The next technique we’ll use to derive the formula for the center deflection of a simply supported beam with a uniform load is the slope-deflection equation:

Let’s start by explaining where all the terms come from. Here’s a beam of length L with arbitrary end supports (could be simple, fixed, free, or sprung) and an arbitrary applied load. We’ll call the left end A and the right end B.

The moment at A is the sum of five terms, which come from the superposition¹ of five conditions. First is the fixed-end moment (FEM), which is the moment that would exist at A if the beam had both ends fixed against vertical displacement and rotation:

The other four terms come from analysis of the unloaded beam when specific geometric end conditions are applied. The end conditions are specified by the clockwise rotation at each end, ${\theta }_A$, and ${\theta }_B$ and the downward defection at each end, $y_A$ and $y_B$.

Each of these shapes comes from applying just one of these end conditions and keeping the others zero. The moment at A that corresponds to each of these shapes is given in the figure.

The general solution for the clockwise moment at A is the sum of these five terms:

Note that we’ve put in some negative signs to account for the counter-clockwise terms.

Let’s now define the span rotation as

This is the clockwise rotation of the straight line connecting points A and B.

Rewriting the third and fourth terms on the right-hand side of the equation using this definition, we get

Pulling out common terms gives us the equation at the top of the post:²

OK, let’s use this to solve our problem. We’ll start with our simply supported beam and label the two ends:

The simple supports mean $M_A=0$ and $\psi =0$ (the straight line connecting A and B stays horizontal through the deflection). Symmetry tells us ${\theta }_B=-{\theta }_A$. And the fixed-end moment for a uniform load is $w{L}^2/12$ (this is another one of those things burned into my brain through repetition).

So

and therefore

which should look familiar.

To get the center deflection, we need to use symmetry in another way. It tells us that the slope at the center of the beam is zero, which means we can treat the left half of the beam as its own problem:

The right end of this half-length beam is guided, which means it’s free to deflect but prevented from rotating. This beam will behave exactly like the left half of our original beam.

For this half-length beam, we know that

where we’ve taken the expression for ${\theta }_A$ from the intermediate solution above. Plugging these into the slope-deflection equation gives us

And therefore, as we’ve seen five times now,

We had to solve two equations to get this result, but they weren’t simultaneous equations, so it wasn’t that much work.

The fourth way we’re going to derive the formula for the center deflection of a simply supported beam with a uniform load is the conjugate beam method. This is probably tied with the moment-area method for the simplest and fastest way to get the formula—at least if you’ve memorized the properties of parabolas.

I wrote about the conjugate beam method last year. In a nutshell, it takes advantage of the similarity of the relationships between bending moment and distributed load,

and between displacement and bending moment,

To get the deflection of the beam, we construct a conjugate beam that’s loaded by the $M/EI$ diagram of the real beam. Calculating the moment at a point of the conjugate beam gives us the deflection at the corresponding point in the real beam.

For our problem, the conjugate beam looks like this:

The supports of the conjugate beam don’t always match the supports for the real beam, but they do for simple supports, so that makes things easy. The intensity of the distributed load at the peak of the parabola is $w{L}^2/8EI$.

To get the bending moment at the center of the conjugate beam, we analyze a free-body diagram of its left half:

By symmetry, the reaction at the left support is $w{L}^3/24EI$ (that’s the area under the parabola, and we’ve done that calculation before). That’s also the resultant of the distributed load. The line of action of the resultant is ⅜ of the way from the center to the left support. Therefore, the moment at the center of the conjugate beam is

so this is the center deflection of the real beam, as expected.

Here’s the first one:

Generally, I’m not a fan of putting stuff in the margins that could be within the body of the plot itself, but at least it’s clear which label goes with which data series. Too bad that’s not the case with Chart 2:

The two series end at almost the same spot, so putting the labels out in the margin means they can’t both be centered on their series. The little leader lines that run from the labels to the series match the colors of the series, but unfortunately, their colors are difficult to distinguish because

1. The two colors are both shades of blue.
2. The leader lines are thin, muting their colors.
3. The leader lines are dashed, further muting their colors.

I’m not saying you can’t tell which is which, but when a chart is showing only two things you shouldn’t have to squint to tell the two apart. And it’s harder if you’re reading the article on a phone.

