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Simply supported beam—dummy unit load method

June 5, 2026 at 7:51 AM by Dr. Drang

We’re in the home stretch of this series, so ANIAT will soon go back to complaining about Apple’s UI choices.¹ Next week’s WWDC keynote should provide some inspiration.

But today’s post covers our eleventh method for deriving the center deflection of a uniformly loaded simply supported beam: the dummy unit load method. When I was an undergraduate, this was the method we used to determine the deflections of truss structures, but it’s more general than that. My favorite explanation of why it works is in Nicholas J. Hoff’s The Analysis of Structures. I have the original Wiley hardcover of this book, but you can apparently get both paperback and hardcover reprints.

Hoff uses the principle of virtual work in his explanation, and I’d like to quote him here, but unfortunately his explanation is split between one section on the analysis of trusses and another on the analysis of beams. I couldn’t figure out a nice way to put them together coherently, so you’re just going to have to trust me.

For a beam, you can get the deflection at a particular point by following these steps:

1. Work out an equation for the bending moment, M, in the beam for the given set of applied loads.
2. Imagine that same beam with those loads removed (this is the dummy structure) and a concentrated unit load placed at the point at which we want the deflection. By definition, the magnitude of a unit load is one. The unit load points in the direction of the deflection we want to calculate.
3. Work out an equation for the bending moment, m, in the dummy beam with the unit load.

4. The desired deflection is

   $$\delta ={\int }_0^L\frac{mM}{EI}dx$$

Let’s apply this simple principle to our problem:

The bending moment is

$$M=\frac{w}{2}(Lx-x^2)$$

where the x coordinate starts at the left end of the beam and increases to the right, as usual.

This same structure with just a unit load at the center is

Its bending moment is

$$m=\frac{1}{2}x\mathrm{\text{for}}0\le x\le \frac{L}{2}$$

and m is symmetric about the center of the beam.

Therefore,

$$\delta =y\left(\frac{L}{2}\right)={\int }_0^L\frac{mM}{EI}dx$$

and we can take advantage of symmetry to say

$$\begin{array}{cc}\hfill y\left(\frac{L}{2}\right)& =\frac{2w}{4EI}{\int }_0^{L/2}(L{x}^2-x^3)dx\hfill \\ \hfill & =\frac{w}{2EI}{[\frac{1}{3}L{x}^3-\frac{1}{4}{x}^4]}_0^{L/2}\hfill \\ \hfill & =\frac{w}{2EI}[\frac{L^4}{24}-\frac{L^4}{64}]\hfill \\ \hfill & =\frac{5w{L}^4}{384EI}\hfill \end{array}$$

You can pretty much do this problem in your head if you’re good at figuring out least common multiples. I’m not, but after writing things out, I did realize that the LCM of $24$ and $64$ is $192$.

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Simply supported beam—finite element method

June 3, 2026 at 8:24 AM by Dr. Drang

The biggest problem I face when writing posts in this series is deciding how much background explanation to put in. If I included only the stuff I do to get the answer, the posts would be very terse: one sketch and a dozen or fewer lines of math. A full explanation, though, could easily turn into several semesters’ worth of structural analysis theory. I’m trying to get close to the former while adding just enough background so readers with engineering experience can follow along. Even if you haven’t used the method of a particular post, you should be able to take what’s in the post and find the whatever further information you need to fill in the inevitable gaps.

This problem is most acute in describing today’s analysis. The finite element method—more precisely called matrix structural analysis or the direct stiffness method when applied to structures made of beam, truss, or frame elements—has many analytical parents: the Rayleigh-Ritz method, the slope-deflection equation, and matrix math. Presenting it without at least some of its antecedents would be a crime. But I will try to keep the introductory material to a minimum.

The fundamental piece of this analysis is the beam bending element. This is a length of beam with no interior loading, and its behavior can be described completely by four parameters, called its degrees of freedom. The DOF are the the displacements and rotations at the two ends (or nodes) of the element, usually numbered and directed as follows:

The DOF are often called generalized displacements, and we’ll refer to them as $u_1$, $u_2$, $u_3$, and $u_4$. The deflected shape of this beam is completely determined by the sum of the products of these generalized displacements and a set of shape functions that look like this:

Each shape function is a cubic curve that has a unit displacement corresponding to one of the DOFs and zero displacement corresponding to the other three DOFs. Apart from the sign convention, these are the same shapes we saw in our discussion of the slope-deflection equation.

