Common Mechanical Measurement System and Transducers

Principle: an applied force deforms an elastic member (a spring, ring, beam or column) within its elastic limit. The deflection or strain is proportional to force ($F = kx$ or $\varepsilon = F/AE$) and is measured by a displacement transducer or strain gauges.

Proving ring

A steel ring loaded along a diameter. Diametral deflection is proportional to force.

        F
        v
      .---.
     /     \    deflection measured by
    |       |   micrometer / LVDT / strain gauges
     \     /
      '---'
        ^
        F

Advantages: high accuracy, used to calibrate testing machines, good linearity. Limitations: bulky, a limited range for each ring.

Strain-gauge column load cell

A solid or hollow steel column carries the force. Four gauges are bonded: two axial (aligned with the load, compression) and two transverse (at $90^\circ$, which strain in tension by Poisson's effect).

        F
        v
     +-----+   Axial gauges: R1, R3
     | R1  |   Transverse gauges: R2, R4
     | R2  |
     +-----+
        ^
        F (reaction)

Derivation of the bridge output

Axial strain (compressive): $\varepsilon_a = -\dfrac{F}{AE}$. Transverse strain: $\varepsilon_t = +\nu\dfrac{F}{AE}$.

Connect axial gauges $R_1$, $R_3$ in opposite arms and transverse gauges $R_2$, $R_4$ in the other two arms. Then

$$V_o = \frac{V_{ex}G}{4}\left(\varepsilon_1 - \varepsilon_2 + \varepsilon_3 - \varepsilon_4\right) = \frac{V_{ex}G}{4}\left[2\varepsilon_a - 2\varepsilon_t\right]$$ $$V_o = \frac{V_{ex}G}{2}\left(-\frac{F}{AE} - \nu\frac{F}{AE}\right) = -\frac{V_{ex}\,G\,(1+\nu)}{2AE}\,F$$

The output magnitude is proportional to $F$ and is $(1+\nu)/2$ times that of a full-bridge. The arrangement also compensates temperature (all four gauges change alike) and cancels bending (the gauges on opposite sides are paired).

Example sensitivity: with $G = 2$, $\nu = 0.3$, $\varepsilon_a = 1000\ \mu\varepsilon$, $V_{ex}=10$ V, $V_o = 13$ mV (1.3 mV/V).

Other elastic members: cantilever beam (bending), S-type, shear-beam and ring-type cells, all with gauges as bridge.

(a) Piezoelectric load cell

Certain crystals (quartz, Rochelle salt, barium titanate PZT) develop an electric charge on their faces when a force deforms them (the direct piezoelectric effect):

$$q = d\,F$$

where $d$ is the charge sensitivity (C/N). The crystal is a self-generating (active) transducer, mounted between two plates in a stiff housing. The output is equivalent to a charge source $q$ in parallel with a capacitance $C$, so the open-circuit voltage is $V = q/C$.

        F
        v
   [=========]  <- top plate
   | crystal |  charge +q / -q on faces
   [=========]
   housing -> output

Why a charge amplifier: the voltage $q/C$ depends on the cable capacitance, so changing the cable changes the reading. A charge amplifier (op-amp with feedback capacitor $C_f$) gives $V_o = -q/C_f$, which is independent of cable length and has a high input impedance.

Why not static: the charge leaks away through the insulation and the amplifier input resistance $R$. With a time constant $\tau = RC$, the output decays for a constant force, so only dynamic (and quasi-static) forces can be measured. Advantages: very high stiffness, high natural frequency, wide dynamic range, small size.

(b) Numerical

Charge generated:

$$q = dF = 2.3\times10^{-12} \times 500 = 1.15\times10^{-9}\ \text{C} = 1.15\ \text{nC}$$

Open-circuit voltage ($C = 1.2$ nF):

$$V = \frac{q}{C} = \frac{1.15\times10^{-9}}{1.2\times10^{-9}} = 0.958\ \text{V}$$

Charge amplifier output ($C_f = 100$ pF):

$$V_o = \frac{q}{C_f} = \frac{1.15\times10^{-9}}{100\times10^{-12}} = 11.5\ \text{V}$$

Lower cut-off frequency ($R = 1$ G$\Omega$, $C = 1.2$ nF):

$$f_c = \frac{1}{2\pi RC} = \frac{1}{2\pi(10^{9})(1.2\times10^{-9})} = 0.133\ \text{Hz}$$

Answer: $q = 1.15$ nC; $V = 0.958$ V; charge-amplifier output $= 11.5$ V; $f_c = 0.133$ Hz.

Both cells convert force into a fluid pressure, which is measured by a pressure gauge or transducer.

