A strain gauge is a device used to measure strain on an object. Strain gauge is a passive resistance transducer which converts the mechanical elongation and compression into resistance change. This change in resistance takes place due to variation in length and cross sectional area of the gauge wire,when an external force acts on it.

The resistance of a conductor is expressed as,

$R\ =\ \dfrac{p \times l}{A}$

Where, p : Specific resistance of the material in 2-m.

1 : Length of the conductor in m.

A : Cross-sectional area of conductor in rn2.

The characteristics of a strain gauge are described in terms of its sensitivity which is called as the gauge factor of the strain gauge.

Gauge factor (G.F) is defined as the unit change in resistance per unit change in the length of the strain gauge wire.

$Mathematically, \ GF(K)\ =\ \dfrac{ \Delta R / R}{\Delta l/l}$

where, k = gauge factor

Δ R= change in the intial resistance in Ω ‘s

R = resistance of the gauge wire (without strain).

Δ l = change I n the length in m.

l = length of gauge wire (without strain).

Strain is defined as the change in length divided by the original length.

$\sigma = \dfrac{ \Delta l}{l}$

σ = strain in the lateral direction.

Equation (22) can be written as,

$K = \dfrac {\Delta R/R}{\sigma}$

The resistance of a conductor with uniform cross-section is,

$R = P \dfrac{l}{A}$

$R = P \dfrac{l}{\pi \chi^2}$

but, $r= \dfrac{d}{2}$, $\therefore r^2=\dfrac{d^2}{4}$

$\therefore R=P \dfrac{l}{\pi d^2/4}$

where, p =specific resistance of the conductor.

l : length of the conductor

d : diameter of the conductor

The tension on the conductor causes a change in its length. The length of the conductor increases by factor Al and in turn the diameter decreases by a factor Ad. Thus, the resistance of the conductor changes to,

$R'= \rho \dfrac{l + \Delta l}{ \pi/4 (d- \Delta d^2)}$

$R'= \rho \dfrac{l + \Delta l}{ \pi/4 (d- 2d\Delta d+ \Delta d^2)}$

∴ Δd is small, hence Δd2 can be neglected.

$\therefore R'= \dfrac{ \rho \ (l + \Delta l)}{ \pi/4 (d^2- 2d\Delta d)}$

$\therefore R' = \dfrac{\rho (l+\Delta l)}{\dfrac{\pi }{4}d^2 \left ( 1-\dfrac{2 \Delta d}{d} \right )}= \dfrac {\rho l \left ( 1+\dfrac{\Delta l}{l} \right )}{{\dfrac{\pi }{4}d^2 \left ( 1-\dfrac{2 \Delta d}{d} \right )}}$