And now it’s all this

I just said what I said and it was wrong
Or was taken wrong

Simply supported beam—fourth-order ODE

May 20, 2026 at 7:28 AM by Dr. Drang

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

$$V=\frac{dM}{dx}\mathrm{a}\mathrm{n}\mathrm{d}q=-\frac{dV}{dx}$$

which can be combined to give

$$q=-\frac{d^{2}M}{d{x}^2}$$

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,

$$M=-EI\frac{d^{2}y}{d{x}^2}$$

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

$$EI\frac{d^{4}y}{d{x}^4}=q$$

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 $x=0$:

$$y=y_0+{\theta }_{0}x-\frac{M_0}{2EI}{x}^2-\frac{V_0}{6EI}{x}^3+y_p$$

(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,

$$y_p=\frac{w}{24EI}{x}^4$$

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

$$y_0=0,M_0=0,V_0=\frac{wL}{2}$$

Therefore,

$$y={\theta }_{0}x-\frac{wL}{12EI}{x}^3+\frac{w}{24EI}{x}^4$$

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

$$y(L)={\theta }_{0}L-\frac{w{L}^4}{12EI}+\frac{w{L}^4}{24EI}=0$$

Solving for ${\theta }_0$ gives us

$${\theta }_0=\frac{w{L}^3}{24EI}$$

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

$$y\left(\frac{L}{2}\right)=\frac{w{L}^4}{EI}(\frac{1}{48}-\frac{1}{96}+\frac{1}{384})=\frac{5w{L}^4}{384EI}$$

which is the formula we were looking for.

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Simply supported beam—second-order ODE

May 19, 2026 at 7:52 AM by Dr. Drang

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:

$$M=-EI\frac{d^{2}y}{d{x}^2}$$

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

$$M=\frac{wL}{2}x-\frac{w}{2}{x}^2$$

Therefore,

$$y″=\frac{w}{EI}(\frac{1}{2}{x}^2-\frac{L}{2}x)$$

where I’ve started using primes for differentiation.

Integrating once gives

$$y\prime =\frac{w}{EI}(\frac{1}{6}{x}^3-\frac{1}{4}L{x}^2)+C_1$$

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

$$y\prime \left(\frac{L}{2}\right)=\frac{w}{EI}(\frac{1}{48}{L}^3-\frac{1}{16}{L}^3)+C_1=0$$

Which means

$$C_1=\frac{w{L}^3}{24EI}$$

Plugging this result in and integrating again gives us

$$y=\frac{w}{EI}(\frac{1}{24}{x}^4-\frac{1}{12}L{x}^3+\frac{1}{24}{L}^{3}x)+C_2$$

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

$$C_2=0$$

and the deflection at the center of the beam is

$$y\left(\frac{L}{2}\right)=\frac{w{L}^4}{EI}(\frac{1}{384}-\frac{1}{96}+\frac{1}{48})=\frac{5w{L}^4}{384EI}$$

which is the answer we were expecting.

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Areas of my expertise

May 18, 2026 at 12:25 PM by Dr. Drang

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

$$\frac{5w{L}^4}{384EI}$$

This is the formula each post is aiming towards.

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

$$\frac{wL}{2}$$

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

$$\frac{w{L}^2}{8}$$

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.

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Unsound

May 11, 2026 at 2:17 PM by Dr. Drang

I mentioned last week that some wording in a news article struck me as odd. A similar thing happened when I read this Scientific American piece (that’s an Apple News link—I don’t have a SciAm subscription). In this case, I didn’t even have to read the article; the oddity is right there in the headline: “A SpaceX rocket booster is on track to hit the Moon at several times the speed of sound.”

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.

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