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PROFESSOR STRANG: OK, so
this is the second last

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lecture on trusses.
 

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Then we've got a
holiday on Monday.

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And then after that we'll
be into Chapter 3.

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I thought I'd write down just
in case it's use to you, the

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four problems that I intend to
include with the next homework.

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That won't be due for quite a
while, a week from Monday.

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So these will be the problems
on trusses that come from

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particular trusses
drawn in the book.

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And then there'll be some
problems from the new material,

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that we do next week.
 

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So trusses and really, there's
two main jobs for today.

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One is to identify this
matrix A, the strained

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displacement matrix or
the stretching matrix.

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How far do the bars stretch?
 

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Everybody remembers A is going
to come in this step if we have

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displacements then of the nodes
like this would be like a u_1,

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this would be a u_1 h and a u_1
v, this would be a u_2 h and a

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u_2 v, so there are four
movements of the ends of the

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truss and of one particular
bar, and then we'll

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00:01:45 --> 00:01:47
stretch that bar.
 

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And the question, is how much?
 

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So that will be one row of A.
 

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So if we follow one bar, you
remember in the matrix A,

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there's going to be a
row for every bar.

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So a row for each, row of A.
 

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For each bar.
 

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00:02:09 --> 00:02:13
And if we track down one
of those rows, we'll

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have the idea.
 

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00:02:15 --> 00:02:19
And then of course at the end
we'd maybe be constructing A

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without, sort of a free-free A.
 

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And then at the end, any fixed
displacements that will

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knock out columns of A.
 

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So that's one job.
 

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And then to see, so the
A is going to be a

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little more messy.
 

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00:02:39 --> 00:02:41
It's because we're
in two dimensions.

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So compared to the network
problems, and and the

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line of springs, now we
have more happening.

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We've got more columns
because every node

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has now two unknowns.
 

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A horizontal and vertical.
 

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So A is kind of bigger.
 

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And therefore A transpose C A,
you might think, it's going to

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be hard to see what's going on.
 

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But you'll see the right
way to look at A transpose

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C A is a bar at a time.
 

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00:03:15 --> 00:03:21
That's the nice fact about
A transpose C A, I might

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focus on that first.
 

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And then comes the fun part.
 

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I'll draw some more trusses,
that may or may not

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have mechanisms.
 

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They may or may not be stable.
 

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And we can try to
identify the mechanisms.

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Actually, as before.
 

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We'll do it by engineering
instinct rather

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than by solving.
 

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I mean, in principle, we could
always use elimination.

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Or ask MATLAB or any other
system to do it, and look

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for the solutions to Au=0.
 

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And decide are the columns
of A independent.

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In that case the
truss is stable.

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This matrix is invertible,
we know all the

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good possibilities.
 

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And then there's the more
interesting possibility, of

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having some solutions to that.
 

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In which case that matrix
will be singular.

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There'll be some modes
in our big system that

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will cause it to fail.
 

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But it's kind of
fun to find those.

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OK, while I've written A
transpose C A, may I remind

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you about a good way to
do that multiplication.

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OK, so imagine I'm just
putting a number.

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Here's going to
be the matrix A.

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So the matrix A will have
a bunch of rows, row

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one, row two, so on.
 

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These rows will correspond
to bar one, bar

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two, and bar three.
 

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OK.
 

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OK, then we have C, so
that's a square matrix.

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That each bar has a spring
constant, so c_1, c_2, c_3,

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and then we have A transpose.
 

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And those rows, or
columns of A transpose.

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00:05:20 --> 00:05:26
So that's the sort of picture
of A transpose C A, for a

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three bar, three bars only.
 

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00:05:30 --> 00:05:33
But the point is
made right here.

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00:05:33 --> 00:05:36
There's a row of
A for every bar.

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00:05:36 --> 00:05:37
Right?
 

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Because our matrix A is m by n.
 

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00:05:40 --> 00:05:48
If there are m bars, a row for
every bar, and it tells us how

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00:05:48 --> 00:05:50
far that bar is stretched.
 

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00:05:50 --> 00:05:53
And we'll figure out
what its entries are.

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00:05:53 --> 00:05:55
That's our main job.
 

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00:05:55 --> 00:05:56
I'm just looking ahead.
 

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Suppose we've got that row.
 

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And that row, and that row.
 

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So a row for every bar.
 

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Now, here I've
taken three bars.

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00:06:04 --> 00:06:08
Now, how do I multiply
those matrices?

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00:06:08 --> 00:06:11
Well, I can do it
different ways.

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00:06:11 --> 00:06:13
But here's a cool way to do it.
 

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Just the way I want to point
out is column times row.

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If you multiply matrices you're
allowed to, the effective

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c_1, c_2, c_3 is going
to be very simple.

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00:06:26 --> 00:06:32
So I'm really paying attention
here to A transpose A.

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If I want to multiply A
transpose A, I can do row

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times column as usual
and get one number.

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Or I can do column times row
and get a whole little matrix.

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And that's the bar one matrix.
 

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It's the element matrix, and
that's how finite elements will

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be assembled, and that's why I
should keep mentioning

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this point.
 

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So the way to do that column
times row thing, and then of

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course that c_1 just multiplies
that row, that'll be c_1.

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Row one, transpose, that's
the column, times row one.

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That's what's coming
from bar one.

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That column multiplies
the c_1 and that row.

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You see how nice, that's the
element matrix associated

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with the first bar.
 

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And then there'll be a second
column times the c_2,

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times the second row.
 

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So plus c_2, row two,
transpose row two.

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That's a matrix again.
 

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Plus c_3, row three transpose.
 

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Row three.
 

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I focus on that because you
don't think of this as a way to

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multiply matrices, but it's
really a nice thing to notice.

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00:08:00 --> 00:08:05
And it's better to notice it
now when we have three bars

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00:08:05 --> 00:08:11
or something, than in a
big finite element code.

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00:08:11 --> 00:08:12
Yeah.
 

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So this is, just if I complete
it, complete this thought, this

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I would call up a
one bar matrix.

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That's the matrix A transpose
A if there's only one bar.

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Actually, one of the problems
at the end of this section

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is find the element
matrix for one bar.

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00:08:35 --> 00:08:39
And I guess it's about what
we're going to get to

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00:08:39 --> 00:08:41
when we do that one bar.
 

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00:08:41 --> 00:08:46
Do you remember what it was
in the, just to connect

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this thought, what was
the little matrix?

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00:08:50 --> 00:08:53
In the case of networks?
 

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00:08:53 --> 00:08:57
So in the case of networks,
there was just one unknown

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00:08:57 --> 00:09:00
for each, not two.
 

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00:09:00 --> 00:09:05
So for networks, just, I'm just
going to put down the, and

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you'll recognize
it immediately.

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The little element matrix
was the c for that.

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And there was an
1, -1, -1 what?

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00:09:17 --> 00:09:21
Do you remember that guy that
was the -1, 1 from a row?

