Safe Working Load of Simple Rectangular Beam

[Maxarmstrong]

Hi guys,

Can anyone help me with the following calc I have to perform (I'm sure you'll find it basic.)

I have a rectangular beam positioned vertically to support a load. The beam is made from AL AL 6082 T6 size: 30mmx60mmx1000mm. The load applied to the top of the beam(30x60 face)is 4500kg.

Can anyone show me how to calculate the buckling load, and derive either a FOS or the failure criteria?

Many thanks guys.
Max.

 

[Lion06]

Is there any bracing? This will effect the buckling strength. Is there any welding? This will effect the yielding strength.
Do you have (or can you get your hands on) a copy of the Aluminum Design Manual? There are provisions for allowable stresses based on the section type and b/t ratios. If you have welding, that knocks the Fy down significantly within a few inches of the weld.

 

[ishvaaag]

Lesser slenderness as doubly pinned is 115 so it has elastic buckling behaviour and (I don't seem to have the data for your exact 6082 T6 material, but it is T6 what I give as follows) an allowable stress of 85000/115^2 or 6.42 ksi gives about 8.96 kips allowable or 4.05 ton. Your load exceeds the allowable by 11% if the formula for the allowable stress in the elastic range is right for your material. Seek that formula in the aluminum specs.

 

[ishvaaag]

Disregard the previus entry;
Will look a correct allowable stress later. No time at this moment. It seems I am taking a section that doesn' is the correct one.

 

[ishvaaag]

It seems section 3.4.7 overall buckling behaviour in the elastic range controls, hence

Fa=PI^2*10100/(SF*(L/ry))=3.86 ksi, where the safety factor is 1.95 and L/ry=115.

Hence the allowable load is 10.76 kips or 4.87 ton.

Your load is just safe.

It seems I not only choose a bad section in the book I was searching but multiplied with error by the previous allowable. Hurries.

 

[Maxarmstrong]

Thanks for the excellent response guys.

I'll be documenting this straight away

 

[Lion06]

ishvaaag-
How are you determining that without knowing the bracing conditions?

 

[swearingen]

Maxarmstrong,

ishvaaag's calcs are OK considering a pin-pin situation without sidesway. If the top of this column is not restrained adequately, your safety factor effectively disappears.

You might want to have a local structural friend of yours take a look...


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[ishvaaag]

Structural EIT, I was just solving a pin pin situation as sweringen noticed, not negating that bracing influences allowable load.

 

[swearingen]

My point is that pins are modeled as infinitely stiff with no moment resistance. I'm curious if his situation really approaches this assumption...


If you "heard" it on the internet, it's guilty until proven innocent. - DCS

http://www.eng-tips.com/supportus.cfm

 

[ishvaaag]

Certainly, if you do not restrain movements at the tips of the bar, we are in a dangerous situation, 4.5 tons on something that can slip. So, again, assuming pinned ends.

 

[msquared48]

I trust that this is not a student problem...

Mike McCann
MMC Engineering

 

[Maxarmstrong]

I wish it was a student problem, unfortuately its a professional problem.

More info:

Basically this bar is used as a stop bar for a periscope during maintenance, when the hydraulics have been shut down the periscope will fall down the well unless a bar is wedged underneath it to stop it falling.

And thats it, the bar is simply wedged underneath. Because the periscope is housed in a well it can it has only vertical freedom.

I am running an ANSYS solution to this but I would like to compare it to a hand calc.

As for the bracing of the rectangular bar, I have been assuming K=0.5 which is both ends fixed - do you agree?

Unfortunately I am all alone here, surrounded by Electrical Engineers so there are no other Mechanical or Structural Engineers to discuss this with :'(

Max.

 

[ishvaaag]

Bracing length ... if the corners of the 30x60 mm section are tightly fitted to the sheat and this is of some bigger stiffness, it is mostly a compression only in your bar as long the sheath is able to brace the buckling forces (continuosly in this case). Your section would be definitely ok but we would have to check the continuous bracing condition of your bar, i,e, if the sheath has enough strength and stiffness to completely restrain the buckling of your bar.

If your bar has clearance laterally within the well and you have a 2000 mm length that actually causes elastically the sidesway buckling, the midlength point of your bar accompanying the buckling will come to fall to one side of the confining well to be supported there by the bracing action of the same, hopefully strong enough at the buckling deflection there. Assuming you can count on enough restraint there to lateral movement (force and stiffness) we can surmise out of simmetry you have there fixity at midlength and a pinned connection atop (assuming clearance). Hence to determine the slenderness we would have to use the fixed-pinned condition with 1000 mm, or K=0.7. However, in more than the 4.5 ton axial load, we would have by then to check it as well with some concurrent moment corresponding to the initial bending standing out of the clearance, the problem has become not the buckling of a straight bar under under an axial load, but one of the stability and strength of some bar on the described conditions. The moment present is as derived for the standig deflection-clearance for the bar stiffness. You may calculate it without going our old Cross notes as the deflection caused by force enough to cause such deflection in your 1000 mm bar in cantilever way, or P·L3/(E·Iy)=Actual Clearance with tip of the bar and midlength points both touching the sheath. Then the Fcr for the axial load would be calculated as above for the 700 mm length and a formula of interaction with flexure from the aluminum code would be required to check the midpoint of your 1000 mm notional bars. So to perform the calculation and check the safety the clearance as described would be needed, then look for the strength in flexure of the aluminum and then verifiy the well can deliver strength and stiffness to confine the overall buckling of the 2000 mm bar.

 

[Lion06]

The OP says a vertical beam, and I think of a "column" with lateral loading only.
What is the actual loading condition?