All right. So we want to look at owl’s tree. Owl wants to put a library

in one of his trees. So I’m going to model owl’s tree

with a circular cross section. OK. So I’m going to keep it fairly simple. There’s my model for owl’s tree. OK. So I’m going to make it a solid

circular– constant throughout. So here’s my cross

section I’m considering. Owl picked a fairly tall tree. OK. So it’s 35 meters tall. OK. I’ve calculated the self weight. So the tree itself

weighs 85 kilonewtons, so that’s acting downward on the tree. OK. I’m going to assume it has

a diameter of 0.5– oops. Not 0.05. 0.5 meters. OK. So it’s got– it’s a

half a meter in diameter. And I want to make sure

this– well, I want to see how much– hopefully,

this is– the tree is fine for just its self weight, but

we want to see how much extra load he can put on for his library. So what is the extra

load that it can handle? OK. So I’m going to start out

again with compression. So I want to see if it’s OK for

compression and how much extra capacity I might have. So again, I’d 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 I’m going to divide by

the cross sectional area. My cross section is circular. So what? If I calculate the area of that,

it would be pi times r squared. And that’s the part– r is the radius. So this is the radius. So it’ll be half the

diameter, so that’ll be 85 kilonewtons over

pi times 0.25 squared. And that’ll give me

the stress in the tree, and that is 425 kilonewtons

per meter squared. OK. Now, I also need to know the allowable

stress of this tree, of this wood. So we went out 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. OK. Which is much, much greater than the

stress that’s applied with the 85, so I’m going to– we’re

OK 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. And this is a tall tree, so 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. So I’m going to need

those different values. E is the modulus of

elasticity of the tree. For E, I’m going to use 7 times 10

to the sixth kilonewtons per meter squared. OK. For moment of inertia, I have

a circular cross section. So I can calculate moment of

inertia for this cross section, and that is going to be

pi r to the fourth over 4. Again, r is the– r is 0.25 meters. This is the diameter. So that value ends up being

0.0031 meters to the fourth. So now, I have the values I need. L is this 35 meters. So if I want to calculate

the critical buckling load for this column, which is owl’s

tree, it’s pi squared times 7 times 10 to the sixth. That’s kilonewtons per meter squared. Times the moment of inertia. 0.0031 meters to the fourth,

and divide that by L squared. It includes the tall column. 35 meters squared. And I get that the critical buckling

load for this tree is 175 kilonewtons. Now, that’s pretty good. OK. This tree– owl’s tree is

much closer to buckling. So they applied load that’s greater

than the applied load of 75 kilonewtons, but that’s just the weight of the tree. So we want to figure out– can

owl put a library in this tree? So that’s the question. So how much additional

capacity do we have? So if we look at the 75

minus 85, so we have– I’m going to just focus on buckling. So an additional load that I can

put on is going to be 175 minus 85, or I can put on an

additional 90 kilonewtons. So I think that owl’s tree can

handle an additional 90 kilonewtons. He’s going to need some

structure for his library, so I’m just going to make an assumption

that the load for the structures– that would be the wood that

we’d use for the library. Let’s assume that is going

to weigh 30 kilonewtons. OK. That’s going to leave 60 kilonewtons. P for books. The allowable load for

books is 60 kilonewtons. OK. So I went and looked it 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. OK. So if I want to figure out how

many books owl can put on his– in his library– so I

want number of books. OK. 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. So I’ve made a fair amount

of assumptions here. 2,400 books is a lot of books,

but owl does love to read. So we’ll see what happens to his tree.