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-1, 1 from a column?
 

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So this was exactly c times the
-1, 1 from the column times

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the -1, 1 from the row.
 

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That's where this simple
little matrix came from.

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And you remember that the, so
what's involved in creating

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00:09:44 --> 00:09:49
this big A transpose C A is
just create all these

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little pieces.
 

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Which are like this, but
they're going to be

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a little bigger.
 

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Fact, in a minute I'm going to
ask you what size they'll be.

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Well, they're really
big matrices.

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00:10:00 --> 00:10:03
There are a whole lot of zeroes
there that I didn't even put.

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00:10:03 --> 00:10:08
Zeroes are there for rows and
columns that aren't touching

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00:10:08 --> 00:10:11
this particular edge.
 

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And again, this matrix.
 

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There'll be all kinds
of zeroes at A.

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00:10:16 --> 00:10:22
Because a typical row of A,
bar one, is going to have

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00:10:22 --> 00:10:27
non-zeroes only for the,
yeah what's the size?

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00:10:27 --> 00:10:32
How many non-zeroes in
a typical row of A?

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00:10:32 --> 00:10:36
Getting the count right first
is like half the battle.

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How many non-zeroes in
a typical row of a?

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00:10:39 --> 00:10:43
This was the network case where
we had a couple of nodes.

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00:10:43 --> 00:10:44
And they were connected.
 

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00:10:44 --> 00:10:47
And we had an unknown
at each end.

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00:10:47 --> 00:10:51
So two unknowns were involved.
 

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00:10:51 --> 00:10:53
The little matrix
was two by two.

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00:10:53 --> 00:10:57
It properly has lots of zeroes
for all the other nodes

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that are not involved.
 

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And then that matrix kind
of gets, assembled is the

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00:11:04 --> 00:11:06
word I think usually used.
 

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All these little guys get
assembled, you know, pasted,

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stamped, I hear the
verb sometimes now.

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00:11:15 --> 00:11:19
Take these little matrices
for this little element.

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And stamp them into the
big A transpose C A.

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00:11:26 --> 00:11:29
This is the c_1, so this gives
the c_1's in the matrix.

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00:11:29 --> 00:11:31
And the c_2's 2 and the c_3's.
 

192
00:11:31 --> 00:11:32
Alright.
 

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00:11:32 --> 00:11:36
Now, just before we, I'm like
doing this preliminary, before

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00:11:36 --> 00:11:40
I write down anything,
the exact row.

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What's the size of,
for trusses, how many

196
00:11:46 --> 00:11:49
non-zeroes in a row of A?
 

197
00:11:49 --> 00:11:50
So that's my question.
 

198
00:11:50 --> 00:11:58
How many non-zeroes in a
typical row, like for that bar,

199
00:11:58 --> 00:12:05
non-zeroes in a row of A?
 

200
00:12:05 --> 00:12:12
So A is the matrix that tells
us how much, that row of A is

201
00:12:12 --> 00:12:17
the row that tells us how much
this bar stretched when this

202
00:12:17 --> 00:12:25
moved along by u H 1, and up by
you u V 1, and this moved along

203
00:12:25 --> 00:12:32
by say, u H 2, and up by u V 2.
 

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00:12:32 --> 00:12:35
Well, I've written
all those in.

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00:12:35 --> 00:12:36
So that you can tell
me this number.

206
00:12:36 --> 00:12:38
How many?
 

207
00:12:38 --> 00:12:40
What's your guess?
 

208
00:12:40 --> 00:12:43
When I tell you,
you'll say of course.

209
00:12:43 --> 00:12:48
How many u's are involved in
the stretching of that bar?

210
00:12:48 --> 00:12:49
Four.
 

211
00:12:49 --> 00:12:50
Four.
 

212
00:12:50 --> 00:12:51
Exactly.
 

213
00:12:51 --> 00:12:54
Instead of one at each end,
we have two at each end.

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00:12:54 --> 00:12:56
So the answer is four.
 

215
00:12:56 --> 00:12:58
How many, the answer is four.
 

216
00:12:58 --> 00:13:03
And now the only remaining
question is, what are they?

217
00:13:03 --> 00:13:06
What are those four numbers?
 

218
00:13:06 --> 00:13:08
The four non-zeroes in the row?
 

219
00:13:08 --> 00:13:13
So let me just answer that.
 

220
00:13:13 --> 00:13:17
They are, so here is that row.
 

221
00:13:17 --> 00:13:22
So we the two non-zeroes
associated with it.

222
00:13:22 --> 00:13:26
Well, the way I've numbered
these nodes one and two.

223
00:13:26 --> 00:13:29
Since I've numbered them one
and two, the non-zeroes are

224
00:13:29 --> 00:13:31
going to come right
at the start.

225
00:13:31 --> 00:13:33
And then a whole lot, then
this is all going to

226
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be zero after that.
 

227
00:13:34 --> 00:13:37
Because those will be nodes
three, four, or five, whatever

228
00:13:37 --> 00:13:40
that don't involve bar one.
 

229
00:13:40 --> 00:13:43
So bar one just connects
node one to node two.

230
00:13:43 --> 00:13:47
Now, what do you think?
 

231
00:13:47 --> 00:13:51
Well, let me put in the
key quantity here.

232
00:13:51 --> 00:13:53
This bar is at an angle.
 

233
00:13:53 --> 00:13:55
It's at an angle theta.
 

234
00:13:55 --> 00:13:56
So there's a theta.
 

235
00:13:56 --> 00:14:00
Angle theta.
 

236
00:14:00 --> 00:14:02
OK.
 

237
00:14:02 --> 00:14:10
And so that angle is going
to enter these things.

238
00:14:10 --> 00:14:13
In fact, here's what you get.
 

239
00:14:13 --> 00:14:18
You get, I think if I put the
one up there and the two

240
00:14:18 --> 00:14:20
down there, let's see.
 

241
00:14:20 --> 00:14:22
What am I thinking now?
 

242
00:14:22 --> 00:14:26
I'm saying if u 1 H is
positive, that's going

243
00:14:26 --> 00:14:28
to stretch the bar.
 

244
00:14:28 --> 00:14:30
That's a positive stretching.
 

245
00:14:30 --> 00:14:34
So I'm expecting a positive u
1 H to give me, I'm expecting

246
00:14:34 --> 00:14:36
that sort of to come
in with a plus sign.

247
00:14:36 --> 00:14:42
Now suppose the bar
is horizontal.

248
00:14:42 --> 00:14:46
Suppose the bar is horizontal,
then how much does

249
00:14:46 --> 00:14:48
the u 1 H stretch it?
 

250
00:14:48 --> 00:14:52
It stretches it by the
whole u 1 H, right?

251
00:14:52 --> 00:14:57
If the bar was horizontal,
so theta equals zero.

