 Now that you know how to determine the internal forces within a beam using equilibrium, you might be asking yourself, isn't there some way to generalize the effect of different loads on the internal forces within a beam to make them easier to determine? We are engineers after all, and engineers like to be lazy. I mean efficient, yeah, yeah, that's the word I was looking for, efficient. Speaking of efficient, this lesson is going to be quite derivation intensive. So I think I'm going to farm this lesson off to some artificial intelligence. Artificial, eh? I'm going to pretend that you didn't say that. But back to the lesson. How can we generalize the internal loading within a beam? Consider a beam in two dimensions with load acting in the ZY plane. This beam can be subjected to distributed loads, point loads, and moment couples along its length, where the location, distribution, and orientation of these loads can all vary. What we would like to do is look at a small element of this beam with an infinitesimal width dz, and observe how different loads acting on the element will affect the variation of internal forces across it. We will annotate this on our element by stating that the internal forces change by increments dn, dv, and dm when traveling in the positive Z direction. To make it easier to visualize, we can break up these generalized loads into components parallel to the beam, which we will call lateral loads, and components perpendicular to the beam, which we will call transverse loads. We include the moments in our transverse loads as the vector direction of the moment couples are indeed perpendicular to the beam. Also note that we are drawing all applied loads acting on the beam in the positive coordinate directions and that the relationships we derive will be dependent on this convention. Now, all that we have to do is examine how each type of load affects the internal forces. We will start by analyzing the lateral loads, and specifically with a lateral point force acting in the center of the element. We can determine its effect on the internal loading by recognizing that if the beam is in static equilibrium, so too must the infinitesimal element of the beam. Starting with force equilibrium in the Z direction, we will choose forces acting to the right as our reference positive direction. As a result, the normal force acting on the left-hand side of the element will be negative, while the applied point force, PZ, will be positive, as well as the N plus DN internal force acting on the right side of the element. Here we can see that N cancels out in our equation, leaving us with the result that the variation in the internal force, DN, is equal to the negative lateral point force, PZ. We can do the same for force equilibrium in the Y direction. Here we obtain that the internal shear force V acting on the left side of the element has to be balanced by V plus DV acting on the right side. This results in the variation of the internal shear force, DV, being equal to zero. If you think about the nature of a lateral force acting on the beam, this is logical, as we expect that it will not cause the beam itself to shear. Finally, we can also evaluate moment equilibrium of the element. We will sum our moments about point A acting at the center of the left-hand side of the element with counterclockwise as our positive reference direction. Most of the forces acting on the element pass through point A, thus we are left with the following for our moment equilibrium. We can see that our moment M will cancel out, and we can also use our earlier result of DV equal zero to further simplify this, leaving us with the result that DM divided by DZ is equal to V. This result is interesting as it is not actually dependent on our lateral force PZ, which is logical for the same reason that DV was equal to zero, but it does indicate that the rate of change of the internal bending moment in the Z direction, DM by DZ, is equal to the internal shear force. We will see for other forces that this result will continuously pop up and is in fact generally true for a beam. Let's quickly summarize what we have observed so far. We have shown that the rate of change of the internal bending moment, DM divided by DZ, is equal to the internal shear force at that point. Furthermore, we saw that the change in internal normal force, DN in the positive Z direction, is equal to the negative of the point force itself. If we take the limits of this differential term as the width of the element approaches zero, we see that this simply becomes delta N is equal to negative PZ. Next, we will examine the influence of a lateral distributed load on the element. Here we see our element with that distributed force. We need to recognize that our element is infinitesimal in width, meaning it is actually a point. So we can treat the distributed force as a constant value acting at that point. With this established, we can repeat our equilibrium analysis of the element. First, looking at the sum of the forces in the Z direction, we see that the normal force N acting to the left must be balanced by N plus DN acting on the right, as well as WZ times DZ also acting to the right. Canceling terms and rearranging, we get the result that DN is equal to negative WZ times DZ. For the sum of the forces in the Y direction, we see that WZ does not act in the Y direction. So this would result in this same equation as for the last load case. So we'll simply write the final result from the previous load case. The same would occur for some of the moments about point A. WZ, N, N plus DN, and V all act through point A. So we would get the same moment equilibrium equation as in the last load case, which would give the result DM divided by DZ equals V once again. Thus we only obtain one new result from this load case. Let's add this case to our summary board before moving on to the next load case. The result for the lateral distributed force gives us a relationship between the internal normal force and the lateral distributed load that is dependent on the width of the element. Thus we can express this as a rate of change of the normal force being equal to the negative of the lateral distributed load using the directions we established as positive for our applied loading. Let's quickly look now at the transverse loading cases. We will examine all of the transverse load cases before summarizing the results. We will start by examining equilibrium for a single transverse point force. Looking at a force equilibrium in the Z direction, we would expect that the internal normal force would not be varying along the width of the element, as there is no applied force acting in this direction. This is indeed what we see when we simplify and see that DN is equal to zero. Looking at force equilibrium in the vertical direction, we obtain a simple expression that when simplified shows that dV is equal to the negative of the applied transverse point force. The next load case is a transverse distributed force. As we reason for the lateral distributed force, the distributed force can be considered as a constant value as the width of the element is infinitesimal. Establishing this, we can once again look at equilibrium of the element. In the horizontal or Z direction, nothing changes from the previous load case, as the distributed force does not act in this direction. So we can simply repeat our result of dN is equal to zero. In the vertical direction, we will have a contribution of the distributed force on equilibrium. It will produce a resultant force of wY times dz. So the expression will be similar to the previous load case, but with the point force pY replaced with the resultant of the distributed force wY times dz. Simplifying this expression, we will obtain the result that dV divided by dz is equal to negative wY. Finally, recognizing that the resultant of the distributed force will act through the center of the element, if we sum moments around the point in the center of the element, we will obtain the same result as the previous load case. Almost done, one more load case to look at. For a point moment, we will consider a positive applied point moment as acting in a counterclockwise direction. For this load case, we do not need to evaluate equilibrium of forces. We can actually reason that an applied moment couple will not change the internal normal and shear forces. Thus dV and dN will both be equal to zero. We thus only need to look at moment equilibrium. Once again, choosing our summation point as being the center of the left side of the element, we can produce the following expression for moment equilibrium using counterclockwise as our positive reference direction. Here, the m terms cancel and we can replace dV with the result that it has to equal zero. Simplifying, we get the expression dM is equal to negative m0 plus V times dz. In order to isolate the m0 term, we can take the limit of this expression as dz approaches zero, whereby dM becomes delta m, which will be equal to the negative of the point moment couple. Let's add the results from all of our transverse loads to our summary board. The new result from our lateral point force was dV is equal to negative py. In the limit as dz approaches zero, this simply becomes delta V is equal to negative py. From our lateral distributed force, we obtained dV divided by dz is equal to negative wY. And finally, from our point moment, we obtained delta m is equal to negative m0. Looking closely at these results, we can see two types of variations. We see that the rate of change of internal forces are caused by distributed forces, while point forces and moments cause step changes in the internal forces and moments. It is good to view the results in this manner, as you will see later, that we can exploit this to very quickly and easily plot the internal force diagrams. However, we will cover that in another video. For now, I would recommend that you take a look at some of the internal force diagrams we have already generated earlier in this course, and see if you can recognize the presence of these relationships within them.