Hello my name is Vladimir. I'm a researcher at Technical University of Denmark and welcome to this lecture about beam theory. After this lecture you will be able to define boundary conditions and loads on beams. Calculate reactions and internal forces in statically determined beam systems. Explain the basics of classical beam theory, beam deformations and beam stresses. Let's first look at what is a wind turbine blade. Here you can see a blade from a ten mega watt, reference wind turbine that has been developed at Technical University of Denmark. This blade is pretty slim and 86 meter long with five meter root diameter and the widest part of the blade is about six meter wide. So, in many, many problems, this blade is considered as a beam. With variable structural properties along its length. And you can also see in this picture the typical bending loads acting at this blade in normal operation. The major goal of the blade is to provide airdynamic loads. Therefore, this kind of load is very important for the blades. And you can notice that the peak of this load comes through closer to the tip of the blade, which is mainly due to higher air speeds at this region. Before we continue, let's first look at the general case of body equilibrium. What is equilibrium? Mathematically speaking, equilibrium is zero total force and zero total momentum acting at the beam. Let's look at the case of a 2D body with some forces, P_i and moments M_j acting in at it. How can we describe equilibrium? We can write down two vector equations for the forces and for the moments. So first comes the forces, the sum of forces is equal to zero. And the second is for the moments, and notice that we have not only moments acting directly at the body, but those are the moments generated by the forces at the position r_i, each. Let's now look at how we can restrict or fix the beams in space, the way of providing support to them. There are many different ways of doing that. But within this lecture, we will look only at three idealized approaches. The first approach is called sliding simple support. It restricts movement of the beam only in vertical direction. So we can write it like this: Zero vertical displacement while non-zero horizontal displacement and rotation. And the reaction force provided by this support, is only non-zero vertical force while the horizontal force and bending moment are zero. The second one is the simple support. In this case the beam movement is restricted in both horizontal and vertical directions, while the bending in rotation is not restricted. So we have two reaction forces here which are non-zero, the horizontal and the vertical one, where the bending moment is zero. And the last one here is called clamped support. In this case, we restrict all three kind of movements of the beam: Displacement in horizontal and vertical direction and beam rotation. While direction forces are now all non-zero, both force components and bending moment. And very often, the blade is considered as a beam, which is clamped at the wind turbine hub. That's why this type of support is very interesting for our case. Let's now look at a beam supported at two ends. We have simple support at the left end A and sliding simple support in the right end B. And this beam is loaded by a force, F, at the distance, d, apart from the point A. Now we know that the simple support provides two reaction forces, horizontal and vertical one. While the sliding simple support provides one vertical reaction force. So we can write down equilibrium equations for this system. The first one will be the sum of horizontal forces acting at this beam, which is only a reaction force provided by the simple support which is obviously zero. The second one will be the sum of the moments acting at this beam around point A. Now we have only two components. One from the force F, and the other one is from the reaction force R_BY. From this equation, we can figure out what is the reaction force and it is dependent on the magnitude of the force F, apparently. The last one is for the vertical forces. We have three components here from both supports and from the force F. And from here we can figure out what is the vertical reaction force R_AY at the left end of the beam. We have now seen what are the direction forces for this beam system, but let's look at what are the internal forces in the beam in this situation. To do that we need to make a virtual cut at the beam at any position and look what are the forces that have to act there to keep the system in equilibrium. So, that's what we do. We cut the beam at position x, and we add three internal forces here. The tensile force N, the sheer force Q, and the bending moment M. What can we do now is we can write down the three equilibrium equations for this part of the beam. The first one will be for the horizontal forces, and now we can see that we have two components, and we have learned before that the reaction force from the simple support is zero, so the tensile internal force N is also zero. The second equation will be for the vertical component and now we have three components here. The support at the point A, the force itself, and the sheer force Q. From this we can learn what is the magnitude of the sheer force Q. And we can see that they're only dependant on force F and not on the position of the cut. And the last equation will be for the moments. We again do it around point A and we have three components again. The first one is the moment produced by the force F. The second one is the moment produced by the sheer force Q. And the last one is the bending moment itself, internal bending moment. And the internal bending moment is the only unknown in this equation and therefore we can solve it for the internal bending moments and we can see that this is actually the only component that is dependent on the position of the virtual cut. Until now we have looked at the beam like it's something rigid, but let's see what's going on inside the beam. What are the stresses in the bending beam? Let's first look at the straight beam and choose a reference axis which is also straight, and some reference point. Let's say the middle of this beam. Now we can choose two cross sections at a distance delta s away from the middle point and the classical Euler-Bernoulli theory now says that the two cross sections initially plain and perpendicular to the reference axis remain plain and perpendicular even after beam is deformed. So if we have two bending moments acting at the beam at its ends. The reference axis is now a part of a circle with a curvature radius R. If we select the reference axis such that it is a neutral axis, that is the line which does not experience the change in its length during deformation of the beam, then the distance between the two cross sections is still the same, delta s from the middle part. Let's now look closer. We have part of the beam, which is bounded by the two cross sections. And we have a reference axis. Let's look at what happens with the material at distance z away from the reference axis which is neutral axis. So we can write down the beam curvature, kappa, which is actually inverse to the radius of the curvature. And now we can write down the strains. The strains is the relative deformation of the stripe of material. That is delta s star means the new length of the material and the delta s is the old one. Now if we substitute from the kappa equation and simplify it a little bit, we'll figure out that strains are actually in linear relation with the position of the material in the cross section. Now if you look at stresses, stresses are linear with respect to the strains. And we can write it down using the constant E, which is the material stiffness, the Young's modulus of the material. And we can also see that stresses are also linear with respect to the position in the cross section. And now we would like to know what is the bending moment in this cross section? And now we can actually figure out what it is. The only thing that we need to do is we need to integrate the stresses along the entire cross section A of the beam, and we can substitute the sigma. The stresses here, and we can see that if we have the same material in the beam cross section, then the stiffness is constant and we can take it out of the integral. So that we have the relation between bending moment and the beam defamation kappa. And it is through the some parameter E_i which is called bending stiffness. And you can notice that the I which is the integral here, it is called the cross section moment of inertia and you can notice that if we have the same material through the cross section of the beam then we can take out the stiffness outside the integral and this parameter I is only a function of the cross sections shape and nothing else. Often, but not always, beam deformations are much smaller compared to the beam length, width, and height. In this case we can use the number of simplifications and establish the following relations for the beam deformations. Beam bending rotation theta is actually the first derivative of the first displacement, while the bean curvature kappa is the second displacement. So we can see that the bending moment, M, is actually related to the beam deformation through the second derivative of the beam deformation. And we can calculate beam deflection as a function of a position capital X, by doing the double integration of the beam curvature. Taking into account boundary conditions. In this lecture you have learned that wind turbine blades can often be treated as beams. You have also learned how to calculate reactions and internal beam loads, how to calculate beam strains and stresses, and also how to calculate beam bending and deformations.
