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Elastic Buckling 1

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LeonhardEuler

Structural
Jun 19, 2017
200
This seems like an elementary question, but I've always wondered about it and have scoured the internet to no avail. Perhaps it is too simple for Mech of materials books to explain.

My question is how does a compression member buckle with a purely axial load. Is this due to inherent initial out of straightness, which means the load actually isn't purely axial and does have eccentricity and moment, or is this due to poissons effect where the axial load creates a lateral stress?
 
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Does no reply mean the answer eludes others as well? There's no practical need to know exactly "why", but I must figure out for sanitys sake!
 
"Does no reply mean the answer eludes others as well?" Not exactly, it is covered in the first year of an engineering degree. Look up euler buckling.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376
 
I understand what euler buckling is; however, no book explains clearly how an axial force causes lateral deflection of the slender element. Typically it just says that it does happen.
 
I have seen that the buckling load is the load at which the "slightest" lateral load causes the column to jump to a new configuration, so I will assume the initial out of straightness of the member is typically to blame for the buckling.
 
The point of euler buckling is that any perturbation can cause a new geometry without any additional energy in the system, so the causes could be many. Load eccentricity is one, side load is another, as is material distribution. I see by your change in username that you have been doing some research. Good.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376
 
In developing the Euler buckling equation, an initial deformation is assumed. This drops out in the derivation and an initial deformation is not part of the equation. It is simply a factor of E, I, and L. There are several derivations on the web... chase one down.

Dik
 
It all depends on the amount, if any of lateral restraint, the length, the type of "member", diameter to length ratio if under axial force from either end, wall thickness and d/t ratio.

There are many many sources of information out there (BTW it's "scoured" the internet not "scowered"). you just need to use the right terms. It's also quite complex given the variance of issues not there is no simple rule, whereas Euler buckling on a long thin member is pretty basic stuff. Essentially in my mind the higher the axial force essentially reduces the initial deformity to the point where it becomes practically impossible not to have such deformation. However there are limits and the L/d ratio is one of them as well as the stiffness.

E.g.


~And if you search for "elephants foot tube buckling", you find things like the pictures below.

If the conditions are right then a tube especially will fail in a compressive circular bulge or bulges. For a simple example think of a Coke can crushed under your foot. Get it right and it simply collapses vertically into a squashed flat, but wrinkled mass.

Buried pipelines and pipes heavily guided on occasion fail in compression like that, sometimes called a mash buckle or in columns they call it elephants foot.

If you have a solid rod e.g. inside a sleeve, but leave a certain short section able to expand sideways but not buckle, it will in the end simply yield sideways in a uniform "bulge"

buckle_sklrl9.jpg


buckle_2_rqtmcf.jpg


Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
 
"In developing the Euler buckling equation, an initial deformation is assumed." ... I don't think so. I think Euler considers a perfectly straight element in perfect compression. Euler buckling does "just happen" as a result of the math. If the element is imperfect (bowed, eccentric load) this is different (non-Euler) problem with a different (secant for example) solution.

another day in paradise, or is paradise one day closer ?
 
my recollection is that it is the solution to a differential equation, of a perfect column.

from wiki ...
Mathematical Derivation - Pin Ended Column[edit]
The following model applies to columns simply supported at each end.

Firstly, we will put attention to the fact there are no (lateral) reactions in the hinged ends, so we also have no shear force in any cross-section of the column. The reason for no reactions can be obtained from symmetry (so the reactions should be in the same direction) and from moment equilibrium (so the reactions should be in opposite directions).

Using the free body diagram in the right side of figure 3, and making a summation of moments about point A:

Sigma M=0 ... M(x)+Pw=0
where w is the lateral deflection.

According to Euler–Bernoulli beam theory, the deflection of a beam is related with its bending moment by:
M=-EI{d^2w/dx^2}

The column is perfectly straight until the critical load is reached and it fails instantly. This is in the ideal mathematical world, the real world is full of imperfections.



another day in paradise, or is paradise one day closer ?
 
I think some of the items change if you don't have a solid rod or member, but a tube or other hollow item.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
 
I had the same kind of questions when the concept of buckling was first introduced to me many years ago. Best explanation was that the strain energy in bending eventually becomes less than the strain energy of cylindrical compression as the axial load is increased, and (nature liking minimum strain energy as it does) the column buckles.
 
solid or hollow tube only changes I; and also the possibility of local buckling of the wall (Euler assumes a stable cross section).

another day in paradise, or is paradise one day closer ?
 
https://www.youtube.com/watch?v=wrdO8hPJGyg

This is a very good video that shows how I affects Euler.

The problem (as I see it) is K. It is a made-up number.

It's all well and good to say that it is 1 for a pinned-pinned column and a lower number for other situations. What is the lower number? You make it up.

Torsion is well defined because it is repeatable and predictable.
Tension is well defined because it is repeatable and predictable.
Compression is defined using a fudge factor.

You can get pretty close to the correct answer, but you will never be exact. You can see why when you look at the images in LittleInches's post. No two failures were the same. When you get to Point A on his graph almost anything can happen, but that is discussing failure.

Elastic region is defined by the gradient of his graph. The gradient is a factor of how "pinned" the ends are.

If you are compressing the strut using pure axial loading (almost impossible), will the strut bend upwards, downwards, or compress as shown in L/D=2 and L/D=4.
 
K is a correction factor for end effects. It is not "made up". It is based on analysis and tests. Sure, many times we'll assume a value in our hand calcs but that is based on experience (which could be interpreted as "made up", but then so it the result).

another day in paradise, or is paradise one day closer ?
 
My final year project looked at the deformation of welded tresses, up to and including buckling. The curve of deflection is non linear as the structure 'softens' as the load increases and approaches buckling. I got very good correlation with my test results using some standard method for investigating the restraint of an elastic buckling member by the surrounding structure. I can't remember the methodology but if it was in the 3rd year of an engineering course it can't be too obscure.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376
 
If you have a sample in a compression machine. As you load it and tap it with a bar... the sound rings flat at buckling due the the stiffness going to zero.

Dik
 
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