beam plastic bending / "drinking straw" failure?

[madvlad]

Hey everybody...
So, I have a square-section tubular beam in pure bending.

Plastic bending theory tells me that my applied moment is below the allowable. However, plastic bending is for stable sections...So I need to show that the tube doesn't cripple, or even if it does, it can still withstand the applied moment.

Bruhn's C4.16 tells me to treat the compression side (of a square tube in bending) as a flat plate, which I'm guessing means equation C5.1, failing as shown in Fig C5.1 (right?)

When I do this, I get allowable stresses that are really high, no matter what pessimistic value I choose for Kc...which tells me it doesn't cripple, and supposedly the tension side will break first. (right?)

But everyone's bent a tube before... first it ovalises a bit, and then flattens out and fails, like when you bend a drinking straw. Most tubes fail like this, even a solid bar fails in this way if it's a ductile enough material.

The scenario examined by C5.1 doesn't really have anything to do with this observed failure mode.

So how do I predict that type of "ovalising" failure? There happen to be allowable curves given for round tubes (which I assume come from test), but not for any other shape.

Thanks in advance
Chris

 

[graemew]

Bruhn Sect C5.1 is for elastic behaviour and you are dealing in the plastic range, right?
I think that if you read Bruhn C5.4 and consider Eq C5.2 and the Ramberg-Osgood approach to plasticity you will get a better answer.
A tangent modulus replacement for E is another way to go but it entails digitizing the tangent modulus vs stress curve and setting up an Excel solver to get a solution.
I have done this a few times but it is only worthwhile if you intend to re-use the curve for that particular material a few times. You need to enlarge the curve, read off data points, do an accurate Excel curve fit (normally involves at least 3 seperate curves and if statements to get the required accuracy), validate the curve, set up the excel equation and either use the Excel solver or iterate to an acceptable solution.
If you are going to do it once only, use the non-dimensional curves in Bruhn Fig C5.7 and C5.8. Assume simple support all around.
The value will be conservative. In the real world the flat panel will curve across the width and along the length.
Consider using a crippling approach as well using Gerard since crippling is plastic.
Good-luck
graemew

 

[GregLocock]

Isn't one likely failure mode short wavelength elastic buckling of the compression face?



Cheers

Greg Locock

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

 

[graemew]

Yes it is. The equations and pictures in Bruhn do not illustrate this very well but the peaky discontinuities in the Kc curves are related to the number of waves in a panel.

Also, elastic buckling isn't a "failure" per se. It can become a limiting stress since a buckled part will not carry a higher load than that calculated. It does however continue to carry the buckling load, which means that it is possible to have the various parts of a complex member with different buckling loads for different sections. If elastic it also reverts to an unbuckled state after the removal of load so it is possible to have to consider that parts limited by elastic buckling are at ultimate load with the limit load only 2/3 of that.
At this point I should point out that when considering plastic buckling as the ultimate state, that the allowable limit load could quite possibly be noticeably less than the load established by elastic buckling calculations.(not all buckled sections will get to a value where they can go plastic).
graemew

 

[rb1957]

one reason a straw ovalises is because it doesn't have corners

what sort of deflection do you have with plastic bending ?

is small displacement theory still valid ??

if you've checked that the compression side is stable (you've checked the face as a panel in compression, i'd've checked the corners for crippling) then i think you've done the job.

what sort of plastic bending allowable are you using ? (a test based modulus of rupture, a K factor, cozzone, a hand calc using the material curve) ? the point is what is your compression allowable stress (fcy, Ftu?) ?

what sort of b/t do you ahve for the sides of your tube ?

 

[dik]

b/t ratios for local stability. Rotation is required for plastic yielding at mechanism developed and amount of rotation is not as pronounced as drinking straw

Dik

 

[madvlad]

Ok, just for nomenclature: a square tube as a beam has a compression face, a tension face, and two 'webs'. I'm talking about a 2.5 square tube with 3/16 wall. But more generally, it could ANY section, even a solid one.

I'm saying that, ok, I have to prove the section is stable in order to claim plastic bending per the Cozzone method. So what are the possible instabilities?

One potential instability is buckling of the compression face. So I check it for plate buckling...and it passes. The critical buckling stress is well above the applied stress in the compression face. (It's also above Ftu)

So the compression face isn't going to buckle, fine, and Bruhn doesn't mention any other instabilities I need to worry about. Fine, my bosses are buying that, and I guess rb1957 buys that too.

BUT! I don't believe the section is stable all the way up to Ftu, because:

There's another possible instability, failure of the webs, which is unaccounted for. (not shear failure). In the failure mode I'm asking about, the webs bulge outwards as the tube deflects, the webs eventually buckle outwards, the compression and tension sides come inwards towards eachother, and the tube flattens out and fails because the section is reduced locally by deformation. (Like a drinking straw)

I just don't automatically trust that a ductile metal tube will ultimately fail by snapping (tension side failure), I've never seen a tube do that, in practice they ovalize and flatten out (or fall over like a parallelogram)

Beam small displacement theory is NOT valid. It's in the plastic regime, and deformation of the section is the crux of the problem.

At first I also thought about using the tangent modulus in the plate buckling equation, but the failure isn't plate buckling anyway.

Needham/Gerard are limited to 0.8 Fcy, so by definition, one cannot use these methods anywhere near the plastic region.

It was asked: I'm using plastic bending allowables per Cozzone method, C3.14, for 6061-T6 (which closely match NASA TM-X-73305 B4.5 page 147)

I'm not worried about the limit load case, and if the section doesn't fail by crippling, the deflection will be ok.

