Chapter 6, Problem 6.9P

To determine

The relationship between change in resistance and strain.

Answer to Problem 6.9P

The relationship between change in resistance and strain is $dR=R\epsilon \left(1+2v\right)$ .

Explanation of Solution

Formula Used:

Write the expression for the resistance of wire..

  $R=\rho \frac{l}{A}$ …… (I)

Here, $R$ is the resistance, $\rho$ is the resistivity, $l$ is the length of wire and $A$ is the area of wire.

Write the expression for the derivative of volume in terms of length and area.

  $dV=Adl+ldA$ …… (II)

Here, $dV$ is the change in volume, $dl$ is the change in length and $dA$ is the change in area.

Write the expression for change in volume.

  $dV=l\left(1+\epsilon \right)A{\left(1-v\epsilon \right)}^{2}lA$ …… (III)

Here, $\epsilon$ is the strain and $v$ is the Poisson’s ratio.

Calculation:

Differentiate equation (I) with $\rho$ as constant.

  $\begin{array}{c}dR=d\left(\rho \frac{l}{A}\right)\\ =\frac{\rho }{A}dl+\rho ld\left(\frac{1}{A}\right)\\ =\frac{\rho }{A}dl-\rho l\frac{dA}{A^2}\\ =\rho \left(\frac{Adl-ldA}{A^2}\right)\end{array}$ …… (IV)

Rewrite equation (III) in expanded form.

  $\begin{array}{c}dV=l\left(1+\epsilon \right)A{\left(1-v\epsilon \right)}^{2}lA\\ =l\left(1+\epsilon \right)A\left(1-2v\epsilon +v^2{\epsilon }^2\right)lA\end{array}$

For small strain, neglect ${\epsilon }^2$ .

  $\begin{array}{c}dV=l\left(1+\epsilon \right)A\left(1-2v\epsilon \right)-lA\\ =\left(lA+lA\epsilon \right)\left(1-2v\epsilon \right)-lA\\ =lA\epsilon \left(1-2v\right)\\ =Adl\left(1-2v\right)\end{array}$ …… (V)

Equate equation (II) and (V).

  $\begin{array}{c}Adl+ldA=Adl\left(1-2v\right)\\ Adl+ldA=Adl-2Avdl\\ ldA=-2Avdl\end{array}$ …… (VI)

Substitute $-2Avdl$ for $ldA$ in equation (IV).

  $\begin{array}{c}dR=\rho \left(\frac{Adl-\left(-2Avdl\right)}{A^2}\right)\\ \frac{dR}{R}=\frac{\rho }{R}\left(\frac{Adl+2Avdl}{A^2}\right)\\ =\frac{\rho Adl}{R}\left(\frac{1+2v}{A}\right)\end{array}$ ……. (VII)

Substitute $\rho \frac{l}{A}$ for $R$ and $\epsilon$ for $\frac{dl}{l}$ in equation (VII).

  $\begin{array}{c}\frac{dR}{R}=\frac{\rho Adl}{\left(\rho \frac{l}{A}\right)}\left(\frac{1+2v}{A}\right)\\ =\frac{dl}{l}\left(1+2v\right)\\ =\epsilon \left(1+2v\right)\\ dR=R\epsilon \left(1+2v\right)\end{array}$

Conclusion:

Thus, the relationship between change in resistance and strain is $dR=R\epsilon \left(1+2v\right)$ .