There is a generalized force associated with each DOF; these are the forces and moments at the two ends of the beam, and we’ll call them $f_1$, $f_2$, $f_3$, and $f_4$. Note that when we multiply a u term by an f term, we get an energy expression. For DOF 1 and 3, that’s a displacement multiplied by a force; for DOF 2 and 4, that’s a rotation multiplied by a moment.

For this element, we can write a matrix equation for the relationship between the generalized displacements and generalized forces,

$$\mathit{𝒌}\mathit{𝒖}=\mathit{𝒇}$$

where

$$\mathit{𝒖}=\left\{\begin{array}{c}\hfill u_1\\ \hfill u_2\\ \hfill u_3\\ \hfill u_4\end{array}\right\},\mathit{𝒇}=\left\{\begin{array}{c}\hfill f_1\\ \hfill f_2\\ \hfill f_3\\ \hfill f_4\end{array}\right\}$$

and

$$\mathit{𝒌}=\frac{EI}{L^3}\left[\begin{array}{rrrr}12& 6L& -12& 6L\\ 6L& 4L^2& -6L& 2L^2\\ -12& -6L& 12& -6L\\ 6L& 2L^2& -6L& 4L^2\end{array}\right]$$

Each column is the set of generalized forces that correspond to that shape function.

You may notice that the terms in the second row of $\mathit{𝒌}$ match the coefficients in the slope-deflection equation before the $y_A$ and $y_B$ terms were combined into the $\psi$ span rotation term.

What should we do when, as in our case of a beam with a distributed load, there is interior loading? We convert that internal loading into what are called consistent loads at the ends. These are equal in magnitude and opposite in direction to the reaction forces and moments that would occur if the interior loads were applied to a beam with both ends fixed: $wL/2$ for DOF 1 and 3 and $w{L}^2/12$ for DOF 2 and 4.

To solve a problem by the finite element method, we assemble the structural stiffness matrix, $\mathit{𝑲}$, and the structural force vector, $\mathit{𝑭}$ from all the individual elements and then solve the structural equation,

$$\mathit{𝑲}\mathit{𝑼}=\mathit{𝑭}$$

for the vector of structural displacements, $\mathit{𝑼}$. As you can see, I tend to use lower-case letters for the element terms and upper-case for the structural terms. That’s a reasonably common, but by no means universal, notation.

In general, this assembly and solution of a matrix equation can be quite complex, which is why the finite element method is typically thought of as a process that requires a computer. But our problem is simple enough to do by hand.

Recall that in both the slope-defection and Myosotis solutions, we took advantage of symmetry and considered just half the beam. We’ll do that again here.

This half-beam has just two DOF, the rotation at the left end and the deflection at the right end. In the assembly process, element DOF 2 matches structural DOF 1 and element DOF 3 matches structural DOF 2. The other element DOF don’t appear in the structural matrix equation because their U terms are zero. Here’s an illustration of the assembly process for the stiffness matrix, accounting for the fact that the beam element has a length of $L/2$:

I took a class in matrix structural analysis (CE 361) from Prof. Jamshid Ghaboussi in the fall of 1981. He used the phrase destination array for the list of structural DOF that the element DOF go to. I’ve always thought that was a particularly good description and have used it ever since, even though I don’t think I’ve ever seen it in any finite element textbook. Note again that DOF 1 and 4 in the element have no destination because those displacements are zero in the structure.

When the assembly of both the stiffness matrix and force vector (which is done in a similar way) are complete, we get this equation:

$$\frac{8EI}{L^3}\left[\begin{array}{rr}{L}^2& -3L\\ -3L& 12\end{array}\right]\left\{\begin{array}{r}{U}_1\\ U_2\end{array}\right\}=\{\begin{array}{r}–\frac{w{L}^2}{48}\end{array}–\frac{wL}{4}\}$$

There are different ways to solve this, but because a 2×2 matrix is easy to invert, that’s how I did it. Premultiplying each side by the inverse of $\mathit{𝑲}$ gives us

$$\left\{\begin{array}{r}{U}_1\\ U_2\end{array}\right\}=\frac{L}{24EI}\left[\begin{array}{rr}12& 3L\\ 3L& L^2\end{array}\right]\{\begin{array}{r}–\frac{w{L}^2}{48}\end{array}–\frac{wL}{4}\\ \}$$

Matrix multiplication gives us

$$\left\{\begin{array}{r}{U}_1\\ U_2\end{array}\right\}=\{\begin{array}{r}–\frac{w{L}^3}{24EI}\end{array}–\frac{5w{L}^4}{384EI}\\ \}$$

The signs are the opposite of what we’ve seen before because of the directions we used in defining the degrees of freedom. Physically, though, these are the same answers we’ve seen several times.

