calculate material for hole to edge rip-out 3

[paintballJim]

OK, this is kind of embarrassing but I haven't dealt with this topic for some time so I wanted to check that I am going at this right. I have a plate of 3/8" A36 steel with a 1" dia. hole centered 1-3/8" from the edge. A load will be hanging from a bar inserted through 2 of these plates. To determine the allowable load to suspend from the bar, do I take a 45 degree angle each way from the center of the hole to the edge and determine the area of material in the triangle? or do I simply measure the distance from hole edge to part edge (.875) mutliply by thickness (.375) to find area then multiply by 36,000 and double for 2 parts?

Thanks.

Jim

 

[charliealphabravo]

A sketch would help. Are you loading it from the edge or the face?

You will have to decide which component or failure mode in the system is the weak link...the rod, local plate rupture, plate span, etc.

Also what is the load?

 

[paintballJim]

I am attempting to determine the failure point in the side plates. Put a 1" hole centered 1-3/8" up from the bottom edge of a 3/8" plate of steel. Place a 1" bolt or bar in the hole and pull straight down. How much load will this hold? I am thinking that simply calculating the area of distance from edge * thickness is the most conservative but I don't think that is right. I seem to remember calculating out at an angle and figuring on strength of more material.

 

[charliealphabravo]

Well a few things...

1. The standard hole for a 1" bolt 1-1/8". Are you designing to a particular code?
2. The limit state for bolt holes should also consider splitting of the plate which may govern in your case in addition to plate tear out.
3. The ultimate strength of the steel (fu) should be used not the yield strength (fy) for bolt .

 

[charliealphabravo]

Also depending on the treatment of the edge you may not meet the minimum edge distance. For a 1" bolt and a sheared edge, the edge distance is 1-3/4" measured from the center of the hole.

Assuming the gross area yielding of the plate is sufficient then the capacity for properly factored loads is based on the net fracture of the plate = 0.75 * fu * Net Area. Net area is the thickness of the material times (the width of the hole plus 1/16"). In other words there is no 45 degree angle, just straight line to the edge.

As I say you will also need to check bolt bearing...and a picture would help

 

[paintballJim]

Unfortunately, I am not "designing" anything. My boss just threw a trolly assembly on my desk and said "calculate if this is strong enough." He has sourced the bearings, had the plates cut, rollers machined, and the whole 9 yards before ever consulting engineering. Typical for the way he likes to do things. Thus I don't have any drawings or any design data however I did do a quick sketch to show why I am not an artist. I hope you can understand what I mean. the second side would be a mirror and be to the viewer. He has 2 side plates with rollers at the top held together with a 1" grade 8 bolt. There is a 3" piece of 1" I.D. pipe clamped between the side plates that the bolt sticks through to keep the sides at the desired width. The trolly is for our paint booth and the projected load is less than 3000 pounds per trolly. Yes, it is over built but I wanted to do the calculations anyway first for a refresher in how to do it, and secondly to show how far over built it is.
I guess I must have confused the bolt clamping area where effective clamping area is 30 from the bolt edge to 1/2 the thickness of the material with the strength of material for the edge of a sheet. If I understand What you are saying, The area used is simply hole width plus 1/16" times the thickness without regard to the distance from the edge? (providing you are at least "X" distance from the edge?) If so how do I determine the area when I am closer to the edge as in this case? Sorry for all the questions but I have dealt with weldments for far to long. I remember covering this in school, but that was many moons ago and I'm not finding it in my books.

 

[rb1957]

easy way is to test it, load to 3,000 lbs *SF.

sketch is not clear to me ...
the rollers are on the lower I beam flange, on both sides of the web.
the brkt bends over the I-beam lower flange and straight down to the load.
the two brkts are spaced 3" apart (like the lower flange width, + clearance).

the load is introduced between the two brkts, on the bolt.

problem i see is the free body of the brkt ...
the brkt load is offset from the reaction onto the I-beam flange (at the wheel contact)
(note also that the wheel contact is offset from the attmt of the wheel to the brkt, so there's a small bending moment, maybe being carried by the wheel axle.)
so i see a moment somewhere near the brkt interface with the bolt.

i'd've put a sleeve between the two brkts, and tightened the bolt against this (maybe you have this ?)

 

[charliealphabravo]

I won't speak for your beam and trolly assembly or for the pipe but it looks like the controlling failure mode for the plate will be fracture along a single line from the center of the bolt to the near loaded edge at the bottom. A single rod will not tear out a block of metal for short edge distances.

