Question

Mechanical lab 1

Answer 1

I) Refer diagram below:

0 2

Let the initial resistance of all the gages and R₃ be R ohms

Let the gage factor br S_g

Let the applied voltage br V volts

From the equation of Wheatstone bridge with strain gages

$\Delta E=\frac{r}{(1+r)^2}\left ( \frac{\Delta R_1}{R_1}- \frac{\Delta R_2}{R_2}+ \frac{\Delta R_3}{R_3}- \frac{\Delta R_4}{R_4}\right ) V$

$r=\frac{R_2}{R_1}=\frac{R_3}{R_4}=\frac{R}{R}=1$

Since R₂ is aligned perpendicular to strain direction

$\epsilon_y=-\nu\epsilon_g$

$\frac{\Delta R_2}{R_2}=-S_g\nu\epsilon_g+S_g\epsilon_{temp}$

Since R₃ is not loaded with strain gage

$\Delta R_3=0$

$\frac{\Delta R_1}{R_1}=\frac{\Delta R_4}{R_4}=S_g\epsilon_g+S_g\epsilon_{temp}$

Hence

$\Delta E=\frac{1}{2^2}\left ( S_g\epsilon_g+S_g\epsilon_{temp}+S_g\nu\epsilon_g- S_g\epsilon_{temp}+ 0-S_g\epsilon _g-S_g\epsilon_{temp}\right ) V=\frac{S_g(\nu\epsilon_g-\epsilon_{temp})}{4} V$

Here

$\epsilon_{temp}=$ strain induced by temperature

If a single strain gage is used

$\Delta E_1=\frac{1}{4}S_g(\epsilon_g+\epsilon_{temp})V$

Hence bridge constant is

$\kappa=\frac{\Delta E}{\Delta E_1}=\frac{\nu\epsilon_g-\epsilon_{temp}}{\epsilon_g+\epsilon_{temp}}$

Neglecting temperature strain

$\kappa=\nu$

The $\Delta E$ which is a measure of strain is affected by room temperature as is evident from equation for $\Delta E$.

II) Refer the diagram below:

The connections of gages have been adjusted for best output

Strain at top surface = $\epsilon_b$

Strain at bottom surface=$-\epsilon_b$

The voltage equation is (for the modified circuit)

$\Delta E=\frac{r}{(1+r)^2}\left ( \frac{\Delta R_1}{R_1}- \frac{\Delta R_3}{R_3}+ \frac{\Delta R_2}{R_2}- \frac{\Delta R_4}{R_4}\right ) V$

$\frac{\Delta R_1}{R_1}=-S_g\nu\epsilon_b+S_g\epsilon_{temp}$

$\frac{\Delta R_2}{R_2}=S_g\epsilon_b+S_g\epsilon_{temp}$

$\frac{\Delta R_3}{R_3}=S_g\nu\epsilon_b+S_g\epsilon_{temp}$

$\frac{\Delta R_4}{R_4}=-S_g\epsilon_b+S_g\epsilon_{temp}$

Hence

$\Delta E=\frac{1}{2^2}\left (( -S_g\nu\epsilon_b+S_g\epsilon_{temp})-(S_g\nu\epsilon_b+S_g\epsilon_{temp})+(S_g\epsilon_b+S_g\epsilon_{temp})-(-S_g\epsilon_b+S_g\epsilon_{temp})\right ) V=\frac{S_g(1-\nu)\epsilon_b}{2} V$

Neglecting thermal strain

$\Delta E_1=\frac{1}{4}S_g\epsilon_bV$

Hence

$\kappa=2(1-\nu)$

Referring to the formula for $\Delta E$ above, the measured voltaga and hence strain is not affected by room temperature.