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Simply supported beam—Castigliano's method

June 1, 2026 at 8:06 AM by Dr. Drang

Continuing our series on the many ways to get the center deflection of a uniformly loaded beam, we come to another energy-based technique: Castigliano’s method.

Castigliano’s second theorem provides a relationship between displacements, forces, and the strain energy in a linearly elastic system. The strain energy, U, is the potential energy of the system due to its deformation. For a generic elastic body, like this,

Castigliano’s second theorem can be as:

If an elastic system is mounted so that rigid-body displacements of the entire system are impossible and certain external point forces $P_1$, $P_2$, … act on the system, in addition to distributed loads and thermal strains, the displacement component ${\delta }_i$ of the point of application of force $P_i$ in the direction of force $P_i$ is determined by the equation

$${\delta }_i=\frac{\partial U}{\partial P_i}$$

I took this from Henry Langhaar’s Energy Methods in Applied Mechanics, a favorite text of mine. I’ve altered the variable names to avoid conflicting with some variable names we’ve used in previous posts in this series.

Looking this over and thinking about it in terms of our problem,

two concerns come to mind:

1. Our problem doesn’t have a point force at the center of the beam, where we want to determine the deflection.
2. Even if we did have a point force at the center of the beam, how do we write an expression for the strain energy in terms of that force?

Let’s tackle the second concern first. Recall from our two Rayleigh-Ritz posts (Fourier and polynomial) that the potential energy associated with beam bending is

$${\int }_0^L\frac{1}{2}EI(y″{)}^{2}dx$$

This is the strain energy, U. Unfortunately, it’s written in terms of deflection, not force, but we can fix that. We saw early on that the curvature, $y″$, and the bending moment, M, are proportional:

$$M=-EIy″\mathrm{\text{or}}y″=-\frac{M}{EI}$$

We substitute the second of these into the expression for U to get

$$U={\int }_0^L\frac{M^2}{2EI}dx$$

Because the bending moment can be written in terms of the applied loads, we’ve solved the second concern.

We solve the first concern with a trick. We can pretend there’s a load, P, at the center of the beam, and write out an expression for the bending moment, M, that includes P.

We then do the integration to get U, take its derivative with respect to P, and—here’s the trick—evaluate the derivative for $P=0$. This may seem like cheating, but it’s perfectly legitimate. Imagine P is very very small—it won’t contribute much to the deflection. So why not just make it zero?

Let’s go ahead and see what happens. The upward support reaction force at end of the beam above is $(wL+P)/2$, so we can draw a free-body diagram for the left portion of the beam like this:

where $x<L/2$. The bending moment in the left half of the beam is therefore

$$M=\frac{1}{2}[(wL+P)x-w{x}^2]$$

and its square is

$$M^2=\frac{1}{4}[(wL+P{)}^2{x}^2-2w(wL+P)x^3+w^2{x}^4]$$

This formula applies only in the left half of the beam, but because of symmetry we know that

$$U={\int }_0^L\frac{M^2}{2EI}dx=2{\int }_0^{L/2}\frac{M^2}{2EI}dx={\int }_0^{L/2}\frac{M^2}{EI}dx$$

So

$$\begin{array}{cc}\hfill U& =\frac{1}{4EI}{\int }_0^{L/2}[(wL+P{)}^2{x}^2-2w(wL+P)x^3+w^2{x}^4]dx\hfill \\ \hfill & =\frac{1}{4EI}{[\frac{1}{3}(wL+P{)}^2{x}^3-\frac{1}{2}w(wL+P)x^4+\frac{1}{5}{w}^2{x}^5]}_0^{L/2}\hfill \\ \hfill & =\frac{1}{4EI}[\frac{L^3}{24}(wL+P{)}^2-\frac{L^4}{32}w(wL+P)+\frac{L^5}{160}{w}^2]\hfill \end{array}$$

Taking the derivative with respect to P gives us

$$\frac{\partial U}{\partial P}=\frac{1}{4EI}[\frac{L^3}{12}(wL+P)-\frac{L^4}{32}w]$$

Therefore the downward deflection at the center of the beam is

$$\begin{array}{cc}\hfill \delta & ={\frac{\partial U}{\partial P}|}_{P=0}\hfill \\ \hfill & =\frac{w{L}^4}{48EI}-\frac{w{L}^4}{128EI}\hfill \\ \hfill & =\frac{5w{L}^4}{384EI}\hfill \end{array}$$

as we’ve now seen many times.

