Strain gauges, load cells and torque sensors

Why it matters

Force and torque are rarely measured directly — they are inferred from the tiny strain they produce in an elastic element. Strain gauges inside load cells weigh trucks and dose powders, and torque transducers on shafts verify motor and gearbox performance, so every mechatronics engineer must be able to design and read a strain-gauge bridge.

Key ideas

Strain gauge principle

• A bonded foil (or wire) grid changes resistance when strained, because its length and cross-section change (geometric effect) and its resistivity changes (piezoresistive effect).
• The gauge factor G = (ΔR/R)/ε combines both: about 2 for constantan foil gauges; 50–200 for semiconductor (silicon, piezoresistive) gauges, which are more sensitive but more nonlinear and temperature-sensitive.
• Typical nominal resistances are 120 Ω and 350 Ω. Strains are small — engineering structures see a few hundred to about 2000 microstrain (µε, 10⁻⁶) — so ΔR is a fraction of an ohm and a Wheatstone bridge is used to read it.
• Gauges measure strain only along their grid axis, averaged over the grid length. Bonding quality, surface preparation and gauge alignment directly affect accuracy.

Bridge arrangements (excitation V_ex, equal arms R, small strains)

• Quarter bridge (one active gauge): V_o ≈ V_ex·G·ε/4.
• Half bridge (two active gauges with opposite strains, e.g. top and bottom of a beam): V_o ≈ V_ex·G·ε/2.
• Full bridge (four active gauges, two in tension and two in compression): V_o = V_ex·G·ε.
• Gauges in adjacent arms subtract; gauges in opposite arms add. This is how bridges cancel unwanted effects: a temperature change or axial load that strains all gauges equally cancels out, while bending (opposite signs) adds.
• Temperature compensation: a dummy gauge on an unstrained piece of the same material in an adjacent arm, or self-temperature-compensated gauges matched to the substrate's expansion. Three-wire connection compensates lead resistance in quarter bridges.

Load cells

• An elastic element (cantilever beam, S-beam, shear beam, column or ring) carries the load; gauges are placed where strain is high and of opposite signs, wired as a full bridge.
• Rated output is quoted in mV/V at rated capacity (commonly 1–3 mV/V). Output at a load W is (W/capacity) × rated output × V_ex.
• Other types: hydraulic and pneumatic load cells (pressure of a trapped fluid), piezoelectric force sensors (dynamic forces only, charge leaks away), capacitive force sensors.

Torque sensors

• A shaft in pure torsion has maximum principal strains at ±45° to the axis, equal in magnitude and opposite in sign. Four gauges at ±45° in a full bridge give the maximum output and cancel bending and axial strain.
• Reaction (static) torque sensors mount between a motor and its frame; rotary (in-line) sensors spin with the shaft and transmit power and signal through slip rings or, more commonly now, a rotary transformer or telemetry. Non-contact magnetoelastic and phase-shift (twist angle between two toothed wheels) methods also exist.

Formulas

G = (ΔR/R) / ε → ΔR = G·ε·R

• G: gauge factor (dimensionless), R: gauge resistance (Ω), ε: strain (m/m).

V_o = (V_ex/4)·(ΔR₁/R − ΔR₂/R + ΔR₃/R − ΔR₄/R)

• General bridge output (V) for small changes, arms numbered round the bridge so that 1 and 3 are opposite. Signs show which arms add.

V_o ≈ V_ex·G·ε/4 (quarter), V_ex·G·ε/2 (half), V_ex·G·ε (full)

σ = 6F·x / (b·h²) ; ε = σ/E

• Surface bending stress (Pa) of a rectangular cantilever at distance x (m) from the load F (N); b: width, h: thickness (m); E: Young's modulus (Pa).

τ = 16T / (π·d³) ; ε_±45 = ±τ / (2G_s)

• τ: surface shear stress (Pa) in a solid shaft of diameter d (m) under torque T (N·m); G_s: shear modulus (Pa). Principal strains at ±45°.

