We want to look at Owl's Tree. Owl wants to put a library in one of his trees. I'm going to model Owl's Tree with a circular cross section. I'm going to keep it fairly simple. There's my model for Owl's Tree. I'm going to make it a solid circular constant throughout. Here's my cross section I'm considering. I will pick a fairly tall tree. It's 35 meters tall. I have calculated the self-weight. The tree itself weighs 85 kilonewtons. That's acting downward on the tree. I am going to assume it has a diameter of 0.5 meters. It's a half a meter in diameter. And I want to make sure, well, I want to see how much, hopefully the tree is fine for just itself weight, we want to see how much extra load he can put on for his library. What is the extra load that it can handle? I'm going start out again with compression. I want to see if it's okay for compression and how much extra capacity I might have. Again, I compute the compressive stress and that is going to be force over cross-sectional area. In this case, my force so far is 85 kilonewtons. That's just from the self-weight of the tree, and then I'm going to divide by the cross-sectional area. My cross-section is circular. If I calculate the area of that, it would be Pi times r squared, r is the radius. This is the radius. We have the diameter. That will be 85 kilonewtons over Pi times 0.25 squared. That will give me the stress in the tree, and that is 425 kilonewtons per meter squared. I also need to know the allowable stress of this tree, this wood. We went on and tested Owl's Tree and we know that the allowable stress for the wood for his tree is 30,000 kilonewtons per meter squared, which is much, much greater than the stress that's applied with the 85. That's okay, in compression. I'm not worried about Owl's Tree in compression. But we also need to check buckling. Let's switch here, and look at buckling. This is a tall tree. I expect buckling might be a bit of an issue, so, I will need to calculate the critical buckling load. That again is Pi squared EI over L squared. I'm going to need those different values. E is the modulus of elasticity of the tree. For E, I am going to use 7 times 10 to the sixth kilonewtons per meter squared. For moment of inertia, I have a circular cross section. I can calculate moment of inertia for this cross section. That is going to be Pi r to the fourth over 4. Again, r is point 25 meters. This is the diameter. That value ends up being 0.0031 meters to the fourth. Now I have the values I need, L is this 35 meters. If I want to calculate the critical buckling load for this column, which is the Owl's Tree, it's Pi squared times 7 times 10 to the sixth, it's kilonewtons per meter squared, times the moment of inertia, 0.0031 meters to the fourth, and divide that by L squared, includes tall column, 35 meters squared. I get that the critical buckling load for this tree is 175 kilonewtons. Now, that's pretty good. Owl's Tree is much closer to buckling, so the applied load that's greater than the applied load of 75 kilonewtons, but that's just the weight of the tree. We want to figure out, can Owl put a library in this tree? That's the question. How much additional capacity do we have? If we look at the 75 minus 85, so we have, I am going to just focus on buckling. An additional load that I can put on is going to be 175 minus 85, or I can put on an additional 90 kilonewtons. I think that Owl's Tree can handle an additional 90 kilonewtons. He's going to need some structure for his library, so I am just going make an assumption that the load for the structures, that would be the wood that we would use for the library. Let's assume that is going to weigh 30 kilonewtons. That's going to leave 60 kilonewtons. P for books, the allowable load for books is 60 kilonewtons. I went and looked up, I tried to figure out how much one book would weigh. One book only weighs, it's very small, 0.025 kilonewtons, is a fairly big book. If I want to figure out how many books Owl can put in his library, so, I want number of books. That is going to be my allowable load, 60 kilonewtons divided by the weight for one book, 0.025 kilonewtons per book. I calculate that he can put 2,400 books in his library. I have made a fair amount assumptions here; 2,400 books is lot of books, but Owl does love to read. We'll see what happens to his tree.