In the text just below Chart 2, Krugman says

The blue line labeled “Euro relative constant prices” supports the Draghi-Smith story of badly lagging European productivity, with Europe starting well above the US level but falling far behind. But the black line labeled “Europe relative current prices” shows Europe holding its own.

This is essentially alt text. It helps but would be better if there actually were a black line and not just a darker blue line.

Better still would be moving the labels so what they’re labeling is obvious. Something like this:

Yes, I should have pushed the upper label a little further to the left.

Charting software can’t judge the clarity of its output, but we can. A little touch-up in a graphics program can work wonders.

Continuing our odyssey through various ways of calculating the deflection at the center of a simply supported beam with a uniform load, today we use the moment-area method.

There are two moment-area theorems used to calculate the slopes and deflections of a beam. Here’s how they’re given in the textbook I used as an undergrad, the 3rd edition of Elementary Structural Analysis by Norris, Wilbur, and Utku.¹

The first theorem is

The change in the slope of the tangents of the elastic curve between two points A and B is equal to the area under the $M/EI$ diagram between these two points.

The second theorem is

The deflection of point B on the elastic curve from the tangent to this curve at point A is equal to the static moment about an axis through B of the area under the $M/EI$ diagram between points A and B.

The “elastic curve” is the curve made by the beam in its deflected position. The “static moment” of an area is the product of that area and the distance from the area’s centroid to the given point. (It can also be expressed as an integral, as we’ll see in a minute). This sort of thing is easier to describe with examples than with words.

The first theorem should be pretty obvious, given that we already know that

so

The second theorem isn’t so obvious, but if we write it in terms of an integral,

you can get to the second moment-area theorem through integration by parts.

But we’re not here to derive the moment-area method—I did that last year—we’re here to use it. For our problem, let’s sketch the deflected shape of the beam and note that the slope at its center must be zero by symmetry.

We’ll call the center A and the support at the left end B. The deflection at A is equal to the deflection of B relative to the horizontal tangent at A, which we’ll call $\delta$.

By the second moment-area theorem, we take the moment diagram, divide it by EI, and multiply the area under the left half of the curve by the distance from B to the centroid of that area.

The area under the shaded part of the parabola is ⅔ of the area of the enclosing rectangle,

The distance from B to the centroid of the shaded area is ⅝ of the distance from B to A, so

which is the answer we were expecting.

How did I know the area under a parabola and the location of its centroid without looking it up? Lots and lots of practice doing problems like this 45 years ago. There’s a diagram with all the values in a post I wrote a few years ago.

One last thing: Because the slope at the center of the beam is zero, the slope at the left end should be equal to the shaded area calculated above. Click back to the previous article in this series, and you’ll see that it does match the value of ${\theta }_0$ calculated there.

In the introductory post to this series, I mentioned that the shear is the derivative of the moment. It’s easy to see that for our specific problem, the simply supported beam with a uniform distributed load,

but it’s also true in general. A further relationship is that the distributed load function—let’s call it q—is the derivative of the shear (with a minus sign according to the usual sign convention). Therefore

which can be combined to give

Again, for our specific problem with a uniform load, $q$ is just the constant value $w$, and it’s easy to see that the slope of the shear diagram is $-w$.

In yesterday’s post, we used the differential relationship between the moment and displacement,

and the fact that we already knew that the moment was a parabola to get a solution quickly. In general, though, people usually put the last two equations together to get

This is a non-homogeneous fourth-order linear differential equation with constant coefficients. The solution will have four constants of integration, which can be written in terms of the initial conditions, i.e., the displacement, slope, moment, and shear at $x=0$:

(For small displacements, the slope and angle are the same, which is why the initial slope is called ${\theta }_0$.)

The final term, $y_p$, is the particular solution to the original equation, i.e., four integrations of q. Since q in our problem is just the constant w,

We know three of the initial conditions right off the bat:

Therefore,

We solve for the fourth initial condition by noting that the displacement at $x=L$ is zero:

Solving for ${\theta }_0$ gives us

Using this, the displacement at the center of the beam is

which is the formula we were looking for.

Here’s the first of the derivations for the center deflection of a simply supported beam with a uniform load.

