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Beam Deflection Methods
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0:00
Thanks to Brilliant for sponsoring this video. Beams are used to carry loads in a whole range of different structures, and it’s important for engineers to be able to predict how much a beam will deflect when loads are applied to it. Allow a beam to deflect too much and it could cause damage to other parts of the structure, feel unsafe to the user, or fail to meet its intended function, even if stresses in the beam are low and there’s no risk of failure.
0:29
Construction codes typically define what the maximum allowable deflection is, and so it’s up to the engineer to figure out how much the beam will deflect. There are quite a few different approaches that can be used to do this. I'll cover five of the more common methods in this video. All of these methods make the same important assumption, which is that deflections are small and stresses remain in the elastic region. For this reason, the deformed shape of the beam is called the elastic curve.
1:03
We can define x as the distance along the axis of the undeformed beam and y as the vertical deflection of the beam. The objective is to figure out how y varies along the length of the beam. We will assume that there’s no displacement along the x axis. We’re also interested in the angle of rotation of the beam, Theta. Let’s take a closer look at the angle. We can define the slope of the beam as dy divided by dx. Since we’re limiting ourselves to scenarios where deflections are small, we can apply
1:42
the small angle approximation, and so we can assume that the angle of rotation is equal to the slope of the beam. In other words, the angle of rotation is equal to the first derivative of y with respect to x.
1:59
We can start thinking about how beams deflect under load by reviewing the moment-curvature equation. It allows us to calculate the curvature of a beam for a given bending moment. M is the bending moment, R is the radius of curvature of the deformed shape, and El is the flexural rigidity of the beam, which defines how much the beam will resist bending due to the stiffness of its material, and due to its cross-sectional shape. The term 1/R is the curvature of the beam.
2:35
The moment-curvature equation is the first step towards being able to calculate the deflection of a beam, since it gives us information about how the beam will deform for an applied bending moment. It turns out there’s a mathematical equation that can be used to calculate the curvature at any point along a curve.
2:59
We can combine this with the moment-curvature equation to get an expression that relates the deflection of the beam and the bending moment along it. But this equation is still quite complicated, so let’s try and simplify it. Since we’re limiting ourselves to elastic deformation and assuming that the slope of the elastic curve is small, the dy/dx term will be small. And since it’s squared it will be even smaller, and so can be neglected.
3:29
This gives us a much simpler differential equation defining the deflection of the beam. All we need to do to get an expression for the deflection along a beam is integrate the equation twice. This is called the double integration method. It’s easiest to understand how the method works using an example. Here we have a cantilever with a load applied at the end. We’ll start by drawing the free body diagram and applying equilibrium to determine the reaction forces.
4:04
Next we want to calculate the bending moment along the beam. If we draw a free body diagram of a slice of the beam we can see that as we move along the beam, the bending moment increases linearly with the distance from the concentrated force P. So the bending moment M is equal to -Px. We can insert this expression for M into the deflection differential equation and apply the double integration method to figure out how the beam will deflect.
4:36
E and I are constant along the length of the beam, so we can pull them outside of the integral, and then integrate the equation for M with respect to x. Remember that whenever you integrate you need to add a constant of integration at the end. We'll call it C1. It’s an unknown for now. Then we integrate a second time, remembering to add a second constant of integration. And so we now have an equation that defines the deflection y of the beam.
5:13
The only problem is that we have two unknown constants. Fortunately we can use what we know about the deflection and the slope of the beam at the support to calculate C1 and C2. At the fixed end the beam cannot rotate, so the slope dy/dx must be equal to zero. This allows us to calculate C1. And at that same location the deflection y must also be zero, which gives us C2.
5:59
And so we've solved the deflection curve for this beam. In this case it's obvious that the maximum deflection will occur at the end of the beam.
6:11
For cases where the location of maximum deflection is less obvious, like for this simply supported beam, we can observe that the slope of the beam will be zero at the location of maximum deflection. So by setting the slope equation to be equal to zero we can figure out the location of ymax. Boundary conditions are key to applying the double integration method. We just saw that at a fixed support the displacement and the slope must both be zero.
