Simply supported beam—the Myosotis method

May 26, 2026 at 8:28 AM by Dr. Drang

The sixth way we’ll derive the formula for the center deflection of a uniformly loaded simply supported beam is the Myosotis method, which I wrote about over a decade ago. This is the method popularized1 by J.P. Den Hartog in his Strength of Materials textbook.

Forget-me-not from Wikipedia

Image from Wikipedia.

Myosotis is the genus of the forget-me-not flower, and the idea behind the Myosotis method is to memorize the following six equations for the tip angle and deflection of a cantilever beam under different loading conditions.

Myosotis excerpt from Den Hartog

Once you have the formulas memorized, you can combine them to generate the solution for almost any beam that’s subjected to point and uniformly distributed loads. I wouldn’t say the Myosotis method is, or has ever been, a practical tool for working engineers, but it’s a great pedagogical tool for teaching engineering students how to take advantage of symmetry, antisymmetry, and superposition. Using it even a few times will get you thinking about how complex structural problems can be broken down into a combination of simpler solutions, and that will stay with you even if you never use the Myosotis method again.

We mentioned in the slope-deflection post that the left half of our simply supported beam behaves like a simple-guided beam. Let’s be more explicit about that. The symmetry of the problem we want to solve,

Simply supported beam with uniform load

means it deflects like two simple-guided beams back to back:

Shape of two half-beams back-to-back

This time, we’ll consider the right half:

Half-beam with guided and simple ends

Statics tells us that the upward reaction at the right support is wL/2.

This is, apart from an overall downward displacement, the same as a fixed-free beam with both a uniform load over its length and an upward load at its tip:

Half-beam with fixed and free ends

So the downward deflection at the center of our full-length simple-simple beam is equal to the left end deflection of our half-length guided-simple beam, which in turn is equal to the upward right end deflection of our half-length fixed-free beam. One of the purposes of a structural engineering education is to get you to see these relationships in a lot less time than it takes to type them out.

Now we can use superposition and two of the Myosotis formulas to get our answer. Here’s a graphical expression of how the superposition works:

Superposition for Myosotis solution

So the upward deflection of the right end of the fixed-free beam is

−w(L/2)48EI+(wL/2)(L/2)33EI=−wL4128EI+wL448EI=5wL4384EI

and that’s the same as the downward center deflection of our original problem, as expected.

  1. “Popularized” may be going a bit far; I’ve never seen the Myosotis method in any other book. Still, thanks to Dover, Den Hartog’s book is still in print, something you can’t say about many other textbooks from 1949. ↩

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:

MA=2EIL(2θA+θB−3ψ)−FEMA

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, θA, and θB and the downward defection at each end, yA and yB.

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:

MA=4EILθA+2EILθB+6EIL2yA−6EIL2yB−FEMA

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=yB−yAL

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

MA=4EILθA+2EILθB−6EILψ−FEMA

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

MA=2EIL(2θA+θB−3ψ)−FEMA

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 MA=0 and ψ=0 (the straight line connecting A and B stays horizontal through the deflection). Symmetry tells us θB=−θA. And the fixed-end moment for a uniform load is wL2/12 (this is another one of those things burned into my brain through repetition).

So

0=2EILθA−wL212

and therefore

θA=wL324EI

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

MA=0,θB=0,θA=wL324EI,ψ=yBL/2,FEMA=w(L/2)212

where we’ve taken the expression for θA from the intermediate solution above. Plugging these into the slope-deflection equation gives us

0=4EIL(2wL324EI−32yBL)−wL248

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

yB=L224EI(wL23EI−wL248EI)=5wL4384EI

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

  1. 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. ↩

  2. 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″=−MEI

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:

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 wL2/8EI.

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 wL3/24EI (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

M=(wL324EI)(L2)−(wL324EI)(38L2)=wL4EI(148−3384)=5wL4384EI

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

  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:

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.

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