Worked examples

Example 1 (standard) — quarter bridge. A 120 Ω gauge with G = 2.1 sees 500 µε; bridge excitation is 5 V. Find ΔR and V_o.

1. ΔR = G·ε·R = 2.1 × 500 × 10⁻⁶ × 120 = 0.126 Ω.
2. V_o ≈ V_ex·G·ε/4 = 5 × 2.1 × 500 × 10⁻⁶ / 4 = 1.3125 × 10⁻³ V.
3. Answer: ΔR = 0.126 Ω, V_o ≈ 1.31 mV — this is why bridge outputs need an instrumentation amplifier.

Example 2 (GATE level) — cantilever load cell. A steel cantilever (E = 200 GPa), 20 mm wide and 5 mm thick, carries F = 50 N. Four gauges (G = 2) sit 100 mm from the load, two on top and two on the bottom, wired as a full bridge with 10 V excitation. Find the strain and output.

1. σ = 6F·x/(b·h²) = 6 × 50 × 0.1 / (0.02 × 0.005²) = 30 / (5 × 10⁻⁷) = 60 × 10⁶ Pa = 60 MPa.
2. ε = σ/E = 60 × 10⁶ / 200 × 10⁹ = 300 × 10⁻⁶ (300 µε): + on top gauges, − on bottom gauges.
3. Full bridge: V_o = V_ex·G·ε = 10 × 2 × 300 × 10⁻⁶ = 6.0 × 10⁻³ V.
4. Answer: ε = 300 µε, V_o = 6.0 mV (0.6 mV/V).

Example 3 (GATE level) — torque shaft. A 25 mm solid steel shaft (G_s = 80 GPa) carries 100 N·m. Four gauges at ±45° (G = 2) form a full bridge at 10 V.

1. τ = 16T/(πd³) = 1600 / (π × 0.025³) = 1600 / 4.909 × 10⁻⁵ = 32.6 MPa.
2. ε = τ/(2G_s) = 32.6 × 10⁶ / 160 × 10⁹ = 203.7 µε.
3. V_o = 10 × 2 × 203.7 × 10⁻⁶ = 4.07 mV.

Common mistakes

• Using V_o = V_ex·(ΔR/R) for a single gauge — a quarter bridge gives one quarter of that.
• Putting two gauges with the same strain sign in adjacent arms, so their outputs cancel.
• Forgetting microstrain is 10⁻⁶; 500 µε is 0.0005, not 0.5.
• Mounting torque gauges along the shaft axis — under pure torsion the axial strain is zero; use ±45°.
• Ignoring temperature: a few °C change can produce apparent strain comparable to the real signal without compensation.
• Confusing mV/V rated output with mV output; multiply by excitation voltage.

For GATE ME

Expect NAT problems on ΔR from G and ε, quarter/half/full bridge output, cantilever or shaft strain feeding a bridge, and load-cell output from mV/V rating. MCQs test gauge factor values, temperature compensation with dummy gauges, ±45° placement for torque, and which bridge arms add or cancel. Revise bending and torsion formulas from strength of materials alongside this topic.

Quick check

1. What is the approximate gauge factor of a constantan foil gauge?
2. A full-bridge load cell rated 2 mV/V is excited at 10 V and loaded to half capacity. Output?
3. Where are gauges placed on a shaft to measure torque, and why?
4. How does a dummy gauge compensate for temperature?
5. Why is a Wheatstone bridge used instead of measuring R directly?

Answers: 1. About 2. 2. 10 mV. 3. At ±45° to the axis, along the principal strain directions of pure shear. 4. It sits in an adjacent arm and sees the same thermal resistance change but no strain, so the thermal effects cancel. 5. ΔR is tiny (fractions of an ohm), and the bridge converts it into a near-zero-based voltage that can be amplified, with built-in cancellation of common effects.