6:42
At a pinned support the beam is free to rotate, so the slope can be non-zero and only the boundary condition for the displacement applies. The same is true for a roller support, which is also free to rotate. The double integration method is quite easy to apply, but even for some quite simple configurations like this one the bending moment won't be defined by a single expression along the full length of the beam. To apply the double integration method in situations like this we need to split the
7:16
beam into sections, and determine separate expressions for the bending moment in each section. We then apply the double integration method to the different sections, and we end up with a lot of different constants of integration. By considering the boundary conditions at the supports we can calculate some of these constants. To solve the remaining constants we need to consider what’s called the continuity condition between the different sections of the beam.
7:44
Since the beam remains continuous across the transition between these two sections, the displacement and the slope must be the same at the right of Section 1 and at the left of Section 2. This gives us a system of equations we can solve to determine the remaining constants of integration. Having to split the beam into sections and solve a system of equations complicates things and makes it more difficult to solve the deflections of the beam, particularly for cases where
8:15
the beam needs to be split into a large number of sections. Fortunately there’s a powerful mathematical trick that allows us to define a bending moment profile like this one using a single expression, so that we can apply the double integration method without having to split the beam into sections. It uses singularity functions to make solving this type of beam much easier. They are defined using this special notation, and follow specific rules.
8:47
When the expression inside the bracket gives a positive number, the output is the expression, but when it is negative, the output is zero. These are called Macaulay brackets, and applying them to calculate the deflection of beams is called Macaulay’s Method. Their behaviour allows us to essentially turn different terms in the bending moment equation on and off as we move along a beam. We can work through the beam we looked at earlier to understand how.
9:18
Let’s see how the bending moment changes as we move along the beam. In the first half of the beam only the reaction force contributes to the bending moment, and the moment arm is equal to x. But once we get past halfway, we need to include the effect of the concentrated load as well, and so the bending moment expression has to include an additional term. The moment arm is equal to x minus half the length of the beam.
9:51
If we replace the brackets in this expression with Macaulay brackets, the second term in the expression will be “switched off” in the first half of the beam, since x - L/2 will give a negative number. And so this expression using Macaulay brackets defines the bending moment along the entire length of the beam. The same thing works for distributed loads, although it’s slightly different. When using Macaulay brackets for a uniformly distributed load, the bracket has to appear
10:23
twice - once in the magnitude of the load and once in the moment arm. The moment arm is the distance to the centre of the distributed load, so is equal to the length over which the load is applied, divided by 2. And so we get this expression, where the Macaulay bracket is squared and the distributed load is divided by 2. This expression is only valid if the distributed load extends all the way to the end of the beam.
10:52
If it doesn't, we have to adjust the expression to balance the part of the load that’s missing. Here's how you can apply Macaulay brackets for a few other load types. Macaulay brackets also have special rules for integration. They are integrated in a similar way to polynomial functions, like this, which makes it easy to apply the double integration method when using them. Let's go back to our beam with a concentrated load and solve the deflections using Macaulay’s method.
11:47
Since we have a single expression that defines M along the full length of the beam, we no longer need to split the beam into sections to solve it, and we can just apply the double integration method as we normally would. The double integration method can be quite time consuming to apply, even when using Macaulay's method. To save us from having to go through the integration process every time we want to calculate the
12:19
deflection of a beam, reference texts often include tables which list the deflection equations for a set of typical load and support configurations. This is more powerful than it might seem at first glance, because the principle of superposition can be used to apply this small set of solutions to a much wider range of problems. It allows us to combine different load configurations on top of one another, and to calculate the
12:47
overall deflection caused by the combined loading by summing up the contribution of each load to the total deflection. This is a really powerful technique that's not just used for the analysis of beams, but also more widely in structural analysis. For the superposition method to be applicable, deflections must vary linearly with the applied loads. And deflections need to be small, since large deflections could mean that the applied loads don't act independently.
13:22
Sometimes when analysing a beam we might only be interested in calculating the angle of rotation or the displacement at a single location, instead of along the entire length of the beam. There’s another method we can use for cases like this, which is called the moment-area method because it’s based on the area under the bending moment diagram. If we take two points, 1 and 2, and we integrate both sides of the deflection differential
13:52
equation between those two points, we end up with this equation. The term on the left is the change in angle between both points, which we can visualise by extending lines tangent to the elastic curve at the two points. And the term on the right is just the area under the bending moment diagram between both points, A, divided by the flexural rigidity EI, assuming that both E and I are constant along the length of the beam.
14:23
So to calculate the change in angle between two points along the elastic curve all we have to do is divide the area under the bending moment diagram between the two points by EI. This is the first moment-area theorem. If we rewind and multiply both sides of the deflection differential equation by x before integrating, we can get another useful equation. This time the integration is a bit more complicated, since we need to use integration by parts.
