Strain Gage

(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 $G = 2.05$.

(a) Bonded and unbonded gauges

Common bonded types:

• Wire gauge: a grid of fine wire (about 0.025 mm) on paper or epoxy; the older type.
• Foil gauge: an etched metal foil grid on a polyimide backing; the most used type, it has better heat dissipation, stability and shape flexibility.
• Semiconductor gauge: silicon, high gauge factor (100-200), strongly temperature-dependent.
• Rosettes: two or three gauges at fixed angles for biaxial strain.

Gauge materials: constantan (Cu-Ni, $G\approx 2.1$, low temperature coefficient), Karma (Ni-Cr), nichrome, isoelastic ($G\approx 3.6$), platinum-tungsten (high temperature).

(b) Selection factors

• Gauge length (short for stress peaks, long for averaging on heterogeneous material such as concrete).
• Gauge resistance (120 or 350 $\Omega$), gauge factor and its tolerance.
• Strain range, fatigue life and elongation limit.
• Temperature range, and self-temperature compensation matched to the specimen material.
• Backing, adhesive and environment (moisture, chemicals).
• Grid pattern: uniaxial, rosette, diaphragm, or shear pattern.

Installation of a bonded gauge

1. Choose the position and direction of principal strain.
2. Prepare the surface: degrease with solvent, abrade with fine emery paper, and clean again with a conditioner and neutralizer.
3. Mark gauge alignment lines with a pencil (not a scriber).
4. Apply adhesive (cyanoacrylate or epoxy), place the gauge using tape, and press with a thumb for about a minute under steady pressure.
5. Cure as the adhesive requires (room temperature or heat).
6. Solder lead wires to the terminals; use a three-wire arrangement for lead compensation.
7. Check insulation resistance (above 500 M$\Omega$) and gauge resistance, and apply a moisture-proof coating.

Given: $R_g = 120\ \Omega$, $G = 2.1$, $\varepsilon = 1500\ \mu\varepsilon = 1.5\times10^{-3}$.

Change in resistance

$$\frac{\Delta R}{R} = G\varepsilon = 2.1 \times 1.5\times10^{-3} = 3.15\times10^{-3} = 0.315\%$$ $$\Delta R = 120 \times 3.15\times10^{-3} = 0.378\ \Omega$$

Quarter-bridge output

All arms are 120 $\Omega$ and $V_{ex} = 5$ V. With one active arm of resistance $R + \Delta R$:

$$V_o = V_{ex}\left[\frac{R + \Delta R}{2R + \Delta R} - \frac{1}{2}\right] = V_{ex}\frac{\Delta R}{2(2R + \Delta R)}$$ $$V_o = 5\times\frac{0.378}{2(240.378)} = 3.931\times10^{-3}\ \text{V}$$

The usual small-change approximation gives

$$V_o \approx \frac{V_{ex}}{4}\frac{\Delta R}{R} = \frac{5}{4}(3.15\times10^{-3}) = 3.94\ \text{mV}$$

        +---R1=120---+---R_g=120+dR---+
        |            |                |
   5 V -+            o V_o            |
        |            |                |
        +---R3=120---+----R2=120------+

Answer: $\Delta R = 0.378\ \Omega$ ($0.315\%$); bridge output $\approx 3.93$ mV (3.94 mV by the approximate formula).

A Wheatstone bridge converts the small resistance changes of strain gauges into a measurable voltage.

Bridge output

Four arms $R_1$ (between A and B), $R_2$ (B-C), $R_3$ (C-D), $R_4$ (D-A). The excitation $V_{ex}$ is applied across A and C, and the output is measured between B and D.

          A
        /   \
      R1     R4
      /       \
     B         D     output V_o between B and D
      \       /
      R2     R3
        \   /
          C   (V_ex across A and C)

$$V_o = V_{ex}\left[\frac{R_1}{R_1 + R_2} - \frac{R_4}{R_3 + R_4}\right]$$

The bridge is balanced ($V_o = 0$) when $R_1R_3 = R_2R_4$.

Let each arm be $R$ initially and change by $\Delta R_i$. For small changes:

$$V_o = \frac{V_{ex}}{4}\left[\frac{\Delta R_1}{R} - \frac{\Delta R_2}{R} + \frac{\Delta R_3}{R} - \frac{\Delta R_4}{R}\right]$$

With $\Delta R/R = G\varepsilon$:

$$V_o = \frac{V_{ex}G}{4}\left(\varepsilon_1 - \varepsilon_2 + \varepsilon_3 - \varepsilon_4\right)$$

Adjacent arms subtract and opposite arms add, which is the basis of compensation.

