SSC JE Mechanical Engineering 2018 Set 3 Paper

Question : 125 of 200

Marks: +1, -0

What will the thermal stress developed in a rod having a diameter of 4 cm and length of 2 m. It experiences heating from temperature 50°C to 200°C. The coefficient of thermal expansion is

α = 10 × 10^{–6}/

°C and young’s modulus is 250 GPa?

Validate

Solution:

Thermal stress

= E∝ ∆ t

= 250 × 109 × 10 × 10^{–6 }× (200 -50)

= 375 N/m

^{2}

or MPa

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