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, 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. ↩

Simply supported beam—fourth-order ODE

May 20, 2026 at 7:28 AM by Dr. Drang

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,

Shear and moment diagrams

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

V = \frac{d M}{d x} a n d q = - \frac{d V}{d x}

which can be combined to give

q = - \frac{d^{2} M}{d x^{2}}

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,

M = - E I \frac{d^{2} y}{d x^{2}}

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

E I \frac{d^{4} y}{d x^{4}} = q

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 $x = 0$ :

y = y_{0} + θ_{0} x - \frac{M_{0}}{2 E I} x^{2} - \frac{V_{0}}{6 E I} x^{3} + y_{p}

(For small displacements, the slope and angle are the same, which is why the initial slope is called $θ_{0}$ .)

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,

y_{p} = \frac{w}{24 E I} x^{4}

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

y_{0} = 0, M_{0} = 0, V_{0} = \frac{w L}{2}

Therefore,

y = θ_{0} x - \frac{w L}{12 E I} x^{3} + \frac{w}{24 E I} x^{4}

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

y (L) = θ_{0} L - \frac{w L^{4}}{12 E I} + \frac{w L^{4}}{24 E I} = 0

Solving for $θ_{0}$ gives us

θ_{0} = \frac{w L^{3}}{24 E I}

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

y (\frac{L}{2}) = \frac{w L^{4}}{E I} (\frac{1}{48} - \frac{1}{96} + \frac{1}{384}) = \frac{5 w L^{4}}{384 E I}

which is the formula we were looking for.

Simply supported beam—second-order ODE

May 19, 2026 at 7:52 AM by Dr. Drang

Here’s the first of the derivations for the center deflection of a simply supported beam with a uniform load.

Beam and moment diagram

We start with the differential relationship between the bending moment, M, and the deflection, y:

M = - E I \frac{d^{2} y}{d x^{2}}

The second derivative of y is the curvature of the beam (for small deflections, which is one of the fundamental assumptions of beam theory), and the negative sign is there to account for the usual sign conventions for moment and displacement.

M is a parabola that passes through 0 at each end of the beam and peaks at $w L^{2} / 8$ at the center. Its formula is

M = \frac{w L}{2} x - \frac{w}{2} x^{2}

Therefore,

y ″ = \frac{w}{E I} (\frac{1}{2} x^{2} - \frac{L}{2} x)

where I’ve started using primes for differentiation.

Integrating once gives

y' = \frac{w}{E I} (\frac{1}{6} x^{3} - \frac{1}{4} L x^{2}) + C_{1}

Symmetry tells us the slope at the center of the beam is zero, so

y' (\frac{L}{2}) = \frac{w}{E I} (\frac{1}{48} L^{3} - \frac{1}{16} L^{3}) + C_{1} = 0

Which means

C_{1} = \frac{w L^{3}}{24 E I}

Plugging this result in and integrating again gives us

y = \frac{w}{E I} (\frac{1}{24} x^{4} - \frac{1}{12} L x^{3} + \frac{1}{24} L^{3} x) + C_{2}

Because the deflection is zero at $x = 0$ ,

C_{2} = 0

and the deflection at the center of the beam is

y (\frac{L}{2}) = \frac{w L^{4}}{E I} (\frac{1}{384} - \frac{1}{96} + \frac{1}{48}) = \frac{5 w L^{4}}{384 E I}

which is the answer we were expecting.

Areas of my expertise

May 18, 2026 at 12:25 PM by Dr. Drang

A few years ago, I wrote a post describing how I asked ChatGPT to solve a couple of elementary beam bending problems and how its answers were persistently wrong, even after I told it the mistakes it had made. For the first problem, determining the deflection at the center of a simply supported beam with a uniform load, ChatGPT gave the correct formula—presumably because the correct formula was part of its training corpus—but couldn’t come up with the correct numerical solution. As I said in the post:

Strictly speaking, this wasn’t a good example of a structural analysis homework problem. Students don’t get asked to just look up formulas and plug in numbers. More likely, they’d be asked to derive the equation that ChatGPT started with by either solving the differential equation for beam deflection or using some simplified technique like the moment-area or conjugate beam method. I didn’t think asking ChatGPT to do something like that would be fair.

This got me wondering how many ways I could derive the formula. A handful of ways came to me immediately, and I kept thinking of other methods over the course of the next several weeks.

Here’s a sketch of the problem:

Simply supported beam with uniform load

where w is the intensity of the load, in units of force per length, L is the length of the beam, E is the modulus of elasticity of the beam’s material, and I is the moment of inertia of the beam’s cross-section. I’m not going to get into the details of these terms or the assumptions implicit in my derivations. Suffice it to say that I’m using the typical definitions and assumptions described in strength of materials and structural analysis textbooks.

I gave myself some rules for the derivations:

They had to be truly different methods; slight variations on the same technique didn’t count.
They had to give an exact formula; no numerical approximations.
They had to be arrived at by hand; no use of a computer. The upshot of this was that although methods that yielded simultaneous equations were OK, there could be no more than three simultaneous equations. No one in their right mind solves more than three simultaneous equations, and frankly, my skills have deteriorated to the point that more than two equations is iffy.

I scratched out the derivations in my notebook, eventually coming up with twelve ways. They were:

Second-order differential equation
Fourth-order differential equation
The moment-area method
The conjugate beam method
The slope-deflection method
The “myosotis” method
Energy minimization with polynomials
Energy minimization with Fourier series
Castigliano’s second method
Finite element analysis
The dummy unit load method
Newmark’s method

I thought about presenting the derivations here, but I dithered over the best way to organize them. Eventually, other parts of my life intruded, and I gave up on the idea. It wasn’t until I wrote about the definition of “kip” a couple of weeks ago that I decided to just do a brain dump of all the derivations, one post for each. That’s what you’ll see here for the next couple of weeks. I know most of you don’t care about this sort of stuff, but I don’t care that you don’t care. Forewarned is forearmed—you’ll know what each post is about from their titles and can skip as you see fit.

For those few who are interested, this post will serve as a table of contents. The items in the list above will be turned into links as the posts are written.

Let me put a couple of things here that will be common. First, the deflection at the center is

\frac{5 w L^{4}}{384 E I}

This is the formula each post is aiming towards.

Second, the upward reaction forces at each end of the beam are

\frac{w L}{2}

which is, as you might expect, half the total applied load.

Third, the shear and moment diagrams for the beam are

Shear and moment diagrams

The moment diagram is very important to many of the derivations. It’s a parabola with a peak value of

\frac{w L^{2}}{8}

We won’t be using the shear diagram¹ directly in any of the derivations, but I tend to draw it whenever I draw a moment diagram. The mathematically inclined might notice that the shear is the derivative of the moment. It passes through zero when the moment is at its peak.

OK, that’s the setup. We’ll start zipping through the derivations next time.

Shear is usually denoted V because it’s a vertical force in most beams. ↩

And now it’s all this

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

Simply supported beam—moment-area method

Simply supported beam—fourth-order ODE

Simply supported beam—second-order ODE

Areas of my expertise

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Simply supported beam—moment-area method

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Areas of my expertise

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