A tensile force of 200 kN is applied on a 0.02-m diameter and 2-m long rod. After applying the load, the diameter of the rod decreases to 0.01998 m and the length increases to 2.01 m. Assuming the material is within the linear elastic region, determine the modulus of elasticity.

We are given the following details about the rod:

- Applied tensile force: \( F = 200 \, \text{kN} = 200 \times 10^3 \, \text{N} \)
- Initial diameter: \( d_0 = 0.02 \, \text{m} \)
- Final diameter: \( d_f = 0.01998 \, \text{m} \)
- Initial length: \( L_0 = 2 \, \text{m} \)
- Final length: \( L_f = 2.01 \, \text{m} \)

We need to determine the modulus of elasticity (Young's modulus), \( E \).

### Step 1: Calculate the axial stress, \( \sigma \)
The formula for axial stress is:

\[
\sigma = \frac{F}{A}
\]

where \( F \) is the applied force and \( A \) is the cross-sectional area of the rod. The cross-sectional area \( A \) for a circular rod is:

\[
A = \frac{\pi d_0^2}{4}
\]

Substituting \( d_0 = 0.02 \, \text{m} \):

\[
A = \frac{\pi (0.02)^2}{4} = 3.1416 \times 10^{-4} \, \text{m}^2
\]

Now, calculate the stress \( \sigma \):

\[
\sigma = \frac{200 \times 10^3}{3.1416 \times 10^{-4}} \approx 636619772.4 \, \text{N/m}^2
\]

\[
\sigma \approx 636.62 \, \text{MPa}
\]

### Step 2: Calculate the axial strain, \( \varepsilon \)
Axial strain is the ratio of change in length to the original length:

\[
\varepsilon = \frac{\Delta L}{L_0} = \frac{L_f - L_0}{L_0}
\]

Substitute \( L_f = 2.01 \, \text{m} \) and \( L_0 = 2 \, \text{m} \):

\[
\varepsilon = \frac{2.01 - 2}{2} = \frac{0.01}{2} = 0.005
\]

### Step 3: Calculate the modulus of elasticity, \( E \)
Young's modulus \( E \) is the ratio of stress to strain:

\[
E = \frac{\sigma}{\varepsilon}
\]

Substitute \( \sigma = 636.62 \, \text{MPa} \) and \( \varepsilon = 0.005 \):

\[
E = \frac{636.62}{0.005} = 127324 \, \text{MPa}
\]

Thus, the modulus of elasticity is approximately:

\[
E \approx 127.32 \, \text{GPa}
\]

### Final Answer:
The modulus of elasticity of the material is approximately **127.32 GPa**.

Would you like to go over any details, or have further questions?

Here are 5 related questions to explore:
1. What is Poisson's ratio and how could it affect the material's behavior?
2. How would the modulus of elasticity change if the material entered the plastic region?
3. What is the relationship between stress, strain, and elasticity in different materials (e.g., metals vs polymers)?
4. How would thermal expansion affect the rod's dimensions and stress in a real-world scenario?
5. How is the modulus of elasticity used in structural engineering design?

**Tip**: When working with modulus of elasticity, always ensure the material remains within the elastic limit to avoid permanent deformation.

Calculate the modulus of elasticity for a rod subjected to a tensile force of 200 kN, given changes in its dimensions. Learn how to use stress, strain, and Young's modulus formulas to solve elasticity problems. Suitable for undergraduate mechanical engineering students.

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