 In this video, we are going to talk about some simplifications we often use for structural loading in engineering structures. In fact, we have already been working with these simplifications within this course, but we'll examine them a bit more deeply in this video, including when such simplifications are dangerously inappropriate. We have already performed some analysis on engineering systems, such as the aircraft shown here. In such analyses, we simplified the system as a rigid body, or even, simply as a single point. And then we looked at the forces acting on it, such as thrust, lift, weight, and drag. But, is this really a realistic representation of the system? Looking at it from just the lift perspective, does a wing really generate lift as a point force? Of course, the answer is no. Hopefully it is intuitive that the lift generated by the wing is distributed over the entire wing, as shown here. But this is less convenient to work with for some analyses, particularly when examining the system using Newton's laws of motion, as we do in many facets of engineering mechanics. In such cases, we often replace the distributed load with its statically equivalent resultant force. So how can we do that? First, we need to consider what a distributed load is. The first type of distributed load we will look at is a surface distribution. Here, the entire force is distributed over an entire surface area, resulting in a pressure distribution, where we use a lowercase p to denote the pressure. And this pressure will have the units of force per unit area, or Newtons per meter squared. If we zoom in on a small patch of that area that we will denote as dA, we see that if the area is small enough that we can consider the pressure p to be constant over that small area, then the resultant force dr acting on it would simply be p times dA. If we wanted to know the resultant force acting over the entire surface, we would need to sum up all of the smaller resultant forces over the surface of the wing, resulting in the integral of p times dA over the entire surface of the wing. Now, don't worry if that went a little fast for you. We'll examine those concepts more closely on the second type of distributed load that you will more commonly find in this course, the line distributed load. Here we have the lift on a wing approximated as a one-dimensional distribution over the span of the wing. It is one-dimensional as it only varies along one dimension, the span of the wing, or x. It is technically still a pressure distribution, so it can have the units of force divided by area, or Newtons per meter squared. However, it is more common to replace the pressure distribution with a line load, where, in this case, the pressure distribution is multiplied by the chord length of the wing at every location x. The resulting distribution, denoted with a lower case w, results in a line load, w of x, with the units of force per unit length, or Newtons per meter. Now that we have defined our generalized line load, let's take a look at some resultant force systems. For this analysis, we will look at one wing of the aircraft where x is the distance from the root of the wing, as shown here. We will start by determining the resultant force couple acting at the root of the wing where x is equal to zero. In order to do this, we will first look at a small infinitesimal force that we will call dr. dr will be acting at a distance x, and it is an infinitesimal force caused by a small slice of our distributed load, dx. So just like previously, where dr was the pressure times the area da for a line load, dr will be our distributed load, w of x, times the width, because it's one-dimensional. The infinitesimal moment due to this infinitesimal force, we will denote as dm sub-zero, and it will be equal to our distance x to our infinitesimal force times dr. We can integrate this infinitesimal quantities along the entire length to get the total resultant force and resultant couple acting at x equals zero or point o. We can perform that integration quite easily, or transform that into an integral form quite easily, where the resultant force r is simply the integral of dr over the entire length where dr was w of x times dx. Similarly for our resultant moment, it is going to be equal to the integral of our infinitesimal moment over the length of the wing, which is the integral of x times w of x dx. So this gives us our two generalized integral equations for our resultant force couple at the root of the wing. Alternatively, we could also analyze the system to find the equivalent resultant force. The magnitude of the force will be the same as in the previous analysis. It will be given by the integral of the distributed line load over the span of the wing. However, this force will be offset from the root of the wing by a distance x bar. As this force system will be equivalent to the previous force couple system, then r times x bar will equal to m sub zero that we previously calculated. Expanding this, we can get that x bar is equal to the integral of x times w dx over the span of the wing divided by the integral of w of x times dx over the span of the wing. This gives us our two generalized integral equations for the resultant force of a distributed line load. With the previous equations, you can reduce a distributed load to an equivalent resultant force couple or a resultant force. This is very useful in a lot of engineering analyses where we are applying Newton's second law and need to sum all the forces acting on a body that we consider to be rigid. However, it is not appropriate for all engineering analyses, particularly when it comes to how load is transmitted through a structure and the required strength and stiffness of a structure. We can see such a case in the structural test of the Boeing 777 aircraft shown here. In order to replicate the structural response of the wing, the lift needs to be applied as a distributed load. This is done with a test fixture known as a wiffle tree that distributes the load over the load carrying portion of the wing. Here is a more simplified version of a wiffle tree for a simple beam. A single force R is applied at the top of the fixture as shown here. And then through a series of levers, this force is progressively distributed into smaller forces F1 through F5 which better replicate the distributed nature of a distributed load. Although this distributed load is needed for the structural response, it still reduces down to that single equivalent resultant force R which is applied at one point. Now don't worry too much if this concept of when a resultant force is appropriate or not is not too intuitive yet. We will revisit it later in the course when we examine internal force diagrams for structures. For now you can focus on being able to determine the resultant force or force couple for a distributed load.