252
00:14:57 --> 00:15:01
I'm just doing these, we
got to sort out this

253
00:15:01 --> 00:15:02
theta angle stuff.
 

254
00:15:02 --> 00:15:04
So here's my thing.
 

255
00:15:04 --> 00:15:08
If the bar happens to be
horizontal, then that

256
00:15:08 --> 00:15:17
stretching by u H 1, will
completely stretch the bar.

257
00:15:17 --> 00:15:20
If the bar happened
to be, yeah yeah.

258
00:15:20 --> 00:15:22
And of course, this way.
 

259
00:15:22 --> 00:15:26
So that for a horizontal bar,
I'll just be back to this step.

260
00:15:26 --> 00:15:32
I'll have a one and a
minus one u, oh yeah,

261
00:15:32 --> 00:15:34
remind me about that.
 

262
00:15:34 --> 00:15:40
Why doesn't u vertical, for a
horizontal bar like this, why

263
00:15:40 --> 00:15:46
does this one not
stretch the bar?

264
00:15:46 --> 00:15:49
You remember that from last
time, that was the little bit

265
00:15:49 --> 00:15:54
of trig that we did when we
were forced ourselves

266
00:15:54 --> 00:15:56
to stay linear.
 

267
00:15:56 --> 00:16:00
So we dropped the second
order correction, that would

268
00:16:00 --> 00:16:03
come from going this way.
 

269
00:16:03 --> 00:16:06
Right?
 

270
00:16:06 --> 00:16:09
I mean you must have
noticed, like walking?

271
00:16:09 --> 00:16:14
Suppose you want to walk from
here to the end of the bar, OK?

272
00:16:14 --> 00:16:17
Well, if somebody moves the end
of the bar forward, you have

273
00:16:17 --> 00:16:19
to take those extra steps.
 

274
00:16:19 --> 00:16:21
The bar really stretches.
 

275
00:16:21 --> 00:16:25
But, if somebody moves the
bar this way, then the extra

276
00:16:25 --> 00:16:28
bit of length is much less.
 

277
00:16:28 --> 00:16:33
In fact, it's zero
to first order.

278
00:16:33 --> 00:16:37
This is like taking shortcuts
when you walk across

279
00:16:37 --> 00:16:40
the courtyard.
 

280
00:16:40 --> 00:16:45
So when the angle's theta,
I'm only expecting a

281
00:16:45 --> 00:16:49
one and a minus one.
 

282
00:16:49 --> 00:16:50
On the horizontal.
 

283
00:16:50 --> 00:16:51
And zeroes on the vertical.
 

284
00:16:51 --> 00:16:52
OK, now I'm ready to write it.
 

285
00:16:52 --> 00:16:56
I think when the angle's theta,
when the angle's theta, any

286
00:16:56 --> 00:17:01
theta, that was when the angle
was zero, I think we

287
00:17:01 --> 00:17:05
get a cos theta.
 

288
00:17:05 --> 00:17:08
Doesn't your instinct
say that this is on the

289
00:17:08 --> 00:17:11
u_1, u horizontal 1.
 

290
00:17:11 --> 00:17:17
And then the u vertical 1,
tell me what these should be.

291
00:17:17 --> 00:17:23
And then we'll make, what do
you suppose is the entry,

292
00:17:23 --> 00:17:27
second non-zero, the one
that corresponds to

293
00:17:27 --> 00:17:29
a vertical movement.
 

294
00:17:29 --> 00:17:32
Here it would be, for a
horizontal bar when theta

295
00:17:32 --> 00:17:35
is zero, I'm going
to see a zero there.

296
00:17:35 --> 00:17:38
But if the bar is at an
angle like this, what

297
00:17:38 --> 00:17:40
am I going to see?
 

298
00:17:40 --> 00:17:42
Everybody's going
to get it right?

299
00:17:42 --> 00:17:44
What do I put in there?
 

300
00:17:44 --> 00:17:45
Sine theta.
 

301
00:17:45 --> 00:17:46
What else could it be?
 

302
00:17:46 --> 00:17:48
Right, OK, sin(theta).
 

303
00:17:49 --> 00:17:53
And now what about the next
guy, the other end of the bar?

304
00:17:53 --> 00:17:59
u H 2 and u V 2, those are
the other two non-zeroes.

305
00:17:59 --> 00:18:03
What your guess
for u 2 H, u H 2?

306
00:18:03 --> 00:18:11
If I move this forward,
What's the change in

307
00:18:11 --> 00:18:15
length of the bar?
 

308
00:18:15 --> 00:18:20
What would your guess be
that goes into there?

309
00:18:20 --> 00:18:21
Say it again?
 

310
00:18:22 --> 00:18:22
-cos(theta).
 

311
00:18:24 --> 00:18:26
-cos(theta), right, yeah.
 

312
00:18:26 --> 00:18:32
The movement of the other end,
like if I move this guy a

313
00:18:32 --> 00:18:34
little bit to this side.
 

314
00:18:34 --> 00:18:37
That will shorten the bar.
 

315
00:18:37 --> 00:18:38
Forget about that one.
 

316
00:18:38 --> 00:18:43
If I move this over, the
bar becomes shorter.

317
00:18:43 --> 00:18:52
And the cosine tells me
the key number there, how

318
00:18:52 --> 00:18:53
much it becomes shorter.
 

319
00:18:53 --> 00:18:58
If the bar was horizontal,
the cos(theta) was one, it

320
00:18:58 --> 00:19:00
counts a hundred percent.
 

321
00:19:00 --> 00:19:05
If the bar is vertical, and
I move it horizontally,

322
00:19:05 --> 00:19:06
it comes zero percent.
 

323
00:19:06 --> 00:19:10
Because the linearity
says there was no

324
00:19:10 --> 00:19:11
first order change.
 

325
00:19:11 --> 00:19:14
And now tell me the
final non-zero entry.

326
00:19:14 --> 00:19:18
And I see I didn't leave much
room for all the zeroes.

327
00:19:18 --> 00:19:24
OK, what's the u 2 V entry?
 

328
00:19:24 --> 00:19:26
-sin(theta), of course.
 

329
00:19:26 --> 00:19:33
And then come all the zeroes,
four whatever other joints are

330
00:19:33 --> 00:19:35
not involved with bar one.
 

331
00:19:35 --> 00:19:40
So let me, maybe to make this
picture best, I should move

332
00:19:40 --> 00:19:43
that over to where it belongs.
 

333
00:19:43 --> 00:19:48
Now, if I add up along the
bar, add up the four numbers

334
00:19:48 --> 00:19:49
there, what do I get?
 

335
00:19:49 --> 00:19:51
Zero.
 

336
00:19:51 --> 00:19:56
You expected that, right?
 

337
00:19:56 --> 00:19:59
In fact, if I add just
that and that I get zero.

338
00:19:59 --> 00:20:01
If I add that and
that I get zero.

339
00:20:01 --> 00:20:04
Just the way I got zero here.
 

