[SOUND] Welcome back to module 21 of Mechanics of Materials, Part I. 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 sheer stresses occur. Last time we talked about the maximum normal stresses. This time we're gonna talk about the maximum sheer stresses. And sometimes as we get involved with these mathematics, I often say we get lost in the trees rather than seeing the forest, and so we're gonna 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. Cuz that could indeed cause the member to fail, or we could look at the performance of the member. And so we'll continue on to try to find now the angle for a known 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 transformation equation equal to 0. That's gonna give us the max and the mins. And so I want you to do that on your own, come on back and see how you did. And so here is the result. If I take the derivative of this first term, the derivative of sin 2 theta is 2 cos theta. So that's gonna cancel out the 2 in the denominator there. When I take the derivative of cos 2 theta I get negative 2 sin 2 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 sheer stress and we found out that this was an expression for the angle where the sheer stress maximum occurs. You'll recall that this was the expression where the theta sub P, the angle, was defined to the principal planes. And you'll notice that these are negative reciprocals. So therefore, we know that 2 theta sub P and 2 theta sub 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. If you look at tangent of 80 degrees, that ends up equaling 5.67. And then if I take the tangent of -10 degrees, that ends up being -0.1768. So if I take the negative, the reciprocal of one of these, so let's take 1 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 2 theta sub P and 2 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 stresses occur are 45 degrees from the principal planes. And so, similar to before with this relationship for the angle to the maximum sheer stress, we can set up sines and cosine functions, substitute them into my sheer 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 sigma sub 2 from sigma sub 1, what I get is this value here. And we can see that that underneath the radical sign, underneath the square root sign, is equal to tau sub MAX up here. And so we find tau sub MAX is sigma sub 1 minus sigma sub 2 over 2. Where the maximum in-plane sheer stress equals one half the difference of the two in-plane principal stresses. A very important result. And so here's a worksheet, and with this worksheet, I'd like you to find the principal stresses. Remember, we're gonna turn, we're gonna find the angle to the principal planes where we have principal stresses. These 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. [NOISE]
