Tag: column buckling

Engineering analysis has its roots deep in mathematics and physics, it is an applied form of it after all. Any student of engineering knows well is that boundary conditions are very important to defining your problem. One can say that the problem isn’t made to represent anything physical until these are specified, a bit of an exaggeration but fairly fitting when analyzing any piece of real structure.

This is well described in an old topic in structural analysis: column buckling.

Here I want to present column buckling in a relatively new way. I have seen aspects of this approach here and there but not to the extent I will present here. As a brief overview I’ll give away the ending now, all simple column buckling modes can be extracted as zeroes of a generalized buckling curve that encompass all possible end conditions. Now the curve we will derive is physically meaningless as derived but only takes its physically meaningful forms when picking certain values of the constants of the displacement curve.

To give an idea where we are heading towards, Figure 1 is a graph of the three separate buckling curves we will be looking at and the zeroes of these curves, denoted by the red points. These zeroes correspond to a buckling modes of the column, one for each curve’s zero. All possible basic buckling modes are contained within these curves as the zeroes.

It’s fascinating, to me at least, and hopefully to the reader, that we can take a curve and ascribe meaning to it in certain conditions which is used in engineering analysis all over the world. But enough pontificating for now and lets get down to our journey into buckling with plenty of stops along the way.

What is buckling and how does it look like?

The best way to describe buckling is that it is a failure of the stiffness of a structure, a column in this case. Usually a column or any structure resists deformation and displacements, like trying to bend a stick. You bend and it exerts an equal and opposite bend, or moment as its termed in mechanics, to your hands. The same can be said for any compressive force/load it is put under, the stick/column pushes right back.

Buckling occurs when the column/structure does not resist the deformations applied to them anymore because its internal strain energy has reached a critical point and cannot increase anymore. So instead of resisting the deformation it just accepts it and buckles.

Figure 2 shows a pinned-pinned column, note that a pinned condition is also referred to as a simple-support in this post because we are primarily concerned with lateral movement only.

Let’s define the types of basic supports you can have at each end of the beams. We are going to look at 4 different types and they are listed as follows.

1. FREE – Fairly self explanatory, the end is free to translate laterally and rotate
2. SIMPLY SUPPORTED/PINNED – The end cannot translate but is free to rotate
3. GUIDED – The end cannot rotate but is free to translate laterally
4. FIXED/CLAMPED – The end cannot translate nor rotate

Here we are treating a simple support and pin as the same since we are only concerned with the lateral motion of the column. This link has a good explanation of the different types of supports the end of beams/columns can have.

Assumptions/Limitations

As any practical engineering analysis problem, we need to place limitations based on the assumptions we will make to generate our mathematical model. So lets list these as well.

1. The column is made of an isotropic material. This implies that the material reacts the same in any direction it is deformed. Most metals fall into this category.
2. The column’s cross-section is the same throughout its length. Like a 2″ by 4″ piece of wood.
3. The compressive force/load is applied directly at the center of the cross-section. This eliminates any off-center moments/rotations caused by a force applied off-center.
4. The displacements are small compared to the length of the column. This assumes a linear behavior in the forces and resulting displacements. This is valid for metals since they will not deform much before they fail.
5. Euler-Bernoulli Beam Theory is applicable, this is the classic way of analyzing beams/columns.

Each of these assumptions can merit itself a stand-alone article, which will happen eventually but for now lets take them as is. So lets take advantage of the great minds before us and use their results.

How does buckling look like?

Lets now take a look at the 10-inch column we are starting out with, a fixed-free column, basically one sticking out of the ground where one end it is restricted from translating and rotating but the other end is free to move around. This is shown in Figure 3 and a compressive force is applied like in Figure 2.

After the applied force reaches a critical value, the column will reach its instability point, where it is neutrally stable, and at this point, the column takes the shape as shown in Figure 4. Note that the units on the lateral deflection, y, has normalized values where at the max value it is equal to one. When the column reaches this point, it is considered to be buckled and any further perturbation will lead to unstable displacements until it collapses.

The Mathematical Model

The mathematical model used to calculate the shape of the elastic curve (column) is Euler-Bernoulli Beam Theory. Like previously mentioned, I won’t derive these equations here, at least not in this post, but just present the results. The internal forces which develop to resist deformation and the applied load can be presented in the free-body diagram shown in Figure 5.

