Buckling explained for beginners (buckling load, deformation mode)

In the previous lesson, we finished explaining beam deflection, which is a highlight of the middle section of mechanics of materials.

I want to read it together

Typical needle deflection that even beginners can understand (actual examples of needle use) Last time, we explained typical second moment of area. This time, we will introduce typical beam deflection. It's not an exaggeration to say that a mechanical structure, in simplified terms, is composed of multiple beams…

Up until now, we have introduced deformations such as pulling, twisting, and bending of materials. Starting from this time, we will introduce deformations such as pressing and crushing, and with this, we will have covered all the major deformations of materials.

このDeformation through pulling, twisting, bending, and crushing (compression)If you understand this, you can break down any structural deformation into its individual components, and it will basically consist of combinations of four types of deformation.

As I will explain further, even with the same material and size, the conditions differ depending on whether the load is applied under tension or compression.Materials are often stronger when subjected to compression. Smart designers often utilize components effectively under compression.

Therefore, it is a very important deformation for efficient design, so I would like you to understand it.

Please note that while deformation due to compression has a long history, it is not fully understood, and therefore, empirical formulas may be used in some cases.

What is buckling?

As usual, let's unify our understanding before explaining buckling.

First, imagine a short, round rod standing on flat ground. What would happen if you pressed down on it with extremely strong force? That's right. It would get crushed and bloated.

Now, what would happen if we pushed a fairly long round bar with extremely strong force?

If you were to press the exact center of a perfectly round cylinder, made with an accuracy of 1μ, or even 0.001μ or more, without any deviation whatsoever, it might just collapse.

But such a thing doesn't exist in this world. If you push a long round rod, it will bend at a certain point where the force is too strong. Depending on the length, it might simply bend into a V-shape, or it might bend in an S-shape.

Yes, thisThinking of a round bar as a long column, the deformation that causes it to bend is called buckling.And when it bends like this...The load is called the critical load..

Furthermore, the points where the shape of a "V" deforms bend are called nodes. The number of nodes is used to describe an S-shape or even more pronounced bend.

The form of deformation will vary depending on the thickness and length of the column and the compressive force applied to it.

Let's consider a more precise image of buckling.

Finally, to illustrate the concept, let's look at a diagram showing what happens when we apply increasing load to a rod that bends into a V-shape. The more the rod deforms into a V-shape, the further apart the load points become.Let the distance be denoted as ω0 (omega) and the load as P.

More and moreWhen a load is applied and the column is bent into a V-shape, it will start to flex beyond a certain point without further load increases (the column flexes and the bending moment increases spontaneously), and depending on the material, it will eventually reach its tensile strength and break. This phenomenon is also called buckling, and the load at this point is called the buckling load Per (Per in the graph).

To put it simply, imagine a long, thin stick standing upright, and if you press down on it, the stick will bend and eventually break.

I think everyone has the experience of bending and breaking a gardening stick or something similar while trying to stick it into hard ground during a gardening class in elementary school. And you probably remember feeling a sudden loss of strength when it bent. (I'm sure everyone has done that as a prank.)

As explained earlier, the deformation due to the deflection is greater than the applied load, causing the rod to suddenly break, resulting in a sudden loss of tension.

This is a typical example of buckling, where the load that causes a gardening pole to break is the buckling load.

I think you now have a grasp of what buckling is like. Next, we'll set up an equation that probably everyone dislikes.

Elastic buckling and Euler's formula

Now, let's set up an example problem as usual to formulate the buckling equation.

First, as a rough example, consider a long column that is deformed by a load P. When considering buckling, the origin of the coordinate system is placed at the left end of the column, not at the center (to account for the eccentricity of the load point P). The coordinate system is shown in yellow in the diagram. Therefore, the load P is at a slightly eccentric point.

To further elaborate on the example, consider a column of length l rising from flat ground, with a load P applied to its tip, causing it to bend at a single point. The coordinate system is defined with the base of the column as the origin, with +x in the vertical direction and +y to the right in the diagram. The load point is the amount of column deflection δ (delta, lowercase Δ), and the eccentricity is denoted as e.

From here on, it's the same as with the cantilever beam deflection we've done so far, and although the position of load P is a bit tricky, we'll solve it using the differential equation for deflection.

As usual, if we ignore the shear force and calculate the bending moment for a sufficiently long beam, it will be -Pe (because it is against the direction of deflection) of the load P and the eccentricity e, but since it has already deflected by δ (delta), strictly speaking the bending moment M at any coordinate (x,y) will be as shown in the following figure.