We start with the differential relationship between the bending moment, M, and the deflection, y:

The second derivative of y is the curvature of the beam (for small deflections, which is one of the fundamental assumptions of beam theory), and the negative sign is there to account for the usual sign conventions for moment and displacement.

M is a parabola that passes through 0 at each end of the beam and peaks at $w{L}^2/8$ at the center. Its formula is

Therefore,

where I’ve started using primes for differentiation.

Integrating once gives

Symmetry tells us the slope at the center of the beam is zero, so

Which means

Plugging this result in and integrating again gives us

Because the deflection is zero at $x=0$,

and the deflection at the center of the beam is

which is the answer we were expecting.

A few years ago, I wrote a post describing how I asked ChatGPT to solve a couple of elementary beam bending problems and how its answers were persistently wrong, even after I told it the mistakes it had made. For the first problem, determining the deflection at the center of a simply supported beam with a uniform load, ChatGPT gave the correct formula—presumably because the correct formula was part of its training corpus—but couldn’t come up with the correct numerical solution. As I said in the post:

Strictly speaking, this wasn’t a good example of a structural analysis homework problem. Students don’t get asked to just look up formulas and plug in numbers. More likely, they’d be asked to derive the equation that ChatGPT started with by either solving the differential equation for beam deflection or using some simplified technique like the moment-area or conjugate beam method. I didn’t think asking ChatGPT to do something like that would be fair.

This got me wondering how many ways I could derive the formula. A handful of ways came to me immediately, and I kept thinking of other methods over the course of the next several weeks.

Here’s a sketch of the problem:

where w is the intensity of the load, in units of force per length, L is the length of the beam, E is the modulus of elasticity of the beam’s material, and I is the moment of inertia of the beam’s cross-section. I’m not going to get into the details of these terms or the assumptions implicit in my derivations. Suffice it to say that I’m using the typical definitions and assumptions described in strength of materials and structural analysis textbooks.

I gave myself some rules for the derivations:

• They had to be truly different methods; slight variations on the same technique didn’t count.
• They had to give an exact formula; no numerical approximations.
• They had to be arrived at by hand; no use of a computer. The upshot of this was that although methods that yielded simultaneous equations were OK, there could be no more than three simultaneous equations. No one in their right mind solves more than three simultaneous equations, and frankly, my skills have deteriorated to the point that more than two equations is iffy.

I scratched out the derivations in my notebook, eventually coming up with twelve ways. They were:

1. Second-order differential equation
2. Fourth-order differential equation
3. The moment-area method
4. The conjugate beam method
5. The slope-deflection method
6. The “myosotis” method
7. Energy minimization with polynomials
8. Energy minimization with Fourier series
9. Castigliano’s second method
10. Finite element analysis
11. The dummy unit load method
12. Newmark’s method

I thought about presenting the derivations here, but I dithered over the best way to organize them. Eventually, other parts of my life intruded, and I gave up on the idea. It wasn’t until I wrote about the definition of “kip” a couple of weeks ago that I decided to just do a brain dump of all the derivations, one post for each. That’s what you’ll see here for the next couple of weeks. I know most of you don’t care about this sort of stuff, but I don’t care that you don’t care. Forewarned is forearmed—you’ll know what each post is about from their titles and can skip as you see fit.

For those few who are interested, this post will serve as a table of contents. The items in the list above will be turned into links as the posts are written.

Let me put a couple of things here that will be common. First, the deflection at the center is

This is the formula each post is aiming towards.

Second, the upward reaction forces at each end of the beam are

which is, as you might expect, half the total applied load.

Third, the shear and moment diagrams for the beam are

The moment diagram is very important to many of the derivations. It’s a parabola with a peak value of

We won’t be using the shear diagram¹ directly in any of the derivations, but I tend to draw it whenever I draw a moment diagram. The mathematically inclined might notice that the shear is the derivative of the moment. It passes through zero when the moment is at its peak.

OK, that’s the setup. We’ll start zipping through the derivations next time.

I know what the writer means, of course. He’s telling us that the SpaceX debris will hit the Moon at several thousand feet per second, 1,000 fps being about the speed of sound we’re used to here on Earth. But the Moon isn’t the Earth, so it’s kind of a weird comparison, don’t you think? It’s not as if the booster is going to create a sonic boom. I’d find this less odd if the story were in a normal newspaper or magazine instead of Scientific American.

