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Elastic transducers and strain gauge load cells

Short answer

Elastic force transducers measure force through elastic deformation; a proving ring measures diameter change. A column load cell with two axial and two transverse gauges in a full bridge gives Vo = -Vex G (1+nu) F/(2 A E), proportional to force and temperature compensated.

Explain how force is measured using elastic force transducers. Describe the proving ring and the strain-gauge column-type load cell, and derive the bridge output of a column load cell with four gauges (two axial, two transverse).

Principle: an applied force deforms an elastic member (a spring, ring, beam or column) within its elastic limit. The deflection or strain is proportional to the force (F=kxF = kx or ε=F/AE\varepsilon = F/AE) and is measured by a displacement transducer or by strain gauges.

Proving ring

A steel ring is loaded along a diameter. The change in diameter is proportional to the force.

        F
        v
      .---.
     /     \    deflection measured by
    |       |   micrometer / LVDT / strain gauges
     \     /
      '---'
        ^
        F

Advantages: high accuracy and good linearity; used to calibrate testing machines. Limitations: bulky, and each ring covers only a limited range.

Strain-gauge column load cell

A solid or hollow steel column carries the force. Four gauges are bonded: two axial (aligned with the load, in compression) and two transverse (at 90∘90^\circ, in tension because of the Poisson effect).

        F
        v
     +-----+   Axial gauges: R1 (front face), R3 (back face)
     | R1  |   Transverse gauges: R2 (front face), R4 (back face)
     | R2  |
     +-----+
        ^
        F (reaction)

Derivation of the bridge output

Axial strain (compressive): εa=−FAE\varepsilon_a = -\dfrac{F}{AE}. Transverse strain: εt=+νFAE\varepsilon_t = +\nu\dfrac{F}{AE}.

Connect the axial gauges R1R_1, R3R_3 in opposite arms and the transverse gauges R2R_2, R4R_4 in the other two arms. Then

Vo=VexG4(ε1−ε2+ε3−ε4)=VexG4[2εa−2εt]V_o = \frac{V_{ex}G}{4}\left(\varepsilon_1 - \varepsilon_2 + \varepsilon_3 - \varepsilon_4\right) = \frac{V_{ex}G}{4}\left[2\varepsilon_a - 2\varepsilon_t\right] Vo=VexG2(−FAE−νFAE)=−Vex G (1+ν)2AE FV_o = \frac{V_{ex}G}{2}\left(-\frac{F}{AE} - \nu\frac{F}{AE}\right) = -\frac{V_{ex}\,G\,(1+\nu)}{2AE}\,F

The output magnitude is proportional to FF and is (1+ν)/2(1+\nu)/2 times that of a full bridge with all four gauges at the axial strain. The arrangement also compensates for temperature (all four gauges change alike) and cancels bending (gauges of each pair are on opposite faces).

Example sensitivity: with G=2G = 2, ν=0.3\nu = 0.3, εa=1000 με\varepsilon_a = 1000\ \mu\varepsilon and Vex=10V_{ex}=10 V, Vo=13V_o = 13 mV (1.3 mV/V).

Other elastic members: cantilever beam (bending), S-type, shear-beam and ring-type cells, all with strain gauges wired as a bridge.

✓ Answer checked by Claude Opus 5.5 (high effort)

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