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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