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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. ↩

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