Question

1. (15) A truck scale is made of a platform and four compression force sensors, one at each corner of the platform. The sensor itself is a short steel cylinder, 22 mm in diameter. A single stain gauge is pre-stressed to 3% strain and bonded on the outer surface of the cylinder. The strain gauges have a nominal resistance (before pre-stressing) of 340 Ohms and a gauge factor of 6.9. The steel used for the cylinders has a modulus of elasticity of 30 GPa. Calculate: a. The maximum truck weight that the scale can measure. b. The change in resistance of the sensors for maximum weight. c. The sensitivity of the scale assuming the response of the strain gauges is linear.

Answer 1

Answer:

a). 139498.24 kg

b). 281.85 ohm

c). 10.2 ohm

Explanation:

Given :

Diameter, d = 22 m

Linear strain, $\epsilon$ = 3%

                        = 0.03

Young's modulus, E = 30 GPa

Gauge factor, k = 6.9

Gauge resistance, R = 340 Ω

a). Maximum truck weight

σ = Eε

σ = $0.03 \times 30 \times 10^9$

$\frac{P}{A} =0.03 \times 30 \times 10^9$

$P = 0.03 \times 30 \times 10^9\times \frac{\pi}{4}\times (0.022)^2$

 = 342119.44 N

For the four sensors,

Maximum weight = 4 x P

                            =  4 x 342119.44

                            = 1368477.76 N

Therefore, weight in kg is $m=\frac{W}{g}=\frac{1368477.76}{9.81}$

                     m = 139498.24 kg

b). Change in resistance

$k=\frac{\Delta R/R}{\Delta L/L}$

$\Delta R = k. \epsilon R$    , since $\epsilon= \Delta L/ L$

$\Delta R = 6.9 \times 0.03 \times 340$

$\Delta R = 70.38$ Ω

For 4 resistance of the sensors,

$\Delta R = 70.38 \times 4 = 281.52$ Ω

c). $k=\frac{\Delta R/R}{\epsilon}$

If linear strain,

$\frac{\Delta R}{R} \approx \frac{\Delta L}{L}$  , where k = 1

$\Delta R = \frac{\Delta L}{L} \times R$

$\Delta R = 0.03 \times 340$

$\Delta R = 10.2$ Ω