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Column buckling analysis

July 3, 2024 at 2:04 PM by Dr. Drang

Last time, we looked at the solutions for six elementary column buckling problems. In this post, we’ll outline the method for getting those solutions and give a couple of detailed examples.

In all of these problems, the column is assumed to be straight and prismatic, which means that the cross-section doesn’t change along its length. The minimum moment of inertia of the cross-section is $I$, and the column buckles in the direction associated with $I$.¹ The column has a compressive load $P$ acting at either end, and the load is aligned with the centerline of the column. The material is linearly elastic with a Young’s modulus of $E$, and there is no yielding of the material at any point.

With these assumptions the governing differential equation of the column is

$$EI\frac{d^{4}y}{d{x}^4}+P\frac{d^{2}y}{d{x}^2}=0$$

where $x$ is the coordinate along the centerline of the column and $y$ is the (small) lateral deflection of the column when it buckles. This is a fourth-order linear differential equation with constant coefficients, which is great for us because it’s one of the few types of differential equation that can be solved analytically.

Here’s the general form of the solution,

$$y=A\sin \lambda x+B\cos \lambda x+Cx+D$$

where $\lambda$ is a constant that combines the coefficients,

$$\lambda =\sqrt{\frac{P}{EI}}$$

and $A$, $B$, $C$, and $D$ are the four constants of integration that appear when solving a fourth-order differential equation. We’ll use the boundary conditions (how the column is supported at either end) to solve for these.

Simple-simple

The column with simple supports at either end is the fundamental problem of column buckling, and it’s pretty easy to work out its solution.

(I’m showing the column horizontally to take up less room on the screen. Because we’re not considering the weight of the column itself, the orientation doesn’t matter.)

We’ll start with the conditions at $x=0$. Here, the column is prevented from moving laterally, so

$$y(0)=B+D=0$$

or

$$D=-B$$

The column is free to rotate at this end, which means there’s no bending moment. Since the bending moment is proportional to the second derivative of the deflection,

$$y^{\prime \prime }(0)=-{\lambda }^{2}B=0$$

where I’m using the prime notation for derivatives. This means $B=0$ and therefore $D=0$, which simplifies things nicely.

Now we look at the other end, $x=L$, where again the lateral defection is zero,

$$y(L)=A\sin \lambda L+CL=0$$

and the bending moment is zero,

$$y^{\prime \prime }(L)=-A{\lambda }^2\sin \lambda L=0$$

Since $\lambda$ isn’t zero, the second equation means

$$A\sin \lambda L=0$$

and combining that with the first equation tells us that $C=0$.

So we’ve eliminated all the constants of integration except $A$. The equation above means either

$$A=0$$

or

$$\sin \lambda L=0$$

The first of these is the solution when the column is straight and $y=0$ throughout its length. It is a solution, but it’s trivial—we know the column can be straight if the load is small.² We’re interested in the nontrivial solution for which the column isn’t straight. That’s the buckling solution.

So when does $\sin \lambda L=0$? When $\lambda L=n\pi$, for any integer $n$. The buckling load will be when $P$ is the smallest positive value that satisfies this equation. In other words,

$$(\lambda L{)}_{\mathrm{cr}}=\sqrt{\frac{P_{\mathrm{cr}}}{EI}}L=\pi$$

or

$$P_{\mathrm{cr}}=\frac{{\pi }^{2}EI}{L^2}$$

which is the answer we saw in the last post for the simple-simple column.

What about $A$? Well, $A$ is undetermined. Remember at the top of the post where I snuck the parenthetical “(small)” into the description of $y$ and didn’t explain it? Here’s where that comes in. Buckling deflections really aren’t small, and to solve for them properly we’d have to start with a nonlinear differential equation. But that doesn’t mean what we’ve done is useless. The value of $P_{\mathrm{cr}}$ is correct, and what we’ve done tells us that the deflected shape of the buckled column,

$$y=A\sin \lambda x=A\sin \frac{\pi x}{L}$$

is one-half of a sine wave. This is useful because it helps us visualize the buckling.