Arrangements

Temperature compensation

Temperature changes the resistance of all gauges equally (the same $\Delta R_T$). Because adjacent arms subtract, this common change cancels. In a quarter bridge a dummy gauge (same type, same temperature, mounted on unstrained material) is placed in the adjacent arm. In half and full bridges the active gauges act as each other's compensators. Strain from bending or axial load can similarly be separated by placing gauges in the correct arms (e.g. on top and bottom of a beam).

Comments

• Output of a quarter bridge is slightly nonlinear for large $\Delta R/R$; half and full bridges are more linear.
• A three-wire connection compensates the lead resistance.
• Balancing and calibration (shunt calibration) are done before use.

Given: $E = 200$ GPa, $b = 20$ mm, $t = 4$ mm, $P = 10$ N, gauge distance $L = 250$ mm from the load, $G = 2.1$, $V_{ex} = 10$ V.

Bending moment and stress at the gauges

$$M = PL = 10 \times 0.25 = 2.5\ \text{N m}$$

Section modulus:

$$Z = \frac{bt^2}{6} = \frac{0.02 \times (0.004)^2}{6} = 5.333\times10^{-8}\ \text{m}^3$$ $$\sigma = \frac{M}{Z} = \frac{2.5}{5.333\times10^{-8}} = 46.9\ \text{MPa}$$

Strain

$$\varepsilon = \frac{\sigma}{E} = \frac{46.875\times10^{6}}{200\times10^{9}} = 2.344\times10^{-4} = 234\ \mu\varepsilon$$

(The same as $\varepsilon = 6PL/(Ebt^2)$.) The top gauges read $+\varepsilon$ (tension) and the bottom gauges $-\varepsilon$ (compression).

Bridge output

With the two tension gauges in opposite arms and the two compression gauges in the other two opposite arms, all four contributions add:

$$V_o = V_{ex}\,G\,\varepsilon = 10 \times 2.1 \times 2.344\times10^{-4} = 4.92\times10^{-3}\ \text{V}$$

        P (load)
         v
  fixed |============================|
  wall  | T1  T2 <- top gauges       |
        | C1  C2 <- bottom gauges    |
        |<-- 250 mm -->|

  Bridge:  T1 (arm 1)   C1 (arm 2)
           T2 (arm 3)   C2 (arm 4)

Answer: strain $= 234\ \mu\varepsilon$, bridge output $= 4.92$ mV (0.49 mV/V).

A ballast circuit is a simple potential-divider circuit in which the strain gauge $R_g$ is in series with a fixed ballast resistor $R_b$ across a supply $V$. The output is taken across the gauge. It is used mostly for dynamic strain, since a slow drift appears as a false signal.

   V o-----+---- R_b ----+----+
           |             |    |
           |            R_g   o--- V_o (across R_g)
           |             |    |
           +-------------+----+

Output change

$$V_o = V\frac{R_g}{R_g + R_b}$$

Differentiate with respect to $R_g$:

$$\frac{dV_o}{dR_g} = V\frac{R_b}{(R_g + R_b)^2}$$

For a small change $\Delta R_g$:

$$\Delta V_o = V\frac{R_bR_g}{(R_g + R_b)^2}\cdot\frac{\Delta R_g}{R_g}$$

Maximum sensitivity

Let $m = R_b/R_g$. Then $\Delta V_o = V\dfrac{m}{(1+m)^2}\dfrac{\Delta R_g}{R_g}$. Setting $d/dm[m/(1+m)^2] = 0$ gives $m = 1$, i.e. $R_b = R_g$. The maximum output change is

$$\Delta V_o = \frac{V}{4}\frac{\Delta R_g}{R_g}$$

(The same as a quarter bridge, but the circuit carries the dc level of $V/2$ so that the amplifier must block it with a capacitor.)

Numerical

$\Delta R_g/R_g = G\varepsilon = 2.0 \times 1000\times10^{-6} = 2\times10^{-3}$, $R_b = R_g$.

$$\Delta V_o = \frac{6}{4}(2\times10^{-3}) = 3\times10^{-3}\ \text{V}$$

Answer: $R_b = R_g$ for maximum sensitivity; $\Delta V_o = 3$ mV.