 

[rb1957]

the point about asking about deflections was that i think other load carrying mechanisms come into play, lke a thin membrane carrying pressure by hoop stress.

i'm looking at a rectangular tube, similar to your dim'ns (6061T6) that we tested ... 'cause i was a little uncomfortable with the analysi i'd done (and it was easy to test). the sides haven't bowed out appreciably, there's a small buckle on the upper face, an indentation where the press applied the load. one of the very intersting things to come out of the test was that the 6061 is so ductile that it was very difficult for the press to apply the maximum moment (in fact it couldn't).

i agree with you, that the section won't fail in tension, that something else will let go alittle before then ... as you say, the sides might buckle under the applied shear. probably the corners will start to yield, which will start the ovalizing that you're (IMHO) alittle fixated on. btw, i don't think your straw analogy is particularly applicable to your pretty darn hefty tube.

another consideration, as this piece of a larger structure starts to yield, the stiffness of the load path decreases ... will the strcuture will redistribute load away from the tube ?

 

[madvlad]

I'm not really fixated, I was just curious about a particular subject that I didn't understand, hope thats ok!

rb1957, you got me on the right track, making me sit down and think, well, what IS the deflection of the beam? Especially mentioning that your test structure never reached a maximum.

I now think that my "failure mode" isn't really a failure mode at all, it's a deformation that occurs after the *real* failure, which is plastic collapse of the beam. (which didn't occur to me until you implied excessive deflections) It's a symptom, not a cause.

Understanding that, I now understand why one of the examples in Bruhn's chooses Fmax (in the Cozzone method) as coming from a particular strain value, rather than an ultimate stress.

It's done because the permitted strain (and beam curvature) is what defines Fmax, otherwise I'm just analyzing what happens in a tube bending machine. I didn't realize that.

Thanks guys and especially rb1957

 

[graemew]

Good on you. I am pleased that you are now viewing this in the correct light.
Here was me thinking that you were dealing with a bit of light structure.
All the best with it.

 

[JZuk]

madvlad,
I was searching for another topic, and I came across this one. Not sure if you are still keeping track of it. I have some knowledge that may help you.

I would limit the compression side to Fcy (34 ksi). If you do a buckling check of the compression side segment of the square tube using the tangent modulus, Et, (k = 4.0), the buckling stress is about 33 ksi. The tangent modulus is the most conservative of all reduced moduli. The tangent modulus can be calculated using equation 9.8.4.2(b) (page 9-191) in MMPDS-01. Thefore, you can conclude that the segment buckles at about Fcy.

The vertical segment also buckles at Fcy. Use K = 24 (from Fig C5.15 from Bruhn - this chart has an error in it - the values on the X axis should go from 0.3 to 2.3 - this is a know typo in Bruhn). Also use the tangent modulus again. This results in a buckling stress of 34 ksi - same as Fcy.

The section is not stable and compact. The b/t ratio is 2.5/0.1875 = 13.33. Although the topic is not coverd in literature (due to the fact that each situation is different). If you had a shape that was more stable and compact, say a solid squae section, or a rectangular section with a b/t of no more than, say, about 8, then you can do plastic bending. There is the "intermediate" range were, yes you buckle at Fcy, but you are not compact and stable such that you can do plastic bending.

If you were to test your 2.5" square, t = 0.1875" tube, you'd find that it probably exhibit some plastic bending capability, but it would be very small. There are two resons for this.

1. Your section is not stable and compact. it is in the intermediate range. It does not buckle below Fcy, but it also does not have the capability to redistribute load in a plastic manner (Cozzone's method) because it is essentially buckling at Fcy.

2. Even if it could re-distribute load per Cozzone (testing will prove that is can - to some very small degree), the cozzone method will buy you very little based on your shape. First you shape factor will be small - there is very little material a the center of the cross section to redistribute load to. But you could take it ot some apparent stress level over Ftu - thus seeming to buy you more strength. But, once you go into the plastic range, you must interact bending and shear stress - thus reducing your load carrying capability.

So in summary, the walls of your tube buckle at Fcy. But the section is not capable of going plastic using conventional methhods. Testing might show some small, small plastic ability - but it will be peanuts.

 

[madvlad]

Hi JZuk,
Respectfully, I'm not sure I agree with you...

Et is a function of stress, and in the vicinity of the compressive yield stress, the tangent modulus is equal to the elastic modulus (by definition).

Ie at Fcy, Et=~E = 10.5x10^6 psi

Solving for (Sigma)cr using C5.1, k=4, v=0.3, and t/b as 0.1875/2.5, you get 214 ksi, which is a long ways from your 33 ksi which you got somehow. There is still more strength to be had, beyond Fcy...

Sure, as one moves beyond Fcy into the plastic region, Et starts falling off, and I guess would eventually converge with the applied stress, resulting in buckling. If you try to bend a tube in a tube bender, eventually you get to a small enough inside radius that the compression face will buckle. For round aluminum tubes, it's ~4D...

But by that time, it makes no sense to call the thing a beam.

The beam doesn't fail by buckling or crippling or rupture. It never really 'fails', it just runs away from the force..the more you push, the more it bends, until it reaches plastic collapse or some other geometric limitation, without ever buckling or rupturing.

RB1957's comment about never being able to apply the 'maximum moment' using a press was the perspective I was initially missing...

The other thing I don't agree with is that I could get a "stress level over Ftu" somehow...that's impossible, by definition, and I don't think that's what you meant...