Fixed-simple

This is the problem that inspired the previous post. It won’t take as long because we can use some of the simple-simple results to speed things up.

Again, we start with the general solution,

$$y=A\sin \lambda x+B\cos \lambda x+Cx+D$$

The boundary conditions at $x=0$ are the same as before,³ so we know that

$$B=D=0$$

The boundary conditions at $x=L$ are partly the same and partly different. The deflection is zero,

$$y(L)=A\sin \lambda L+CL=0$$

and the slope is zero,

$$y^{\prime }(L)=A\lambda \cos \lambda L+C=0$$

The slope condition means

$$C=-A\lambda \cos \lambda L$$

Substituting this into the deflection boundary condition gives

$$A\sin \lambda L-A\lambda L\cos \lambda L=0$$

or

$$A(\sin \lambda L-\lambda L\cos \lambda L)=0$$

Once again, there’s the trivial solution, $A=0$, and the nontrivial solution,

$$\sin \lambda L-\lambda L\cos \lambda L=0$$

Divide this through by $\cos \lambda L$ and move one term to the other side of the equation to get

$$\tan \lambda L=\lambda L$$

As we saw in the previous post, the smallest positive solution is

$$\lambda L=4.4934\dots$$

and therefore

$$P_{\mathrm{cr}}=\frac{{4.4934}^{2}EI}{L^2}=\frac{20.19EI}{L^2}$$

Once again, we can’t solve for $A$, but we can see that the buckled shape is

$$y=A(\sin \frac{4.49x}{L}-\frac{4.49x}{L}\cos \frac{4.49x}{L})$$

which isn’t any sort of standard trig function, but it is what I plotted for the fourth column in this image:

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All the other column problems in the image above can be solved the same way. The boundary conditions are different, but the solution process is the same. Notice that all the columns other than the fixed-simple buckle into shapes that are portions of sine waves. As we showed above, the simple-simple column is half of a sine wave. As for the others:

• The fixed-free column is a quarter of a sine wave.
• The fixed-fixed column is a full sine wave.
• The fixed-guided column is half of a sine wave.
• The simple-guided column is a quarter of a sine wave.

The effective length is based on the number of half sine waves the buckled shape has.

This simple visualization of the buckled shapes allows you to work out the buckling loads for all four of these columns based on the simple-simple column. The only other thing you have to remember is 4.49.

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Tangents and columns

July 1, 2024 at 2:35 PM by Dr. Drang

Continuing with John D. Cook posts, I should write up a little thing that I meant to post about a month ago. It was in response to this post of Cook’s in which he considers solutions to the equation

$$\tan x=x$$

Cook talks about the solution in terms of the sinc function and Bessel functions, but my immediate response upon seeing that equation was a single number: 4.49. I first saw this equation over 40 years ago and the solution—or at least one of the solutions—has been stuck in my head ever since.

As Cook shows in his post, the best way to think about solutions to this equation is to consider two separate equations,

$$y=\tan x$$ $$y=x$$

and plot them both on the same graph. Wherever we see intersections will be solutions to $\tan x=x$.

As you can see, there are an infinite number of solutions. The obvious one, which you probably didn’t need the graph to figure out, is $x=0$. After that are solutions that are relatively close to the asymptotes of the tangent function at the odd multiples of $\pi /2$ (excluding $\±\pi /2$).

While the graph helps you figure out where the solutions are, you’ll still have to use some numerical technique to get the value. You might think that direct iteration,

$$x_n=\tan x_{n-1}$$

would work, but if you try it, you’ll soon find yourself diverging instead of converging. The reasons are given in this blog post from a few years ago. A direct iteration formula that will converge is

$$x_n={\tan }^{-1}{x}_{n-1}+\pi$$

where the added $\pi$ term accounts for the fact that the inverse tangent function returns values between $-\pi /2$ and $\pi /2$. If you start with a number just under $3\pi /2$, this formula will quickly converge to the solution that’s circled in the graph.

The circled intersection is the lowest postive solution. That’s 4.4934094579…, or just 4.49 for short, which is the solution that matters in one of the standard problems of column buckling and is the reason I’ve remembered it all these years.