14:56
But we end up with this expression, which is the second moment-area theorem. It tells us that the vertical distance between point 2 and the tangent line at point 1 is equal to the first moment of the area of the bending moment diagram, divided by EI. The first moment of area is just the area under the diagram, multiplied by the horizontal distance from point 2 to the centroid of the area This vertical distance is called the tangential deviation.
15:26
It might not seem that useful, since it doesn’t directly give us the deflection at point 2. But if we calculate the tangential deviation at the other end of the beam we can then use trigonometry to figure out the actual deflection at point 2. The moment-area method is particularly useful when there’s a point along the beam where we know that the slope of the elastic curve will be equal to zero, since in that case calculating the tangential deviation gives us the deflection directly.
15:58
This is why the moment-area method is often applied for cantilever problems. Let's look at an example. We’re interested in the deflection and angle of rotation at the free end, so let’s define point 1 at the fixed support and point 2 at the free end. Because the angle and deflection are zero at point 1, the moment-area method will give us the deflection and angle at point 2 directly. We can define x as the horizontal distance from the free end, and draw the bending moment diagram.
16:34
To apply the moment-area method we need to calculate the area under the bending moment diagram between points 1 and 2, and the distance between point 2 and the centroid of the area. It’s easy to do this by splitting the diagram into two triangles and a rectangle. The areas are easy to calculate. And for the centroids we just need to remember that the centroid of a right angled triangle is one third of the way down and one third of the way across from the right angle.
17:15
Then we can just plug the numbers into the first and second theorems to calculate the angle and deflection at point 2. There's one last approach to analysing the deflection of beams that I’d like to cover, which is an energy approach that uses the concept of strain energy. When a body is deformed elastically, energy is stored within it in the form of strain energy. For a beam deformed by bending the strain energy is a function of the bending moment, and it can be calculated using this equation.
17:50
It’s common to neglect any shear deformation and assume all of the strain energy is coming from pure bending. Castigliano’s theorem states that the deflection at the location of an applied load along the line of action of the load is equal to the partial derivative of the strain energy with respect to that load. The deflection can be a displacement or an angle of rotation. Because the bending moment term is squared in the strain energy equation, it’s usually
18:21
easier to move the partial derivative inside of the integral, like this.
18:32
Let’s look at an example. The bending moment along the beam shown here is defined by this expression. We can calculate the partial derivative of M with respect to P. Using Castigliano’s theorem, and integrating between zero and L, we get an expression for the displacement at point A.
19:05
The powerful thing about Castigliano's thoerem is that the load we consider doesn't even have to be a real load. If we want to determine deflections at a location where no load is applied, we can just apply a fictitious force. We include the fictitious force in the bending moment equation, and then once we’ve calculated the partial derivative of the strain energy with respect to that force, we can set it to zero to obtain an equation for the deflection at that location.
19:35
The fictitious load can be either a moment or a force. Applying a moment allows us to calculate the angle of rotation, and applying a force allows us to calculate the deflection y of the beam. Applying a fictitious moment at Point A allows us to calculate the angle of rotation at that point. This time we take the partial derivative of M with respect to M_A. Once we have an expression for the angle of rotation we can set the fictitious moment M_A to be equal to zero.
20:13
Although it might be slightly more difficult to conceptualise than the other methods we’ve looked at, Castigliano’s theorem can be very powerful when applied properly, and it has uses that extend well beyond the analysis of beams. With experience you'll be able to select the method that will be easiest or most appropriate to apply, based on the specifics of the situation you are analysing. But one inescapable fact when it comes to applying any of these methods is that they
20:42
rely heavily on having a solid understanding of calculus. Mastering the fundamentals of calculus is the best way to make beam deflection problems feel much less daunting, and this video's sponsor Brilliant is a math and science learning platform that will help you do exactly that. It has all of the steps covered, from the basics of what differentiation and integration actually are, all the way up to complex topics like partial differential equations and Laplace transforms.
21:13
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21:40
So, to take the first step towards calculus mastery, and support this channel at the same time, head over to brilliant.org/efficientengineer or click the link in the description, and sign up for free. The first 200 people to sign up using that link will get 20% off the annual Premium subscription. And that’s it for this look at the deflection of beams. Thanks for watching!
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