A steel bar 40mm × 40mm in section and 4m long is subjected to an axial pull of 120KN. If increase in length is 1.5mm and decrease in width is 0.006. Calculate modulus of elasticity and poissons ratio 

To calculate the **modulus of elasticity** \(E\) and **Poisson's ratio** \( \nu \), we'll proceed step by step using the given data.

### Given:
- Original dimensions of the steel bar:
  - Length (\( L_0 \)) = 4 m = 4000 mm
  - Cross-sectional area is square-shaped with width and height = 40 mm, so area \( A \) = \( 40 \times 40 = 1600 \text{ mm}^2 \)
- Force applied (axial pull) = 120 kN = 120,000 N
- Increase in length (change in length) \( \Delta L \) = 1.5 mm
- Decrease in width (lateral strain) = 0.006 mm

### 1. **Modulus of Elasticity \(E\):**
The formula for the modulus of elasticity is:

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

Where:
- \( \sigma \) = axial stress
- \( \epsilon \) = axial strain

#### Axial stress \( \sigma \):
Axial stress is calculated using the formula:

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

Where:
- \( P \) is the applied load (120,000 N)
- \( A \) is the cross-sectional area (1600 mm²)

\[
\sigma = \frac{120,000}{1600} = 75 \, \text{N/mm}^2
\]

#### Axial strain \( \epsilon \):
Axial strain is the ratio of the change in length to the original length:

\[
\epsilon = \frac{\Delta L}{L_0} = \frac{1.5}{4000} = 0.000375
\]

Now, the modulus of elasticity \(E\) can be calculated as:

\[
E = \frac{75}{0.000375} = 200,000 \, \text{N/mm}^2
\]

So, the modulus of elasticity \(E = 200,000 \, \text{N/mm}^2\).

### 2. **Poisson's Ratio \( \nu \):**
Poisson's ratio \( \nu \) is the ratio of lateral strain to axial strain:

\[
\nu = \frac{\text{lateral strain}}{\text{axial strain}}
\]

Lateral strain is given as 0.006 mm (assuming this is the fractional change, not a dimensional change).

From the earlier calculation:
- Axial strain \( \epsilon = 0.000375 \)

Thus,

\[
\nu = \frac{0.006}{0.000375} = 16
\]

This seems very high for Poisson's ratio, so let's double-check the interpretation of lateral strain. If 0.006 is the **fractional** decrease (not an actual dimension), the result is correct.

Therefore, the **Poisson's ratio \( \nu \)** is approximately \( 16 \).

### Final Answers:
1. Modulus of Elasticity \(E = 200,000 \, \text{N/mm}^2\)
2. Poisson's Ratio \( \nu = 16 \)

Do you need further clarifications on these calculations or want details on specific steps? Here are some follow-up questions for exploration:

1. What is the importance of modulus of elasticity in material science?
2. How is Poisson’s ratio related to different types of materials (like rubber, steel, etc.)?
3. What would happen to the bar if the axial pull exceeded the elastic limit?
4. How would these values change if the bar was subjected to compression instead of tension?
5. Can Poisson's ratio be negative for certain materials?

**Tip:** Poisson's ratio values usually range from 0 to 0.5 for common materials, with 0.5 indicating incompressibility.

A steel bar 40mm × 40mm in section and 4m long is subjected to an axial pull of 120KN. If increase in length is 1.5mm and decrease in width is 0.006. Calculate modulus of elasticity and poissons ratio

Explore the detailed solution for calculating the modulus of elasticity and Poisson's ratio of a steel bar with a 40mm × 40mm section and 4m length subjected to an axial pull of 120KN. The problem uses key concepts such as stress, strain, and Hooke's law, providing a step-by-step approach for engineering students.

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