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Torque about a Point

A composite t...

Question

A composite tube is made by striking a thin steel tube on a brass tube. If $A_{S}$ and $A_{B}$ are the respective sectional areas of the steel and brass tubes and $Y_{S}$ and $Y_{B}$ are their Young's moduli. Find the Young's modulus of a single tube of the same length and total sectional area, which would behave in the same fashion as that of the composite tube.

\frac{A_{S} Y_{S} + A_{B} Y_{B}}{A_{S} + A_{B}}

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\frac{A_{B} Y_{S} + A_{S} Y_{B}}{A_{S} + A_{B}}

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\frac{Y_{S} + Y_{B}}{2}

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Y_{S} + Y_{B}

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Solution

The correct option is A $\frac{A_{S} Y_{S} + A_{B} Y_{B}}{A_{S} + A_{B}}$
$Method 1:$ Let the composite tube be subjected to an axial (tensile) force $F$ and $δ l$ be the corresponding elongation.

The stress produced in the steel tube,

$σ_{S} = \frac{F Y_{S}}{A_{S} Y_{S} + A_{B} Y_{B}}$

But we know, $Y_{S} = \frac{σ_{S}}{ε_{S}} = \frac{σ_{S}}{(\frac{Δ l}{L})}$

$Δ l = \frac{σ_{S} L}{Y_{S}} = \frac{F L}{(A_{S} Y_{S} + A_{B} Y_{B})} . . . . (i)$

If $Y$ be the required Young's modulus of the tube which behaves in the same fashion as that of the composite tube, then

$Δ l = \frac{F L}{(A_{S} + A_{B}) Y} . . . . (i i)$

Comparing eqs. $(i)$ and $(i i)$ , we have

$Y = \frac{A_{S} Y_{S} + A_{B} Y_{B}}{A_{S} + A_{B}}$

Hence, $(A)$ is the correct answer.

$Method 2:$

We can write equivalent force constact of spring as,

$k_{e q} = \frac{1}{L} (A_{S} Y_{S} + A_{B} Y_{B}) . . . . (i i i)$

If we replace the composite rods into a single rod of length $L$ and area $(A_{S} + A_{B})$

Here $k_{e q} = \frac{(A_{S} + A_{B}) Y}{L} . . . . (i v)$

Comparing eqs. $(i i i)$ and $(i v)$ , we have equivalent Young's modulus as

$Y = \frac{A_{S} Y_{S} + A_{B} Y_{B}}{(A_{S} + A_{B})}$

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