Column buckling is governed primarily by the restraints, if any, at the column ends. Four types of ends are considered standard:

1. A simply-supported end is prevented from moving laterally but has no restrictions on rotation.
2. A fixed end is prevented from both lateral movement and rotation.
3. A free end has no restrictions on either lateral movement or rotation.
4. A guided end is prevented from rotating but has no restrictions on lateral movement.

The six standard column types come from combinations of these four end conditions. Here’s a graphical representation of the columns and the shapes they take on when they buckle. Sorry about the size of the formulas—you can zoom in to get a better view.

From left to right, they are

1. Simple-simple
2. Fixed-free
3. Fixed-fixed
4. Fixed-simple
5. Fixed-guided
6. Simple-guided

Under each column is its elastic buckling load, denoted as $P_{\mathrm{cr}}$, where $E$ is the elastic modulus of the material, $I$ is the moment of inertia of the cross-section, and $L$ is the length. The simple-simple column is considered the most fundamental, probably because it has the least complicated and easiest to remember formula. The other column types are related to the simple-simple column through the concept of effective length, $L_{\mathrm{eff}}$. If you take the effective length of any of the other columns and plug it into

$$\frac{{\pi }^{2}EI}{L_{\mathrm{eff}}^2}$$

you’ll get the buckling load for that type of column.

So where does 4.49 come into this? Well, in solving for the buckling load of the fixed-simple column, you run into the equation

$$\tan \lambda L=\lambda L$$

where

$$\lambda =\sqrt{\frac{P}{EI}}$$

The buckling load is the lowest non-zero value of $P$ that satisfies this equation.

$$\lambda L=\sqrt{\frac{P_{\mathrm{cr}}}{EI}}L=4.4934\dots$$ $$\frac{P_{\mathrm{cr}}{L}^2}{EI}={4.4934}^2$$ $$P_{\mathrm{cr}}=\frac{20.19EI}{L^2}$$

And if we put this in the form

$$P_{\mathrm{cr}}=\frac{{\pi }^{2}EI}{L_{\mathrm{eff}}^2}$$

we get $L_{\mathrm{eff}}=0.699L$, which we can round to $L_{\mathrm{eff}}=0.7L$ with little error.

Variations on the image above are found in many structural engineering books. For example, here’s the table of basic columns from Johnson, Lin, and Galambos’s Basic Steel Design, the textbook used in my first steel design class.

The authors have ordered the columns in this table according to their effective length. I put the columns in the order they’re typically taught.

I took my first steel design class my senior year, so that’s when I learned that the effective length of a fixed-simple column is $0.7L$. But I didn’t learn how to derive that solution until I was in grad school the next year. It’s pretty simple, which should give you a hint about the next blog post.

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Plenty of polygons

June 28, 2024 at 1:48 PM by Dr. Drang

John D. Cook wrote a fun geometry post yesterday, and I wanted to use Mathematica to generate my own versions of his images.

Start with a circle of unit radius and draw an equilateral triangle with sides that are tangent to the circle.

It wouldn’t take much to turn this into the Sign of the Deathly Hallows, but we have other things to do.

Draw a circle that passes through the vertices of the triangle and a square whose sides are tangent to the second circle.

Note that we drew the square with a corner at the 12:00 position. We’ll do that with all the polygons.

Continue by drawing a circle that passes through the corners of the square and a pentagon whose sides are tangent to that circle.

The radii of the circles have the following relationship to one another:

$$r_n=\frac{r_{n-1}}{\cos \frac{\pi }{n}}$$

where $r_n$ is the radius of the circle that passes through the vertices of the n-sided polygon.

If we redo this drawing without showing the circles and continue adding sides to the circumscribing polygons (up to 30), we get this,

which is pretty cute.

Normally I wouldn’t repeat as much of someone else’s post, but I wanted to replicate all of Cook’s drawings myself in Mathematica.¹ The images above were generated using this code:

The list of radii, saved in the variable r, is generated using the RecurrenceTable function in a straightforward translation of the formula above. The only thing you might be wondering about is why I added a term of zero to the front of the list. It’s because I wanted the indices of the r list to match the subscript in the formula.

