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Figure shows two rods $A$ and $B$ of same length $L$ and same cross-sectional area $S$ but of different material having coefficient of linear expansion $α_{1}$ and $α_{2}$ respectively. They are clamped between two rigid walls, separated by a distance $2 L$ . This all refers to temp $t^{\circ} C$ . Find the tension in each rod at temp $2 t^{\circ} C$ (Take the young’s modulus for the two rods to be $Y_{1}$ and $Y_{2}$ respectively).

A

\frac{S Y_{1} Y_{2} (2 α_{1} + α_{2}) t}{(2 Y_{1} + Y_{2})}

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B

\frac{S Y_{1} Y_{2} (α_{1} + 2 α_{2}) t}{(Y_{1} + 2 Y_{2})}

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C

\frac{S Y_{1} Y_{2} (α_{1} + α_{2}) t}{(Y_{1} + Y_{2})}

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D

\frac{S Y_{1} Y_{2} (α_{1} + 2 α_{2}) t}{(Y_{1} + Y_{2})}

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Solution

The correct option is C $\frac{S Y_{1} Y_{2} (α_{1} + α_{2}) t}{(Y_{1} + Y_{2})}$
Since the clamps are fixed, the amount of thermal expansion should be equal to the compression.

Thermal expansion of the first rod $= α_{1} L Δ t$

Thermal expansion of the second rod $= α_{2} L Δ t$

Let compression of first rod be $Δ L_{1}$ and compression of second rod be $Δ L_{2}$

Net expansion = net compression
$\Rightarrow α_{1} L Δ t + α_{2} L Δ t = Δ L_{1} + Δ L_{2}$ ----- (1)

Since the two rods are connected, tension across them will be equal.

$T_{1} = T_{2}$

$\Rightarrow \frac{T_{1}}{S} = \frac{T_{1}}{S} \Rightarrow σ_{1} = σ_{2} \Rightarrow ϵ_{1} Y_{1} = ϵ_{2} Y_{2}$

$\Rightarrow \frac{Δ L_{1}}{L_{1}} Y_{1} = \frac{Δ L_{2}}{L_{2}} Y_{2}$

$\Rightarrow Δ L_{1} = Y_{1} = Δ L_{2} . \frac{Y_{2}}{Y_{1}} . \frac{L_{2}}{L_{1}}$ ------ (2)

Solving (1) and (2),

$Δ L_{2} = \frac{(α_{1} + α_{2}) L t . Y_{1} L_{2}}{Y_{1} L_{2} + Y_{2} L_{1}}$

Therefore tension $T = σ_{2} . S = ϵ_{2} Y_{2} . S = \frac{Δ L_{2}}{L_{2}} . S Y_{2}$

$= \frac{S . Y_{1} Y_{2} . (α_{1} + α_{2}) t}{Y_{1} L_{2} + Y_{2} L_{1}}$

$= \frac{Y_{1} Y_{2} t (α_{1} + α_{2}) S}{Y_{1} (1 + α_{2} t) + Y_{2} (1 + α_{1} t)}$

Taking $1 + α_{1} t = 1 + α_{2} t \approx 1$

$T = \frac{Y_{1} Y_{2} t (α_{1} + α_{2}) S}{Y_{1} + Y_{2}}$

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