A free-body diagram represents the internal reactions to an externally applied force or displacement. In the figure, the force P is reacted by an equal and opposite force at any point along the beam. However, the rotation resulting from this force is reacted by a moment which varies along the length depending on much the column is bent/curved.

The expression for the internal force and moment can be summarized as follows.

Now lets define some of these terms before we go on, time for another list.

1. EI – Flexural/Bending Rigidity: A measure of how much is required to bend a structure by one unit of curvature. The higher the value, the harder it is to bend.
2. EA – Extensional/Axial Rigidity: A measure of how much is required to stretch/compress a structure by one unit of distance. Again the higher the value, the harder it is to stretch.
3. M – Bending Moment: the bending “force” that is resisting the deformation of the column trying to return it to its initial shape. This can also be thought of as a rotational spring.
4. P – Axial force: The force that is resistance the vertical deformation of the column trying to stretch it out to its original length. Think of a spring, the more you compress it, the more it pushes back.
5. κ(x) – The curvature of the column: This measures how “bent” a structure is. The higher value the more bent it is, e.g. a bobby/hair pin has high curvature where it is bent and straightening the bobby pin out reduces the curvature up to a value of zero when the bobby pin is straight
6. ϵ(x) – Total strain along the length of the column (x-direction): Includes contributions from both the change of vertical displacement, u'(x), and the rotation of the cross-section ,y'(x).
7. u(x) – centroid vertical displacement: This is how much the column moved down from its initial position.
8. u'(x) – The extensional strain: Rate of change of the vertical displacement
9. y(x) – The displacement along the y-direction: This is how much the column moves laterally
10. y'(x) – The rotation of the column: This measures the angle which an initially horizontal cross-section rotates, think of what angle the face of the column makes with the ground.
11. y”(x) – alternative way to write κ(x): Shows its relation to the displacement y(x).

The governing differential equation of a column under a compressive force like this has the form of a homogenous fourth-order differential equation with constant coefficients.

Which has a fairly textbook solution and has an elastic curve of the following form.

Where we can define the wave-number, k

Since we have an isotropic and constant cross-section column, the flexural rigidity, EI, is constant so this wave-number is dependent only on the magnitude of the applied compressive force, P.

Now from here we still need to determine the unknown constants a, b, c, and d. This is usually done by using the boundary conditions to calculate the prescribed values of the displacement, rotation, bending moment, and shear of the beam at these ends. Once these constants are determined, the final shape of the column is determined. This can be tedious and requires a decent amount of calculations for all six configurations we will be looking at.

Why do these ends matter?

We already defined the four different types of supports we will be looking at for the ends of the column but now its time to look at the different configurations of these supports. These are nicely tabulated in Figure 6. which comes from a textbook on Advanced Mechanics of Materials by Boresi. However, these are standard so they could easily be found by a quick search online.

This kind of gives away the ending, from Figure 6 we can see what the critical buckling load is and what the buckled shape will be like. However, this will serve as a nice sanity check to make sure this method works and its always nice to have some affirmation.

Also note from the critical buckling loads in Figure 6 that this critical load is always of the form:

Or to write it in terms of the wave-number, k.

We are on the start of our journey to generating the curves we saw in the beginning and the next couple of stops are into the world of stability analysis and calculus of variations.

The fancy fundamental principle

Earlier I described the expressions for the internal force and moment along the length of the column, for this next step we will need to know the strain energy these produce in the column. Again, similar to a spring, when you compress a spring, you are transferring the work you are performing on the spring into strain energy stored within the spring. This means its potential energy increases which it will release as soon as you let go and it… well… springs back.

The internal strain energy of the column is given by the following integral over its entire length, L.

The energy definition we will be using is the product of the force and its displacement. As in, the applied compressive force does work on the column, which is the product of the force and the vertical displacement.

These two terms make up the total potential energy of the column, which is expressed by an uppercase Pi.

The total potential energy encodes all the information we need to derive the governing differential equation for equilibrium and we can also use it to check the stability. This requires the use of one of the most important concepts in engineering and physics: The Principle of Minimum Potential Energy. This concept can be defined in the following way:

“Any admissible displacements which satisfy the equilibrium equations require that the first variation of the total potential energy to be zero”

So what does this mean exactly? Well, it says any solution to the governing differential equation of the elastic curve of the column also has the added requirement that it sets this “first variation” to zero. Lets take a closer look at this variation business.

