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  • Impact Loads....Shear Modulus v.s Tensile Strength

    Discussion in 'The main mechanical design forum' started by samstreet101, Feb 18, 2013.

    1. samstreet101

      samstreet101 New Member

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      Hi all,

      New to the forum. I have a problemI'm tying to solve with regards to an impact load acting against a small peice of a safety catch. The scenario is this...

      I have a safety catch welded to some box section. This safety catch is made from mild steel and is there to stop an arm (made from 1/2" tube) from swingning out and hitting someone in the face. The arm will therefore hit a section of this safety catch, causing an impact load (see pitcure below for detail).

      [​IMG][​IMG]

      The section that the arm hits is 6mm thick. I've used the formula below (a simple impact load/shock load forumla) to work out the stress such a load will cause.

      [​IMG]
      In this forumla V=velocity in m/s, m = mass in kg, E = Young's modulus in GN/m[SUP]2 [/SUP]or GPa, A = area in m[SUP]2 [/SUP], and l = length in m.

      The arm weighs 1.5Kg, the catch is made from mild steel (E=220 GPa), the area that the arm will hit is 0.00092 m[SUP]2 [/SUP]and the length of that section is 6mm (0.006m). I've chosen velocity as 3 m/s as a benchmark, as this is a relatively slow speed and I cannot imagine the arm breaking this piece of catch at that speed. However, when using all these figures, the stress result is 733.5 MPa.

      My issue then is that the tensile strength of this mild steel is approx. 400 MPa, meaning that the arm at this velocity, will break the catch...I find that damn near impossible to believe. However, I did think that because of the way that the arm is striking the catch, it may cause shear stress. In this case, is the shear modulus (which is around 80 GPa) what determines whether or not the catch will fail in this situation?

      Any help would be greatly appreciated.
       
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    3. itsenzothebaker

      itsenzothebaker New Member

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    4. AndrewNew

      AndrewNew Well-Known Member

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      Well your equation appears to be dimensionally consistent and I get the same answer as you do when I run your numbers. Do you have any more information on the source of the equation? It may be that your interpretation of which dimension to use as l is at fault....
       
    5. samstreet101

      samstreet101 New Member

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      L could be af fault indeed, I've looked at that Roymech page before and actually the forumla I'm using is the same one as on that page (Linear kinnetic imact loading I think is the sub-heading). Any other formulas anyone can recommend?
       
    6. Alfred Putter

      Alfred Putter Member

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      Impact Stress

      I have calculated the stress to be 73,35 N/MM[SUB]2
      [/SUB]not 733.5 N/MM[SUB]2[/SUB]

       
    7. Alfred Putter

      Alfred Putter Member

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      I get 73,35 N/MM2 not 733,5 N/MM2
       
    8. samstreet101

      samstreet101 New Member

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      Wrong units

      Hi Alfred,

      Are you sure? I've ran those numbers several times and others have and got the same result. Sounds like a decimal point gone missing somewhere.
       
    9. Alfred Putter

      Alfred Putter Member

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      sorry a calc mistake by me
       
    10. AndrewNew

      AndrewNew Well-Known Member

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      If you have access to FEA you can use an energy equivalence approximation to determine the stresses in your structure under impact. You know the kinetic energy of your arm (0.5 x m x v^2), assume that is all converted to strain energy in your catch at impact, then back-calculate the load that approximates the impact conditions and gives the same strain energy. The stress distribution under that load can be used in the usual way to predict whether failure will occur. The difficulty is often getting the strain energy out of the FEA package - providing you can do that it's relatively straightforward and gives conservative (safe) if not always totally accurate results.
       
    11. srdfmc

      srdfmc Well-Known Member

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      Don't forget to evaluate wich part of teh E is really absorbed by the material.

      During a impact the main governing eq. is m1v1=m2v2 with m1 and m2 the mass of the objects ant v1 and v2 the respect. velocity.

      Tyou'lle then see that the inertia of the receptive solid play it's part absorbing the E and then only the component's material.

      try to imagine a fly impacting a moving glass. If you consider the glass being static, it's obvious that you do not hve any material def to study. But then what about the fly being static... Yeah huge def indeed.

      At first glance, this is here what gets you out of proportions here. Just a wild guess.

      Best rgeards,

      SRDFMC
       

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