In that case the nominal strength against properly factored loads is the fracture capacity of that line of steel = phi*fu*An where phi is 0.75 for tension fracture, fu is 58,000psi and An is the thickness of the plate times the edge distance minus half the hole diameter.

For a 1-1/16" hole that works out to 13.2 kips which matches the tabulated value in my old 1994 AISC LRFD manual (Table 8-47a).

You should factor your actual loads by anywhere from 1.2 to 2.0 for normal impact type conditions.

 

[paintballJim]

Other than to put the component into production we really don't have any way of testing items here. Sorry about the sketch. the intent is to place the rollers on both bottom flanges of the I-beam. the flange is 4" wide thus the sides are bent to allow the rollers to be wide enough to fit. There is a pipe for a sleeve between the sides where they are clamped by the 1" grade 8 bolt. I have followed the path and have found moments at the rollers and shafts. For a 3000# load this bracket is clearly over built. The closest thing to a concern is the 4 roller bearings rated at 1330# static load each. I am currently using this exercise to help me remember how to determine the bracket capabilities before I need to do it for real some time. My question really regards the bolt tear out on the edge of the material and how to calculate the strength there. I know I did it in class over 10 years ago, but until you do it for real a few times it doesn't mean much and I forget how it was done. I am trying to find the process in my text books but so far have been unsuccessful.

Thank you

Jim

 

[WillisV]

Bolt bearing and tearout for structural bolts is covered explicitly by Section J3.10 equation J3-6a or b of the 2010 AISC Specification located for free here: www.aisc.org/2010spec

 

[paintballJim]

Charlie,
Thank you. Your quick glance at the drawing is the same as I came up with. That was why I wanted to know the correct way to determine the edge strength. Figuring on being conservative I would have figured basically the same as you said but I would have used the yield rather than the intimate and would have missed the tension fracture. Being that it is 13.2 kips X 2 for 2 plates and the load is 3kips I would have been ok but I wanted to know the right way. Again, Thank you

Willis,
Thank you for the link. It is knowledge of resources like this that I miss being in a one horse operation.

Jim

 

[dhengr]

PaintballJim:
Your and Charlie’s calc. for ultimate shear cap’y of the side pl. looks about right under the bolt, assuming nicely finished pl. edges, no nicks, notches, etc., no stress raisers. But, this is going to be loaded many times, thus some pin or pinhole wear; and there is a very high bearing stress in the side pl. immediately under the pin (a Hertz Stress problem). And, the high tri-axial stress picture there is the point at which a failure will start. The failure is actually a mix of high shear and tensile stresses. And, as WillisV suggested AISC 360-10, I wonder if the more appropriate equation isn’t J3-6c, a slightly lower available bearing strength. I’d have to look a little deeper to try to understand how AISC accounts for the Hertz stress problem in their approach to the pin/bolt bearing problem. A FoS of 3 or 4 (your 3k load to calc’ed. 13.2k cap’y.) is reasonable for this type of equipment. There is some good discussion on this problem in various literature and text books related to lifting pin-plates and eye-bars, and the like.

I’ll add something more for you to think about..., since the flange on the I-beam is sloped, and I assume the treads on the rollers are sloped to match, what prevents the two side plates from spreading apart under load and with movement down the beam? Draw a free body diagram of the trolley looking down the beam axis, and study this problem a bit. Can the bolt and side plates resist this tendency for the rollers to try to move outward? Or, should you have pipe spacers and through bolts btwn. the side plates, immediately under the I-beam bot. flg.?

 

[charliealphabravo]

Equation J3-6c gives the same value that I calculated. Equation 6a and 6b are increased by a coefficient 1.2 and 1.5 respectively. In view of the impact loading, the single rod/bolt, and the small edge distance I would go with the 6c as well. It is also important that the rod remain perpendicular to the rail so that no prying action is occurring.

I think that AISC assumes/allows some initial self-limiting yielding (deformation) of the plate where it contacts the bolt so that the high contact stresses are equalized for mild steel. I don't think you will see the high contact stresses dhengr mentioned unless you are working with hardened plates where the stresses are within the elastic limit of both materials. It is important that the rod/bolt not be allowed to rock or wallow in the hole though.

 

[paintballJim]

I would agree with both of you that there will be a slight deformation initially as the bolt "seats" into the hole. Otherwise we would technically have a point load with a 1" bolt in a 1-1/8" hole. The point about the I-beam flanges being sloped is a good point. The roller surface is sloped to match it but I may suggest another bolt and sleeve higher up to counter the tendency to spread.

Thank you both for your help.