Determining force required to rip or tear sheetmetal

  • Thread starter Spoolx
  • Start date
  • #1
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Main Question or Discussion Point

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?

Thanks
 

Answers and Replies

  • #2
38
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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?

Thanks
 
  • #3
2,155
568
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.
 
  • #4
38
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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.
 
  • #5
OldEngr63
Gold Member
732
51
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.
 
  • #6
38
0
I would say no because the sheet metal is tearing at the bottom of the bracket.
 
  • #7
2,155
568
You still have not shown us a picture, so it is impossible to make any meaningful comments on this matter.
 

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