 Now that we've looked at idealized structures using the sheet-boom approximation, let's look at some useful simplifications and shortcuts that we can use when considering Q as a constant on these sheets between booms. Let's first look at the resultant shear force on an idealized sheet. Here you can see a generalized curved sheet between 1 and 2 with a boom at 1 with x coordinate x1 and y coordinate y1 within our xy coordinate frame given here and between boom 2 with a coordinate x2 y2. So this is an idealized plate section and because it's idealized we know that the Q12 value shown here is constant and this is a result of our idealization. We moved all of the area of our sheet to the two booms. So now what we want to do is find out what the resultant shear force on the idealized sheet is. To do this let's look at the two components, the x and y component. First looking at the resultant force in the x direction, let's consider a small element delta s which is at angle phi with respect to our xy coordinate system. Now the resultant force acting on element delta s will just be Q1 delta s but we want the x component so we would have to multiply that by the cosine of phi. So we get Q12 delta s cosine phi and we would have to integrate that over the entire segment 1, 2. I can write that down as here except now I've moved the cosine phi before the delta s but the result is the same. So we get Q12 times delta s is the resultant on our element delta s multiplied by cosine phi and we integrate that from boom 1 to boom 2. What we can recognize is cosine phi is actually just delta x. It's the change in x coordinate along delta s. So we can modify this to the constant shear flow Q12 times the integral from 1 to 2 of dx which will just be Q12 times x2 minus x1. We can do the exact same analysis in the y direction and we see that indeed the shear force, resultant shear force in the y direction is just Q12 times the length or the distance between booms 1 and 2 in the y direction. Now if we want to figure out what the total resultant force is we can then use Pythagorean theorem with the x and y component and we get that v is the square root of vx squared plus vy squared. Looking at this we see that the shear flow is constant in both vx and vy and we essentially get the square root of the x distance and y distance squared which will be the linear distance between 1 and 2. So combining these results we get that the x component of the shear flow is just the shear flow times the change in x coordinate from 1 to 2. The y component is the shear flow times the change in y coordinate and the total resultant shear force is just Q12 times the linear distance between booms 1 and 2. Now let's look at the resultant moment of an idealized sheet. We have to do this sometimes when we're doing analysis of closed sections and we have to calculate the moments about a point or for finding shear center in open sections. So we have to do this quite a bit. So we still have constant Q12 due to our idealization but now we want to find out what the moment is due to this shear flow acting between booms 1 and 2. We will again define an element ds and instead of showing Q12 I'm going to remove Q12 and show the resultant force Q12 times delta s acting on this element. This is because for looking at moments we have to multiply the resultant force on that element by a moment arm so it just helps us with the visualization but Q12 is still acting on this entire segment. Now to look at the moments we have to pick a point and so I'll pick an arbitrary point O that we will sum the moments about. So this could be any point but we'll illustrate it here to the right of segment 1, 2. The moment due to the resultant force acting on ds will be the resultant force Q12 delta s times the moment arm which is the perpendicular distance between that resultant force and point O denoted as rho here. So our moment due to Q about point O will be the integral from 1 to 2 of Q12 times this distance rho ds. And rho is going to be variable as we are at different points. The distance is actually changing. However, this integral of rho ds around a curved surface we saw this in torsion of closed wall sections. In fact, we use that to calculate the area enclosed and we use that rho ds is 2 times the enclosed area because the swept area is this triangular area and if we do the distance times the length we'll get a rectangular area. So the resultant of that integral is 2 times the swept area along ds. So if we utilize this what we can actually see is the resultant moment is going to be related to the total enclosed area between whatever point we're taking the moments about and our generalized curved surface. So our total moment due to that shear flow Q is 2 times that enclosed area times the shear flow. And that's it. You can utilize these two simplifications in order to help you with determining resultant forces and resultant moments due to constant shear flows on generalized curved surfaces in a boom sheet approximation of a structure.