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Question
A steel rod is clamped at its two ends and rests on a fixed horizontal base. The rod is unstrained at 20°C.
Find the longitudinal strain developed in the rod if the temperature rises to 50°C. Coefficient of linear expansion of steel = 1.2 × 10–5 °C–1.
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Solution
Given:
Temperature at which rod is resting on a fixed horizontal base without any strain, T1=20 °C. Then the rod is heated to temperature, T2 = 50 °C
So change in temperature,ΔT =T2-T1=30°C
Coefficient of linear expansion of steel, α = 1.2 × 10–5 °C-1
Let L be the length of the rod without heating and L' be the length of the rod on heating.
Let longitudinal strain developed in the rod be S.
We know that
L' =L(1+∝ΔT)
⇒ ΔL =L∝ΔT
Strain, S = `(ΔL)/L`
`=(L∝ΔT)/L`
=αΔT
⇒ S =1.2 × 10-5 ×(50-20)
=1.2 × 10-5 ×30
=1.2 × 10-5 × 30
=36 × 10-5
S = 3.6 × 10-4
The strain of 3.6 × 10-4 will be opposite to the direction of expansion.
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