Variations aren’t real but don’t tell them

A variation can be described as a small “virtual” perturbation to any variable of a function, the main use of these is to see how a function changes to small changes to its independent variables. Again this is not going to serve as a primer on the Calculus of Variations, which is the subject these variations live in. Instead I’m going to summarize three important points about them. you knew another list was coming, right?

• Variations act and operate like derivatives. Which makes sense to an extent since they both deal with changes of a function.

• Variations and derivatives commute. For independent variables, you can take its variation before or after taking its derivative and get the same result. This is analogous to how the order of the mixed partial derivatives of a smooth function does not matter.

• The integral of a variation is also equal to the variation of the integral.

With these tools under our belt we can calculate variations of the total potential energy. One important thing to note however is that we are looking at the variation within a bound, in this case over the length of the beam. Due to this, we require there be no variation at each end since we can determine what values they have due to the boundary conditions. We can illustrate this with a simple but classic analogy.

What is a straight line?

Think of two points in space and we want to trace out the shortest possible path between point one and point two. It is fairly obvious that the answer is a straight line but say we didn’t know that and we traced out a few different candidates as shown in Figure 7.

Note that the curves begin and end at the same place, this is what is meant by there being no variation at the boundary. What happens in-between is a different story, the optimal and correct path is the straight line any difference between that straight line and any of the other curves is the variation from the optimal path. Any curve can become a straight line by reducing its variation to zero and that is what it means that the equilibrium for our column occurs when the first variation is set to zero.

Okay, but is the ball stable?

We saw that equilibrium happens occurs when the first variation of the total potential energy is zero but how do we know its a stable equilibrium, well for that we will need the second variation. Similar to how we can analyze curves using its derivatives, like how a minimum/maximum is related to the first derivative and using the second derivative to determine which one it is. Here we are going to use the second variation to determine the stability of the column.

To see how this works lets look at another classic analogy, a ball resting at the bottom of a ditch. Here we can say that the height is stable because if you give the ball a slight nudge, it will roll back to the bottom.

However, if the ball is at the peak of a hill instead; a slight nudge will not return it to the top. Instead it would continue to roll down, this is an unstable system.

The last case, we have is if the ball is on flat ground. Here if you nudge the ball slightly, it will just roll to one side slightly and stop; this is called neutrally stable. The neutrally stable condition is what we are after because it is the transition point between stable and unstable.

Figure 8 illustrates the transition of a ball when its on a stable, neutral, and unstable surface.

Now we know why the second variation shows the stability of an equilibrium configuration. More specifically, when it is zero, the system is neutrally stable and any further perturbations will cause it to become unstable. So lets calculate the second variation of the total potential energy.

The first variation

Finally we can start off by taking the first variation of the total potential energy, to limit

For the full details lets substitute the expressions for the bending moment M(x), P(x), and the strain ϵ(x) from the beginning into the strain energy. While the work done by the external

Now we can distribute the variation among all the deformations, the integrand include squares of deformations so we can note that when taking the variation the 1/2 outside the integral is canceled with the square coming to the front. Just like taking the derivative of x squared.

The strain energy due to the internal axial force

Before we take the second variation we can simplify the above expression, specifically the second term in the integrand. Where the strain in the x-direction can be split into the two components described before.

This variation leads to the following:

The integral can be split into two terms.

We are going to take a step back and bring back the definition of the internal axial force, P(x), The (x) is actually a bit misleading because the actual magnitude of P is constant along the column, it only pushes back with the same strength of the applied force P. That is why I labelled them the same. The total strain is dependent on location in a way that the combination of the two results in the magnitude of P.

Since the magnitude of P is the same everywhere along the length of the column, we can treat it as a constant and bring it out of the integral.

Above, the identity for the variation of an integral was used to shift the variation after we integrate. The parenthetical term should also look familiar to first year calculus students, its the integral of a derivative. Here we can employ the Fundamental Theorem of Calculus that essentially says that the integral of the derivative of a function is the original function.

The vertical strain component is equal to the work done by applied force, which is to be expected because that work is converted into strain energy of the column. So we can put everything back together now.

The second variation

Using the result we just came up with above we can re-write the first variation in a simpler form since we can cancel the work done by the applied force.

Remember we are only considering the second variation in terms of the first variation of the independent variables. so any variations of variations of deformations are neglected. This leads to a fairly simple form of the second variation, which for the condition for stability we defined, is equal to zero.

