[SOUND] [MUSIC] Hi, and welcome back. In today's module, we're going to continue going through Transverse Shear Stress. And we'll work through an example. The learning outcome for today's module is to understand how to calculate transverse shear stress in a fairly simple object with a straight forward cross section. Again this is intended to be a review, so if you've never seen transverse shear stress before, you can go back and check out these modules by Dr. Whiteman here. Okay so last time we learned that the formula for transfer shear stress is tau is equal to VQ divided by ib, and now we're going to apply that formula to this example problem. You can see here we have this rod OA which is attached to a rod AB and what we're trying to do is find the transverse shear stress at this point, C, which we're assuming is at the very front of the beam. Acting at equal distance from the top and the bottom of the rod, along the neutral axis which would be in the XZ plane. Let's write out our formula, so our transfer share stress is equal to tau which is equal to VQ over Ib. And transverse shear is typically from a load that causes bending. In this case we have this axial load F that's not going to contribute to our transverse shear. And then we have this load here P that is pushing down on AB which causes the rod AB to push down on rod OA. And so, in the module where we discussed on bending stress, we drew a free body diagram for the whole system and we figured out that rod OA essentially has a force P from rod AB pushing down here, and then a reactionary force up at the wall, and a moment. And then we go forward and draw the shear diagram, the shear force diagram, what we see is our shear force throughout the beam is equal to 500 newtons. We know in this equation that V is going to be 500 newtons, and now we just need to figure out the other components in this equation. Let's give me some more space to work with here. We have V is equal to P, which is 500 newtons. And then I, I is going to be the moment of inertia of this rod or cylinder, which is going to be pi D to the 4th, over 64. I'm going to draw the cross section of the rod OA and here is our point C which is acting along this neutral axis. And if you recall B is the width of the beam at the point of interest, which will be right here. B Is equal to D. And now we have to figure out now is Q, which was, A prime, y prime. Where A, is the area above, the point at the level at the point of interest. A is going to be this cross sectional area, right here, which is going to be one half of pi d squared divided by 4 or pi d squared over 8. And y prime is going to be the distance between, if this is our centroid of that cross sectional area here, it's going to be the distance between the centroid and the neutral axis, and if you, you can go ahead and either derive the centroid if you want, or you can look it up, and what you'll find is the centroid of the half circle is 4d over 6 pi. If we plug all of this into our equation, what we'll see is tau is equal to V [SOUND], times Q which is pi d squared over 8, times 4d over 6 pi. [INAUDIBLE] [SOUND] Which is over I which will if we take the inverse of I will be 64 divided by pi d to the 4th. And if we take the inversive of P, we get one over d. [SOUND] And so when we break this down, what we end up with is, tau is equal to 4 times 64 times v, divided by 8 times 6 times pi d squared. And what we can break this out into is actually combined pi D squared and this 4 and that will give you the inverse of the area, so we can end up with 64 divided by 48 times V over A which boils down to 4 3rds V over A [SOUND]. And if you go into any machine design textbook, you'll often see this shortcut equations for cross sectional areas, or I'm sorry, for transverse shear at the maximum point. At point C our transverse shear is going to be maximum, because we're right at that neutral axis right here. this is actually one of those shortcut equations where if you're at the neutral axis, your transverse shear in a cylinder is 4 3ds V over A. If we go ahead and we calculate four thirds V over A what we will find is 4 times 500 Newtons divided by 3 times pi times 0.04 divided by 2 squared which is going to be right around a half of megapascal. And the distribution for this [SOUND] stress would look something like this. If we're right here at this cross section, what we would see is, you would see really low stresses at to zero at the exterior, and then you'd see really high stresses at the neutral axis and it ends up looking something like that. There's also a short cut for a rectangle. If you're looking at the maximum stress or transverse shear at the neutral axis, what you'll see, again, your stress is going to be max at the neutral axis. You end up with this almost circular shape on the transverse shear at the neutral axis the max is 3V over 2A which you guys could derive out on your own time for fun, [SOUND]. Okay so that's it for today, and that concludes the module on transfer shear. I'll see you next time. [SOUND]