} \text{ mm} $ $ \Delta L \approx 0.76527 \text{ mm} $
4. Round to Appropriate Precision: Rounding the result gives 0.77 mm.

Result

The resultant elongation of the copper piece is approximately 0.77 mm.