# Diameter of steel rod fixed between two rigid supports is 20 mm. Find the stress in the rod when the temperature increases by 80°C. (Take Young's modulus, Y = 2 × 10^11 N/m^2 and thermal expansion coefficient, a = 12 × 10^-6/^°C) A) 150 MPa B) 192 MPa C) 250 MPa D) 100 MPa

## Question

Diameter of steel rod fixed between two rigid supports is 20 mm. Find the stress in the rod when the temperature increases by 80°C. (Take Young's modulus, Y = 2 × 10^11 N/m^2 and thermal expansion coefficient, a = 12 × 10^-6/^°C)

A) 150 MPa B) 192 MPa C) 250 MPa D) 100 MPa

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## Answer

**Question:**

The diameter of a steel rod fixed between two rigid supports is 20 mm. Find the stress in the rod when the temperature increases by 80°C. (Take Young's modulus, $Y = 2 \times 10^{11}$ N/m² and thermal expansion coefficient, $\alpha = 12 \times 10^{-6}$/°C)

- A) 150 MPa
- B) 192 MPa
- C) 250 MPa
- D) 100 MPa

**:**

The correct option is **B) 192 MPa**

First, recall the formula for the **Modulus of Elasticity**:

$$ Y = \frac{\text{Stress}}{\text{Strain}} $$

From this equation, we can express **Stress** as:

$$ \text{Stress} = Y \times \text{Strain} $$

Next, we need to calculate the **Strain** due to thermal expansion. The strain is given by:

$$ \frac{\Delta L}{L} = \alpha \Delta T $$

Substitute the given values:

$$ \alpha = 12 \times 10^{-6}/°C \quad \text{and} \quad \Delta T = 80°C $$

Calculate the strain:

$$ \frac{\Delta L}{L} = 12 \times 10^{-6} \times 80 = 9.6 \times 10^{-4} $$

Now, substituting the strain into the stress equation:

$$ \text{Stress} = 2 \times 10^{11} \times 9.6 \times 10^{-4} $$

Hence, the Stress is:

$$ \text{Stress} = 192 \text{ MPa} $$

Therefore, the stress in the rod when the temperature increases by 80°C is **192 MPa**.

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