[MUSIC] Hi, and welcome to module 13 of Mechanics of Materials part 1. Today's learning outcomes are to define and calculate something we're gonna call Poisson's ratio. And then to define homogeneous materials and isotropic materials. So, Poisson's ratio was named after Simeon Poisson who was a French mathematician and physicist and it applies for uniaxial stress and strain. This is the affect of Poisson's ratio or the Poisson effect if you will. Here's a material that's got very little stiffness and with very little stiffness we can see the effects much easier because I can apply forces that make the effect take place. And so when I put a tensile force on this, we see that the specimen becomes thinner, okay. Not as wide. And likewise, if I take a very soft material And I put it into a compressive force. We can see that it goes ahead, and it gets wired. And so that's Poisson's effect. Here's a graphic of Poisson's effect. And Poisson's ratio. What we look at is the lateral strain which is the change in the lateral dimension over the initial lateral dimension. In this case I am showing it as width. And the longitudinal strain is the change in the longitudinal direction or delta longitudinal divided by the original. And then Poisson's ratio is given the symbol nu and it's equal to minus the lateral strain over the longitudinal strain. And there's a couple of assumptions here we make. One is that it's homogeneous material, or the same material throughout. And the the other is is that it's isotropic which means it has the same material properties in all directions, okay? An anisotropic material would be something like carbon fiber which would be stronger in the directions of the fiber than against. Or grains in wood cause wood to be anisotropic in certain directions. And maybe another common example is a firehose. If you've ever looked at a firehose it's got fibers in it which make the material properties different in one direction than other directions. So, let's go ahead and take Poisson's ratio now and let's look at an example. And so this is the campus recreation center here at Georgia Tech. And let's say that we took a structural member of that building and we want it to go ahead and test it and find out what Poisson's ratio is. It's going to be a rectangular steel alloy and so it's going to have a cross section of 25 millimeters by 50 Millimeters. And it's going to be 2 meters long. And so this is not to scale, but this dimension for the length would be 2 meters. And we're going to calculate Poisson's ratio, so Poisson's ratio is defined up here. We have nu equals now minus the lateral strain. And for the lateral strain, we're told that it's subjected to a tensile force of 40 kilo Newtons. And the test section's gonna elongate by 2 millimeters. And it's gonna narrow in the width direction by 0.014 millimeters. And so for the lateral dimension it's going to be, since it's shrinking or becoming thinner, it's going to be negative 0.014 millimeters. Over the original width, which was 50 millimeters. And that's gonna be divided by the longitudinal strain, which is 2 millimeters over the original length, which was two meters or 2,000 millimeters. And if we calculate that out it becomes 0.28 is our Poisson's ratio. Let's proceed now with part two, which is to find the new height of the cross section. We found Poisson's ratio based on what happened to the width and the length so now let's use the information given to find the change in height. So, the Poisson's ratio here was found to be 0.28 and that's equal to the lateral strain which is minus delta lateral over the original dimension, for the lateral dimension, we had used width, but now we're going to use height, which is 25 millimeters. And again, we're gonna divide that by the longitudinal strain which was two millimetres. Over 2,000 millimeters. And we can solve then for D lateral. And it's not D lateral, delta lateral. And delta lateral becomes minus 0.007 millimeters. And which that now means that we can find the new height. We know that the height shrinks by .007 millimeters because it's a negative number. And so, the original height was 25 millimeters. We subtract the .007 millimeters, and our new height then is 29 Excuse me, 24.993 millimeters. And that's our answer. Okay, so that's an example. Plus ounce ratio. I'd like you to try it on your own now. I have another worksheet for you. And this is a typical bench vice that many of you may be familiar with. And we're gonna put a section of magnesium alloy in it and test it. I give you the dimensions. I give you the compressive force now instead of a tensile force. I tell you what Poisson's ratio is, .35. And I want you, from that information, to find the average normal stress at the instant when the load is applied and to find the height and width of the cross section after compression. And the solution is in the module handouts, and we'll see you next time. [MUSIC]