The key point here is that, unlike with beams, we consider the bending moment by taking into account the deformation amount δ.

Bending moment M = P(δ + ey)

All that remains is to substitute this into the partial equation for deflection. Substituting it gives us the following equation.

$ \frac{d^2y}{dx^2}=\frac{P}{EI}(δ+ey) $

If we set $ \frac{P}{EI}=α^2$ which is a constant, then we get the following.

$ \frac{d^2y}{dx^2}+α^2y=α^2(δ＋e) $

We need to solve this differential equation, but simply integrating as we have done before won't work because we have a variable y.

Here, we'll use mathematical techniques to solve it. While the aim of this site is to explain things without using mathematical techniques, solving differential equations, like factorization, differentiation, and integration, requires special techniques, so it's unavoidable. We'll explain differential equations a little later on this page, so for now, let's move on.

Now let's solve it. The general solution to the differential equation above is expressed by the following equation, where A and B are constants.

$ y=Asin(αx)+Bcos(αx)+δ+e $

$ \frac{dy}{dx}=Aαcos(αx)-Bαsin(αx) $

It can be calculated using this method.

Here, the boundary condition is that the lower end is fixed.

$ At x=0, y=0, and the angle of deflection θ\frac{dy}{dx}=0 $

Therefore, A=0 and B=-(δ+e), and the differential equation becomes:

$ y=(δ+e)(1-cos(αx)) ・・・① $

On the other hand, the upper end of the long column is

$x=l and y=δ$

And so the respective constants are,

$ A=0, B=-e\frac{1}{cos(αl)}, δ=e(\frac{1}{cos(αl)}-1) $

Substituting these into the solution to the differential equation,

$ y=\frac{e(1-cos(αx))}{cos(αl)} $

Becomes

This will tell us how much the column is deflecting.

Next, let's consider the meaning of this equation.

The key point is the denominator of the equation; when cos(αl) becomes 0, the deflection becomes infinite. If the deflection is infinite, the column will completely collapse.

So, cos(αl) becomes 0 when, looking at the positive side of the graph below, θ is $ \frac{π}{2}, \frac{3π}{2}...$.

When cos(α1) is 0, the numerator is odd, so if we let the odd number be m+1,

$ αl=(2m+1)\frac{π}{2}, (m=0,1,2,3,….) $

Under these conditions, the deflection becomes infinite and the column breaks.

Next, let's consider the case where the eccentricity e is 0.

If we consider that the column will bend (not collapse and become thicker) even when the eccentricity e is 0, then the intermediate equation for the deflection mentioned earlier, ①$ y=(δ+e)(1-cos(αx))$, holds true, and if we set this eccentricity e to 0,

$ y=δ(1-cos(αx)) $

Therefore, the deflection at the top of the long column (x = l) is δ.

$ δcos(αl)=0 $

Since δ is the deflection at the top, it cannot be zero, so cos(αl) becomes 0.

So, in the end, even if the eccentricity e is 0, cos(αl) is destined to become 0, and the deflection at that time will of course become infinite according to the formula we derived earlier, leading to damage.

Now let's find out what kind of load occurs when the condition αl=0 causes the column to break.

When α is returned to its original state

From $ \frac{P}{EI}=α^2 $ and $ αl=(2m+1)\frac{π}{2}, (m=0,1,2,3,….) $

$ P=(2m+1)^2\frac{π^2EI}{4l^2}, (m=0,1,2,3,……) $

$ P1=\frac{π^2EI}{4l^2}, P2=\frac{9π^2EI}{4l^2},…..$

Becomes

In other words, the load calculated using the above formula results in infinite deflection of the column, making it a load that causes failure.

The smallest solution to this equation, $P1=\frac{π^2EI}{4l^2}=Per$, is called the buckling load Per, where one end is fixed and the other end is free.

Looking at the formula, you should be able to understand the characteristics of the buckling load.As the length increases, the load decreases by the square of the length, meaning it buckles easily. Higher elastic modulus and second moment of area result in a higher buckling load, meaning it becomes stronger.

Conversely, it has the interesting characteristic of failing regardless of the eccentricity and the strength of the material itself (its strength is determined geometrically).

If you make good use of thisBy using short columns with a large second moment of area, a strong structure can be created regardless of the type of material. This technique, when using only weak materials, allows for the design of incredibly strong structures.This is where the design skills come into play.

You might be wondering why there are multiple loads, m=0, 1, 2, 3... To understand what this means, please look at the following diagram.

