Question

A steel rod is clamped at its 2 ends and rests on afixed horizontal base. The rod is unstrained at 25 degree Celsius . Find longitudinal stress developed in the rod if the temperature rises to 60 degree . Coefficient of li ear expansion of steel = 1.5×10^-5K-. Youngs modulus of steel = 20×10^10Nm-

Answer 1

Answer:

Stress in the rod is given as

$stress = 1.05 \times 10^8 N/m^2$

Explanation:

As we know that ratio of stress and strain is known as Young's modulus

So we will have

$stress = Y ( strain)$

here we know that

$strain = \frac{\Delta L}{L}$

so we know that due to thermal expansion of rod strain in it

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

$\frac{\Delta L}{L} = (1.5 \times 10^{-5})(60 - 25)$

$\frac{\Delta L}{L} = 5.25 \times 10^{-4}$

now we have

$Stress = (20 \times 10^{10})(5.25 \times 10^{-4})$

$stress = 1.05 \times 10^8 N/m^2$

#Learn

Topic : thermal Expansion and thermal stress

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