Hydraulic load cell

The force acts on a piston that bears on a liquid (oil) held in a chamber and sealed by a rolling diaphragm. The fluid pressure rises:

$$p = \frac{F}{A}$$

and the pressure gauge, calibrated in force units, shows the load.

      F
      v
   [=======] <- loading plate
   |piston |
   |-------|  diaphragm seal
   | oil   |---> pressure gauge / transducer
   |_______|

• Almost frictionless because of the diaphragm.
• Needs no electric power; suitable for hazardous areas.
• Useful for heavy loads (tank weighing, bridge testing, truck scales) and remote reading.
• Temperature changes affect the oil volume and pressure.

Pneumatic load cell

Force is balanced by air pressure acting on a diaphragm. A flapper-nozzle detects the position of the diaphragm. Air supply pressure rises until the balancing force $pA = F$ is reached, and the back pressure is read.

   Air supply --[restrictor]--+--> gauge
                              |
                      nozzle  |
                     ======== flapper moved by F
                        F v
                   [diaphragm]

• Intrinsically safe, clean and inexpensive.
• Insensitive to temperature, unlike hydraulic cells; no fluid hazard.
• Typical for food, pharmaceutical and weighing of small loads (up to about 20 kN), with high accuracy.
• Limited by the response lag (compressible air, tubing volume).

Comparison with strain gauge cells

(a) Classification of dynamometers

A dynamometer measures torque, and with speed the shaft power of an engine, motor or turbine.

Absorption types are used for engine testing, and transmission types are used for in-line measurement on shafts.

(b) Rope brake

Brake torque is the net tangential force times the effective radius. The rope has finite thickness, so the effective radius is measured to the rope's centre line:

$$r_e = \frac{D + d}{2} = \frac{1.0 + 0.025}{2} = 0.5125\ \text{m}$$

Net load on the brake: $W - S = 600 - 50 = 550$ N.

$$T = (W - S)\,r_e = 550 \times 0.5125 = 281.9\ \text{N m}$$

Brake power:

$$P = \frac{2\pi N T}{60} = \frac{2\pi \times 300 \times 281.9}{60} = 8855\ \text{W} = 8.86\ \text{kW}$$

        drum (rotating)
         .---.
        ( r_e )----- rope ends: W (dead load) below
         '---'                  S (spring balance) above

Answer: brake torque $= 281.9$ N m, brake power $= 8.86$ kW.

Torque measurement by strain gauges

A shaft in torsion has pure shear at its surface. The principal stresses are $\pm\tau$ along directions $45^\circ$ to the axis (tension on one diagonal, compression on 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 ignores axial load and bending (they strain all gauges equally) and compensates temperature. The bridge supply and output are carried to the rotating shaft by slip rings, or a telemetry/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 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.

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

$$\varepsilon_{top} = \varepsilon_a + \varepsilon_b, \qquad \varepsilon_{bottom} = \varepsilon_a - \varepsilon_b$$

Solving:

$$\varepsilon_a = \frac{\varepsilon_{top} + \varepsilon_{bottom}}{2}, \qquad \varepsilon_b = \frac{\varepsilon_{top} - \varepsilon_{bottom}}{2}$$

Calculation

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

$$\varepsilon_a = \frac{600 - 200}{2} = 200\ \mu\varepsilon, \qquad \varepsilon_b = \frac{600 + 200}{2} = 400\ \mu\varepsilon$$

Area: $A = 0.02 \times 0.01 = 2\times10^{-4}$ m$^2$.

$$F = EA\varepsilon_a = 200\times10^{9}\times 2\times10^{-4}\times 200\times10^{-6} = 8000\ \text{N} = 8\ \text{kN}$$

Section modulus about the bending axis:

$$Z = \frac{bh^2}{6} = \frac{0.02 \times (0.01)^2}{6} = 3.333\times10^{-7}\ \text{m}^3$$ $$M = EZ\varepsilon_b = 200\times10^{9}\times 3.333\times10^{-7}\times 400\times10^{-6} = 26.7\ \text{N m}$$

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

Wiring to separate the effects

• For bending moment only: use gauges on the top and bottom faces in adjacent arms of a half bridge (or two on each face in a full bridge). Axial strain is equal on both faces, so it cancels in adjacent arms, and the bending strain adds. Output $= V_{ex}G\varepsilon_b/2$.
• For axial force only: use gauges on the top and bottom faces in opposite arms of a half bridge (or in a full bridge with Poisson gauges). Bending strain is opposite in sign and cancels, and axial strain adds.

 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)

Both bridges can use the same two gauges, with separate bridge completion resistors and two separate readout channels.