# Stabilization against buckling.

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#### proppastie

##### Well-Known Member
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Crippling from Bruhn. You probably should read the whole chapter though.

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#### Autodidact

##### Well-Known Member
Thanks, proppastie. You did spur me into finding my old copy of Bruhn (wasn't sure I had one). The subject matter you posted is actually in a completely different chapter from my old 1949 copy (smells slightly funky!), but mine might not have the Gerard method. Will look at this more in the next day or so.

#### wsimpso1

##### Super Moderator
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OK, lets suppose you apply compression to the column to the point that it is considered buckled as per Euler, is it then a simple trig problem, not unlike calculating the forces in a truss joint? In other words, draw a line from the mid-point of the buckled column to one end, and find that line's angle from the column's original axis, if the angle is zero then the lateral force is zero, and if the angle is 10°, then the lateral force is compression times sin10°; but that can't be right because even in it's buckled state the column is capable of supporting a certain amount of compressive force, hence the problem is statically indeterminate.

I'm sure there is a way to solve this, but I'd hate to go through the statically indeterminate calculations. Thanks for helping me think this through, although if anyone can suggest a simpler method of solving I would appreciate it.
Autodidact,

First, Chapter 10 of Mechanics of Materials by Timoshenko and Gere covers columns, and Chapter 15 of Roark's covers elastic stability. Go there, be prepared to plow on past the math while reading, and get the gist of it.

This is kind of complicated territory. let's start by knowing your original Fcr for the unsupported case. If your jury strut were infinitely stiff, and firmly attached to the column, the restoring force developed in the strut when you get to Fcr is vanishingly close to zero. If the strut has real stiffness, that is it has measurable deflection vs load, the force developed in the jury strut at the Fcr will be measurable but still small if the stiffness in the strut is high, and the force in the jury strut will get higher as the stiffness is reduced. Now, open a clearance in the collar of the jury strut connection to the column (you know, the clamp is loose), now the column can move some before the jury strut starts to restrain it, and forces developed in the jury strut get even higher at Fcr, maybe high enough to buckle the jury strut. Makes you want to look close at your jury strut connections before every flight, doesn't it?

What force is developed in the tight stiff jury strut? Well, we can approximate it. Each column has its Fcr that is close to what is calculated by Euler, etc. They vary, some, but let's stay with Fcr for this one. At Fcr minus one pound, there is very little deflection - tough to ever measure - let's just say zero. Try to get to Fcr, and lateral deflection occurs WITH NO UPPER BOUND. This is buckling, the thing can take off to the side and collapse. How to prevent the collapse? Well, if the column load times the lateral deflection drops rapidly, you might stop it. If some other elements took the load at some point, you might save the day. But if you place a constant load on it that is as big as Pcr, it will try to collapse and very quickly too. Now this moment that is trying to bend the column is the load on the column times the lateral deflection. And the lateral deflection of the jury strut times its axial stiffness (its own EA/L) is the lateral load that prevents buckling. Now the force in the jury strut times the distance along the column to its mounts is also a moment. In detail, the jury strut forces go into the two ends of the column using lever rules, so apportion the load that way, then apply the force on each end by the lengths to the ends... So, for any column and jury strut, you can play the deflection back and forth until moment from the column and the moment from the jury strut balances, and that is restoring force needed. Bleagh! You want to make it closed form, you gotta use energy methods, which Timoshenko and Gere also cover but do not perform for columns - the extension, as the professors put it, is straightforward. Yeah, maybe not easy, but by being careful in applying algebra and calculus, you can do it. Or really screw it up. Roark's chapter 15 does get into the whole column solution with formula. Have good time...

One other comment about jury struts, they too are columns under compression and subject to offset loads from little things like g's and aero drag. Design them conservatively. Or look at other airplanes that use them and work OK.

Billski

#### wsimpso1

##### Super Moderator
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I've read several different descriptions of crippling, given by people from different engineering disciplines, and the details of the descriptions seem to be driven by what that person learned about their specific area of engineering, but they all involve exceeding the materials yield point. Crippling as I understand it is where the column is short enough that it undergoes plastic deformation (yield) due to compression forces before it can/or as it displace(s) laterally. An analogy would be to step on a coke can and crush it, that's crippling, and if you could stretch the coke can so that it was 12 feet long and apply compression, it would bow in the middle laterally and below a certain amount of force it would not deform plastically but only elastically and that's Euler buckling.

