Answer: 

• The strain Gauge may be defined as the passive resistive transducer based on the principle of conversion of mechanical displacement into the resistance change. The strain Gauge characteristics depensds on gauge sensitivity, accuracy, frequency and the enviromental condition.

Working of Strain Guage :- 

• The basic principle of operation of an strain gauge is that the resistance of the wire changes as a function of strain, increasing with tension and reducing with compression. The change in resistance is measured with a Wheatstone bridge.
• The strain gauge is attached to the specimen and hence the gauge is subjected to the same strain as that of the specimen under test.
• The materials used for fabrication of electrical strain gauges must have some basic qualities to achieve high accuracy, excellent reproducibility, good sensitivity, long life and ability to operate under the required environmental conditions.
• Some of these qualities are attained by selecting materials with high specific resistance, low temperature coefficient of resistance, constant gauge factor, and constant strain sensitivity over a wide range of strain values.
• The bonding cement should have high insulation resistance and excellent transmissibility of strain, and must be immune to moisture effects.

• The most common materials used for wire strain gauges are constantan alloys containing Nickel$45\%$ and$55\%$Copper, as they exhibit high specific resistance, constant gauge factor over a wide strain range, and good stability over a reasonably large temperature range$\mathrm{(from\space 0 ^oC \space to\space 300 ^oC)}$. For dynamic strain measurements, Nichrome alloys, containing$80\%$ Nickel and$20\%$ Chromium are used. They can be compensated for temperature with platinum. Improper bonding of the gauge can cause many errors.

   

Derivation of Gauge Factor

• The Gauge factor is defined as the unit change in resistance per unit change in length. Generally it is denoted by$S$. Thus the below figure shows:-  **Fig Deformed resistance wire** * $\mathrm{S=\dfrac{\Delta R/R}{\Delta l/l}}$ Where $S$= Gauge factor $R$= Gauge wire resistance $\Delta R$= Change in wire resistance $l$= length of the Gauge wire $\Delta l$= Change in the length of the Gauge wire * Let the resistance wire is in tensile stress and it is deformed by$\Delta l$. Thus $\rho$= Specific resistance in $\mathrm {\Omega.m}$ $l$= Length of the wire in $\mathrm m$ $A$= Area of cross sections in $\mathrm{m^2}$ * When uniform stress$\sigma$is applied to the wire along the length, the resistance$R$ changes to$\mathrm{R+\Delta R}$ * Hence $\sigma=\mathrm{Stress}=\mathrm{\dfrac{\Delta l}{l}}$ $\mathrm{\Delta l/l}$=per unit change in length $\mathrm{\Delta A/A}$= per unit change in area $\mathrm{\Delta\rho/\rho}$ = per unit change in resistivity * Now, we know that $\mathrm R=\mathrm {\dfrac{\rho l}{A}}$ $\therefore \space\mathrm{\dfrac{dR}{d\sigma}=\dfrac{d(\dfrac{\rho l}{A})}{d\sigma}}$ $=\mathrm{\dfrac{\rho}{A}.\dfrac{\delta l}{\delta \sigma}-\dfrac{\rho l}{A^2}.\dfrac{\delta A}{\delta \sigma}+ \dfrac{l}{A}.\dfrac{\delta \rho}{\delta \sigma}}$ * Thus $\mathrm{\dfrac{\delta}{\delta\sigma}(\dfrac{l}{A})}=\mathrm{-\dfrac{l}{A^2}.\dfrac{\delta\rho}{\delta\sigma}}$ * Multiply both sides by$\mathrm{\dfrac{1}{R}}$ we obtain $\therefore \space\mathrm{\dfrac{1}{R}\dfrac{dR}{d\sigma}}=\mathrm{\dfrac{\rho}{RA}.\dfrac{\delta l}{\delta \sigma}-\dfrac{1}{R}\dfrac{\rho l}{A^2}.\dfrac{\delta A}{\delta \sigma}+ \dfrac{l}{RA}.\dfrac{\delta \rho}{\delta \sigma}}$ * Using$\mathrm R=\mathrm {\dfrac{\rho l}{A}}$, we obtain $\therefore \space\mathrm{\dfrac{1}{R}\dfrac{dR}{d\sigma}}=\mathrm{\dfrac{1}{l}.\dfrac{\delta l}{\delta \sigma}-\dfrac{1}{A}\dfrac{\delta A}{\delta\sigma}+ \dfrac{l}{\rho}.\dfrac{\delta \rho}{\delta \sigma}}$ * Canceling$\mathrm \delta\sigma$ from both sides from above equation , we obtain $\therefore \space\mathrm{\dfrac{dR}{R}}=\mathrm{\dfrac{d l}{l}-\dfrac{dA}{A}+ \dfrac{\delta \rho}{\rho}}$ $\mathrm {i.e. }=\mathrm{\dfrac{\Delta R}{R}=\dfrac{\Delta l}{l}-\dfrac{dA}{A}+\dfrac{\Delta \rho}{\rho}}........(1)$ * Thus for finite stress, total change in resistance is due to fractional change in length , are and resistivity * Thus for a circular wire, $\mathrm {A}=\mathrm{\dfrac{\pi}{4}d^2}$ $\therefore \space \mathrm{\dfrac{\delta A}{\delta s}=\dfrac{\pi}{4}(2d)\dfrac{\delta d}{\delta s}}$ $\therefore \space \mathrm{\dfrac{1}{A}\dfrac{\delta A}{\delta s}=\dfrac{1}{A}\dfrac{\pi}{4}(2d)\dfrac{\delta d}{\delta s}}$ $\therefore \space \mathrm{\dfrac{1}{A}\dfrac{\delta A}{\delta s}=\dfrac{1}{d^2}(2d)\dfrac{\delta d}{\delta s}}$ * Canceling$\delta s$ from both sides from above equation , we obtain $\therefore \space \mathrm{\dfrac{\delta A}{A}=\dfrac{2}{d}{\delta d}}$ * Hence $\therefore \space \mathrm{\dfrac{\Delta A}{A}=\dfrac{2\Delta d}{d}}$ * Now the Poisson's ratio$\mathrm {\mu=-\dfrac{\Delta d/d}{\Delta l/l}=\text{Poisson's ratio}}$ $\therefore \space \mathrm{\dfrac{\Delta d}{d}=-\mu(\dfrac{\Delta l}{l})}...........\mathrm {Equation 3}$ * Using 2 and 3 in 1 we obtain $\mathrm {i.e. }=\mathrm{\dfrac{\Delta R}{R}=\dfrac{\Delta l}{l}-\dfrac{2\Delta d}{d}+\dfrac{\Delta \rho}{\rho}}$ $=\mathrm{\dfrac{\Delta R}{R}=\dfrac{\Delta l}{l}-2[-\mu\dfrac{\Delta l}{l}]+\dfrac{\Delta \rho}{\rho}}$ $\mathrm{=\dfrac{\Delta l}{l}[1+2\mu]+\dfrac{\Delta \rho}{\rho}}$ * Neglecting piezoelectric effect,$\dfrac{\Delta\rho}{\rho}$ can be neglected $\mathrm{=\dfrac{\Delta l}{l}[1+2\mu]}$ * $\bf{\mathrm{Guage\space Factor}\space\mathrm{S=\dfrac{\Delta R/R}{\Delta l/l}}=1+2\mu}$