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Concentrated Force on HSS

Concentrated Force on HSS

Concentrated Force on HSS

(OP)
I have a connection I am analyzing for a fixture or sorts (NOT a building connection)where a 2.375" dia. round HSS is slid through holes of the same diameter drilled thru both walls of a 6" HSS rectangular tube.

The 2.375" round is loaded as a cantilever so it will have concentrated forces resulting from the moment where it bears on the rectangular HSS.
I am trying to check this condition using Chap K in AISC 360-05 and in particular equation K1-1.
K1-1 refers you to K2-1 for determining the factor Qf.
I am getting an extremely low value for Qf killing my connection capacity.

Is Qf even applicable here?

ConnectEgr - you out there?
 

RE: Concentrated Force on HSS

Have you checked the commentary with reference to the calculation of Qf.  They do have some helpful diagrams.

Wouldn't this be similar to a "thru" plate?

RE: Concentrated Force on HSS

(OP)
Yes I read the commentary but it's not much help.

What are your thoughts SteelPE?  

RE: Concentrated Force on HSS

(OP)
Essentially yes

RE: Concentrated Force on HSS

Then I am not sure you are using the correct equations.  I would design the example as if it was a through plate using the method described in HSS connection manual.  

RE: Concentrated Force on HSS

(OP)
I dont think I agree...all this is is a 6x6 tube with a hole drilled in it with  2.375" HSS round slid through the hole.

The 6x6 is a column and the 2.375" round is loaded as a cantilever.

The reactions where the round HSS passes through the walls of the 6x6 tube are essentially concentrated forces transverse to the longitudinal axis on the 2.375" round HSS.

NO?  

RE: Concentrated Force on HSS

If you look in the design example they give you an example of a longitudinal plate and a transverse loaded plate.  Both of those examples are different how your "outrigger" tube is loaded.

In the HSS connection manual they end up with a resultant shear in each face of the tube.  This shear is what you design your weld for.

I may be wrong but more diagrams should be given in the AISC manual to make this a little more clear.

RE: Concentrated Force on HSS

(OP)
this "outrigger" is not welded.  

RE: Concentrated Force on HSS

The outrigger is not even welded a little?  What stops it from falling out?

If that was the case, I would use the same equation and then use a modified bolt bearing equation to resist Vf and Vr.  How much force are we talking about?

RE: Concentrated Force on HSS

It's just a propped cantilever.  If the pipe is good in bending, the tube is good in bearing, and the tube is still structurally adequate with the hole, what's the problem?

RE: Concentrated Force on HSS

(OP)
Checking the tube bearing is the problem. I was trying to check it is if it were a concentrated force according to AISC chap. K.

I will also need to check the bearing on the hole itself.
Also not sure what to apply here.


The tube is held in place with a collar and a set screw

RE: Concentrated Force on HSS

I think you are using the wrong equations.  I always thought the equations in Chapter K were for tubes/plates welded to the face of a HSS.  Your connection is different.

Bearing on the hole would be the same as bearing for a bolt.  What's the difference?   Bearing on the tube would be the same as well.

RE: Concentrated Force on HSS

Toad -
Are you concerned with crushing of the round HSS or with bearing on the wall of the HSS 6x6 column?

As far as crushing of the round HSS I would need to look at DG24 more or I would look at roark's rings section.

For bearing on the HSS square column unless there is something more specific in DG24 I think checking bearing as is done with bolts (as SteelPE suggested) should be a reasonable solution  

EIT

RE: Concentrated Force on HSS

(OP)
Yes.
I am concerned with the round HSS wall crushing as it bears on the Rectangular HSS wall.
I am also concerned with bearing on the thin wall of the Rectangular HSS

 

RE: Concentrated Force on HSS

(OP)
So are you suggesting that I simply use Chapter J.8?

Seems to simplistic
 

RE: Concentrated Force on HSS

(OP)
Actually, how does concentrated forces on HSS not apply here?

The reactions of the pipe on the tube walls are concentrated forces on the pipe, no?  

RE: Concentrated Force on HSS

The pipe is fully restrained.  It can fail in bending, but arching action will prevent any other mode of failure.

RE: Concentrated Force on HSS

OOOOHHHH, you are concerned about the concentrated force acting transverse to the 2.375" dia pipe.  For some reason I was thinking you were concerned about the HSS6x6 square tube.

Kind of an odd duck here.  If you are having problems, why can't you just increase the size of your 2.375" dia pipe?

RE: Concentrated Force on HSS

(OP)
yes- I was wondering why no one agreed.

The connection is existing.  

RE: Concentrated Force on HSS

(OP)
I am determining the capacity of the connection.
In using K1-1 and subsequently K2-1, the Qf value I get is only 0.03.
Nonsense.  

RE: Concentrated Force on HSS

The lowest I could see Qf getting to is 0.4.  That would be at U=1.  Otherwise your pipe should be overstressed no (or at least close to it)?  Something isn't right if your Qf=0.03 that means your U = 1.366.

RE: Concentrated Force on HSS

(OP)
Steel PE-
Here is the issue.

In calculating U, Mr = the full available moment because I am determining the capacity.
Pa in my case seems to be = 0 (no required axial strength)
So U = Mr/S*Fc
Yes, you are correct. When I calc U, I get 1.366
 

RE: Concentrated Force on HSS

And the reason why you are getting 1.366 is because the shape factor for a circle is about equal to 1.366.  Interesting, then I would say you need a larger outrigger or don't add any more load to the existing outrigger.

