8

If you are looking at the column as being supported at the ends, you are correct that the n=1 mode gives the lowest buckling load. The other modes (n=2,3,...) aren't useless though. Long columns are often braced at regular intervals to reduce the unbraced length of the column. For a given length of column, these braces force the column to buckle under a ...


7

Whether or not buckling modes with $n>1$ exists depends on how you look at the structure. As @hazzey notes in his answer, columns with bracing may display buckling modes with $n>1$. These buckling modes, however, are simply equivalent to the $n=1$ modes of the individual segments that compose the column. To be clear, this doesn't mean that the ...


7

Yes, buckling is a sudden change in shape. A unevenly made column that bows as a result of increasing stress would just experience deformation. Or a metal bar bent by a "strongman." Buckling is different from regularly occurring plastic deformation.


7

In a wing the normal situation is that the aerodynamic force is upwards (resisting gravity). You are right that there is some shear and a fair amount of torsion, but the result is that: the top side of the wing is in compression, while the bottom is in tension. So in essence what happens is the opposite from the picture below (the force is applied upwards, ...


5

Defining buckling as "a sudden change of shape" seems to be mixing up cause and effect IMO. What happens is that the load-deflection path of the structure bifurcates when the load reaches a critical value, and the structure then follows whichever branch requires the least energy. Since the stiffness of the two possible paths are usually several orders of ...


5

Shear buckling of beam web happens when the shear at a section of the beam under consideration surpasses the controlling combination of factored shear. The following three equations are the LFRD method for: no web instability Inelastic web buckling Elastic web buckling These depend on the ratio $h/t$ of the web. If this ratio is bigger than 260 web ...


3

Real structures do suffer from buckling. Yes, real structures are never perfect. Columns aren't perfectly vertical, cross-sections aren't perfectly consistent throughout the span, the material isn't perfectly homogenous, and the loads aren't perfectly centered. That is all true. However, for loads lower than the true buckling load (which is much smaller ...


3

It means exactly what it says. The geometric tolerances in the beam are exactly zero. The material is perfectly isotropic and homogeneous. The end load is distributed across the surface of the end of the beam in a way which is perfectly consistent the Euler beam theory and does not produce any local stress concentrations at the end. Etc, etc... These "...


3

The basic Euler buckling formula is unrealistic because it implies perfect geometry and perfect alignment of the loads. Therefore a large empirical safety factor is required. It is also a potentially catastrophic failure mode, in the sense that the buckled column supports no load at all, unlike plastic failure in compression for example. There are ...


3

To answer the specific question: (a) No, that equation is irrelevant and (b) Yes. see below. The question isn't about Euler buckling. It says the bar is rigid. Euler buckling only applies to a flexible bar - the Euler formulas include Young's modulus, and also the moment of inertia of the bar which is not mentioned in the question. One way to solve the ...


3

I found this formula here: $$p_{crit}=\frac{2\,E\,t}{D}\left(\frac1{(n^2-1)\left(1+\left(\frac{2\,n\,L}{\pi\,D}\right)^2\right)^2}+\frac{t^2}{3(1-\nu^2)D^2}\left(n^2-1+\frac{2\,n^2-1-\nu}{\left(\frac{2\,n\,L}{\pi\,D}\right)^2-1}\right)\right)$$ So you should set $p_{crit}$ to about 3 atm to be safe, then find $E$ and $\nu$ for your aluminum, plug those in ...


3

What you want to look for is Shear buckling or Web buckling. It is quite an advanced subject though. In order to determine mode of failure, critical loads etc, you need to specify explicity the boundary conditions, and then solve the problem for the dimensions. In case that you want to determine the permissible load of the member outside the context of a ...


3

For Euler Buckling the critical load is: $$ P_{cr} = \left(\frac{\pi}{K L}\right)^2 E I $$ where: $P_{cr}$ is the critical load $E$ is Young's modulus (for steel assume 200 [GPa]) $I$ is the second moment of area. ($\frac{\pi}{64}(d_o^4- d_i^4)$ $d_o, d_i$ is the outer and inner diameter respectively $L$ is the length of the rod $K$: is a parameter that ...


2

In principle the axial load in an element is related to its natural frequency. Think of it like a guitar string - increasing the tension in the string raises the pitch (frequency). It may be possible to attach an accelerometer and then strike the element with a hammer to get it vibrating and measure the natural frequency (or even measure with the laser - ...


