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

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

and m is symmetric about the center of the beam.

Therefore,

and we can take advantage of symmetry to say

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 $192$.

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,

where

and

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,

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:

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

Matrix multiplication gives us

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.

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

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:

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

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

and its square is

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

So

Taking the derivative with respect to P gives us

Therefore the downward deflection at the center of the beam is

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.

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,

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,

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:

Here’s the form we’ll start with:

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:

The second derivative is

Since $y″(0)=0$,

which means

and therefore, after some more algebraic cancellation and rearranging,

and

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

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

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:

and

Therefore,

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

which we’ve seen as an intermediate solution before.

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

Our old friend comes to visit again.

I just learned that people are listening to music pitched slightly down because it makes them feel better. Instead of the A above middle C being set at 440 Hz, they have it tuned down to 432 Hz.

This strikes me as odd, but how you feel is how you feel. Do whatever you want, as long as it doesn’t hurt anyone. I was interested, though, in the math behind this pitch change.

In the equal-tempered scale, the frequency ratio of a semitone, which I’ll call $r_s$ is the twelfth root of two:

This is the ratio of frequencies of adjacent piano keys.

The ratio of 440 Hz to 432 Hz is

so the pitch difference you get from moving down to 432 Hz is distinctly less than a semitone. How can we characterize that difference?

Small differences in pitch are measured in cents. There are 100 cents in a semitone, so the frequency ratio of one cent, $r_c$, is

To get the number of cents we move down in going from 440 Hz to 432 Hz, we solve this equation for n:

Taking the base-2 logarithm of both sides yields

and therefore

So going from A440 tuning to A432 tuning means going down about 32 cents or about a third of a semitone. Not a lot, but you (probably) can hear it.

Here’s two seconds of A440:

And here’s two seconds of A432:

It’s easier to hear the difference when they’re played simultaneously because the beat frequency is distinct:

Given the name of this blog, I would be remiss if I didn’t mention the famous splice in “Strawberry Fields Forever.” There were two takes that John Lennon liked: a slower version in a lower key and a faster version in a higher key. He wanted the final song to have part of one and part of the other. Right. As luck would have it, though, producer George Martin and engineer Geoff Emerick learned that adjusting the tape speeds to bring the two tempos together also put them in the same key.

Continuing our trip through various methods to derive the equation for the center deflection of a uniformly loaded simply supported beam, today we’re going to do the first of two solutions using the Rayleigh-Ritz method.

Of all the possible shapes a beam can deform into, the shape it will deform into is the one that minimizes the potential energy of the system, the system being the beam and the load. The equation for the potential energy for our beam, $\mathrm{\Pi }$, is

where the first term comes from the bending of the beam (note its similarity to the formula $\frac{1}{2}k{x}^2$ for a spring) and the second term comes from the load acting through the deflection. The first term is positive because the potential energy of the beam increases as the beam bends; the second term is negative because the potential energy of the uniform load decreases as the load moves down with the beam.

Minimizing an expression like this with respect to the displacement function, y, is what the calculus of variations was invented to do. But Lord Rayleigh and Walther Ritz came up with a way to avoid the calculus of variations. Instead of considering all possible shapes for y, we can consider only certain shapes governed by a set of associated parameters. We then express the potential energy in terms of these parameters and solve for the parameter values that minimize it.

Let’s demonstrate with a simple example. We’ll assume y is of this form,

and find the value of a that minimizes $\mathrm{\Pi }$

This sine function is a good choice because it meets all the boundary conditions of the simply supported beam: both it and its second derivative are zero at the two ends of the beam, i.e.,

(Recall that the moment is proportional to the second derivative of the displacement—since the moment is zero at a simply supported end, so is the second derivative.)

Given our choice for y, we can say that

Therefore,

The first integral works out to be $L/2$ and the second to $2L/\pi$. So

To find the value of a that minimizes this, we take its first derivative with respect to a and set it equal to zero:

which means

and

Compare this with our previous solution,

and we see that the one-term Rayleigh-Ritz approximation is awfully close to the exact solution.

But our goal wasn’t to get awfully close; it was to get the exact solution. To do that, we need to take not a single sine term, but the sum of an infinite number of sine terms, like this:

This is called a Fourier series, and you may recall seeing somewhere that a Fourier series can be fit to any function. In general, a Fourier series will have both sine and cosine terms, but for our problem the cosine terms drop out to meet the boundary conditions.

It may seem that we’ve just assigned ourselves an infinite amount of work, given that our expression for potential energy is now

But there are some features of the sine function that we can take advantage of. First and foremost, that nasty integral in the first term is actually quite simple:

And the second term can be simplified, too:

So we end up with

We minimize with respect to the $a_m$ by setting

for all m. Solving for $a_m$ we get

Plugging these results into our series expression for y and evaluating it at $x=L/2$ gives us

The sine term inside the sum alternates between 1 and –1, so we could write this as

At this point, I could get the sum from Mathematica with this expression,

Sum[(-1)^((m - 1)/2)/m^5, {m, 1, Infinity, 2}]

but that would be breaking my self-imposed rule against using computers in the derivation. Luckily, I have a book, An Introduction to the Elastic Stability of Structures by George Simitses, that discusses using infinite series in Rayleigh-Ritz solutions, and it includes this table of closed form solutions for infinite sums:

We can use the last entry in this table with $x=\pi /2$ to get

And through the magic of cancellation,

I’m not suggesting this is the best way to derive this formula, but it’s nice to know you can do it. And when you don’t need an exact answer, the Rayleigh-Ritz method can give you a good approximation without much work.

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.

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.

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,

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

This time, we’ll consider the right half:

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:

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:

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

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

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:

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.

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:

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, ${\theta }_A$, and ${\theta }_B$ and the downward defection at each end, $y_A$ and $y_B$.

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:

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

Let’s now define the span rotation as

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

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

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

The simple supports mean $M_A=0$ and $\psi =0$ (the straight line connecting A and B stays horizontal through the deflection). Symmetry tells us ${\theta }_B=-{\theta }_A$. 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).

So

and therefore

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:

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

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

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

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

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,

and between displacement and bending moment,

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

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

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.

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

so

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

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,

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

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.

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

which can be combined to give

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,

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

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 $x=0$:

(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,

We know three of the initial conditions right off the bat:

Therefore,

We solve for the fourth initial condition by noting that the displacement at $x=L$ is zero:

Solving for ${\theta }_0$ gives us

Using this, the displacement at the center of the beam is

which is the formula we were looking for.