MethodWhat is calculated?
Pure torsion produces principal normal stresses at ±45° to the axis equal in magnitude to the shear stress, with opposite signs. Two perpendicular gauge pairs at 45°, wired alternately into a full bridge, add all four strain magnitudes and are insensitive to axial force, bending and temperature. Keil's example: hollow shaft 24/14 mm gives W_p = 2400 mm³; at 400 N·m τ = 167 N/mm² and ε = 1054 µm/m.
Equations
τ = M_t/W_p (Keil Gl. 10.35)
W_p = π/(16d_a)·(d_a⁴ − d_i⁴) (Gl. 10.36)
ε₁ = −ε₂ = τ·(1+ν)/E (Gl. 10.38, 10.39)
U_M/U_B = k·ε₁ (Vollbrücke, Gl. 10.41)
ProcedureStep by step
- Enter torque, diameters and material data.
- Choose the full bridge with two 45° pairs 180° apart to compensate bending.
- Match the signal with the amplifier range and rated strain (≈ 1000 µm/m).
Typical mistake
Using longitudinal strain instead of 45° strain or computing τ with the bending section modulus.
PracticeApplication and limits
Torque measuring shafts, drivetrain measurements and retrofit instrumentation of shafts, including slip-ring or telemetry transmission.
Circular or annular section without keyways, Saint-Venant torsion; misalignment of the 45° gauges reduces the signal and couples in bending.
SourceTechnical basis
Keil, Dehnungsmessstreifen, 2nd ed. 2017, sec. 10.2.3 torsionally loaded objects, eqs. (10.35)–(10.41); sec. 4.5, eq. (4.18).
The source supports the equation structure and worked examples; this calculator does not replace calibration of the measuring chain.
FAQFrequently asked questions
Why ε = τ(1+ν)/E?
From Hooke's law with σ₁ = −σ₂ = τ, ε₁ = (σ₁ − νσ₂)/E = τ(1+ν)/E, equivalent to τ/(2G).
Is one gauge pair enough?
As a half bridge yes, but bending is then not compensated; Keil recommends two opposite pairs.
How do long rotor leads matter?
Keil's fig. 10.20 shows leads should be placed in adjacent bridge arms so their resistance changes cancel.
In the same cluster
All strain-gauge calculators