A column will not buckle under pure tension and is much stronger in tension, so a fully metal skinned wing would not have this trouble most likely since in the fore/aft lateral directions the stabilizing sheet (treated as a thin, wide column) would always be in tension. A good example of the problem area I'm thinking of would be the Ercoupe's wing (which is not a "problem", and I do not suggest here that it is!), which has a Warren truss rib arrangement with a fairly large distance between rib attachments to the rear side of the spar. On the front side it has that partial D-tube with false (or partial) ribs, and that would stabilize the spar unless the upper cap elastically displaced laterally forward with enough force to cripple the partial D-tube skin, but if the D-tube skin/false rib structure is strong/stiff enough, the upper cap will be stabilized from buckling and the ultimate failure of the wing will be crippling in the upper cap as it reaches the materials yield point without elastic buckling failure first. I'm just trying to think of a simple yet conservative way to do this w/o FEA or other analytically difficult methods. I suppose that most fabric covered wings have a small enough distance between ribs that this is not a problem, but with a Warren truss rib arrangement, there is more distance between the ribs (note: I am not saying that the Ercoupe wing is problematic; it has a nice wide spar cap and would need minimal stabilization if any):

View attachment 68038

The Hurricane has apparently (proprtionally) shorter distances between Warren elements (not actually ribs in this case) and those large tubular caps and may not need the extra stabilization that the Ercoupe wing has:

View attachment 68039

I think the simplified method that Dana expounded upon in post #17 may be a step toward a workable procedure. It would probably be mainly just a check, since with minimum gauges any LE skin is very possibly strong enough, especially with a partial rib, but we're always railing against TLAR, and I think something like this needs to be quantified in some way as long as it's conservative and not too inefficient weight-wise.

Thanks for everyone's thoughts on this.
We were talking about column buckling, which is described nicely by Euler formula for long columns and transitions nicely to Johnson and others for shorter columns and applies nicely to large scale buckling while the section EI's of the columns are maintained (the column shapes are maintained, at least early in process - once collapse is proceeding, shapes may go to hell in a handbasket) and everything is very sensitive to length.

Now crippling is being brought up. Crippling is independent of length. Crippling is changes in the cross section, and is small scale buckling of the structure. If the column failure starts before crippling, the eventual shape changes that occur later in the collapse are irrelevant. If the column is short and crippling starts before the column failure begins, it is relevant. Yep, for thin sections, crippling should be checked carefully too. If you are working in FEA with the buckling tools turned on, all of these modes are covered down to buckling at a scale of the element sizes. Want to look closer, tighten your node spacing, or go with Roark's and/or Bruhn on crippling and Elastic Stability.

By the way, the soda can trick works is crippling, but not the way you are describing it. Poking the sides removes a bunch of the can wall from load carrying, leaving only a little. It may also have had its shape changed enough that buckling stress is also reduced by the pokes, but mostly, you just removed a bunch of wall from load carrying...

Billski

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#### wsimpso1

##### Super Moderator
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I am not sure I understand how there is failure of a column before the crippling psi is reached unless we are talking about deflections short of failure which cause other problems.
It can and it does. Long slender elements will go into buckling at very low stresses. Most extreme examples are cables. They have strength in tension, but none in compression, and they will fold up without any yielding at all if you try to push on them. So will chains. When you make elements to carry some compression, like streamline tubes used as wing struts, streamline tubes used as jury struts supporting the middle of wing struts, thinwall tubes used as pushrods for control systems, and so on, you can compute Critical Force (Fcr) from Euler, Johnson, etc. That is where the column will deform out of plane, load carrying capability drops, and collapse begins. Stresses in the column right up to Fcr can be quite low up to and after the column commences collapse. Now, when you get to the smoking hole or pile of rubble, you may find the column or pushrod plastically deformed, but it most likely got to that state later in the collapse.

Want some demos? Got a drill press? Grab a foot long piece of 0.063 welding rod, hit the end with the torch to put a ball on it, put a piece of scrap wood on the table, chuck up the straight end, bring the chuck down until the round end is denting the wood, then gently bring the handle down until the rod just begins to deflect out of plane. It reached Fcr for the case of one end fixed, one end free, that length, and has elastically buckled. Bring the chuck back up, and the rod is as straight as it ever was. Bring the handle back down until the rod again deflects out of plane, and then bring it a little bit further, then let up. If you are careful about only making little steps past buckling, the rod will come out straight. Make a big enough step, and the rod will take a set in the middle - you got to yield strength there.

Now, drill a tight hole, press a straight chunk of rod into the whole, chuck the other end, and repeat. Deformed shape will be different, Fcr will be higher at same exposed lengths, this is both ends fixed. Use your torch to put nice balls on both ends, set one in a dimple in the table, the other end in the little dimple with the chuck closed down, and Fcr will be lower at same exposed length, this is both ends free.

And in all of these, if you somehow avoid letting the deformation go very far, only elastic deformation. Sometimes very low stresses. Make the column short enough and yeah, it will be tough to just get elastic buckling. Put a couple undersized columns underneath a porch roof, start adding snow to it, and when you get to Fcr on one column, you get a rubble pile with snow on top. Yeah, the columns are bent in two, but at the very instant that buckling commenced, it was at modest stress, out of plane and the crash was already in motion. The load stayed on the parts all of the way down, the columns first deformed out of column, then further and further and eventually reached yield strength.

Billski

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#### Aesquire

##### Well-Known Member
Back in the day when Boeing started cranking out 747s the price of aluminum went up enough that recycling beer & pop cans paid off.