340
00:20:04 --> 00:20:07
In the incidence matrices.
 

341
00:20:07 --> 00:20:14
The column of all ones is
certainly going to solve Au=0.

342
00:20:14 --> 00:20:17
 
 

343
00:20:17 --> 00:20:22
Unless the supports
remove those, of course.

344
00:20:22 --> 00:20:28
If the supports don't allow all
ones because some have to stay

345
00:20:28 --> 00:20:31
at zero, then I could
have a stable truss.

346
00:20:31 --> 00:20:34
OK, that's a typical bar.
 

347
00:20:34 --> 00:20:35
A typical row.
 

348
00:20:35 --> 00:20:37
That's a typical row.
 

349
00:20:37 --> 00:20:43
OK, and now maybe while I'm
on the same subject, what

350
00:20:43 --> 00:20:52
is the size, what is this
thing look like now?

351
00:20:52 --> 00:20:54
This is in A.
 

352
00:20:54 --> 00:21:00
This is in a matrix A, and now
I want to ask you before I even

353
00:21:00 --> 00:21:05
come back to all this stuff,
what about in A transpose A?

354
00:21:05 --> 00:21:10
In A transpose A, A transpose
C, it the whole deal.

355
00:21:10 --> 00:21:17
The element for the little
matrix, the element matrix,

356
00:21:17 --> 00:21:18
can I call it that?
 

357
00:21:18 --> 00:21:25
Or the one bar matrix, call
it the one bar matrix.

358
00:21:25 --> 00:21:31
Will be, is what?
 

359
00:21:31 --> 00:21:37
So I want this, it would
be typical. c_1, row one.

360
00:21:37 --> 00:21:41
Transpose row one,
that's the typical guy.

361
00:21:41 --> 00:21:48
And how many
non-zeroes in that?

362
00:21:48 --> 00:21:51
Multiplying a row.
 

363
00:21:51 --> 00:21:56
Sorry, multiplying a column
that has four non-zeroes

364
00:21:56 --> 00:22:00
times a row that has four
non-zeroes times a number,

365
00:22:00 --> 00:22:01
which is just fine.
 

366
00:22:01 --> 00:22:07
How many non-zeroes are going
to sit in this element

367
00:22:07 --> 00:22:12
matrix, this one-bar matrix?
 

368
00:22:12 --> 00:22:14
16.
 

369
00:22:14 --> 00:22:18
16 non-zeroes.
 

370
00:22:18 --> 00:22:24
And they're going to be, I have
cosine, sine, minus cosine,

371
00:22:24 --> 00:22:30
minus sine, multiplying cosine,
sine, minus cosine, minus sine.

372
00:22:30 --> 00:22:32
And all multiplied by c_1.
 

373
00:22:33 --> 00:22:34
So that's the matrix.
 

374
00:22:34 --> 00:22:40
And you see what it looks
like. c squared, cs, so on.

375
00:22:40 --> 00:22:43
16 guys.
 

376
00:22:43 --> 00:22:49
So we have four squared as
our element matrix, where

377
00:22:49 --> 00:22:51
here we had two squared.
 

378
00:22:51 --> 00:23:00
And in finite elements, when
you get to elasticity, and

379
00:23:00 --> 00:23:07
you've got triangles, you've
got triangular elements, then

380
00:23:07 --> 00:23:09
there are three nodes involved.
 

381
00:23:09 --> 00:23:12
So you're up to higher numbers.
 

382
00:23:12 --> 00:23:14
But this gives you the idea.
 

383
00:23:14 --> 00:23:18
And remember that this four by
four, the way I've done it, the

384
00:23:18 --> 00:23:22
way I've numbered it, one,
two, happens to sit up in

385
00:23:22 --> 00:23:24
the upper left corner.
 

386
00:23:24 --> 00:23:26
Of A transpose C A.
 

387
00:23:26 --> 00:23:31
But can you sort of imagine how
the code would be written?

388
00:23:31 --> 00:23:36
The code would be written,
take each bar, and what do I

389
00:23:36 --> 00:23:38
have to know about the bar?
 

390
00:23:38 --> 00:23:41
Just imagine a code
that would do trusses.

391
00:23:41 --> 00:23:46
Actually, the final problem
that I'm not assigning in

392
00:23:46 --> 00:23:49
this section says what
would the code be like?

393
00:23:49 --> 00:23:55
Can we just have a think about
what the code would look

394
00:23:55 --> 00:23:57
like if we were to write it.
 

395
00:23:57 --> 00:24:03
What would the
input have to be?

396
00:24:03 --> 00:24:08
For each bar, what
input do I need?

397
00:24:08 --> 00:24:14
For this bar, I need to know,
and for the whole truss.

398
00:24:14 --> 00:24:18
What do I have to tell, what's
the information that I

399
00:24:18 --> 00:24:20
need for the whole truss?
 

400
00:24:20 --> 00:24:25
I have to know the positions
of all of the joints, right?

401
00:24:25 --> 00:24:28
So I'd have to know the
coordinates of that, x

402
00:24:28 --> 00:24:31
y, the coordinates of
this one, x_1, y_1.

403
00:24:31 --> 00:24:33
 
 

404
00:24:33 --> 00:24:36
For joint one, x_2, y_2.
 

405
00:24:36 --> 00:24:39
So I'd have to have a
little list of what

406
00:24:39 --> 00:24:41
would that be? m by two?
 

407
00:24:41 --> 00:24:44
I don't know. n. n by two.
 

408
00:24:44 --> 00:24:50
I have n, what do I have now?
 

409
00:24:50 --> 00:24:53
Think of what information
do I have to report

410
00:24:53 --> 00:24:54
about this truss?
 

411
00:24:54 --> 00:24:58
I guess I have N,
capital N, joints.

412
00:24:58 --> 00:25:02
And I need two coordinates,
x, y for each position.

413
00:25:02 --> 00:25:04
So that's N by two this.
 

414
00:25:04 --> 00:25:09
OK, and then for every bar,
what do I need to tell it?

415
00:25:09 --> 00:25:12
What do I need to put in the
code for a typical bar?

416
00:25:12 --> 00:25:19
I certainly have to put
in the c for that bar.

417
00:25:19 --> 00:25:21
And what else do
I need to know?

418
00:25:21 --> 00:25:26
I need to know which
joints it's connected.

419
00:25:26 --> 00:25:27
Right?
 

420
00:25:27 --> 00:25:32
I have to tell the system
that this bar is between

421
00:25:32 --> 00:25:34
two and one, one and two.
 

422
00:25:34 --> 00:25:36
I have to tell it which pair.
 

423
00:25:36 --> 00:25:43
So I guess I have a list of m
bars, and for each bar I must

424
00:25:43 --> 00:25:49
tell the system the two
node numbers, and the

425
00:25:49 --> 00:25:51
c, the stiffness.
 