The above represents a neutrally stable system, the point between stable and instable. Setting this equal to zero, rearranging, and representing the curvature in terms of the displacement leads to the equation we are looking for after taking the constant terms out of the integrals.

This can be rearranged again in the following way to represent a familiar term, the wave-number k.

Here is the condition for stability for any column, regardless of boundary conditions or end support types. This encodes the entire family of buckling curves into a single expression. A fraction made up of the variation of the first and second derivatives of the displacement curve. So now the next question… How do we calculate it?

Solution to the general buckling problem

What do we mean by taking the variation of the displacement and its derivatives? We can actually represent the displacement like a wave, deformations do behave like waves after all to some extent. This representation includes the product of an amplitude, displacement magnitude, and a function for the shape for the curve.

The variation of this then only affect the amplitude not the shape of the displacement. This is because we are trying to find out if the current configuration is in equilibrium and/or stable, so the amplitude is varied to see how the overall system changes. The important distinction here about why this doesn’t include a derivative of the shape function is that the displacement is the independent variable while the potential function depends on this deformation. So we have to take its derivative with respect to the deformation not with respect to x. Also note that the amplitude is a constant.

Where f(x) is the displacement curve normalized by the max displacement which does not affect our solution since we don’t care about what the displacement is at the onset of buckling since it will quickly become unstable and collapses. We can substitute this expression for the derivatives of the displacement curves and cancel the amplitudes.

We can finally take a look at the solution for the general buckling problem by taking the expression for the shape of the column, f(x), to be equal to the displacement curve, y(x), with a unit displacement. Now since we are looking at the shape of the curve only, it will be best to express the wave-number, k, not in terms of P or EI but in terms of the length L.

This form of the displacement is what we will use to integrate the equation we developed.

I’ll spare the painful details of carrying out the integration and simplifying the results but instead show the end result.

The functions N, M, and R are a bit messy but they do have a fairly aesthetic appearance.

You might have noticed the leading factor of this expression is the square of the wave-number. Note that the term in parenthesis in R(m) has the form of a displacement with d equal to -b.

So we end up that the parenthetical term has to be equal to one, all that messiness simplifies to one. We can also rearrange it so we are looking at the zeroes of the function g(m).

Let’s reflect on how we got to this point, to get this equation which governs all buckling solutions we had to go through various physics principles, solving differential equations, regular calculus, and calculus of variations. All these branches of mathematics and physics lead to this point where we can represent reality and practical engineering equations.

Well… almost, this general buckling curve is dependent on the choice of the constants a, b, c, and d. Until then, this is a very interesting but also very abstract equation. Lets see how these constants relate to the boundary conditions of the column.

Making the math real

One thing you may have noticed is that the constant d does not make an appearance in g(m), this is because its an initial displacement which vanishes when taking derivatives of y(x). However, it is important in determining the shape of the buckled column, so lets determine what it should be.

We do this by noticing that all of our supports in Figure 6 require at least one end to be fixed in translation, meaning no lateral displacement in y. If we take this to be the end when x = 0, the displacement curve can be solved for d.

Since the sine function is zero and the cosine is one at zero, this leads to the following expression.

We can employ this in our definition of y(x) and g(m) but also note that this came out naturally earlier in the function R(m).

Now while we can determine the other constants in a similar way, that is not how I want to do it here. Instead we are going to pick values for the remaining constants and see what comes out. Though what comes out will be the buckling shapes for all the supports in Figure 6. This is how we will make g(m) represent real column configurations.

Pinned supports are simple

This first case we will set b = c = 0 and see what form g(m) takes, so after plugging in and simplifying we get the following.

The plot of this buckling curve is shown in Figure 9.

This has an infinite amount of zeroes but we are only interested in the first two for this curve, subsequent zeroes correspond to higher order buckling modes which require more energy to initiate so they aren’t of interest to us here. We are only interested in what is the smallest force P which will initialize buckling. Lets take a look at these zeroes and what they tell us.

m = 1/2

The wave-number associated with this is the following:

The shape of the displacement curve associated with this is the following:

The buckled shape of the column is shown in Figure 10. Notice how at the bottom of the column, has rotated by a small angle while at the top, it has only moved laterally with no rotation.

The associated critical buckling force to initiate this buckling mode is the same as shown in Figure 6.

To give an idea of the magnitude of numbers we are working with here, lets use some real world values for these. Let’s assume we have a titanium 1/2-in square column that is 10 inches long.

