A specimen of aluminum having a rectangular cross section 10 mm
x 12.7 mm (0.4 in.
x 0.5 in.) is pulled
in tension with 35,500 N (8000 lbf) force, producing only elastic deformation. Calculate the resulting strain

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.

A specimen of aluminum having a rectangular cross section 10 mm x 12.7 mm (0.4 in. x 0.5 in.) is pulled in tension with 35,500 N (8000 lbf) force, producing only elastic deformation. Calculate the resulting strain.

Learn how to calculate the strain in an aluminum specimen with a rectangular cross section under a tensile force of 35,500 N. This solution involves stress-strain relationships and Young's modulus, providing a detailed step-by-step calculation. Ideal for undergraduate engineering students.

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