426
00:25:51 --> 00:25:54
The constant for Hooke's Law.
 

427
00:25:54 --> 00:25:56
Right, do you see this picture?
 

428
00:25:56 --> 00:26:02
Just sort of visualizing,
creating a code here.

429
00:26:02 --> 00:26:07
And then the code would do all
this, oh, have I given enough

430
00:26:07 --> 00:26:09
information to find theta?
 

431
00:26:09 --> 00:26:13
Or do I have to
import theta also?

432
00:26:13 --> 00:26:14
No.
 

433
00:26:14 --> 00:26:18
I told you the positions, so
it'll figure out cos(theta)

434
00:26:18 --> 00:26:18
theta and sin(theta).
 

435
00:26:19 --> 00:26:22
It actually won't figure out
theta, that's always a dumb

436
00:26:22 --> 00:26:25
thing to do find the actual
angle. cos(theta) and

437
00:26:25 --> 00:26:27
sin(theta) is the
quantities we want.

438
00:26:27 --> 00:26:32
So given that position, x, y
and this position x_2, y_2,

439
00:26:32 --> 00:26:35
it would know cos(theta)
and sin(theta).

440
00:26:36 --> 00:26:40
And having drawn his picture
allows me to make once

441
00:26:40 --> 00:26:44
more the key point about
small displacements.

442
00:26:44 --> 00:26:49
What's the angle of the
bar after it's moved?

443
00:26:49 --> 00:26:51
After it's displaced?
 

444
00:26:51 --> 00:26:54
It was theta before it
was displaced, and the

445
00:26:54 --> 00:26:58
angle after is theta.
 

446
00:26:58 --> 00:27:02
To first order, the
angle doesn't change.

447
00:27:02 --> 00:27:07
Because these are little
tiny movements of the ends.

448
00:27:07 --> 00:27:10
I've drawn them much bigger
than they should be drawn.

449
00:27:10 --> 00:27:14
They're little, tiny movements
of the ends so that the angle

450
00:27:14 --> 00:27:18
is not significantly changed.
 

451
00:27:18 --> 00:27:23
Otherwise we're into geometric
nonlinearity and that stuff,

452
00:27:23 --> 00:27:27
that makes the problem
much, much harder.

453
00:27:27 --> 00:27:30
OK, are you seeing
sort of the picture?

454
00:27:30 --> 00:27:36
I guess what I haven't
completely, I've really

455
00:27:36 --> 00:27:41
depended more on your intuition
than on a calculation to say

456
00:27:41 --> 00:27:45
that these are the
four non-zeroes.

457
00:27:45 --> 00:27:47
What did I ask you to do?
 

458
00:27:47 --> 00:27:51
I asked you to check that that
was right in the extreme cases,

459
00:27:51 --> 00:27:55
like if theta is zero, the bar
is horizontal, then we just

460
00:27:55 --> 00:28:00
have a one, zero minus one,
zero vertical isn't happening.

461
00:28:00 --> 00:28:04
If the bar is vertical so that
the angle is 90 degrees then we

462
00:28:04 --> 00:28:08
would have a zero, one, zerom
minus one, everything's

463
00:28:08 --> 00:28:09
vertical.
 

464
00:28:09 --> 00:28:15
And the book draws a little
picture, and computes.

465
00:28:15 --> 00:28:20
Computes delta l from these
four small movements.

466
00:28:20 --> 00:28:24
And takes the leading term
and sure enough it produces

467
00:28:24 --> 00:28:30
that row of the matrix.
 

468
00:28:30 --> 00:28:33
Gosh, I talk real fast.
 

469
00:28:33 --> 00:28:38
But do you think you could
now create the stiffness?

470
00:28:38 --> 00:28:45
If you had a real truss,
you could create the

471
00:28:45 --> 00:28:48
matrix A for it?
 

472
00:28:48 --> 00:28:51
C is simple, it's given to you.
 

473
00:28:51 --> 00:28:54
You could create
A transpose C A?

474
00:28:54 --> 00:28:58
You might just want to
write the command as

475
00:28:58 --> 00:29:03
A'*C*A or something.
 

476
00:29:03 --> 00:29:06
And let MATLAB do the thinking.
 

477
00:29:06 --> 00:29:12
But I wanted to just see what
these, how this four by

478
00:29:12 --> 00:29:21
four piece appears in this
product from each bar.

479
00:29:21 --> 00:29:26
The 16 non-zeroes will appear
in different positions and

480
00:29:26 --> 00:29:30
you told the code what
those positions are.

481
00:29:30 --> 00:29:35
You had to give the code a
local to global picture.

482
00:29:35 --> 00:29:37
This is the local picture.
 

483
00:29:37 --> 00:29:39
Watch one bar.
 

484
00:29:39 --> 00:29:46
Then it has to fit in this big
n by n matrix, and that means

485
00:29:46 --> 00:29:52
you have to know what joints
was that bar connecting.

486
00:29:52 --> 00:30:05
So which positions do these
16 non-zeroes assemble into?

487
00:30:05 --> 00:30:11
That's some time devoted to
a job that I actually don't

488
00:30:11 --> 00:30:14
plan to require you to do.
 

489
00:30:14 --> 00:30:18
Creating this truss problem.
 

490
00:30:18 --> 00:30:23
What I think is kind of more
fun and that's these homework

491
00:30:23 --> 00:30:30
problems would deal with it,
is part two of the lecture

492
00:30:30 --> 00:30:33
going back to mechanisms.
 

493
00:30:33 --> 00:30:42
And now thinking about
more complicated trusses.

494
00:30:42 --> 00:30:47
We now in principle could find
the solutions to Au=0 because

495
00:30:47 --> 00:30:52
we now have constructed A, and
we could get MATLAB to do the

496
00:30:52 --> 00:30:55
work or Python or whoever.
 

497
00:30:55 --> 00:31:01
But can I go to part two now
and draw a truss and ask

498
00:31:01 --> 00:31:05
you about the mechanisms?
 

499
00:31:05 --> 00:31:10
Let's see.
 

500
00:31:10 --> 00:31:16
I guess somewhere in the
problem set, but not one

501
00:31:16 --> 00:31:25
of the assigned ones is,
start with those six bars.

502
00:31:25 --> 00:31:33
And six joints, so
these are six joints.

503
00:31:33 --> 00:31:42
And OK, as it stands how many,
that's a good question.

504
00:31:42 --> 00:31:47
As it stands, what's the
shape of the matrix A?

505
00:31:47 --> 00:31:50
How many rows has it got?
 

506
00:31:50 --> 00:31:57
So as it stands, so I'll call
it A_0, for no supports

507
00:31:57 --> 00:31:58
have been added.
 

508
00:31:58 --> 00:32:01
A_0, just the full matrix.
 

509
00:32:01 --> 00:32:06
Is what shape? six by 12.
 

510
00:32:06 --> 00:32:06
Good.
 

