Shaft torque measurement using strain gauges

Torque measurement by strain gauges

A shaft in torsion is in pure shear at its surface. The principal stresses are $\pm\tau$ along directions at $45^\circ$ to the axis (tension along one diagonal, compression along the other). Four gauges are bonded at $\pm45^\circ$, two on each diagonal, and connected as a full bridge so that all four outputs add. This arrangement is insensitive to axial load and bending (their strains cancel in the bridge) and compensates for temperature. The bridge supply and output are carried to and from the rotating shaft by slip rings, or by a telemetry or rotary-transformer system.

   shaft axis ------------------------>
        /  \      gauges at +45 and -45 degrees
       / R1 \     R1, R3 : +45 (tension)
      / R2   \    R2, R4 : -45 (compression)

Strain at $45^\circ$ in pure shear:

$$\varepsilon_{45} = \frac{\gamma}{2} = \frac{\tau}{2G}$$

The torque is then found from $\tau = 16T/(\pi d^3)$.

Calculation

Given: $d = 40$ mm, $G = 80$ GPa, $\varepsilon = 400\ \mu\varepsilon$, $N = 1500$ rpm.

Shear stress:

$$\tau = 2G\varepsilon = 2(80\times10^{9})(400\times10^{-6}) = 64\times10^{6}\ \text{Pa} = 64\ \text{MPa}$$

Torque:

$$T = \frac{\pi d^3\tau}{16} = \frac{\pi(0.04)^3(64\times10^{6})}{16} = 804.2\ \text{N m}$$

Power:

$$P = \frac{2\pi N T}{60} = \frac{2\pi(1500)(804.2)}{60} = 126.3\times10^{3}\ \text{W} = 126.3\ \text{kW}$$

Bridge output (all four arms active):

$$V_o = V_{ex}\,G_f\,\varepsilon = 10 \times 2.1 \times 400\times10^{-6} = 8.4\ \text{mV}$$

Answer: $\tau = 64$ MPa, $T = 804$ N m, $P = 126.3$ kW, bridge output $= 8.4$ mV.