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Full-bridge cantilever load cell output

Short answer

Bending moment is 2.5 N m, section modulus 5.333e-8 m^3, stress 46.9 MPa, strain 234 microstrain. Full bridge output is Vex G epsilon = 10 x 2.1 x 2.344e-4 = 4.92 mV.

A steel cantilever beam (E = 200 GPa) of width 20 mm and thickness 4 mm carries a 10 N load at its free end. Two strain gauges (gauge factor 2.1) are bonded on the top surface and two on the bottom surface at 250 mm from the load, connected as a full bridge with 10 V excitation. Calculate the strain at the gauge location and the bridge output voltage. Draw the gauge positions and the bridge connection.

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

Bending moment and stress at the gauges

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

Section modulus:

Z=bt26=0.02×(0.004)26=5.333×10−8 m3Z = \frac{bt^2}{6} = \frac{0.02 \times (0.004)^2}{6} = 5.333\times10^{-8}\ \text{m}^3 σ=MZ=2.55.333×10−8=46.9 MPa\sigma = \frac{M}{Z} = \frac{2.5}{5.333\times10^{-8}} = 46.9\ \text{MPa}

Strain

ε=σE=46.875×106200×109=2.344×10−4=234 με\varepsilon = \frac{\sigma}{E} = \frac{46.875\times10^{6}}{200\times10^{9}} = 2.344\times10^{-4} = 234\ \mu\varepsilon

(This is the same as ε=6PL/(Ebt2)\varepsilon = 6PL/(Ebt^2).) The top gauges read +ε+\varepsilon (tension) and the bottom gauges −ε-\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:

Vo=Vex G ε=10×2.1×2.344×10−4=4.92×10−3 VV_o = V_{ex}\,G\,\varepsilon = 10 \times 2.1 \times 2.344\times10^{-4} = 4.92\times10^{-3}\ \text{V} 
  fixed |                                  P (load)
  wall  |                                  v
        |==================================|  <- free end
        |     T1 T2  (top gauges)
        |     C1 C2  (bottom gauges)
              |<---------- 250 mm -------->|

  Bridge (going round the four arms):
     arm 1: T1   arm 2: C1   arm 3: T2   arm 4: C2
  (T1, T2 in opposite arms; C1, C2 in opposite arms)

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

✓ Answer checked by Claude Opus 5.5 (high effort)

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