511
00:32:06 --> 00:32:10
Six by 12 is six by 12.
 

512
00:32:10 --> 00:32:12
OK, of course it's not stable.
 

513
00:32:12 --> 00:32:13
We know that.
 

514
00:32:13 --> 00:32:16
We haven't supported anything.
 

515
00:32:16 --> 00:32:24
So in a typical case, how many
solutions to A, so I'm going

516
00:32:24 --> 00:32:28
to ask you how many
solutions to Au=0?

517
00:32:28 --> 00:32:32
 
 

518
00:32:32 --> 00:32:35
And what's your guess?
 

519
00:32:35 --> 00:32:37
Six.
 

520
00:32:37 --> 00:32:42
Got six equations, we've
got 12 u's, 12-6, so this

521
00:32:42 --> 00:32:43
is going to be 12-6.
 

522
00:32:44 --> 00:32:49
And of course six solutions
to that equation.

523
00:32:49 --> 00:32:51
It's what I would expect.
 

524
00:32:51 --> 00:32:57
There could be, it could be
possible that the six equations

525
00:32:57 --> 00:32:59
are not independent.
 

526
00:32:59 --> 00:33:02
If they really dropped to
five then this would bump

527
00:33:02 --> 00:33:05
up to seven, I don't think
it's going to happen.

528
00:33:05 --> 00:33:10
Now, can you describe those six
solutions, not with numbers,

529
00:33:10 --> 00:33:13
just with, tell me.
 

530
00:33:13 --> 00:33:17
I hope you can, because
I can't right now.

531
00:33:17 --> 00:33:19
OK, three of them we know.
 

532
00:33:19 --> 00:33:31
So with no supports at all,
what are three rigid motions?

533
00:33:31 --> 00:33:33
And what are they?
 

534
00:33:33 --> 00:33:38
The whole truss could move to
the right, the whole hexagon.

535
00:33:38 --> 00:33:41
It could all move up, it could
all rotate about one point.

536
00:33:41 --> 00:33:47
All three of those would be
movements, displacements that

537
00:33:47 --> 00:33:48
don't stretch anything.
 

538
00:33:48 --> 00:33:51
OK, three rigid motions.
 

539
00:33:51 --> 00:33:59
Across, up, and rotate.
 

540
00:33:59 --> 00:34:03
OK, and I can get rid of those
by supporting some nodes.

541
00:34:03 --> 00:34:07
But let me see, I don't know
what's going to happen.

542
00:34:07 --> 00:34:14
When I describe this topic as
the fun one in 18.085, it's

543
00:34:14 --> 00:34:16
more fun for you than for me.
 

544
00:34:16 --> 00:34:21
Because I draw something like
that and I start worrying can

545
00:34:21 --> 00:34:26
I think of three, how many
mechanisms to look for?

546
00:34:26 --> 00:34:26
Three.
 

547
00:34:26 --> 00:34:31
That's a big number.
 

548
00:34:31 --> 00:34:36
I bet you I can find one but
you guys have got to, alright,

549
00:34:36 --> 00:34:37
tell me some mechanisms.
 

550
00:34:37 --> 00:34:40
Let me try to draw them.
 

551
00:34:40 --> 00:34:44
What would be one mechanism?
 

552
00:34:44 --> 00:34:47
Collapses, yeah, somehow.
 

553
00:34:47 --> 00:34:50
How shall I make it collapse?
 

554
00:34:50 --> 00:34:53
Can squeeze in, yeah, maybe
that's the first one.

555
00:34:53 --> 00:35:05
This guy comes in, this
guy, what does that do?

556
00:35:05 --> 00:35:09
I've got 15 minutes here, I
could pull that board down

557
00:35:09 --> 00:35:10
and draw another one.
 

558
00:35:10 --> 00:35:13
What is this one here?
 

559
00:35:13 --> 00:35:16
Let's see, if that comes in, do
these guys have to go up a bit?

560
00:35:16 --> 00:35:20
Yeah, because that
angle's no 90 degrees.

561
00:35:20 --> 00:35:24
So we've got a first
order change in this.

562
00:35:24 --> 00:35:29
So this comes in a little,
this goes up a little, this

563
00:35:29 --> 00:35:35
guy maybe stays straight.
 

564
00:35:35 --> 00:35:40
Would you go for this,
I mean please say yes?

565
00:35:40 --> 00:35:41
Something happens there, right?
 

566
00:35:41 --> 00:35:50
These things go in
and those go out.

567
00:35:50 --> 00:35:56
Could you create the, I mean is
this was A equal side, regular

568
00:35:56 --> 00:36:03
hexagon, you could put in all
the numbers for all 12, I would

569
00:36:03 --> 00:36:05
be looking for 12 numbers.
 

570
00:36:05 --> 00:36:11
Six joints and they each have
two u's, so that wouldn't be so

571
00:36:11 --> 00:36:15
simple but you could do it.
 

572
00:36:15 --> 00:36:20
So that would be one.
 

573
00:36:20 --> 00:36:23
Looking for number two.
 

574
00:36:23 --> 00:36:24
What would be another one?
 

575
00:36:24 --> 00:36:32
So sort of squeezing
in like whatever.

576
00:36:32 --> 00:36:36
What do you think?
 

577
00:36:36 --> 00:36:40
Any others?
 

578
00:36:40 --> 00:36:42
Maybe that's possible.
 

579
00:36:42 --> 00:36:49
Maybe I just look at it,
you think that would work?

580
00:36:49 --> 00:36:52
We could hope those were
independent, but I wouldn't

581
00:36:52 --> 00:36:54
put my life on it.
 

582
00:36:54 --> 00:36:59
I have this squeeze in, this
squeeze in and this squeeze in,

583
00:36:59 --> 00:37:03
I would worry a little bit.
 

584
00:37:03 --> 00:37:06
I can see another one.
 

585
00:37:06 --> 00:37:09
AUDIENCE: Half, for--
 

586
00:37:09 --> 00:37:10
PROFESSOR STRANG:
Fold it in half.

587
00:37:10 --> 00:37:16
AUDIENCE: Along, I guess that
sort of gets into 3-D, but.

588
00:37:16 --> 00:37:17
PROFESSOR STRANG: Yeah,
we've got to stay in

589
00:37:17 --> 00:37:24
the plane, right.
 

590
00:37:24 --> 00:37:28
I have instead of what, OK.
 

591
00:37:28 --> 00:37:31
AUDIENCE: [INAUDIBLE]
 

592
00:37:31 --> 00:37:34
PROFESSOR STRANG:
Squeeze that out.

593
00:37:34 --> 00:37:36
Yeah.
 

594
00:37:36 --> 00:37:39
Good.
 

595
00:37:39 --> 00:37:42
Number two.
 

596
00:37:42 --> 00:37:47
And then I can see, here's one,
here's an easy one to think of.

597
00:37:47 --> 00:37:51
Leave these three alone and
just rotate these guys.

