# Testing Bridge

For testing a bridge, there are used four strain gauges ($R_1-R_4$) glued to it. The bridge is made of 30 ft. long I beams, each having a height of 3 ft. $R_2$ and $R_4$ are glued on top while the other two are placed on the bottom. A strain gauge is a 1 inch long resistor (like a stamp) that can be glued to the surface to be tested. The resistance depends on the elongation as $R=R_0(1+(Delta L/L))$ where $R_0=600 Omega$. The resistors placed under the beam $R_1$ and $R_3$ expand while the other two resistors on top contract. If the deflection of the beam is $x$ please find the dependence of the voltage $V_x$ on $r$ if the voltage source of the bridge is $E=18 V$. What is the gain for a reading of $1 Volt/1 foot$? For a 10 A battery, what is the time before it gets drained?

By considering A as ground (zero reference) one has

$V(R_1 )=E*R_1/(R_1+R_4)$ and $V(R_2 )=E*R_2/(R_2+R_3)$

So that the voltage on the diagonal of the bridge is

$V_{diag}=V(R_1)-V(R_2)=E*(frac{R_1}{R_1+R_4}-frac{R_2}{R_2+R_3})$

One has

$R=R_0 [1+(ΔL/L_0)]$

Since $R_1$ and $R_3$ extend and $R_2$ and $R_4$ contract and all have the same initial resistance $R_0$ then

$V_{diag}=E*left [frac{1+(ΔL/L_0)}{1+(ΔL/L_0) +1-(ΔL/L_0)}-frac{1-(ΔL/L_0)}{1-(ΔL/L_0) +1+(ΔL/L_0)}right ]=E*frac{2ΔL/L_0}{2}=E*frac{ΔL}{L_0}$

$V_x=(E/L_0) *x$

The deflection is 1 foot for 30 foot long beam. Same deflection is found on the gauges (see above figure). Therefore

$ΔL/L_0 =1/30$ and $V_{diag}=E*(ΔL/L_0) =18/30=0.6 V$

For 1 volt calibration the amplifier gain is

$G=1/0.6=1.67$

If $R_0=600 ohm$ the total resistance of the bridge is (2 resistors series and these two groups in parallel)

$1/R_{tot} =1/(R_0+R_0 )+1/(R_0+R_0 )=1/R_0$ so that $R_{tot}=R_0$

Current from source is

$I=E/R_{tot} =18/600=0.03 A$

Assuming the current stays the same with time a $10 Ah$ battery will last of

$t=(10 Ah)/(0.03 A)=333.33 hours$