Since this is based on Castigliano’s second theorem, you may be wondering what kinds of problems we use his first theorem to solve. It’s a good question, but one I can’t answer. In the 45 years since I first learned of Castigliano’s theorems, I’ve never used his first one.

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Simply supported beam—energy minimization with a polynomial

May 30, 2026 at 2:09 PM by Dr. Drang

Today we’ll use the Rayleigh-Ritz method again, but this time we’ll avoid dealing with an infinite sum. In case you’ve forgotten, this is our problem:

We’ll express the shape as a polynomial. The form of the governing differential equation,

$$EI{y}^{iv}=w$$

tells us that our solution won’t have any terms of higher power than $x^4$. That gets rid of the infinity problem, but a generic fourth-order polynomial still has five parameters,

$$y=a_0+a_{1}x+a_2{x}^2+a_3{x}^3+a_4{x}^4$$

which is more than we want to deal with. Let’s cut down further on the number of parameters by taking advantage of some other things we know. First, the solution will have to be symmetric, so we can express it with symmetric terms right from the start. Second, we know the solution will have to meet these boundary conditions:

$$y(0)=y(L)=y″(0)=y″(L)=0$$

Here’s the form we’ll start with:

$$y=ax(L-x)+b{x}^2(L-x{)}^2$$

This is a fourth-order polynomial. We know it’s symmetric about the center of the beam because switching the $x$ and $L-x$ terms (which is like flipping the beam around) yields the same equation. And it’s clear that $y(0)=y(L)=0$, so we meet two of the four boundary conditions. Now we’ll take on the other two boundary conditions.

We’ll use the product rule and a little algebraic rearranging to give us the first derivative with respect to x:

$$y\prime =a(L-2x)+2b[x(L-x{)}^2-x^2(L-x)]$$

The second derivative is

$$y″=-2a+2b[(L-x{)}^2-4x(L-x)+x^2]$$

Since $y″(0)=0$,

$$-2a+2b{L}^2=0$$

which means

$$b=\frac{a}{L^2}$$

and therefore, after some more algebraic cancellation and rearranging,

$$y=a[x(L-x)+\frac{1}{L^2}{x}^2(L-x{)}^2]$$

and

$$y″=-\frac{12a}{L^2}x(L-x)$$

Because we started with a symmetric function, its second derivative is also symmetric and $y″(L)=0$.

Now it’s time for Rayleigh-Ritz. Recall that the potential energy, $\mathrm{\Pi }$, is

$$\mathrm{\Pi }={\int }_0^L\frac{1}{2}EI(y″{)}^{2}dx-{\int }_0^{L}wydx$$

Plugging in our expressions for $y$ and $y″$ gives us

$$\Pi =\frac{72EI{a}^2}{L^4}{\int }_0^L{x}^2(L-x{)}^{2}dx-wa{\int }_0^{L}x(L-x)dx-\frac{wa}{L^2}{\int }_0^L{x}^2(L-x{)}^{2}dx$$

where the first term is the potential energy of the bent beam and the second two terms are the potential energy of the load as it moves down with the beam.

The integrals are fairly easy to work out. Just expand the polynomials and integrate term-by-term. Here are the results:

$${\int }_0^L{x}^2(L-x{)}^{2}dx=\frac{L^5}{30}$$

and

$${\int }_0^{L}x(L-x)dx=\frac{L^3}{6}$$

Therefore,

$$\mathrm{\Pi }=\frac{72EI{a}^2}{L^4}\left(\frac{L^5}{30}\right)-wa\left(\frac{L^3}{6}\right)-\frac{wa}{L^2}\left(\frac{L^5}{30}\right)=\frac{12EIL}{5}{a}^2-\frac{w{L}^3}{5}a$$

We minimize this by taking the derivative with respect to a, setting it to zero, and solving for a. That gives us

$$a=\frac{w{L}^2}{24EI}$$

which we’ve seen as an intermediate solution before.

Finally, after plugging in this expression for a, we get

$$\begin{array}{cc}\hfill y\left(\frac{L}{2}\right)& =\frac{w{L}^2}{24EI}(\frac{L}{2}\frac{L}{2}+\frac{1}{L^2}\frac{L^2}{4}\frac{L^2}{4})\hfill \\ \hfill & =\frac{w{L}^4}{24EI}(\frac{1}{4}+\frac{1}{16})\hfill \\ \hfill & =\frac{5w{L}^4}{384EI}\hfill \end{array}$$

Our old friend comes to visit again.

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