To calculate the strain, we first need to determine the stress applied to the aluminum specimen, and then use Young's modulus to relate stress and strain.

Step 1: Calculate Stress

Stress ($\sigma$) is defined as force divided by the cross-sectional area: $\sigma = \frac{F}{A}$ Where:

• $F$ is the applied force (35,500 N)
• $A$ is the cross-sectional area

The cross-sectional area $A$ is calculated by multiplying the width and height of the specimen: $A = 10 \, \text{mm} \times 12.7 \, \text{mm} = 127 \, \text{mm}^2 = 127 \times 10^{-6} \, \text{m}^2 = 1.27 \times 10^{-4} \, \text{m}^2$

Now calculate the stress: $\sigma = \frac{35,500 \, \text{N}}{1.27 \times 10^{-4} \, \text{m}^2} = 279,527,559 \, \text{Pa} = 279.53 \, \text{MPa}$

Step 2: Use Young's Modulus to Find Strain

Strain ($\epsilon$) is defined as the stress divided by Young's modulus ($E$): $\epsilon = \frac{\sigma}{E}$

For aluminum, Young's modulus $E$ is approximately $70 \, \text{GPa} = 70 \times 10^9 \, \text{Pa}$.

Now, calculate the strain: $\epsilon = \frac{279.53 \times 10^6 \, \text{Pa}}{70 \times 10^9 \, \text{Pa}} = 0.003993 \, \text{(dimensionless)}$

Result:

The resulting strain is approximately $0.00399$, or 0.399%.

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Would you like further details or have any questions?

Here are 5 related questions:

1. How would the strain change if the force applied were doubled?
2. What is the significance of elastic deformation in materials?
3. How is the concept of stress related to tensile strength?
4. What factors affect Young's modulus in different materials?
5. How does strain differ in elastic and plastic deformation?

Tip: Always check the material’s properties before calculating strain to ensure that deformation remains in the elastic region.