SOP for flattening a can for us was to carefully stand on one foot on a can, which would support my 200 pounds, then lean over, balancing, and tap the sides of the can, lightly, with fingertips on both sides, QUICKLY pulling the fingers out of the way to keep from trapping the fingers in the crushed can. A test of balance, speed, and the best example of column failure you could ask for. At my weight, it was tricky to not prematurely crush the can, and a quick weight bounce did the job. But the "one leg back lean over & tap" technique was the "cool" approach.

#### Autodidact

##### Well-Known Member
At my weight, it was tricky to not prematurely crush the can, and a quick weight bounce did the job. But the "one leg back lean over & tap" technique was the "cool" approach.
Heh, I always did the quick bounce and that would crush the can nicely - I'm not sure I ever had good enough balance to do the side poke trick and I'm not sure I ever even heard of it until you guys brought it up!

Log Member
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#### Autodidact

##### Well-Known Member
Hmmm the missing phrase from the conversation is "Beam Column". The restraining load is going to be very dependent on any eccentricity between the axis of the force and the column axis, using beam column calculations would allow the lateral loads be calculated. The applicable "beam column" model would be one with end moments & a centre point load.

Now, though, there is the complication that a spar cap that is part of a shear-web style beam, will have progressively higher compression loads toward the root end, and it will not buckle like a simple column. The average of the compression forces at each end might still be a reasonable approximation since in reality (in the actual spar) there will be some end fixity factor.

##### Well-Known Member
There is a page of beam column calculations in bruhn, these are a selection of those listed in Niles & Newell.

If none of those do the job & you have now abandoned "simple"and a sucker for punishment, the next best thing Newmarks paper Numerical procedure for computing deflections, moments and buckling loads,
https://engineering.purdue.edu/~ce371/Docs/Newmark_calculating deflections moments buckling loads_1942.pdf
There is nice worked examples in BDM–6255 (Boeing Design Manual) if you happen to find a copy on the web.

The average of the compression forces at each end might still be a reasonable approximation since in reality (in the actual spar) there will be some end fixity factor.
That would be very hard to tell if it is conservative. Most of it could probably be broken down reasonably rationally without too much extra effort which would give you handle on it being conservative. .

#### wsimpso1

##### Super Moderator
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Looking at Table C7.1 Post 21 the max allowable crippling stress PSI is .8 Fcy for tubes. Are the "yield strength" above in the quote the yield psi (Fcy) of the metal at the point of yield?

http://www.homebuiltairplanes.com/forums/attachment.php?attachmentid=68047&d=1513395893
Yes. I know, you look at the part after either type of failure, and the parts are all bent up.

In general, buckling (whether large scale or in crippling) initiates elastically, the distortion then adds bending moments to the stress picture and will increase stresses up into permanent distortion areas. While it can happen in a short time period, at initiation, stresses are below yield. You have to prevent the initiation to prevent either long element buckling or short element crippling.

Billski

#### proppastie

##### Well-Known Member
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The calculations for an element with corners (not a tube) often yield numbers much less than Fcy. For tubes I use the tube allowable charts Bruhn C4. I do not think we are far apart here.

#### ragflyer

##### Well-Known Member
autodidact, lots of good answers here. In general determining stiffness of supports to prevent buckling can be done by hand using energy methods (very complicated for an amateur) or FEA (again not suitable for most amateur) . Both of these methods are over kill and fraught with uncertainty. I would avoid this and use typical spar/rib spacing/skin dimensions combined with conservative assumptions/calculations.

1. First if this is a cantilever spar, Euler column buckling is irrelevant. The spar is a pure beam. All you need to make sure (in terms of compression) is the flange does not cripple. As a simple (conservative) rule for aluminum: if the b/t ratio (width of cap to thickness) is less than 5, an Al flange can be assumed to be good for 0.7 times yield stress. For other b/t ratios use Bruhn or Perry.

2. Theoretically, if very slender, the spar could fail by torsional buckling ( twists when subject to pure bending). This is very unlikely with typical spar dimensions and rib spacing. Further, in D cell wings, if the D cell is capable of taking flight torsional loads of the wing then it is of ample thickness to prevent torsional buckling of the spar.

3. If your spar is braced, then the entire spar (both flanges and web) is subjected to compression due to strut and in addition the flange in the compression side is subjected to an additional compressive load due to bending . You need to make the following checks:

a. Check the flange for crippling as above but take both factors into account: load due to pure compression induced by strut and the load due to pure bending.

b. Check to see it the entire spar (Euler) buckles due to compression induced by the strut. The support rendered by skin and ribs are hard to factor in. Ignore them and assume the beam is the entire length from the strut attachment to the root. Keep in mind MOI (I) is the entire cross-section of the spar not just the flange. In general this case is not often critical- If the spar can take the bending it will take the compression due to strut, but check it.

c. Finally you have to consider the beam-column case. The pure bending of the spar results in the compressive strut adding a secondary bending moment due to eccentricity. Perry has a simple approximation to help with this. You will need to plug in the numbers from a. and b. into the formula to calculate this.