Tearout stress calculation on hole near edge 2

[tnteng]

I am calculating the shear stress of a pin tearing out of a hole that is near the edge of a plate. Does anyone know if the area used in the tearout shear stress calculation would be based on 2 times the distance from the center of the hole to the edge of the plate times the thickness of the plate (as is done with padeyes)? Or would it be based on 2 times the distance from the edge of the plate to the radius of the hole times the thickness of the plate?

Also would there be a limit as to how close the hole could be to the edge without causing a problem with tensile stresses on the section that is between the hole and the edge of the plate?

Thanks in advance for the help,

Tony Billeaud
Mechanical Engineer
Franks Casing Crew

 

[desertfox]

HI tnteng

The formula for shearing out of a fixing to the edge of the plate is :-

F= allowable shear stress in plate * 2*e*t


where F is the force to cause failure

e is distance from fixing centre
to edge of plate

t is the plate thickness

It is normally reccomended that fixings should be at least
(1.5 * "diameter of fixing") away from plate edges.


regards desertfox

 

[CoryPad]

[blue]tnteng[/blue],

[blue]desertfox[/blue] provided the answer for tearout analysis, which agrees with your first item. Your second item is for plate fracture, which may be the limiting condition for your item. It is necessary to check both of these (and bearing yield and pin shear) to avoid failure.

Regards,

Cory

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

 

[tnteng]

Thanks to all who have helped with this question.

Cory,

How would you check for plate fracture?

Thanks,

Tony Billeaud
Mechanical Engineer
Franks Casing Crew

 

[CoryPad]

S < S_f

S = F/[(W-d)t]

where

d = hole diameter
F = applied force
S = plate stress
S_f = plate fracture stress
t = plate thickness
W = plate width

This is a simple analysis that ignores stress concentration.

Regards,

Cory

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

 

[tnteng]

Thanks again for the help.

Desertfox,

What if the hole was very large in order to accept a large pin that was required to resist bending. Would the rule of thumb of placing the center of the hole 1.5*D from the edge still apply in that case? It seems that it would be too large a distance?

Thanks,

Tony Billeaud
Mechanical Engineer
Franks Casing Crew

 

[desertfox]

Hi tnteng

The formula I posted yesterday is based only on tensile loading of plates held together with rivets.
If your encountering bending loads then the situation is different. I need more detail about what your application is and how the load is transfered to the pin.



regards desertfox

 

[tnteng]

Desertfox,

I have a large pin in a hole near the edge of a plate and the force on the pin is pulling the pin towards the edge of the plate. I was only suggesting that the 1.5*D distance to the edge of the plate could be much larger than needed if the hole was very large based on the pin needing to be large for other reasons.

Your thoughts?

Tony Billeaud
Mechanical Engineer
Franks Casing Crew

 

[desertfox]

Hi tnteng

If the pin is trying to shear out through the plate then I would stick to the 1.5*Diameter of the pin.


regards Desertfox

 

[John2004]

Hi everyone,

Regarding the comment by Desertfox,

>It is normally reccomended that fixings should be at least
(1.5 * "diameter of fixing") away from plate edges.<

Is this distance from the center of the round pin to the plate edge, or is it the distance from the OD or quadrant of the round pin to the plate edge ?

If I have a .125" OD pin ...

.125 X 1.5 = .1875" from edge of plate to center of pin ?

Thanks
John

 

[IFRs]

The center of the hole should be 2x the pin diameter from the edge of the metal. If your material has grain structure from rolling or extruding operations, there may be need for more edge distance to prevent tearout.

 

[JStephen]

Check the AISC steel design manuals. The approach there is not a stress analysis, but gives the minimum edge distance based on loading.

 

[John2004]

Hi everyone,

Regarding the formula given by desertfox,

>F= allowable shear stress in plate * 2*e*t

where F is the force to cause failure

e is distance from fixing centre
to edge of plate

t is the plate thickness<

Do you just use the shear stress in PSI, and this gives you the tear out load force in pounds ?

If I have a rectangular plate 0.1181" thick, .1875" wide, and say 3" long, and I have some .125" OD dowels running lengthwise down the center of the plate, with the plate having an allowable shear stress of say 20,000 PSI, I calculate...

20,000 PSI * 2 * .09375 * .1181 = 442.875 pounds to cause the dowel to tear through the plate across the .1875" width ?

Is this the correct approach based on the formula ?

I am also interested in learning more about the method from the AISC design manual if possible, I don't have the manual for reference.

Thanks
John

 

[desertfox]

Hi John

Your understanding of the formula is correct, however if you follow the reccomended edge distance to the dowel fixing centre you won't need to worry to much about a tear out failure.

regards

desertfox