hi, please can you help me out! A strain gauge has a unstrained resistance of 120 ohms, guage factor of 2.0 is connected to steel girder so it experieces tensile stress. If strained resistance of gauge is 120.13 ohm. How do i calculate the tensile stress value? I have looked all over the internet for equations relating but have not succeded, any help would be great
Take a look at the formula in this wikipedia.org article on gauge factor. The formula for gauge factor has all you need I believe. Just plug in the formula and solve for ε. hgmjr
As HGMJR says the Wiki formula will give you the strain at the location of the gauge. You can then calculate the stress from the relation Youngs Modulus = stress/strain This article also gives suitable values for your steel girder. http://en.wikipedia.org/wiki/Young's_modulus You can also calculate the theoretical stress from normal materials engineering formulae involving the value of the loads and the geometry of the setup.
i hate wiki as it always makes things much more complicated as they actaully are. so the formula is Y = F*L /(dL*A) where , Y=Young's modulus in units of pressure (F/A) F=force A=cross-section area L=length dL=change in length surely all i know is youngs module which is 2x10^11 (provided in question- youngs module of steel). we will not be given any tables of that kind of stuff in the exam!
What kind of exam? I asked because I'd like to know if you are studying Physics, General Engineering or some purely electrical discipline. then I can answer appropriately. All you need to solve this problem is Young's modulus, the resistance change and the gauge factor. All of which you say you are given. You are quoting the wrong formula.
So it's not unreasonable for you to have little or no knowledge of mechanical engineering, Or is it? I'm not getting at you, but Electrical Engineers need to have a good appreciation of the physical world. Electrical Engineering is an intensely physical subject. Transducers act as go between from the electrical world to the mechanical one. Enough preaching, to your problem. As I noted, Young's M = Stress / strain Now look again at your equation. stress = force /area and is measured in the same units as pressure. There are two forms of stress normal stress of which pressure is an example and tangential or shear stress, of which friction is and example. Both tyoes are measured as force/area ie N/sqmm or Pascal or whatever. So you have F/A = stress in your equation, (which incidentally is correct for stretching a wire) So using the equation strain = gauge factor x change of resistance/original resistance you can obtain the strain. Finally you can calculate stress = strain x YM Good luck with your finals. Post again for more help
thanks studiot. so am i right in sayin strain is 2.1 x (0.13 / 120) = 2.275 x 10^-3 tensile stress value = (2 x 10^11) x (2.275 x 10^-3) = 455 x 10^6? if thats the answer ^^ it seems abit easy for 8 marks of the paper :s cheers
Yes correct, except for the units which are 455 MN/sqm or MPascal. Always state units. That's a very strong steel by the way.
good , do you know much about wheatstone bridge? hope you dont mind me taking up your time but this is important . http://people.sinclair.edu/nickreeder/EET150/PageArt/circuitWheatstoneLoaded.gif if you look at the diagram and change the RL to Vout (cant find appropiate diagram). what is the condition at which the output voltage will be zero? in terms on resistance? correct me if im wrong but is it R4/R1 = R3/R2?
http://people.sinclair.edu/nickreede...toneLoaded.gif if you look at the diagram and remove the resistance RL and change it to Vout (cant find appropiate diagram). Question is what is the condition at which the output voltage will be zero in terms on resistance? correct me if im wrong but is it R4/R1 = R3/R2? i have one more question after this just to get it off my chest! thanks for help
The ouput will be zero in your example when the potential at A is the same as at B. Since {R1 & R3 } and {R2 & R4 } form potential dividers from a common voltage source Surely a third year EE should be able to work out that R1R4 = R2R3 You must be pulling my leg.
nope, just :s. final question. R1 and R2 are 2 guages with resistance of 100 ohm. A steel bar has diameter 50mm and modulus of steel bar is 161GN/m^2 and guage factor of 2. How would I calculate Vout when a load of 40kN is applied? i know the Vout equation.
How about you make a stab at the calculation and post it here. Then we can give you corrective feedback if it is needed. hgmjr
thats my problem. i do not know were to start. once i have all the information i can easily apply it. its just extracting the information. so if i could be guided in the right direction. thanks
Post14******************************************************************* First where did your diagram come from? It is not the normal (simple ) application of a strain gauge. In normal operation R2 is strained with the force and R1 provides a reference and is not strained.It is usually bonded to a spearate piece of material in the same environment as R2. In your diagram R1 is shown bonded at to the strained material, at right angles to R2. This arrangement would allow measurement of Poisson's ratio, but I think that is beyond your scope. If both gauges were bonded in the same direction to the same test sample, the strained resistance changes would indeed cancel each other out, so there would still be no output from the bridge. Please confirm what you actually mean. End of post 14 question******************************************************* To derive the relation I gave earlier in post 13, note the common potential dividers and equate the potential at A and B. V * R1/(R1+R3) = V * R2 /(R2+R4) Cancel V and cross multiply R1 (R2+R4) = R2 (R1+R3) R1R2 + R1R4 = R2R1 + R2R3 R1R4 = R2R3 Are you happy with this derivation?