 Today, we're going to look at one of the most important concepts in structural design, the concept of a load path, and how we can visualize it using internal force diagrams. To do this, let's start with a bit of a thought experiment. Are these beams equivalent? Intuitively, we may say that we expect these beams to behave quite differently, but what happens if we start to analyze them using the methods we have learned so far in this course? We have learned about making free body diagrams, so let's remove the supports and replace them with possible reaction forces. We also learned that we can calculate what these reaction forces are using equilibrium. If we evaluate equilibrium, we will find that the reactions are the same for both of these beams. As this is a simple load case, I'll leave it up to you to verify this result, but the result remains. The reactions are the same. Does this mean that the beams are the same? Well, not entirely. We would say that the loading on the beams is statically equivalent. They produce the same resultant force, which leads to them having the same reaction forces and moments. But we intuitively know that something must be different. We can validate our intuition by looking at how we test the wing of an aircraft. Here you can see the loading and unloading of an aircraft wing during testing. Where a very complicated load introduction system, known as a wiffle tree, is used to distribute the load over the entire wing. If a point load really would be equivalent, then we wouldn't go through all the trouble of using a wiffle tree. So what is the difference? The answer to this is related to the concept of a load path. So let's start by defining this. The function of a structure is to carry and transmit loads. Where you can view a structure as a connection between externally applied loads and supports or boundary conditions. The load path refers to how that load is transmitted through the structure. You can think of this as if the structure was a maze. The load will flow through the structure, depending on its design, until it reaches the boundary condition. We have actually already looked at this concept indirectly in our analysis of trust structures. Producing the internal forces in the trust members is the way in which we analyze the load path in trust structures. Take for example the following simple structure. We can analyze two separate load cases for the structure which cause it to behave as a trust structure as the loads are applied at the joints. We can easily see from the symmetry of the problem that both load cases are statically equivalent producing the same reaction forces at the removed supports in our free body diagrams. The calculation of those reaction forces is quite straight forward so I'll leave it up to you to verify. To analyze these load cases we would first identify the zero force members. If I fade away these members you are left with a partial view of the load path through the structure. Next we would calculate the internal forces in the remaining members using the method of joints or sections for which I will only show the final results. From this view we can clearly see that although the load cases are statically equivalent they have very different load paths. This can have a significant impact on the design of a structure in terms of deciding on the necessary strength of each of the trust elements for all the relevant load cases. But this is for a trust structure that is comprised solely out of two force members. How could we go about analyzing and visualizing the load path in more complicated structures such as beams and frames? Let's see how this would work for our uniformly loaded cantilever beam. We have already drawn the free body diagram and determined the reaction forces acting at the built-in support so we'll start from there. What we want to do is examine how the distributed load flows through the structure to the support reactions. To do this we need to cut into the beam to expose the internal forces and moments. We will do this at an arbitrary distance zed along the beam. Since we cut into the beam we need to think about the possible internal reaction forces that could be exposed at the cut surface. You can approach this the same way we did for determining the reaction forces at supports by examining the degrees of freedom of the beam that are constrained. As the two sides of the beam cannot translate or rotate relative to each other we need to add a normal force, shear force and bending moment in order to properly constrain our segment of the beam. As you will learn later on I am drawing my internal forces and moments in a particular direction but don't worry about that for the time being. With our new free body diagram complete we can now calculate these internal loads in the beam using equilibrium. Summing the forces in the zed direction we see that the internal normal force has to be equal to zero for the body to be in equilibrium. Summing the forces in the vertical or y direction we get the following expression. We need to remember that our distance zed is not a fixed value in this expression so if we rearrange it we can obtain an expression for the internal shear force within the beam as a function of zed. We need to be careful of the units in our expression so I'll make a little note here that the results will be in kilonewtons when zed is in meters. Finally we can sum the moments about the section plane taking the counterclockwise direction as our positive reference direction. By choosing this point to sum our moments the internal normal and shear forces will not contribute to the summation. Thus we end up with negative mx as it acts clockwise not counterclockwise being summed with the moment caused by the resultant force of the distributed load 2 times 10 minus z acting at a moment arm distance of 10 minus z divided by 2. If we rearrange this we can then obtain an expression for the internal bending moment as a function of z which will have the units of kilonewton meters when zed is in meters. We can take these three functions we have calculated and plot them as functions of zed as shown here. We call these diagrams the internal force diagrams and they help us visualize how the externally applied loads are transmitted through the structure in other words the load path through the structure. We can see the different demands placed on different parts of the structure which could influence the design and sizing of that part of the structure. We can visualize this better by looking at the difference between a point force and a uniform distributed force acting on a wing. Here we have the internal force diagrams for a uniform distributed load we just determined on the left as well as the internal force diagrams for statically equivalent point force on the right. By comparing the two sets of diagrams it becomes clear why we can't simply test a wing of an aircraft using a point force. In the point force case the free end of the beam is completely unloaded while the internal shear and bending moment carried by the clamped half of the beam differs from the distributed case except for precisely at the fixed boundary condition. The demands on the structure are quite different in these two load cases and would have consequences on how the structure would be designed and reinforced along its length. Hopefully this illustrates how important visualizing the internal loads within a structure can be in the design of engineering structures. You will see throughout your engineering career that a lot of time and effort is spent in understanding the load path through a structure and we will certainly be spending a lot of time in this course practicing how to do so.