--- title: "Full-bridge cantilever load cell output" subject: "Instrumentation and Measurement" chapter: 6 key: "semester_4/instrumentation_and_measurement:cantilever-full-bridge-output-numerical" source: https://www.bctengineer.com/q/semester_4/instrumentation_and_measurement/cantilever-full-bridge-output-numerical site: BCT Engineering --- # Full-bridge cantilever load cell output Instrumentation and Measurement · Chapter 6: Strain Gage - Asked in: year not given, 8 marks - Video: [Full (Wheatstone) Bridge – Bending.](https://www.youtube.com/watch?v=ISMqdfvHKrM) — Micro-Measurements - a VPG Brand - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/instrumentation_and_measurement/cantilever-full-bridge-output-numerical - Whole chapter: https://www.bctengineer.com/md/semester_4/instrumentation_and_measurement/chapter-6 ## Question 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. ## 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. ## Answer 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 $$ \varepsilon = \frac{\sigma}{E} = \frac{46.875\times10^{6}}{200\times10^{9}} = 2.344\times10^{-4} = 234\ \mu\varepsilon $$ (This is 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: $$ V_o = V_{ex}\,G\,\varepsilon = 10 \times 2.1 \times 2.344\times10^{-4} = 4.92\times10^{-3}\ \text{V} $$ ```text 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 $= 234\ \mu\varepsilon$, bridge output $= 4.92$ mV (0.49 mV/V).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the bending moment at the gauge location, 250 mm from the 10 N load? A. 2.5 N m B. 25 N m C. 0.4 N m D. 2500 N m Answer: A. M = P L = 10 x 0.25 = 2.5 N m. (Key point: Bending moment) 2. Which expression gives the section modulus of the rectangular section (width b, thickness t)? A. b t / 6 B. b t^2 / 6 C. b^2 t / 6 D. b t^3 / 12 Answer: B. Z = b t^2 / 6 = 5.333e-8 m^3 for 20 mm x 4 mm. (Key point: Section modulus) 3. What is the strain at the gauges? A. About 234 microstrain B. About 469 microstrain C. About 23.4 microstrain D. About 1170 microstrain Answer: A. sigma = M/Z = 46.9 MPa, and strain = sigma/E = 46.9e6/200e9 = 2.34e-4. (Key point: Strain value) 4. With two tension and two compression gauges in a full bridge, the output is: A. Vex G e / 4 = 1.23 mV B. Vex G e / 2 = 2.46 mV C. Vex G e = 4.92 mV D. 2 Vex G e = 9.8 mV Answer: C. All four contributions add, so Vo = 10 x 2.1 x 2.344e-4 = 4.92 mV. 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