Question: Explain in detail - Resistance Strain Gauges.
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 modified 18 days ago by Abhishek Tiwari ♦♦ 50 written 3.3 years ago by Ramnath • 3.8k
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The strain gauge is one of the most widely used strain measurement sensors. It is a resistive elastic unit whose change in resistance is a function of applied strain.

$$\frac{dR}R=S.ε$$

whereR is the resistance, εis the strain, and S is the gauge factor.

Among strain gauges, an electric resistance wire strain gauge has the advantages of lower cost and being an established product. Thus it is the most commonly used type of device. Other types of strain gauges are acoustic, capacitive, inductive, mechanical, optical, piezo-resistive, and semi-conductive.

A wire strain gauge is made by a resistor, usually in metal foil form, bonded on an elastic backing. Its principle is based on fact that the resistance of a wire increases with increasing strain and decreases with decreasing strain, as first reported by Lord Kelvin in 1856. Consider a wire strain gauge, as illustrated above. The wire is composed of a uniform conductor of electric resistivity with lengthl and cross-section area A. Its resistance R is a function of the geometry given by

$$R=\frac{ρ.l}A$$

The resistance change rate is a combination effect of changes in length, cross-section area, and resistivity.

$$dR=\frac{ρ}A dl-\frac{ρl}{A^2} dA+\frac{l}A dρ \\ \; \\ →\frac{dR}R=\frac{dl}{l}-\frac{dA}A+\frac{dρ}ρ$$

When the strain gauge is attached and bonded well to the surface of an object, the two are considered to deform together. The strain of the strain gauge wire along the longitudinal direction is the same as the strain on the surface in the same direction.

$$ε_l=\frac{dl}l$$

However, its cross-sectional area will also change due to the Poisson's ratio. Suppose that the wire is cylindrical with initial radius r. The normal strain along the radial direction is

$$ε_r=\frac{dr}r=-v.ε_l=-v \frac{dl}l$$

The change rate of cross-section area is twice as the radial strain, when the strain is small.

$$\frac{dA}A=(1+ε_r )^2-1=2ε_r+ε_r^2≈2ε_r \\ =-2v \frac{dl}l$$

The resistance change rate becomes

$$\frac{dR}R=\frac{dl}l-\frac{dA}A+\frac{dρ}ρ=(1+2v) \frac{dl}l+\frac{dρ}ρ \\ =(1+2v) ε_l+\frac{dρ}ρ$$

For a given material, the sensitivity of resistance versus strain can be calibrated by the following equation.

$$S≜\frac{\frac{dR}R}{ε_l} \\ =1+2v+\frac{\frac{dρ}ρ}{ε_l}$$

When the gauge factor S is given (usually provided by strain gauge vendors), the average strain at the point of attachment of the strain gauge can be obtained by measuring the change in electric resistance of the strain gauge.

$$ε_l=\frac{\frac{dR}R}S≈\frac{∆R}{SR}$$

 written 3.3 years ago by Ramnath • 3.8k