[MUSIC] Okay, let us look one more example, the example of frame where we assume that we know the actual shear and bending moment diagrams. In the neighborhood of a node, of a joint, first examine the neighborhood of the joint. So here for example, we can see that the axial force along this member is -10. Along the vertical column, that is -125. We do not know the value on the, The right beam. And that's our task to deduce what this value is. Similarly for the shear force we assume along the horizontal. Remember the value just before the joint is -50 along the column. Just before the joint is -80. We do not know what the value is for the right beam. That means the value right here. And similarly for the bending moment -220, the bending moment at this point. -65, the value of the bending moment at the top of the column. And we do not know, we will deduce, we will find, we will calculate what is the value of the bending moment just to the right of this joint. Now to do that all we need to do is draw a free body diagram, Of the joint. Of this joint. And we are going to show clearly based on the information that's already given here, the axial, the shear and bending moment acting at these two points, At the point of the beam just before the joint and the point of the column just before the joint. And we'll calculate the values at this point. All right, so let's start with the actual force, what is given? It is given that the axial force is -10. So it means, minus implies that we have a compressive force of 10. Similarly, along the column we have a axial force of minus 125. So, we'll draw here 125 compressing. This value is unknown, okay? So we assume that we have some value here, N. Let's assume it's positive, that we do not know. We will calculate that. Let's draw all the shear forces. The shear force at this point is minus 50. Minus 50 means negative. Shear force means on the right face which is this, the force is downward so we are a force of 50 like this. Along this member it is -80. So it means on the right face it is again downward, so we have force of 80. And here we assume that we have some sheer force we do not know. We will calculate that and we assume also that this positive, positive on the left face, should be drawn downwards. So I will draw v like this. And finally we look at the bending moments. The value of the bending moments is -220. But we don't pay attention to the sign. We don't care that the sign is minus because we said for bending moments the sign actually depends how we look at the beam. For beams it's not so much of a problem but for vertical or for inclined members it is a problem. So the important thing is that we have plotted at 220, value of bending moment, on the top side of the beam. That means the top side of the beam is under tension and that implies that the bending moment here along this beam is in this direction pulling the top. So -20 means a bending moment that pulls the top side of the beam. Now for this vertical member we see the value is minus 65. Again, we don't care about the sign because in this case we will look at the beam like this. But the important thing is that the value is 65. The diagram has been plot on this side. It means that this side of the beam is under tension. It means that the bending moment must be 65 in this direction. And finally assuming that we have a positive bending moment I will draw the bending moment here, As M. Now we simply need to apply the equilibrium equations for this joint, which clearly is in equilibrium. And what we get is that from the equation we get that 10 + 80 + N, these three horizontal forces should add to 0. So from here we get that N is minus 90. We get from the vertical of Y, we get that, -50 + 125 upwards, -v downwards, should be equal to 0. From here we get that V = +75. And finally from the equilibrium of moments, sum of the moments about this point, this distance assumed to be very small. That has been exaggerated here so there is no moment due to this distance, this is zero. So the sum of the bending moments are +220, +65, And +M, All in the counterclockwise direction is equal to 0. And so from here, we get it M = -285. Now, with this values, we go back to the original drawing. Drawing of these diagrams N is minus 90. So it means we must plot it on the bottom side. But the important thing is that we write here the value -90, okay. We know the value for sure at this point. We do not know how it will vary along the member. Let's assume it is a constant since in general, usually there are no horizontal distributed laws or laws I think along the beam. In this case, in the case of the shear force, the shear force was +75. So as we're looking at this beam, I'm going to plot +75 on the top. And of course what will happen after that is unknown. We thus can deduce the value at this point, but let's say it's something like this. And in this case, in the case of the bending moment, we saw that the bending moment was, -25, was in this direction. So it was pulling the top side. So we must plot something on this side. Maybe something like this, where this value is -285. So the bending moment is plot on the top side because the bending moment is in this direction pulling the top side. So we can see that when it comes to calculate the axial force, the axial force can be only obtained considering both the axial force diagrams and shear force diagrams since in the horizontal reaction we have an axial force contributing and a shear force contributing to the equation sigma fx = 0. And therefore the value of n depends on both the axial force of the beam and the shear force on the column. Similarly for the value of the sheer force it depends on the shear force of these beams and the axial force along this column. While in the case of bending moments the bending moments depend only on the bending moments of the contributing members. So once we know the bending moment at two of the members and we are missing, let's say, the bending moment at the remaining member we can just simply consider the equilibrium. Sum of the moments to be equal to 0 to deduce the value of the missing bending moment at this point.