Again, I realize that the writer is just giving his readers a point of comparison, but is the speed of sound (on Earth) really something most people have an intuitive feel for? I remember learning as a kid to count off the seconds between a lightning flash and the resulting thunder to get the distance to the lightning strike in thousands of feet, but that doesn’t mean I have a strong sense of the speed of sound. Other than “really fast,” which isn’t all that helpful.

It’s mentioned in the body of the article that the speed of the booster at impact, which will be on August 5, is estimated to be about 5,400 mph. That’s about 100 times the posted speed limit on a two-lane highway, which is something most Americans do have a feel for.

By the way, if you have any interest in the problem of space debris coming back and striking the Earth (where the speed of sound might be relevant), you should follow Sam Lawler on Mastodon. She’s a professor of astronomy at the University of Regina and also has a keen interest in the clogging of near-Earth orbital space by the huge number of satellites launched in recent years. Her Mastodon feed is also a great source of farm animal photos, most recently a crop of baby goats.

Earlier this week, I saw this article in Apple News. It’s from the San Francisco Chronicle and discusses a recent report on the safety of the Golden Gate Bridge. The report is one of several reports spurred by the Francis Scott Key Bridge collapse a couple of years ago.

Image from Wikipedia.

Spoiler: the report finds the Golden Gate Bridge safe—quite unlikely to suffer damage from the impact of a ship. This has to do with the Golden Gate’s two towers:

• The south tower (in the foreground) is surrounded by a reinforced concrete protective structure that a ship would have to strike and destroy before coming in contact with the tower and affecting the bridge’s integrity. We discussed the lack of protective structures around the piers of the Key Bridge.
• The north tower is right up against the Marin County shoreline, where the water is shallow. A large vessel would run aground before striking the tower, scrubbing away most, if not all, of its kinetic energy. A ship that wouldn’t run aground before hitting the tower would be too light to bring it down.

A striking omission from the Chronicle story is a link to the report itself. But that appears to be the fault of Apple News. I can’t read the story on the Chronicle’s website because I’m not a subscriber, but this reprint on Yahoo! News includes the report embedded as a PDF and available to download. So it looks like Apple removed the report itself, which is pretty poor form.

The nice thing about finding the actual report, written by HDR, Inc., is that it confirmed some suspicions I had regarding this paragraph in the Chronicle story:

The resilience of the Golden Gate Bridge partly comes from sheer strength. The south tower, on the San Francisco side, is described in the report as a “robust structural feature like no other” and is surrounded by a reinforced concrete protective shell up to 28 feet thick. It can withstand about 50,000 kips of force, or roughly 25,000 tons. In many cases, engineers found, a ship would crumple and absorb its own impact energy before it could seriously damage the structure.

First, it’s not “sheer strength,” it’s “shear strength.” HDR calculated the strength of the protective structure around the south tower to be at least 50,000 kips when that force is attempting to shear through the reinforced concrete wall. They specifically mention shearing capacity, shearing interfaces, and shearing area when discussing this calculation on pp. 59–60 of the report. While it’s true that engineers tend to be crummy writers, we definitely know the difference between “shear” and “sheer.”

Second, I suspect the writer, Brooke Park, doesn’t know what a kip is. When writing for a general audience¹ most people wouldn’t use the word “kip” without saying it’s short for “kilopound.” Which is to say,

There’s no “roughly” about it.²

Still, this oddball paragraph doesn’t affect the overall story and provides the engineers who read it some amusement. Too bad about Apple’s redaction of a link to the report itself, though. That’s sheer incompetence.

(My first thought was that this would be an update to yesterday’s post, but I soon realized it was going to be too long for that.)

The first method starts with a swipe to the left on the message whose sender you want to block. This brings up three options:

The options are

• Trash, which deletes the message.
• Move…, which lets you file the message in a folder.
• More, which brings up the following set of commands:

You may need to swipe up to see the whole list.

Two of these commands repeat what was in the previous screenshot, but a little redundancy never hurt anyone. What we’re interested in is the Block Contact command at the bottom, which does exactly what we want.¹

This is a pretty efficient way to block the sender if you’re in list view and can tell that that’s what you want to do from just the first couple of lines of the message. The next method, which I didn’t know about before Ron’s email, gives you a chance to see more of the message before deciding to block.

