You have a solid 2 inch square copper ground support, that is 4 inches tall and

Question

You have a solid 2 inch square copper ground support, that is 4 inches tall and it is loaded axially with 1600 pounds (force). You have instrumented one rectangular flat of the support with 2 strain gages. The strain gages are made from modified Karma - Alloy K, that has a nominal resistance R_0 = 3500, and that has a gage factor GF = 2.1. (First, you must look up the Modulus and Poisson's Ratio of Copper.) What are the two possible changes in resistance of a strain gage under these conditions? If the bridge excitation is V_EXC = 10 volts, what are the possible AV outputs of the Wheatstone bridge? If you used a Full Bridge design (four strain gages), what is the AV output of the bridge circuit? the strain gages in the bridge to give a - nabla V (minus delta V) output for compression

Explanation / Answer

a) Poisson's Ratio of copper is defined as tangential stress by longitudinal stress, For copper value is 0.355 and Young's modulus is defined as pressure devided by longitudinal stress. For copper young's modulus is 17*10^3 in PSI(pound per squre inch).

b) Gage factor is given 2.1 and it is the ratio of (change in resistace by resistance) to longitudinal strain

we have to find change in resistance and we have R=350 ohm, youngs modulus= 17*10^3.

Youngs modulus is Force per unit area by strain. Therefore strain=(1600/4)/17*10^3=0.0235

So change in resistance=2.1*strain*350=17.294 ohm.

c) Under no stress condition wheatstone bridge will remain balanced (Assumed because no data is provided regarding wheatstone bridge).

So voltage will be zero but under stress condition (stress is compressive in nature) resistance od copper will decrease therefore resistance under stress condition is 350-17.294=332.70 ohm

Lets assume all arms having resistance of 350 ohm. then at balance condition voltages at measurement point in wheatstone bridge is 5V. when resistance will decrese then voltage at mid point in one of the arm will decrease. The voltage will V=10-(10/350+332.7)*350=4.873 V.

So change in voltage will be 0 volt to (5-4.873)=0.126 V.

d. Need layout to answer part d.

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