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Determining force required to rip or tear sheetmetal

  1. Jan 22, 2015 #1
    Hi All.
    I am working on a problem at work that is basically sheetmetal tearing due to a force being applied.

    We have a sheetmetal door which we added a gas cylinder to help lift, the problem is the sheetmetal is tearing at the mounting location of the gas cylinder.
    I know that the stress is causing the failure and that increasing the thickness of the mounting location will solve the problem but i want to wrap some formulas around it to make sure I am going big enough.

    Any help or guidance on what formulas to apply here?

  2. jcsd
  3. Jan 22, 2015 #2
    I do the basic calculation of Fmax using stress = F/A
    Assuming Yield stress = 100ksi, the sheet metal being 1/6" thick and the flange being 2" tall
    100ksi=F/(2*.0625) -> F=100ksi*.125 -> Fmax =12500lb

    This seems way wrong considering the gas shock only applies a few pounds of force. Am I calculating it wrong? If not how do I apply it to the fatigue failure?

  4. Jan 23, 2015 #3
    How is the gas shock attached to the panel? Weld, screws, or what? This will make a lot of difference.

    The calculation you showed almost certainly included too much area supporting the load.
  5. Jan 24, 2015 #4
    It's bolted to the plate with 4 10-32 screws.
    I assume it is almost line to line contact and has a huge stress riser. Just not sure how to calculate.
  6. Jan 24, 2015 #5


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    Then is it not true that the critical issue is the pull-out strength of the 4 screws? It would seem that you might estimate the load required to rip out one screw, and then figure that the four screw combination will support no more than 4 times the individual screw load.
  7. Jan 24, 2015 #6
    I would say no because the sheet metal is tearing at the bottom of the bracket.
  8. Jan 24, 2015 #7
    You still have not shown us a picture, so it is impossible to make any meaningful comments on this matter.

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