The shape of the deformation changes when it is destroyed. This is called the buckling deformation mode, and m = 0, 1, and 2 are called the 0th, 1st, and 2nd order modes, respectively.

Generally, failures often occur at m=0, so that's usually sufficient. However, due to the cross-section and length of the long column, it can surprisingly withstand m=0 and fail at m=1 or m=2.

In my case, I often look for similar destructive tests from the past, examine the results, and then determine the value of m.

Furthermore, when buckling calculations need to be performed on parts that are used regularly, most reputable companies should have data on which order of mode to use for the calculation.

If not already done, it is advisable to conduct a buckling test to investigate the buckling modes. If it were me, I would start with an experiment.

Because if you calculate everything using zero-order deformation mode loads, you'll end up designing with weak loads, resulting in a huge and heavy design. Such a design wouldn't be competitive in the market and would be costly, so it's not a good idea.

Differential equations for a two-circuit system of deflection

Although unrelated to buckling in mechanics of materials, I'll briefly explain the general method for solving $ \frac{d^2y}{dx^2}+α^2y=α^2(δ＋e)$ since we're on the subject.

Since $ \frac{d^2y}{dx^2}+α^2y=α^2(δ＋e)$ has both a general solution and a particular solution, let's find them (a particular solution exists because there are constants other than y'', y', and y).

First, let's set aside $α^2(δ+e)$ on the right side and solve $ \frac{d^2y}{dx^2}+α^2y=0$ to find the general solution.

To save time, let's rewrite it as $ y'' + α^2y = 0 $.

I might get scolded by math teachers or experts for saying this, but when dealing with differential equations that contain y, y', and y'', you can often solve them successfully by making a temporary substitution for y as follows (because y'=y basically results in an exponential function like $y=e^(x+c)$).

$ y=e^λx, y'=λe^λx, y''=λ^2e^λx $

Substituting this into $ y'' + α^2y = 0 $

$ λ^2e^λx+α^2e^λx=0, (λ is appropriate) $

Solving the equation, the exponential function is not zero, so we get $λ^2+α^2=0$. Now λ has two solutions, $λ=±αi(i: imaginary number)$, so the solutions to $y''+α^2y=0$ are as follows.

$ y = C1e^{αx} + C2e^{-αx}, (C1 and C2 are integration constants)$

Using Euler's theorem $e^ix=cosx+isinx, e^-ix=cosx-isinx$, we can express it as follows:

$ y=(C1+C2)cosαx+(C1-C2)isinαx $

Here's a less-than-ideal method, but since the solution will never be an imaginary number, we can manipulate C1 and C2 to eliminate the imaginary number i. We make C1 and C2 conjugates.

$ C1 = A + Bi, C2 = A - Bi $

So y is

$ y = 2Acosαx - 2Bsinαx $

Next, we will solve the particular solution.

The particular solution follows the form of $α^2(δ+e)$ and has no variables, so we assume it is a constant y=C.

Substituting this into the original differential equation,

$ \frac{d^2y}{dx^2}+α^2y=α^2(δ+e) , α^2C=α^2(δ+e), C=(δ+e) $

With this, the differential equation can be solved from the general solution and particular solution as follows.

$ y=2Acosαx+2Bsinαx+(δ+e) $

It can be solved like this.

We've encountered many important keywords such as linear differential equations, general solutions, particular solutions, imaginary numbers, and Euler's theorem, but since this is about mechanics of materials, we'll skip over them for now. These are all important concepts, so please wait for the explanations on a separate page in the field of industrial mathematics.

Incidentally, this form of differential equation, $ \frac{d^2y}{dx^2}+Ay=B$, is important because it is frequently used in industry.

Summary

This concludes our explanation of the basics of buckling.

Solving the intermediate differential equations requires some slightly unusual techniques, but for now, just skim over them and focus on the results.

Summary

Next time, we will determine the buckling load for cases where both ends are fixed and cases where both ends are free. Ideally, we would also like to introduce the experimental formulas.

I want to read it together

Buckling stress explained for beginners: Euler's formula and empirical formula Now, I think you understand what buckling is in the previous explanation. This time, I will explain buckling stress based on the buckling load from last time. Furthermore, buckling affects the entire length of the column…

To those who found this article helpful in understanding design:

While there is basically no textbook covering this content, and it is my own original work, I will introduce the textbook that I have been using since I was a student.

Basic Mechanics of Materials (Revised Edition)

created by Rinker

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