A steel rod is 0.02 m in diameter and 2.5 m in length. The modulus of elasticity of the 
material is 210 GPa, with a Poisson’s ratio of 0.2. Calculate the change in diameter after 
applying a tensile force of 150 kN

To calculate the change in diameter of the steel rod due to the applied tensile force, we can follow these steps:

### Step 1: Given Data
- **Diameter of the rod, \(d_0 = 0.02 \, \text{m} = 20 \, \text{mm}\)**
- **Length of the rod, \(L_0 = 2.5 \, \text{m}\)**
- **Tensile force, \(F = 150 \, \text{kN} = 150 \times 10^3 \, \text{N}\)**
- **Modulus of elasticity, \(E = 210 \, \text{GPa} = 210 \times 10^9 \, \text{Pa}\)**
- **Poisson’s ratio, \( \nu = 0.2 \)**

### Step 2: Calculate the axial stress
The axial stress, \( \sigma \), is calculated using:

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

where \( A \) is the cross-sectional area of the rod, and for a circular cross-section, the area is:

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

Substitute the given diameter:

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

Now, calculate the axial stress:

\[
\sigma = \frac{150 \times 10^3}{3.1416 \times 10^{-4}} \, \text{Pa}
\]

### Step 3: Use Hooke's Law to find axial strain
Axial strain, \( \epsilon \), is related to axial stress by Hooke's Law:

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

Now substitute \( \sigma \) and \( E \):

\[
\epsilon = \frac{\sigma}{210 \times 10^9}
\]

### Step 4: Calculate the lateral strain
The lateral strain, \( \epsilon_{\text{lateral}} \), is related to axial strain by Poisson’s ratio:

\[
\epsilon_{\text{lateral}} = - \nu \times \epsilon
\]

### Step 5: Calculate the change in diameter
The change in diameter, \( \Delta d \), can be calculated using the lateral strain:

\[
\Delta d = \epsilon_{\text{lateral}} \times d_0
\]

Finally, we can compute the change in diameter once we calculate the axial strain and lateral strain. Let me perform the calculations.The change in the diameter of the steel rod due to the applied tensile force is approximately \(-9.09 \times 10^{-6} \, \text{m}\) or \(-0.00909 \, \text{mm}\). The negative sign indicates a reduction in diameter, as expected due to Poisson's effect in a tensile load.

Do you need further details or have any questions?

Here are some related questions:
1. How does Poisson's ratio affect the change in diameter?
2. What would happen if the rod were subjected to a compressive force instead?
3. How is the modulus of elasticity related to material stiffness?
4. What are typical values of Poisson’s ratio for different materials?
5. How would the result change if the rod's length were different?

**Tip:** Poisson's ratio measures the material's tendency to expand or contract in directions perpendicular to the applied load.

A steel rod is 0.02 m in diameter and 2.5 m in length. The modulus of elasticity of the material is 210 GPa, with a Poisson’s ratio of 0.2. Calculate the change in diameter after applying a tensile force of 150 kN.

Learn how to calculate the change in diameter of a steel rod due to a tensile force. This example uses key mechanical concepts such as stress, strain, Poisson's ratio, and modulus of elasticity, making it ideal for undergraduate engineering students.

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