Simply supported beam—slope-deflection equation

May 23, 2026 at 7:51 AM by Dr. Drang

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:

M_{A} = \frac{2 E I}{L} (2 θ_{A} + θ_{B} - 3 ψ) - {F E M}_{A}

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.

Beam with arbitrary load and supports

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:

Fixed-fixed beam with arbitrary load

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, $θ_{A}$ , and $θ_{B}$ and the downward defection at each end, $y_{A}$ and $y_{B}$ .

End moments for end displacements

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:

M_{A} = \frac{4 E I}{L} θ_{A} + \frac{2 E I}{L} θ_{B} + \frac{6 E I}{L^{2}} y_{A} - \frac{6 E I}{L^{2}} y_{B} - {F E M}_{A}

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

Let’s now define the span rotation as

ψ = \frac{y_{B} - y_{A}}{L}

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

M_{A} = \frac{4 E I}{L} θ_{A} + \frac{2 E I}{L} θ_{B} - \frac{6 E I}{L} ψ - {F E M}_{A}

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

M_{A} = \frac{2 E I}{L} (2 θ_{A} + θ_{B} - 3 ψ) - {F E M}_{A}

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

Simple-simple beam

The simple supports mean $M_{A} = 0$ and $ψ = 0$ (the straight line connecting A and B stays horizontal through the deflection). Symmetry tells us $θ_{B} = - θ_{A}$ . 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).

0 = \frac{2 E I}{L} θ_{A} - \frac{w L^{2}}{12}

and therefore

θ_{A} = \frac{w L^{3}}{24 E I}

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:

Simple-guided beam

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

M_{A} = 0, θ_{B} = 0, θ_{A} = \frac{w L^{3}}{24 E I}, ψ = \frac{y_{B}}{L / 2}, {F E M}_{A} = \frac{w (L / 2)^{2}}{12}

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

0 = \frac{4 E I}{L} (2 \frac{w L^{3}}{24 E I} - 3 \frac{2 y_{B}}{L}) - \frac{w L^{2}}{48}

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

y_{B} = \frac{L^{2}}{24 E I} (\frac{w L^{2}}{3 E I} - \frac{w L^{2}}{48 E I}) = \frac{5 w L^{4}}{384 E I}

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

There are other ways to explain the slope-deflection equation. I decided to explain it using superposition after getting this Mastodon reply from Chris Huck. ↩
Most texts define the FEM as positive in the clockwise direction, so it has a positive sign in the slope-deflection equation. Since the FEM at the left end of a beam under most loading conditions is counter-clockwise, I prefer to define it that way and use a negative sign in the equation. ↩

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}{E I}

To get the deflection of the beam, we construct a conjugate beam that’s loaded by the $M / E I$ 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:

Conjugate beam with parabolic load

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 $w L^{2} / 8 E I$ .

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

Free-body diagram of conjugate beam

By symmetry, the reaction at the left support is $w L^{3} / 24 E I$ (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{matrix} M & = (\frac{w L^{3}}{24 E I}) (\frac{L}{2}) - (\frac{w L^{3}}{24 E I}) (\frac{3}{8} \frac{L}{2}) \\ = \frac{w L^{4}}{E I} (\frac{1}{48} - \frac{3}{384}) \\ = \frac{5 w L^{4}}{384 E I} \end{matrix}

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

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:

Krugman Chart 1

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:

Krugman 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

The two colors are both shades of blue.
The leader lines are thin, muting their colors.
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:

Krugman Chart 2 improved

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.

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 $M / E I$ 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 $M / E I$ 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}{E I}

y' (x_{A}) - y' (x_{B}) = \int_{x_{A}}^{x_{B}} \frac{M}{E I} d x

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

\int_{x_{A}}^{x_{B}} \frac{M}{E I} (x - x_{b}) d x = \int_{x_{B}}^{x_{A}} y ″ (x - x_{b}) d x

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.

Beam deflection with tangent line

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 $δ$ .

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.

Moment diagram with shaded half

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

\frac{2}{3} (\frac{w L^{2}}{8 E I}) (\frac{L}{2}) = \frac{w L^{3}}{24 E I}

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

δ = (\frac{w L^{3}}{24 E I}) (\frac{5}{8} \frac{L}{2}) = \frac{5 w L^{4}}{384 E I}

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 $θ_{0}$ calculated there.

Commonly known as “the MIT book” because that’s where the two senior authors taught. ↩

And now it’s all this

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

Simply supported beam—slope-deflection equation

Simply supported beam—conjugate beam method

Lousy labels

Simply supported beam—moment-area method

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Simply supported beam—slope-deflection equation

Simply supported beam—conjugate beam method

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Simply supported beam—moment-area method

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I just said what I said and it was wrong
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