Starting again in list mode, long press on the message whose sender you think you want to block. This will bring up two windows, a larger one with a more complete view of the message and a small one with a list of commands.

If you were unsure whether to block the sender from the list view’s abbreviation, you’ll probably decide after seeing this. Tapping on the small list of commands makes it bigger and reveals the Block Contact command.

Why Block Contact is red in this command list and not in the others is one of the wonderful mysteries of Apple interface design.

These blocking methods (thanks again, Ron!) prove that Apple can make reasonably efficient interfaces when it wants to. Too bad it didn’t put the same effort into the blocking method you have to use from message view.

Yesterday, for example, I was going through my mail on my iPhone and came upon a piece of spam. I put it in the Junk folder with the hope that future emails like it would stay out of my inbox, but I also wanted to do something stronger. I wanted to block all future emails from that sender. I’ve done this with other senders many times before, so I know the steps to take, but I decided this time to document the process because it’s so stupid.

Tapping anywhere in the header turns the various fields blue, suggesting that a further tap on any of them will perform some other action. I’m not sure why these fields aren’t blue to start with, but that’s not my real complaint, so we’ll pass over that.

Since I want to block the sender, the natural thing is to tap on the From field. Indeed, a menu pops up with a set of commands associated with that person/address. You might think that one of them would start a new message (as opposed to a reply), but no, which I find kind of weird. More to the current point, though, is that a command named Block Contact or Block Address is also missing.

I know perfectly well that I can block contacts from my phone, so how do I do it? The Copy and Search commands are clearly wrong, and View Contact Card seems even more wrong, as I have no contact card for this person. Nor do I want one—he’s a spammer. But because I once tapped View Contact Card, possibly by mistake, I now know that that’s the choice to make because this is what appears:

It doesn’t show an existing contact card for the sender; it shows a potential contact card, one that I could add to my Contacts app. But also included in the list of things I can do with this potential contact is block him. Which is what I did.

But this process makes no sense. View Contact Card should not be the path you need to take to block someone you have no intention of turning into one of your contacts. It’s not just an extra step (like tapping the header to turn all the fields blue), it’s a step in the wrong direction. No reasonable person who has not already gone through this process would think that View Contact Card is the command you choose to ensure that you never see an email from this spammer again.

The natural place—the correct place—for a Block Contact or Block Address command is on the popup menu you see when you tap the From field. There’s plenty of room for it. Hiding it behind a command whose primary purpose is something else isn’t a matter of taste, it’s an error.

Apple used to think about things like this and put commands where they made sense. I know that Apple has many more products than it used to, but it also has many many more employees and much much much more money. Simple things like this shouldn’t be falling through the cracks.

My favorite picks were the oddballs, the products that weren’t Macs, iPhones, iPads, iPods, or Apple IIs. In other words: the accessories. I was particularly pleased with Jason’s picks of the LaserWriter, the Apple Disk II, the Apple Watch Sport Band, and the second generation Apple Pencil. I confess I was a little disappointed in Myke’s choice of the first generation Pencil, but he more than made up for it by later choosing the Magic Trackpad.

Those of you who weren’t around in the 80s and 90s may think Jason went overboard in putting the LaserWriter in as his third pick, but you’d be wrong. It was both a great product and incredibly important to Apple. Similar comments apply to the Apple Disk II. I never had one—I never owned an Apple II—but I did have its successor, the Integrated Woz Machine, in all of my early Macs.

My oddball entry would have been the AirPort Express. This is not in the “I can’t believe you didn’t pick” category¹ because it’s an oddball even among oddballs, but for a short period of time for a specific subset of users, it was a great accessory.

If you were a business traveler during those few years in the mid-00s when hotels had wired internet access in their rooms but hadn’t yet outfitted themselves with WiFi, the first generation AirPort Express was one of the best things you could pack. It was about the same size as the wall wart power supply that came with your Apple notebook, and it set up a little WiFi network that gave you the freedom to work (or play) anywhere in your room. Even after hotel WiFi became common, I still packed my AirPort Express because it gave me a faster and more reliable wireless network.

I should also mention that “AirPort” was one of Apple’s best product names. Too bad they don’t have any reason to bring it back.