The typical value for the Young’s Modulus for Ti-6Al-4V titanium from the Metallic Materials Properties Development and Standardization text (MMPDS) is about 16 million pounds per square inch,

The second moment of the are for a square of side length a, is given by the standard formula.

Substituting all these values gives the critical buckling load to be over 2000 pounds.

m = 1

The wave-number associated with this is the following:

The shape of the displacement curve associated with this is the following:

The buckled shape of the column is shown in Figure 11. Now both ends are fixed but have rotated by some amount.

The associated critical buckling force to initiate this buckling mode is the same as shown in Figure 6.

Substituting all the same values as before gives the critical buckling load of over 8000 pounds.

The Pinned-Pinned configuration is also known as the Fundamental Case of Buckling for a Prismatic Beam/Column. This due for two reasons, it has been historically been used the most to estimate a buckling load for a beam/column and because there are no additional factors in the calculation of the critical load. It is the “cleanest” looking one out of the bunch.

These are all the buckled shapes we can extract from this curve, so lets move on to the curve which houses all the fixed supports.

Fixed supports are not that simple

For this next case we will set a = c = 0 and see what form g(m) takes, so after plugging in and simplifying we get the following.

This is very similar to the first case aside from one term switching signs. The plot of this buckling curve is shown in Figure 12.

This also has an infinite amount of zeroes but now we are only interested in three zeroes for this curve, all other zeroes again correspond to higher order buckling modes which require more energy to initiate so they aren’t of interest to us here. We are only interested in what is the smallest force P which will initialize buckling. Lets take a look at these zeroes and what they tell us.

m = 1/2

The wave-number associated with this is the same as the first case.

The shape of the displacement curve associated with this however is different.

The buckled shape of the column is shown in Figure 13. Here one end is fully restricted in motion and the other is free to displace and rotate.

The associated critical buckling force to initiate this buckling mode is the same as shown in Figure 6.

Substituting all the same values as before gives the same for a pinned-guided column.

m = 1

The wave-number associated with this is the following:

The shape of the displacement curve associated with this is the following:

The buckled shape of the column is shown in Figure 14. Now both ends do not rotate and only one end displaces.

The associated critical buckling force to initiate this buckling mode is the same as the fundamental case.

Substituting all the same values as before gives the critical buckling load of over 8000 pounds.

m = 2

The wave-number associated with this is the following:

The shape of the displacement curve associated with this is the following:

The buckled shape of the column is shown in Figure 15. Now both ends remain fixed in place.

The associated critical buckling force to initiate this buckling mode is the same as the fundamental case.

Substituting all the same values as before gives the critical buckling load of approximately 32,900 pounds.

These are all the buckled shapes we can extract from this curve, so lets move on to the black sheep of the curve family.

There is always a black sheep in the family

For this next case we will set b and c to non-zero values, specifically to the following:

For b, we approximate the tangent function by its argument which is valid as long as the rotations are small, which is an assumption we made at the beginning of our journey. For d, we set it to zero this time, setting it to -b as usual just has the effect of shifting the entire column by b. We aren’t interested in rigid translations of the columns so we can justify setting it to zero. Plugging these values and evaluating leads to a not so nice looking result.

Approximations can lead to messy results but we aren’t after exact values in this case. The plot of this buckling curve is shown in Figure 16.

This time we don’t get a nice values for the zeroes but we are only interested in the first one which we can find to be almost 1.5.

The wave-number associated with this is then

The shape of the displacement curve associated with this much different than the others.

The buckled shape of the column is shown in Figure 17. Here one end is fully restricted in motion and the other is free to rotate but not translate.

The associated critical buckling force to initiate this buckling mode is the same as the fundamental case.

Substituting all the same values as before gives the critical buckling load of approximately 16,826 pounds.

This covers all basic buckling shapes. Time to wrap up this post

Concluding thoughts

Just as we set out to do, we illustrated how all the basic buckling modes can be derived from a single buckling curve with fine-tuned constants. In Figure 1, these curves were represented with their corresponding zeroes which led to each buckling mode. However, getting to that point was not trivial but it demonstrates how much engineering and mathematics overlap and how the math can illuminate some patterns that the typical engineering methods obfuscate.

Hopefully you got a deeper understanding of how buckling is analyzed and what the methods are to develop the equations seen in standard texts. Until the next post, be sure to avoid your critical buckling load.