Calculating how much weight a tree can hold (compressive stress and buckling)


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.

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