A steel wire and a copper wire of equal length and equal cross-sectional area are joined end to end and the combination is subjected to a tension. Find the ratio of (a) the stresses developed in the two wires and (b) the strains developed. Y of steel = $2 \times 10^{11} N m^{- 2}$ . Y of copper = $1.3 \times 1011 N m^{- 2}$ .

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Solution

Given that, $L_{s t e e l} = L_{C u}$

and $A_{s t e e l} = A_{C u}$

(a) $\frac{stress of Cu}{Stress of Steel} = \frac{F_{C u} A_{S t e e l}}{A_{C u} F_{S t e e l}}$

= $\frac{F_{C u}}{F S t e e l} = 1$ [because $F_{C u} = F_{P} S t e e l$ ]

(b) $Strain = \frac{Δ L_{S t e e l}}{Δ L C u}$

$= \frac{F_{S t e e l} L_{S t e e l} A_{c u} Y_{c u}}{A_{S t e e l} Y_{S t e e l} F_{c u} I_{c u}}$

$[L_{c u} = L_{S t e e l} : A_{c u} = A_{S t e e l}]$

$\Rightarrow \frac{Y_{c u}}{Y_{S t e e l}} = \frac{1.3 \times 10^{11}}{2 \times 10^{11}}$

$\Rightarrow \frac{Δ L S t e e l}{Δ I_{c u}} = \frac{13}{20}$

$\Rightarrow \frac{Δ L_{c u}}{Δ L S t e e l} = \frac{20}{13}$

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A steel wire and a copper wire of equal length and equal cross-sectional area are joined end to end and the combination is subjected to a tension. Find the ratio of (a) the stresses developed in the two wires and (b) the strains developed. Y of steel = 2×1011Nm−2. Y of copper = 1.3×1011Nm−2.

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