598
00:37:51 --> 00:37:54
Bring this down, right?
 

599
00:37:54 --> 00:38:03
Just bring these three
vertically down, or something.

600
00:38:03 --> 00:38:11
You see why it's sort of, did
you like that one alright?

601
00:38:11 --> 00:38:16
It seems simple to me, looking
at it, just leave these guys in

602
00:38:16 --> 00:38:19
rotation, let this turn down.
 

603
00:38:19 --> 00:38:22
This turn down, let's
say, and this go down.

604
00:38:22 --> 00:38:27
Maybe these would all
drop by the same amount.

605
00:38:27 --> 00:38:33
Maybe.
 

606
00:38:33 --> 00:38:38
So anyway, whatever.
 

607
00:38:38 --> 00:38:42
Let's put some
supports on them.

608
00:38:42 --> 00:38:45
And get these numbers down.
 

609
00:38:45 --> 00:38:50
So let's support, as
usual, the bottom guy.

610
00:38:50 --> 00:38:54
OK, so different problem
now I won't call that

611
00:38:54 --> 00:38:57
A_0, I'll call it A.
 

612
00:38:57 --> 00:39:01
I'll ask myself, is it
stable or unstable?

613
00:39:01 --> 00:39:06
The matrix is now six by what?
 

614
00:39:06 --> 00:39:13
Eight, because I've taken away,
I have four reaction forces,

615
00:39:13 --> 00:39:15
two at each support,
horizontal and vertical.

616
00:39:15 --> 00:39:20
I've got four free nodes.
 

617
00:39:20 --> 00:39:23
And six by eight.
 

618
00:39:23 --> 00:39:28
Let me put in, so six by eight,
what am I expecting now?

619
00:39:28 --> 00:39:31
Any rigid motions?
 

620
00:39:31 --> 00:39:34
No, no rigid motions now.
 

621
00:39:34 --> 00:39:38
How many solutions
am I expecting?

622
00:39:38 --> 00:39:39
Two, I think.
 

623
00:39:39 --> 00:39:41
Probably two.
 

624
00:39:41 --> 00:39:42
How many mechanisms?
 

625
00:39:42 --> 00:39:44
Well, no rigid motions.
 

626
00:39:44 --> 00:39:47
So probably two mechanisms.
 

627
00:39:47 --> 00:39:49
Now, can we find
two mechanisms?

628
00:39:49 --> 00:39:51
Alright, this is like
more reasonable.

629
00:39:51 --> 00:39:55
We can see whether whether we
get two mechanisms and whether

630
00:39:55 --> 00:39:57
they're really different.
 

631
00:39:57 --> 00:39:59
OK, what are the
mechanisms now?

632
00:39:59 --> 00:40:01
These guys are fixed.
 

633
00:40:01 --> 00:40:13
So forget my little sketch
here, and think again.

634
00:40:13 --> 00:40:14
What do you see?
 

635
00:40:14 --> 00:40:18
Alright, let's have
one mechanism.

636
00:40:18 --> 00:40:23
What would one
mechanism be now?

637
00:40:23 --> 00:40:25
There have to be two.
 

638
00:40:25 --> 00:40:28
What do you think?
 

639
00:40:28 --> 00:40:32
Sit on it.
 

640
00:40:32 --> 00:40:35
Alright, bring these guys down,
and then these guys will

641
00:40:35 --> 00:40:38
go out, is that it?
 

642
00:40:38 --> 00:40:43
OK, so a number, mechanism
number one, sit on truss.

643
00:40:43 --> 00:40:45
OK.
 

644
00:40:45 --> 00:40:45
Alright.
 

645
00:40:45 --> 00:40:49
Now, I don't know what
number two is, that's

646
00:40:49 --> 00:40:51
why I'm taking time.
 

647
00:40:51 --> 00:40:56
AUDIENCE: [INAUDIBLE]
 

648
00:40:56 --> 00:40:57
PROFESSOR STRANG: Yeah.
 

649
00:40:57 --> 00:40:59
Or could we do this, could
we bring these guys

650
00:40:59 --> 00:41:02
in and let's go up?
 

651
00:41:02 --> 00:41:03
It's the same thing.
 

652
00:41:03 --> 00:41:10
OK, so squeeze truss.
 

653
00:41:10 --> 00:41:11
Squeeze sides.
 

654
00:41:11 --> 00:41:13
Is that the same thing
that I had, number

655
00:41:13 --> 00:41:15
one the same as two.
 

656
00:41:15 --> 00:41:18
Oh jeez, ok.
 

657
00:41:18 --> 00:41:24
Is that what everybody
is agreeing with this?

658
00:41:24 --> 00:41:25
OK.
 

659
00:41:25 --> 00:41:28
So I didn't get, my
number two was no good.

660
00:41:28 --> 00:41:29
OK.
 

661
00:41:29 --> 00:41:33
What's a better number two?
 

662
00:41:33 --> 00:41:34
Hold an edge?
 

663
00:41:34 --> 00:41:35
Like that one.
 

664
00:41:35 --> 00:41:37
I'm just doing what
you say, I'm not.

665
00:41:37 --> 00:41:44
AUDIENCE: [INAUDIBLE]
 

666
00:41:44 --> 00:41:47
Just this guy, rotates like so.
 

667
00:41:47 --> 00:41:50
And this guy will
rotate, and this guy.

668
00:41:50 --> 00:41:56
That looks pretty good to me.
 

669
00:41:56 --> 00:41:59
Good, is that correct?
 

670
00:41:59 --> 00:42:01
Say that one again?
 

671
00:42:01 --> 00:42:03
So the first one was when
these two came down

672
00:42:03 --> 00:42:04
and these went out.
 

673
00:42:04 --> 00:42:07
Right, OK.
 

674
00:42:07 --> 00:42:10
And now your suggestion is,
you picked on this guy

675
00:42:10 --> 00:42:12
and held it fixed.
 

676
00:42:12 --> 00:42:15
And then this one came
down a little bit.

677
00:42:15 --> 00:42:17
It'll, of course,
how will it move?

678
00:42:17 --> 00:42:21
It will move
perpendicular, right?

679
00:42:21 --> 00:42:22
Small movement.
 

680
00:42:22 --> 00:42:24
The bar is not going to
change length, that's

681
00:42:24 --> 00:42:25
the whole point, right?
 

682
00:42:25 --> 00:42:27
The bar is not changing length.
 

683
00:42:27 --> 00:42:32
So the movement must be, it
must be a simple rotation.

684
00:42:32 --> 00:42:33
Around here.
 

685
00:42:33 --> 00:42:35
OK.
 

686
00:42:35 --> 00:42:39
Right, and of course again you
might say well, the bar really

687
00:42:39 --> 00:42:42
did change length because
that's not quite

688
00:42:42 --> 00:42:43
the same as that.
 

689
00:42:43 --> 00:42:46
But then again that's
my second order basis.

