Shear tearout from a hole to close to the edge

[TylerS]

I am currently trying to figure out the tear out strength of an A36 steel pad-eye. The hole in the pad-eye that it will be lifted from is 2" in diameter and 2.5 inches from center to the nearest edge. The problem is that the pin that will be lifting it is 1.38" in diameter and i was unsure how this effected the tear out strength? I referenced the ASME BTH-1-2011 Design of Below the Hook Lifting Devices but it wasn't helpful. I was hoping someone could give some insight on the subject. I am assuming I cant use the standard tear-out calculation which is F = Shear strength*distance from fixing center to edge*thickness I believe.

 

[rb1957]

"I am assuming I cant use the standard tear-out calculation" ... why not?
"which is F = Shear strength*distance from fixing center to edge*thickness" ... if "distance from fixing center to edge" means the minimum distance from the edge of the hole to the edge of the part*2. this is, in my experience, conservative.

the other thing to check is bearing, with e/D < 1.5.

Quando Omni Flunkus Moritati

 

[fegenbush]

The reference I use for these calculations is the AISC Steel Construction Manual (14th Edition is current). Chapter J is the location of bolted connections.

 

[TylerS]

Thank you rb1957, I have the Tear out strength now. I currently do not own the AISC Steel Construction Manual so i cant reference it. I would think the bearing strength would be lower with a pin diameter that is .62" in diameter smaller then the hole. Is there an equation for this?

 

[rb1957]

P/(d*t)

Quando Omni Flunkus Moritati

 

[tbuelna]

Roark's will give you a basic equation for bearing stress of a 1.38" dia cylindrical pin with line contact against the surface of a 2" dia x ?" wide lug bore. Whether the bearing stresses at this contact are acceptable or not depends on the material properties, analysis factors and loads your analysis requires.

 

[boo1]

Design of Machine Elements, M. F. Spotts, 6th ed, pp 443
Contact stress between two cylinders , fig 9-9
Max Compressive Stress= Po
P1 load per axial inch on cylinder, R1 radius pin, E1 Modulus pin, R2 radius plate hole, E2 modulus plate
Po=0.591*sqrt((P1*E1*E2)*(E1+E2)*(1/R1-1/R2))
Contact zone
a=1/2 width of contact area, P1=load/axial inch
a=1.076*sqrt((P1*R1*R2/(R2-R1)*(1/E1-1/E2))

also see
ASME BTH-1–2008
3-3.3.1 Static Strength of the Plates.

 

[dhengr]

TylerS:

Take a look at thread507-360145, in the Struct. Forum. It is on the same topic and has some more info. and discussion.