---
title: "Derivation of strain gauge factor"
subject: "Instrumentation and Measurement"
chapter: 6
key: "semester_4/instrumentation_and_measurement:strain-gauge-factor-derivation"
source: https://www.bctengineer.com/q/semester_4/instrumentation_and_measurement/strain-gauge-factor-derivation
site: BCT Engineering
---

# Derivation of strain gauge factor

Instrumentation and Measurement · Chapter 6: Strain Gage

- Asked in: year not given, 4+2 marks
- Video: [strain gauge | derivation for a gauge factor | EMI|ushendra's engineering tutorials](https://www.youtube.com/watch?v=B-Wus2HsDfI) — ushendra's engineering tutorials
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/instrumentation_and_measurement/strain-gauge-factor-derivation
- Whole chapter: https://www.bctengineer.com/md/semester_4/instrumentation_and_measurement/chapter-6

## Question

(a) Derive the expression for the gauge factor of a metallic resistance strain gauge, $G = 1 + 2\nu + \dfrac{d\rho/\rho}{\varepsilon}$.

(b) For a gauge material with Poisson's ratio 0.3, find the gauge factor if the resistivity does not change with strain. If the measured gauge factor is 2.05, what part of it is due to the change of resistivity?

## Short answer

Gauge factor is fractional resistance change per strain. Differentiate R = rho L/A, use lateral strain -nu epsilon, giving G = 1 + 2nu + (drho/rho)/epsilon. For nu = 0.3 and constant resistivity, G = 1.6; measured 2.05 gives resistivity contribution 0.45, or 22%.

## Answer

#### (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 $G = 2.05$.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The gauge factor G of a strain gauge is defined as:
   A. Strain divided by fractional change in resistance
   B. Fractional change in resistance divided by strain
   C. Change in resistance divided by change in length only
   D. Resistivity divided by strain
   Answer: B. G = (dR/R) / strain. (Key point: Gauge factor definition)

2. From R = rho L / A, differentiating the logarithm gives dR/R equal to:
   A. drho/rho - dL/L - dA/A
   B. drho/rho + dL/L - dA/A
   C. drho/rho + dL/L + dA/A
   D. dL/L - drho/rho
   Answer: B. R is proportional to rho and L and inversely to A, so dR/R = drho/rho + dL/L - dA/A. (Key point: Derivation step)

3. Why does the geometric part of G equal 1 + 2v?
   A. Length gives 2 and Poisson contraction gives 1
   B. Length change gives 1 and the area reduction from lateral contraction (dA/A = -2v strain) gives 2v
   C. Both come from the resistivity change
   D. The area increases by 2v times the strain
   Answer: B. dL/L = strain gives 1, and dD/D = -v strain gives dA/A = -2v strain, so -dA/A adds +2v. (Key point: Geometric term)

4. With v = 0.3 and no resistivity change, G = 1.6. If the measured G is 2.05, what share is due to resistivity change?
   A. 0.45, about 22% of G
   B. 0.60, about 29% of G
   C. 1.6, about 78% of G
   D. 2.05, all of G
   Answer: A. 2.05 - 1.6 = 0.45, and 0.45/2.05 is about 22%. (Key point: Numerical G)

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