2

It is not necessarily a typo, but you wouldn't normally call a thin walled cylinder 8mm high and 100mm diameter a "pipe". However, the formulas for Euler buckling are most definitely not valid for a structure with those dimensions. The appendix of this rather old report http://www.dtic.mil/dtic/tr/fulltext/u2/a801283.pdf gives some basic theory and graphs ...


2

Yes, it is better to design for yield than buckling because Euler buckling is a result of instability as opposed to overloading force. Gamma f is determined from Eurocodes which are from extensive research and materials testing. This means yield is safer than buckling because buckling is not predictable nor easily counteracted in design. It also would mean ...


2

In your example 2, Nastran is scaling both loads as part of the buckling load. In your first case, you have buckling with $90\times 1 = 90$N axially plus $90\times 1=90$N laterally. In the second, you have buckling with $9.5\times 10 = 95$N axially plus $9.5\times 1 = 9.5$N laterally. From example 1 you have a buckling load of 100N axially plus zero ...


2

The real world doesn't know anything about approximations you make in mathematical models of it. If always behaves in a fully 3-dimensional way. For Euler buckling of a column, you are really studying perturbations about the steady stress condition, which is just an axial direct stress in the column. You want to find the end load at which a small ...


2

I briefly glanced this article. Seems interesting but I haven't read it yet - hopefully I'll still be able to help. First, lets say a few words regarding the non-linear buckling analysis. It is important to realize that contrary to linear analysis, it takes into account the following: Geometrical non linearity - the model shape changes according to the ...


2

Euler buckling is a perfectly elastic behavior: if a beam buckles, it takes a sinusoidal shape of arbitrary (possibly infinite) amplitude. If you then remove the applied force, the beam will return to its original, perfectly-straight shape. Real-world beams, however, don't behave this way: if you buckle a beam, it gets destroyed. After all, Euler buckling's ...


2

One of the most fundamental principles in the eurocodes is the distinction between the serviceability limit state (SLS) and the ultimate limit state (ULS). An element that has large, visible deformations but still is able to support the load is ugly and potentially scary, but not dangerous. Therefore, initial buckling is considered an SLS failure, but not ...


2

This is just wrong. The notion of a "critical load" in Euler buckling is a very simplified mathematical idealization of the situation. In the real world, buckling behaviour always involved nonlinear dynamics, and if you apply a big enough load, the structure will always buckle - though if you apply the load fast enough, it may buckle into a different shape ...


2

Wings are designed as a complex structure of spars and ribs, clad with aluminum, titanium, or new composite sheathing, or in the fabric, in the early planes. The Frame has been designed to support all the loads, bending moment of the wings, torsion, shear, compression, and tension. It has been designed to never let the surface buckle because it would be ...


2

First of all, buckling is an instability failure mode. I.e. the structure resists up to a critical load, and beyond that the failure is very rapid. In most textbooks, buckling is considered in single column members. (Obviously, buckling can occur in structures like the one you have.) In an example with two columns (see below), if you had two columns with ...


1

It's worth remembering that buckling is caused by the bending moment generated by axial loads due to (possibly infinitesimal) imperfections. Given that, it becomes simple to understand what the text is saying. The linear buckling analysis will show you the shape the column or structure will take upon buckling. The particular shape this gives you is, by ...


1

Because it's a failure. Once it starts, it weakens the structure and, assuming the load is consistent, the strength is reduced and it can now buckle easier in a feedback cycle. Longer spans of wood can 'buckle' and not be a strength issue. Just yesterday I answered the question of what sized piece of wood to use to span between two posts and that would not ...


1

A column can be considered "failed" if it buckles. The equation for the critical buckling load for a slender column contains the material's elastic modulus but not its yield strength. In this sense, the load-to-buckle is independent of the yield strength.


1

I think that your first link covers it. Stress or strain virus the -strength- of the materiel is not part of the buckling equation. This equation provides the upper limit, it is going to buckle at this point. The strength can also be based on the yield strength if the material is stressed to that point before reaching the Euler buckling point. This is the ...


1

A column or, a bar, even an empty can of soda may have imperfections and or a nonsymmetrical loading and carry a load and deform without buckling. Buckling happens when the load exceeds the critical loading and the deformation is sudden and large and can continue even without increasing the load anymore. A column may and usually does have defects, such as ...


1

There is no buckling under torsion in a beam, there is lateral web buckling under vertical loads under certain circumstances. A beam under pure torque will gradually twist due to St Venant and warping torsion until it yields with no sudden loss of strength as in buckling under axial load, however, under combined axial stress and torque it will buckle and ...


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