690
00:42:46 --> 00:42:50
So that one came down, and
what did this one do?

691
00:42:50 --> 00:42:52
Came down the same.
 

692
00:42:52 --> 00:42:58
OK, and this one it also moves.
 

693
00:42:58 --> 00:43:03
What is that?
 

694
00:43:03 --> 00:43:07
OK, so what am I going
to call this one?

695
00:43:07 --> 00:43:09
Fixed one.
 

696
00:43:09 --> 00:43:10
Yeah.
 

697
00:43:10 --> 00:43:13
Fixed one node.
 

698
00:43:13 --> 00:43:14
And that makes sense.
 

699
00:43:14 --> 00:43:16
Fixed one joint, yeah.
 

700
00:43:16 --> 00:43:19
And then, and rotate the rest.
 

701
00:43:19 --> 00:43:21
I think that would be possible.
 

702
00:43:21 --> 00:43:23
Yep.
 

703
00:43:23 --> 00:43:33
OK, a small prize for anybody
who, maybe handwritten, a

704
00:43:33 --> 00:43:37
picture of two really
nice mechanisms.

705
00:43:37 --> 00:43:42
Somehow this one seems a little
um-symmetric in a problem that

706
00:43:42 --> 00:43:45
so symmetric, so I would guess
that somewhere along the line

707
00:43:45 --> 00:43:50
we could find a kind of
more some symmetric one.

708
00:43:50 --> 00:43:55
But I don't see what
it is right now.

709
00:43:55 --> 00:44:03
Can the whole thing
rotate a little?

710
00:44:03 --> 00:44:10
Could that rotate, could
the whole thing rotate?

711
00:44:10 --> 00:44:12
Yeah, maybe it could.
 

712
00:44:12 --> 00:44:15
This guy would go up,
maybe that's possible.

713
00:44:15 --> 00:44:20
That's somehow got
everybody into the action.

714
00:44:20 --> 00:44:23
So I'll put or rotate.
 

715
00:44:23 --> 00:44:27
OK, so you see what
questions you get into.

716
00:44:27 --> 00:44:30
May I just draw a
different truss?

717
00:44:30 --> 00:44:38
So those homework questions
are other trusses.

718
00:44:38 --> 00:44:46
Here's one that I drew
in the book itself.

719
00:44:46 --> 00:44:52
Yeah, may I draw this, I
called it a treehouse.

720
00:44:52 --> 00:45:01
OK, so I have, so here's one
that's actually in the book.

721
00:45:01 --> 00:45:07
And it's got a couple of bars
going up, and one over.

722
00:45:07 --> 00:45:09
So that's the start.
 

723
00:45:09 --> 00:45:14
Then it's got a
diagonal and that one.

724
00:45:14 --> 00:45:17
And then here comes
the treehouse.

725
00:45:17 --> 00:45:23
OK, right.
 

726
00:45:23 --> 00:45:28
Well, just to get, let's
again get the count right.

727
00:45:28 --> 00:45:34
So what's the matrix A
for this treehouse?

728
00:45:34 --> 00:45:37
A is how many by how many?
 

729
00:45:37 --> 00:45:41
How many bars are
you seeing here?

730
00:45:41 --> 00:45:46
One, two, three, four, five,
six, seven, eight, eight bars.

731
00:45:46 --> 00:45:51
And how many unknown
displacements?

732
00:45:51 --> 00:45:53
Ten.
 

733
00:45:53 --> 00:45:56
We got five joints that are
not supported, and each

734
00:45:56 --> 00:45:59
one has two unknowns.
 

735
00:45:59 --> 00:46:00
So A is eight by ten.
 

736
00:46:00 --> 00:46:06
So I expect two mechanisms.
 

737
00:46:06 --> 00:46:15
OK, so again I'm looking
for two mechanisms.

738
00:46:15 --> 00:46:19
OK, what's one?
 

739
00:46:19 --> 00:46:20
AUDIENCE: [INAUDIBLE]
 

740
00:46:20 --> 00:46:23
PROFESSOR STRANG: It's what?
 

741
00:46:23 --> 00:46:29
This guy just falls, right.
 

742
00:46:29 --> 00:46:32
This looks unfortunately very
much like the treehouses

743
00:46:32 --> 00:46:36
that I built for my kids.
 

744
00:46:36 --> 00:46:44
Well, so linear algebra
sentenced them to fall, right?

745
00:46:44 --> 00:46:47
OK, that's one.
 

746
00:46:47 --> 00:46:51
I probably propped it up with
one more bar, but of course

747
00:46:51 --> 00:46:53
that wouldn't be enough,
because it's got two

748
00:46:53 --> 00:46:57
mechanisms, so if I make it
nine by ten I haven't

749
00:46:57 --> 00:46:58
saved the kids.
 

750
00:46:58 --> 00:47:06
OK, with eight by ten, what's
the other mechanisms?

751
00:47:06 --> 00:47:10
The whole thing could turn
the, nothing preventing

752
00:47:10 --> 00:47:11
turning here.
 

753
00:47:11 --> 00:47:13
They can't move but
they could turn.

754
00:47:13 --> 00:47:19
So the whole thing could go
over, right, the whole thing

755
00:47:19 --> 00:47:22
could just, that would be a
horizontal movement

756
00:47:22 --> 00:47:24
of all five nodes.
 

757
00:47:24 --> 00:47:27
The horizontal of
all five nodes.

758
00:47:27 --> 00:47:31
And again, slightly downwards,
but that's a second

759
00:47:31 --> 00:47:34
order effect.
 

760
00:47:34 --> 00:47:38
OK, so that's the second truss.
 

761
00:47:38 --> 00:47:40
OK.
 

762
00:47:40 --> 00:47:47
So this is really like practice
for discrete problems, for the

763
00:47:47 --> 00:47:50
problems of plane elasticity.
 

764
00:47:50 --> 00:47:57
And the point is that there are
two unknowns for each point.

765
00:47:57 --> 00:47:59
If we have differential
equations.

766
00:47:59 --> 00:48:01
So the differential equations
of plane elasticity

767
00:48:01 --> 00:48:04
are not really simple.
 

768
00:48:04 --> 00:48:05
They're not really simple.
 

769
00:48:05 --> 00:48:08
And 3-D elasticity even more.
 

770
00:48:08 --> 00:48:16
Because the points are physical
points and they can move

771
00:48:16 --> 00:48:22
three ways, and it gets
quite interesting.

772
00:48:22 --> 00:48:26
And those are the major
problems of computational

773
00:48:26 --> 00:48:28
mechanics.
 

774
00:48:28 --> 00:48:33
OK, let's say, holiday time
and I'll see you next

775
00:48:33 --> 00:48:36
Wednesday for Chapter 3.
 

776
00:48:36 --> 00:48:40
Which moves to partial
differential equations.

777
00:48:40 --> 00:48:40