The only thing I would say is that with Pr= 0 you don't really have a chord.  The definition of a chord member = For HSS, primary member that extends through a truss connection.  So I am don't think these equations even apply.   

RE: Concentrated Force on HSS

(OP)
Looking at Chapter G, maybe my concerns are covered with the shear bucking provisions Chap G6.  

RE: Concentrated Force on HSS

What's the wall thickness of the 6" HSS?   

THAT area is going to resist the force of the round member ("up" on back side, "down" on the front side); but only about 1/3 of the bottom arc and 1/3 of the top arc should be counted as effectively resisting the force.

(There's no real resistance to the force on the tube at the 90 degree of the pipe/round HSS wall.)

If the round beam does not fail by bending, then the failure mode will be collapse (kinking) of the round pipe on the bottom at front face.     

RE: Concentrated Force on HSS

(OP)
1/4" thickness.

"(There's no real resistance to the force on the tube at the 90 degree of the pipe/round HSS wall.)"

Not sure I agree with this.  

RE: Concentrated Force on HSS

How could there be resistance to a vertical force, when the round HSS is parallel to the round holes in the square 6 inch HSS member?  

IF the round HSS were welded, then you can take credit for resistance to movement through the fillet weld's 360 degree circumference.  The (two ?) fillet welds will resist tension and compression at both front and back faces of the 6 inch HSS.  But for a "loose" member (one not welded), the only resistance to a vertical force will be the "somewhat horizontal" lower front and upper back.  

If your member is fillet welded between the two members, then the whole problem changes:   Your loads are NOT spread through the wall thickness only, but through the entire leg width of the fillet.  Your bending moment is resisted through a wider lever arm as well (6 inches across the HSS plus 1/2 of each fillet leg on each side, rather than 6 inches across the HSS minus half the vertical HSS wall thickness on each side.)  IF - and this doesn't often happen - the penetrating member is also using a slide fit (a very snug hole) then the walls of the vertical HSS will also assume part of the load as well: Gives you even more surface area to absorb the force.   Your bending force will still be maximized at the front face of the assembly though.  


If there is a problem, then replace the round small HSS member with a thicker wall "pipe": I'm working today with a 1-1/2 diameter pipe (1.90 OD) with almost 7/16 wall.  

What's your max load?

RE: Concentrated Force on HSS

(OP)
Losing ya here....

Its a propped cantilever as Hokie said.
As the cantilever pipe is loaded, there will be a reaction at both faces of the rectangular tube. Vertical reactions in opposite directions.  

RE: Concentrated Force on HSS

interesting problem..probably no exact solution....
in this case I would usually bound the problem.
Calculating the longitudinal bending effects on the pipe are pretty straghtforward.
The effects of the shear loading on the pipe at the rectangular
member is where I would have to do some digging
The round HSS member is somewhat restrained from deforming in the plane of the face of the rectangular member, as hokkie pointed
out, but just outside the face of the rectangular memb, the round HSS member may still be influenced by this shear loading and start to deform.
Rectangular HSS: I might look at it similar to a lifting lug and
                 calc max bearing stress accordingly...this is a
                 very conservative approach as the round pipe is
                 not as rigid as a solid pin..if I can not live
                 with these results, I would back-off and say the
                 pipe will deform locally at the bottom of the
                 hole, say a 60 to 90 degrees included angle and
                 calculate the bearing stress based on that arc
                 length.
                 
Round HSS: use appropriate case out of Roark for a parial line
           load on a pipe or ring...this could be a varying or
           constant load..get max bending and shear stresses.
           combine these with the lonitudinal bending stress and
           get max  resultant stress.
To somewhat validate my approach, I would consider the case where the pipe would actually be welded to the rectangular memb and try and visualise the differnt behavior between the two cases.
What about buckling?..ofcourse the longitudinal bending should be checked for this alone. For the buckling effects of shear loading at the support I am assuming that the constraint provided by the hole minimizes this and one is left with only the concern of max local combined stresses in the round memb at the support.
I am sure there is a more accurate analysis out there, but this what I got off the top of my head.  
 
 

RE: Concentrated Force on HSS

Toad...
Is there a sketch of you condition in this thread somewhere?  Is this ta the top of the column?  Is the pipe welded around the circumference to the tube wall?  Is all the moment resisted by the column, or is a counter force on the far side?

http://www.FerrellEngineering.com

RE: Concentrated Force on HSS

I think I would do as Sail has suggested -

For the round HSS you have shear stress and bending stress due to cantilever the cantilever then I would look at Roarks rings equations to find the moment or shear for the 'crushing or kinking' force. I would provide more direction on this but I have not yet read that section.

For the rectangular if your concerned with bearing I don't see how checking this as if it were a large bolt would be unconservative. So if it works for that process it would be ok, no?

 

EIT

RE: Concentrated Force on HSS

The problem is similar to the design of saddle supports for a pipe.  The following link may be helpful, but in your case the saddle width is only 1/4", much less than the minimum required for a ductile iron pipe (I'm not sure about a steel pipe).

http://www.dipra.org/pdf/pipeOnSupports.pdf

BA

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