And now it’s all this

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

Simply supported beam—conjugate beam method

May 22, 2026 at 8:01 AM by Dr. Drang

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,

$$M″=-q$$

and between displacement and bending moment,

$$y″=-\frac{M}{EI}$$

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

$$\begin{array}{cc}\hfill M& =\left(\frac{w{L}^3}{24EI}\right)\left(\frac{L}{2}\right)-\left(\frac{w{L}^3}{24EI}\right)\left(\frac{3}{8}\frac{L}{2}\right)\hfill \\ \hfill & =\frac{w{L}^4}{EI}(\frac{1}{48}-\frac{3}{384})\hfill \\ \hfill & =\frac{5w{L}^4}{384EI}\hfill \end{array}$$

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

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Lousy labels

May 21, 2026 at 6:13 PM by Dr. Drang

In Paul Krugman’s post today, he includes two charts. One is fine, the other isn’t.

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.

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Simply supported beam—moment-area method

May 21, 2026 at 8:22 AM by Dr. Drang

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

$$y″=-\frac{M}{EI}$$

so

$$y\prime (x_A)-y\prime (x_B)={\int }_{x_A}^{x_B}\frac{M}{EI}dx$$

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

$${\int }_{x_A}^{x_B}\frac{M}{EI}(x-x_b)dx={\int }_{x_B}^{x_A}y″(x-x_b)dx$$

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,

$$\frac{2}{3}\left(\frac{w{L}^2}{8EI}\right)\left(\frac{L}{2}\right)=\frac{w{L}^3}{24EI}$$

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

$$\delta =\left(\frac{w{L}^3}{24EI}\right)\left(\frac{5}{8}\frac{L}{2}\right)=\frac{5w{L}^4}{384EI}$$

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.

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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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