Mechatronics Lab – strain gage rosette experiment analysis discussion

this figure illustrates the apparatus used in our strain gauge rosette experiment we have an aluminum tube with a given cross-section and we have an offset load where we apply mass hanging from the end of an arm the rosette is oriented on the bar where X is the axial direction it's oriented like so three gauges are 45 degrees apart and they're numbered one two and three this figure illustrates the state of stress on the top of the bar we have Sigma X in the axial direction do the bending of the end load and we have torsion in this direction do the twisting of the unload that's tau XY the bending stress is given by MC over I torsion by TC over J and we work through all the details for the two cross section we get expressions for each of these stresses and we can see that the bending stress is this factor times the torsional stress and for our experiment to be is much larger than a so bending is going to be the dominant factor in terms of stress being experienced by the rosette these figures show a qualitative analysis of the strain being experienced by the rosette first we're showing the effects due to bending so when an original element of material is stressed in the axial direction this dashed rectangle illustrates in an exaggerated way how the material deforms and in this case the central gauge will be in tension and will be stretched the most and the other two gauges they will also be stretched slightly but in summary the strain experienced by the center gage is going to be larger than the other two the other two will be equal because this is symmetric all will be positive because they're all attention due to torsion again we're showing an exaggerated view of how the material will deform as a result of the tours in this direction because gage 2 is in the center of the rosette and then the center of this deforming element it experiences almost no strain whereas gage one up here is oriented mostly in the shoe or stretching direction it will experience the most strain it will be in tension gauge 3 because it is oriented against the stretching direction will actually be compressed slightly and it will have a negative strain so by combining these effects assuming that bending is dominant the strain engaged – should be larger than the strain engage 1 which in turn should be larger than the strain engaged 3 we can also do a qualitative analysis of the stress situation knowing the directions of the bending and torsion 'el stresses we can see that the plane of Max and principle stress is going to be about in this direction because the resultant of all of these arrows points in that direction and it would be a tensile maximum principle stress given the strains in the three gauges the strain gauge rosette equations allow us to calculate the maximum and minimum principle stresses along with the direction of those principal stresses the maximum puzzle stress is called Sigma a the minimum personal stress is called Sigma B and the angle from gauge one to the direction of maximum principle stress is called theta P note that gauge one is 45 degrees away from gauge 2 therefore the angle fee from the axial direction of the tube is going to be 45 degrees minus theta P that we calculate from the strain gauge rosette equations Morris circle allows us to visualize all of the gauge directions the maximum minimum principal stresses and all of the angles involved in this analysis first we can see the gauge directions because they are already in at 45 degrees apart on the top of the bar on Mohr circle all angles are twice physical angles so you consider nine degrees apart on Mohr circle the strain gauge rosette equations allow us to calculate Sigma a and Sigma B the maximum and minimum principle stresses and we calculate that angle fee that tells us where the principal stress is located relative to gauge 2 which is aligned with the axis of the 2 since we know the principal stresses from the rosette equations and the direction to the maximum principle stress we can use that information to calculate the stresses in the direction of the L so Sigma X tau XY these are the equations that allow us to compute the desired quantities from Mohr circle first we find the stress at the center of Mohr circle at Sigma average just the average of the two principal stresses we can then find the radius of Mohr circle this is also the maximum shear stress it's a difference between the maximum and minimum principal stresses divided by 2 we can use this information to figure out the desired Sigma X and tau XY again they're computed directly from the circle using the radius and the angle now I'm going to summarize the procedure that's carried out in lab to utilize readings from the strain gauge rosette we first apply a mass to the end of the tube that creates a state of stress on the top of the tube which for measuring with a strain gauge rosette so we measure the three voltages coming from the strain gauge rosette one voltage for each gauge we calculate the strains corresponding to those voltages from the strain gauge fridge equation using the gains proper length then we calculate the maximum and minimum principal stresses from the strain gauge rosette equations and the angle theta P from gauge 1 to the maximum principle stress we then calculate the integral fee which is a direction from the maximum principle stress to strain gauge two because strain gauge 2 is aligned with the x axis we then calculate Sigma X and tau XY corresponding to an element oriented in the X direction which is the direction of gauge 2 we then compare us those values calculated from the gauge readings to the theoretical equations for stress and shear stress this allows us to calculate the mass so what we've done here is actually built a fairly sophisticated scale that measures a mass based on strain gauge readings not on the top of a tube

One Comment

  1. hemda lama said:

    pls upload clear video

    July 12, 2019

Leave a Reply

Your email address will not be published. Required fields are marked *