SeanStone
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| posted on 3/11/10 at 05:07 PM |
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Any engineers out there?
Need some help with a stress analysis question.
The question has a flat rectangular plate, with a hole in the middle.
I am given the height and thickness of the plate and the diameter of the hole, the yield stress of the plate and the tensile load it is under.
How do I go about finding the stress on the top edge of the hole?
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Liam
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| posted on 3/11/10 at 05:44 PM |
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Erm... diagram? Dont really understand from your description how stress is applied to the plate and where exactly you want to measure stress near the
hole.
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britishtrident
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| posted on 3/11/10 at 05:47 PM |
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Think about it look at what little information you are given and why (whats Yield stress got to do with it ?) what you more important what you are
not given (ie not Young's Modulus )
The answer could be looked at either as being very simple or very complex.
Hint: where is the plate actually likely to fail under tensile load.
[I] “ What use our work, Bennet, if we cannot care for those we love? .”
― From BBC TV/Amazon's Ripper Street.
[/I]
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SeanStone
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| posted on 3/11/10 at 06:02 PM |
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quote:
Originally posted by Liam
Erm... diagram? Dont really understand from your description how stress is applied to the plate and where exactly you want to measure stress near the
hole.
http://www.emeraldinsight.com/content_images/fig/1820170201021.png
Much like that.
We are asked 'Determine the stress on the top edge of the hole (Answer in MPa).'
[Edited on 3/11/10 by SeanStone]
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britishtrident
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| posted on 3/11/10 at 06:23 PM |
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If that is the question then the real world answer is
field stress * kt where kt is the stress concentration factor (SCF)
Field stress is the field stress at that area of the plate.
[Edited on 3/11/10 by britishtrident]
[I] “ What use our work, Bennet, if we cannot care for those we love? .”
― From BBC TV/Amazon's Ripper Street.
[/I]
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bitsilly
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| posted on 3/11/10 at 06:51 PM |
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Presume there is no strain before failure So would the stress not be evenly distributed over the cross sectional area of the plate above and below
the hole?
It certainly would be if there is CURRENTLY no strain ( you should be given Youngs mod).
If that is right then just divide cross section by stress?
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bitsilly
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| posted on 3/11/10 at 06:54 PM |
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Shouldn't that be correct if it comes out below the yield stress?
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britishtrident
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| posted on 3/11/10 at 07:12 PM |
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quote:
Originally posted by bitsilly
Presume there is no strain before failure So would the stress not be evenly distributed over the cross sectional area of the plate above and below
the hole?
It certainly would be if there is CURRENTLY no strain ( you should be given Youngs mod).
If that is right then just divide cross section by stress?
By Hookes Law if the plate carries any load it must experience strain.
There will be a stress concentration under any load except the zero load case.
If you start considering yield conditions you get local plastic deformation around the hole and that redistributes the stress.
Link to photoelastic picture of plate under vertical tension.
http://faculty.rmc.edu/dwoolard/public_html/isochromatic.jpg
[Edited on 3/11/10 by britishtrident]
[I] “ What use our work, Bennet, if we cannot care for those we love? .”
― From BBC TV/Amazon's Ripper Street.
[/I]
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bitsilly
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| posted on 3/11/10 at 07:49 PM |
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A good point well made!
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