[MUSIC] This is Module 39 or Mechanics of Materials I. Todays Learning Outcome is to show that the Young's modulus, the Shear modulus and Poisson's ratio are related for isotropic material. I have said this in earlier modules and I told you that we would prove it. And today is the day when we'll show that this is actually true. And so recall the case of pure shear from earlier in the course. And by equilibrium we found that the shear stress xy was equal to shear stress yx. And so the shear stress all around was the same. And we found that Hooke's Law in Shear was valid in the linear elastic region and it was tau equals, tau being the shear stress, is equal to the modulus of rigidity or the shear's modulus times the shear strain. This is completely analogous to the normal stress equal Young's Modulus times the normal strain. And so again, G is define as the module of rigidity or the shear modulus. And so let's now, for this case of pure shear, draw more circle for stress. And so for the horizontal phase, I've got a clockwise rotation. So that's going to be 0 and positive tau xy. For the vertical phase, I've got 0 for my normal stress and negative tau xy for my shear stress. And so this is my Mohr's circle, and I see for this case of pure shear that a 45 degrees rotation, 90 degrees rotation on Mohr's circle or 45 degrees rotation on my stress block, I will end up with my principle stress of one which I'll call sigmus of P. And minus sigmus of P 0 for the second principle stress. And so they're equal. So what I see for this case, for the pure shear case, is that tau xy is equal to my max sheer stress, and it's the same magnitude as my principle stress. And so here is Mohr's Circle for stress. If I turned 45 degrees, my stress block, I would end up with my principle stresses as shown. I can now draw Mohr's Circle for stain for this stress block. And so for Mohr's Circle for strain I've got, let's call this the new X direction, I've got the principle stress sigma sub P. It's going to give us a principle strain epsilon sub P. I've got a negative epsilon sub P, 180 degrees or 90 degrees on my stress block, and then I can draw Mohr's Circle and I find out that my max sheer strain divided by two is shown here and here. And I find out that the max sheer strain divided by two has the same magnitude as the maximum, or the principle, normal strength, epsilon sub p. So here are our consolidated results, Mohr's circle for stress, Mohr's circle for strain. This is Hooke's Law for shear. If I put in values of max, I get maximum shear stress is equal to the shear modules times maximum shear strain. Or the shear modulus is times maximum shear strain is equal to two times the maximum normal strain. And so, let's go back and recall Hooke's Law, generalized Hooke's Law for a Biaxial Stress-Strain that we developed last time. Here was the relationship for sigma sub x, and let's called sigma sub x sigma sub P, and so we end up with this relationship. But we know that sigma sub P is the same as tau MAX. And we also know that tau MAX is equal to 2G epsilon sub P. On the left-hand side, I've done a little bit of a mathematical manipulation. In the numerator, I've factored out epsilon sub p, in the denominator I've take one minus the Poisson's ratio squared and factored it into one minus Poisson's ratio times 1 plus Poisson's ratio. Then then 1 minus Poisson's ratio cancel in the numerator and the denominator. And epsilon sub ps also cancel and I get a relationship that Young's Modulus is equal to 2 times 1 plus Poisson's ratio times the shear modulus. And so they are not independent. All of these are related for isotropic materials. And we did this for the case of pure shear, but we know on Mohr's circle we can rotate for any other condition of stress and strain. And so for isotropic materials, this always holds true. E, G, and Poisson's ratio are related and not independent. And so that's it for this time. And we'll see you on the next module. [NOISE]