how does thickness affect the design

[kamal11]

I put a sheet of following dimensions on floor. Assume the material is glass.

1) thickness 1.2 mm diameter 2"
2) thickness 1.5 mm diameter 2"
3) thickness 1.8 mm diameter 2"
4) thickness 2 mm and diameter 2"

And I apply same compressive load on all these sheets
Let us assume that i transfer the load with a rod of 2" OD diameter and ID of 1".


The compressive stresses on these would be load/(pi*(2*2-1*1)/4)


Since compressive stress is same. As i start increasing the load i know the glass sheet with minimim thickness will break first.

Assuming that this is not a bucking case, how do i deterimine at which load the 1.2 mm (minimim thickness) sheet will break
In the calculation of the stress that includes thickness what is the area i need to use for calculation.
Is it the area i need to consider for calculating the stress is (D*t) or (pi*D*t). what should be the value of D
Is it the outer diameter of the rod or MEan diameter of the ID of the rod.

 

[BigInch]

I believe you should do this calculation using a punching shear analysis. As you have probably seen, glass generally fails in punching shear leaving a cone shape originating from the point of load.

I would therefore take a the surface area of a cone with a slope of 45º and a height equal to the thickness of the glass and calcuate the shear load as the point load divided by that area. The glass will tend to break when the shear stress equals 1/2 the ultimate tensile strength of the glass.

It is also conceivable to use the lateral area of a truncated cone for shear analysis, with the top diameter of the cone = diameter of your rod. The bottom diameter would = the top diameter + 2 * glass_thickness * 1.4142

I don't think the stress in the immediate vicinity directly under the point load is important, unless you need to consider a local crushing failure. I believe punching shear would occur first, unless the glass thickness is relatively great.



Going the Big Inch!

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[kamal11]

I am not doing glass analysis. I have taken glass so that it gives easy visualisation for any one. I wanted to determine the sheet thickness for a sheet metail.

 

[BigInch]

Good, so much the better. Crushing via bearing stress is unlikely. Check punching shear, which I would do for steel using only the lateral area of a right cylinder, unless the sheet was unusually thick. There's probably some thickness criteria as to when you change from a cylinder to a cone, but if there is, I've forgotten what it might be.

Going the Big Inch!

http://virtualpipeline.spaces.msn.com

 

[jedward]

failure of glass is almost almost assured to be a tensile failure and depends primarily on the presence of small localized defects. it seems that you have assumed the surfaces to all be polished or freshly molded without defects.

the glass will also tend to deform elastically and the stress will not be uniform as your calculation requires.

failures of windows almost always occur from edge defects and not from the location where a simple model predicts the maximum stress.


google search under John Pepi, Keith B. Doyle and the website of Schott glass (look for a documet with TIE-31)

Corning also has some usefull information on mechanics of glass

 

[handleman]

I am not doing glass analysis. I have taken glass so that it gives easy visualisation for any one. I wanted to determine the sheet thickness for a sheet metail.

You can't ask us to consider glass as the material and then apply the replies you get to sheet metal. The two materials will behave in entirely different ways! If I read your original post correctly, you are fully supporting the sheet behind the rod. If the material is glass it will shatter violently at some point. There is a defined failure criteria. With sheet metal there is no such point. What do you consider failure? Any plastic deformation at all? Or shearing through the material altogether?

 

[rb1957]

i think the key thing is that you're trying to drive a 2" dia rod through a sheet; so you're shearing the sheet along the OD of the rod.

 

[handleman]

rb1957,

I may be wrong, but the way I read the OP is that a disk of 2" OD is placed on a solid surface and is compressed by a rod of 2" OD and 1" ID. That's why a clear definition of failure is needed. This wouldn't be a case of shear, but ultimately some sort of crushing, ripping failure.

 

[dimjim]

If you punched the parts as stated above
I would think the cone effect will play
a part at the far end of each thickness.
I would expect that you would have a different
size burr or amount of extra material deformed
on each of the different thicknesses.

If you stacked them in the reverse manner as
above, I think you would see even great distortions
at the far ends.

Most of the parts we have with stamped holes in them
are ss and probably less than 1mm thick and even
they have a burr or distorted material on the
far side. I do not know if they muti stack these
or not. We use them for shims. The entry side
is always smooth and probably slightly indented
while the other side is rough or burred typically.

 

[chicopee]

Because of the thickness(1/2") of the punch , I would also be interested in the crushing strength of glass under the area of contact. See if you can get data on crushing strength of glass.