[MUSIC] Welcome back to module 21 of Mechanics and Materials part one. Today's learning outcome is to find the maximum in-plane shear stress. And so we knew from our stress transformation equations, if we had a loading condition that we could find. We could find the normal and shear stress on any plane. As we turn angles at that particular point. And we said that for all structural machine design it's necessary to know what planes and angles the maximum normal stresses and shear stresses occur. Last time we talked about the net maximum and normal stresses. This time we're going to talk about the maximum shear stresses. And sometimes as we get involved with these mathematics, I often say we get lost in the trees rather than seeing the forest. So we're going to step back here for a second. Remember I have some engineering body, some device or whatever. And I have an external loading condition on it. And I want for any point in that body, on any plane, at any angle to find out where do we have the maximum normal and shear stresses? because that could indeed cause the member to fail or to look at the performance of the member. And so we'll continue on to try to find now the angle for unknown stress block loading condition where the maximum shear stress occurs. And so my question is for you, how should we proceed in doing that? And what you should say is, well, we should set the derivative of the shear stress, a transformation equation equal to zero. That's going to give us the max for the mins, and so I want you to do that on your own. Come on back and see how you did. So, here's the result. If I take the derivative of this first term, the derivative of sin two theta is to cos theta. So that's going to cancel out the two in the denominator there. When I take the derivative of cosine two theta I get negative two sine two theta, and so that's our result. We can express that as a tangent function again, and here it is. And this is the angle now, theta sub x, which is where the maximum shear stress occurs. So here again is our expression for the transformation of shear stress. And we found out that this was an expression for the angle to the, where the shear stress maximum occurs. You'll recall that this was the expression where the theta of p, the angle was defined to the principle planes. And you'll notice that these are negative reciprocals. And so therefore, we know that two theta of p and two theta of s are 90 degrees apart. And just to show that, I've drawn the tangent function here. If I take just a quick example. Let's say we have tangent of 80 degrees. Okay, if you look at tangent of 80 degrees that ends up equaling 5.67. And then if I take the tangent of minus ten degrees that ends up being -0.1768. So if I take the negative, the reciprocal of one of these. So let's take one over 0.1768. We'll take the negative reciprocal of that. And what we find out, is that's equal to 5.67. And you'll see that these angles are 90 degrees apart. That occurs over and over again with a tangent function. And so since two theta sub p and two theta sub s are 90 degrees apart. That means that theta sub s and theta sub p are 45 degrees apart. And so the planes on which the maximum in-plane shear stress has occur are 45 degrees from the principal planes. And so, similar to before with this relationship for the angle to the shear. Maximum shear stress we can set up sine and cosine functions, substitute them into my shear stress transformation equation. And that gives you this expression for the shear stress. And so the maximum shear stress is going to be the absolute value of this or what I've shown here. And so here's the maximum in-plane shear stress. You'll recall from last module, the principal stresses which were the max normal stresses. And if I now take from this expression and I subtract sigmas of two from sigmas of one. What I get is this value here. And we can see that underneath the radical sign, the square root sign is equal to tao sub max up here. And so we find tao sub max is sigma sub one minus sigma sub two over two. With a maximum in-plane shear stress equals one half the difference of the two in-plane principle stresses. A very important result. And so here's a worksheet. And with this worksheet, I'd like you to find the principal stressors. Remember, we're going to find the angle to the principle planes where we have principle stresses. This are the normal stresses where shear stresses go away. And then, you're also going to find the maximum in-plane shear stress. And when you do that I've got the solution in the module handouts. [MUSIC]