The radii stored in r work perfectly as arguments to the RegularPolygon function. In Mathematica, the “radius” of a polygon is the radius of the circle that passes through the vertices, which matches our definition above. The only unusual thing needed in the calls to RegularPolygon is an angle argument of Pi/2 to orient the polygons with a vertex at the 12:00 position.

That pretty much does it for the geometry. The rest of the code consists of:

1. Tweaks to the drawing commands to set colors, line thicknesses, and opacity.
2. Export commands to create the PNG files you see above.

I’m still getting myself comfortable in Mathematica. Fun problems like this are a good way to extend myself.

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Moving reminders

June 27, 2024 at 3:12 PM by Dr. Drang

If I told you to go read Stephen Hackett’s post about a problem with Apple’s Reminders app, you might think I’ve suffered a blow to the head and am living in the past. But I’m not talking about his multipart complaint in 2019, his badge complaint in 2021, his continuing badge complaint in 2022, or his task-ordering complaint in 2023. No, I’m talking about his task-moving complaint that surfaced just yesterday.

You should go read the whole thing, because Stephen backs up his complaint with a nice image, but in a nutshell he prefers the iOS way of moving a task from one list to another to the Mac way of doing that.

On the iPhone and iPad, it’s super easy, as there’s a drop-down [that lets you choose the list to move the task to] right in the Inspector. Mac users, however, don’t have that UI, and are left to drag and drop tasks between lists.

That’s fine if you know about drag and drop and have a small number of lists. However, it can be error-prone and confusing to many users.

I’m going to start by disagreeing about drag and drop. Stephen’s a nice guy, and he’s worried that some users won’t know about dragging a task to a different list over in the sidebar. I’m not a nice guy, and I have no sympathy for those users, assuming they even exist. Drag and drop has been part of the Mac since there was a Mac. It is a fundamental—if not the fundamental—action in a GUI. If you don’t know how to drag and drop, you have more problems than being unable to reorganize your reminders.

However, Stephen is absolutely right when he says that drag and drop is fine only when you have a small number of Reminders lists. As your number of lists grows, some of them will inevitably end up below the visible portion of the sidebar and you’ll find yourself starting a drag and drop only to learn mid-drag that the place you want to drop is hidden. You’ll have to cancel the action, scroll down to make the list you want to move to visible, and the start the drag and drop over again.

For years, this was the problem users of Apple Mail had with organizing messages in folders (they’re also called mailboxes, but I’m going to call them folders here). I know this problem particularly well because I had it. My consulting work was divided into hundreds of projects, dozens of which were active at any given time. I had to keep my email messages segregated by project, which meant creating a mail folder for each project and moving messages into the appropriate folder as they arrived or were written.¹ It was a pain in the ass precisely because I had to scroll through the long list of folders before dragging and dropping messages.

For a time, there were plugins for Mail that helped with the organization of messages into folders. They tended to be fragile and would often break when Apple updated OS X. This got worse when the updates became yearly.

Eventually, though, Apple added an option that allowed you to move selected messages to a folder via a drop-down menu that you could put in your toolbar. The menu is called Move and its icon is a little folder.²

If you have the Move tool in your Mail toolbar, you can select all the messages you want to move, click on the Move tool, and select the appropriate folder from the drop-down menu. Because the menu is long, this may seem no easier than dragging and dropping, but it is. The key is that once the drop-down menu is active, you can start typing the name of the folder and it will be selected. Boom.

Reminders could add this same Move tool to its toolbar. I guess Reminders doesn’t have a true toolbar because there’s no way to customize it, but it does have a small set of tools up at the top of the window and there’s plenty of room up there for a Move tool.

Adding a Move tool to Reminders would be more Mac-like than mimicking what we see on iOS. And it would make the Reminders UI more consistent with Mail. Remember when consistency between apps was thought to be important? That goes back as far as drag and drop.

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