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Sagot :
Answer:
a). 139498.24 kg
b). 281.85 ohm
c). 10.2 ohm
Explanation:
Given :
Diameter, d = 22 m
Linear strain, [tex]$\epsilon$[/tex] = 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 = [tex]$0.03 \times 30 \times 10^9$[/tex]
[tex]$\frac{P}{A} =0.03 \times 30 \times 10^9$[/tex]
[tex]$P = 0.03 \times 30 \times 10^9\times \frac{\pi}{4}\times (0.022)^2$[/tex]
 = 342119.44 N
For the four sensors,
Maximum weight = 4 x P
              =  4 x 342119.44
              = 1368477.76 N
Therefore, weight in kg is [tex]$m=\frac{W}{g}=\frac{1368477.76}{9.81}$[/tex]
           m = 139498.24 kg
b). Change in resistance
[tex]k=\frac{\Delta R/R}{\Delta L/L}[/tex]
[tex]$\Delta R = k. \epsilon R$[/tex] Â Â , since [tex]$\epsilon= \Delta L/ L$[/tex]
[tex]$\Delta R = 6.9 \times 0.03 \times 340$[/tex]
[tex]$\Delta R = 70.38 $[/tex] Ω
For 4 resistance of the sensors,
[tex]$\Delta R = 70.38 \times 4 = 281.52$[/tex] Ω
c). [tex]$k=\frac{\Delta R/R}{\epsilon}$[/tex]
If linear strain,
[tex]$\frac{\Delta R}{R} \approx \frac{\Delta L}{L}$[/tex] Â , where k = 1
[tex]$\Delta R = \frac{\Delta L}{L} \times R$[/tex]
[tex]$\Delta R = 0.03 \times 340$[/tex]
[tex]$\Delta R = 10.2 $[/tex] Ω
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