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Combined force and moment measurement

Short answer

Top/bottom strains give epsilon_a=200 microstrain and epsilon_b=400 microstrain. Axial force=EA epsilon_a=8 kN tension, bending moment=EZ epsilon_b=26.7 N m. Wire top and bottom gauges adjacent to measure bending only, or opposite arms to measure axial only.

A rectangular steel bar (E = 200 GPa), 20 mm wide and 10 mm deep, carries both an axial force and a bending moment about the 20 mm-wide axis. Strain gauges on the top and bottom surfaces read +600 microstrain and -200 microstrain. Find the axial force and the bending moment, and show how the gauges can be wired so that each effect is measured separately.

A bar loaded by an axial force FF and a bending moment MM has a surface strain that is the sum of an axial part εa\varepsilon_a (the same on all faces) and a bending part ±εb\pm\varepsilon_b (opposite on the top and bottom).

εtop=εa+εb,εbottom=εa−εb\varepsilon_{top} = \varepsilon_a + \varepsilon_b, \qquad \varepsilon_{bottom} = \varepsilon_a - \varepsilon_b

Solving:

εa=εtop+εbottom2,εb=εtop−εbottom2\varepsilon_a = \frac{\varepsilon_{top} + \varepsilon_{bottom}}{2}, \qquad \varepsilon_b = \frac{\varepsilon_{top} - \varepsilon_{bottom}}{2}

Calculation

Given: εtop=+600 με\varepsilon_{top} = +600\ \mu\varepsilon, εbottom=−200 με\varepsilon_{bottom} = -200\ \mu\varepsilon, E=200E = 200 GPa, b=20b = 20 mm, h=10h = 10 mm.

εa=600−2002=200 με,εb=600+2002=400 με\varepsilon_a = \frac{600 - 200}{2} = 200\ \mu\varepsilon, \qquad \varepsilon_b = \frac{600 + 200}{2} = 400\ \mu\varepsilon

Area: A=0.02×0.01=2×10−4A = 0.02 \times 0.01 = 2\times10^{-4} m2^2.

F=EAεa=200×109×2×10−4×200×10−6=8000 N=8 kNF = EA\varepsilon_a = 200\times10^{9}\times 2\times10^{-4}\times 200\times10^{-6} = 8000\ \text{N} = 8\ \text{kN}

Section modulus about the bending axis:

Z=bh26=0.02×(0.01)26=3.333×10−7 m3Z = \frac{bh^2}{6} = \frac{0.02 \times (0.01)^2}{6} = 3.333\times10^{-7}\ \text{m}^3 M=EZεb=200×109×3.333×10−7×400×10−6=26.7 N mM = EZ\varepsilon_b = 200\times10^{9}\times 3.333\times10^{-7}\times 400\times10^{-6} = 26.7\ \text{N m}

Answer: axial force =8= 8 kN (tension), bending moment =26.7= 26.7 N m.

Wiring to separate the effects

  • For bending moment only: put a top gauge and a bottom gauge in adjacent arms of a half bridge (or two on each face in a full bridge). The axial strain is equal on both faces, so it cancels in adjacent arms, while the bending strain adds. Output =VexGεb/2= V_{ex}G\varepsilon_b/2.
  • For axial force only: put a top gauge and a bottom gauge in opposite arms of the bridge, with dummy gauges or fixed resistors in the other two arms (or use a full bridge with Poisson gauges). The bending strain is opposite in sign and cancels, while the axial strain adds. Output =VexGεa/2= V_{ex}G\varepsilon_a/2.
 Bending bridge:  R_top (arm 1), R_bot (arm 2)  adjacent
 Axial bridge:    R_top (arm 1), R_bot (arm 3)  opposite
 (other arms: fixed resistors or dummy gauges)

In practice two gauges are bonded on each face: one top-bottom pair is wired into the bending bridge and the other pair into the axial bridge, each with its own excitation and readout channel (or a switch box connects the same pair to each circuit in turn).

✓ Answer checked by Claude Opus 5.5 (high effort)

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