Derivation of strain gauge factor

(a) Gauge factor

The gauge factor $G$ is the ratio of the fractional change in resistance to the strain:

$$G = \frac{\Delta R/R}{\varepsilon}$$

For a conductor of length $L$, cross-sectional area $A$ and resistivity $\rho$:

$$R = \frac{\rho L}{A}$$

Taking logarithms and differentiating:

$$\frac{dR}{R} = \frac{d\rho}{\rho} + \frac{dL}{L} - \frac{dA}{A}$$

For a circular wire $A = \pi D^2/4$, so $dA/A = 2\,dD/D$. The axial strain is $\varepsilon = dL/L$ and the lateral strain is $dD/D = -\nu\varepsilon$ (Poisson's ratio $\nu$). Then $dA/A = -2\nu\varepsilon$, and

$$\frac{dR}{R} = \frac{d\rho}{\rho} + \varepsilon + 2\nu\varepsilon$$ $$G = \frac{dR/R}{\varepsilon} = 1 + 2\nu + \frac{d\rho/\rho}{\varepsilon}$$

The term $1 + 2\nu$ is the geometric effect and $(d\rho/\rho)/\varepsilon$ is the piezoresistive effect. For metals $G \approx 2$ (constantan $\approx 2.0$, platinum alloys up to 4), and for semiconductors $G$ is 100-200 because the piezoresistive term dominates.

(b) Numerical

If resistivity is constant, $d\rho = 0$:

$$G = 1 + 2(0.3) = 1.6$$

For the measured $G = 2.05$:

$$\frac{d\rho/\rho}{\varepsilon} = 2.05 - 1.6 = 0.45$$

The fraction due to resistivity change is $0.45/2.05 = 22\%$.

Answer: $G = 1.6$ (geometry only); resistivity contributes $0.45$ (22%) of $G = 2.05$.