So let's draw a picture.
And some of you may be able to tell what I'm drawing.
I'm not an artist.
So perhaps you can tell that's supposed
to be a spring scale, like a luggage scale,
or a fish scale, something like that.
Perhaps you have used one of those.
And inside, there's a piece of metal.
And it's a metal spring.
And it's wound up, usually, and it stretches out.
And the reason that this works is
because most metals have a very simple relationship
between force and displacement.
I could draw the same concept as you have in the spring.
Now, it could illustrate, as many people do in high school
physics, they take a spring, and I've
drawn in just a block of material,
and it's loaded with a force.
And what you do is measure the length of that spring,
or that piece of material, as you
apply different loads to it.
And if you did that, you find that most materials have
a straight line relationship.
That is, it's linear.
There's a linear relationship between force and displacement.
If you measure the slope, we have use the letter K for that.
Try that with the slope.
So it goes through the origin, as the form
of-- the equation that you could fit
to that is the equation for a straight line, which is y
equals mx plus b.
It's a common way that it's written.
Of course, it's just force in this case
equals K, the slope, or the proportionality constant, times
x, the distance, the length of the material.
That's actually kind of a handy equation,
and it's got a name that's commonly attributed to it.
It's called Hooke's Law.
Perhaps you've heard of that.
And it's just a straight line relationship
between force and displacement.
The problem with that is if you were
to say, take two materials now, or two samples, rather,
of the same material-- so I could take
sample A of aluminium, say, at 60,
61T6 aluminium, exactly the same alloy,
and then I take another sample of exactly that same material,
call this one B. But as I said, let's say
they're the same material.
Same material, that's important, same material.
And I load both of those up with a force, and I then take that
and I plot now force versus distance for those two samples,
A and B. I'd find that A had a higher slope,
and had a higher constant of proportionality,
or sometimes that K is called the spring constant.
And A would have an apparently higher spring constant
than B would.
And in fact, B would fail.
It would break at a lower value of force.
And the reason that that's-- well, that's a bit confusing,
is because it should be the same material.
And so really as engineers, we'd like
to have a material property that we could use that would
be independent of sample size.
And so the solution to that is actually just
to divide by the cross sectional area.
So let me show you what we do.
So here's another sample, again.
We're going to pull axially, uniaxially, straight
up and down.
And there's the force, and this is the cross sectional area.
Actually, I'm going to actually just change
this figure up, just a little.
So there we go, all that stuff's gone.
So I'm going to draw in this area again, without the force,
because what I want to do is I want to call that area,
I want to call that the initial area-- cross
sectional area, actually.
Oh, man, I'm all over the place here, today.
Cross sectional area.
And then what happens when you apply a load to this-- so I'm
going to draw with the yellow is it elongates.
And there's this new area, which we'll consider later.
So this is-- I'm exaggerating, hopefully
obviously, the amount by which this is extending and getting
narrower.
But this is longer now, and narrower.
But what we're concerned with at this point
is this initial cross sectional area,
before we had a load on it.
And if we take that, and use that,
put that in the denominator below the force,
we can define this thing called the stress.
So this is the stress, and that's the Greek letter, sigma.
Have to put the equations into little red boxes,
just so you can identify that they're
in fact an important equation.
And that is, again, A sub 0 is the initial cross sectional
area.
Now actually, I'm not completely through here,
because this-- although we often just say
stress, technically this is in fact the engineering stress.
And that will become obvious, or important a little
bit later, when we want to in fact
account for this reduction in cross sectional area
that you see there.
That will be accounted for with something
called the true stress.
But for now, the engineering stress
is actually very, very useful.
It's very commonly used, and it's just
the force over the initial cross sectional area.
Similarly, we can define the engineering strain,
the Greek letter epsilon as the changing
length over the initial length.
So this would be the initial length.
And then there's this little elongation here at the bottom,
and this elongation at the top.
Now, because it has translational equilibrium,
it's not moving if that half of the elongation
at the top, and half of the elongation at the bottom.
If I fixed it on the surface, it would all be on one side.
But just to define what that change in length
is-- there you go-- we put it into our equation,
and that defines our engineering strain, delta l over l sub 0.
And so the nice thing about this is at the end of that,
if we plot now stress versus strain,
we find that for both A and B, that
is, these two different size samples we had up here,
the curve is the same.
It's independent of sample size.
So now what we see that the twos curves are
independent of sample size-- they're
the same-- we can actually plot that Hooke's Law again.
So remember, we had Hooke's Law involving
force and displacement as f equals kx.
So now if we plot stress versus strain-- and you'll notice,
I apologise I didn't actually introduce this properly,
but down here on the x-axis, I plotted strain
where I would formerly have plotted force and displacement.
Displacement x is replaced with strain,
just as we said force was not really sufficient,
because force depended on the size of the sample,
and we normalised by cross sectional area to get strain.
Well, length isn't enough.
This length is insufficient.
It depends on sample size.
So what we do is we normalise by the length of the sample.
So we take that change in length,
the little bit extra length that we get if we're stretching it
out, divide by the total length over which that elongation is
occurring, and that gives us this quantity, which
we call strain, or the engineering strain,
specifically.
So I've plotted engineering stress versus engineering
strain, and we see that at least the initial portion
of the stress strain curve for most materials is straight,
and regardless of size of the sample, the curve is the same.
And in fact, the slope here, where we saw previously
the slope of the straight line, f equals kx, had a slope of K.
Now, the slope of the stress versus strain curve
here, the linear portion, is the modulus.
And so we can rewrite Hooke's law
as stress equals this constant of proportionality, which
is actually called the-- new terminology,
I like to use yellow for.
So that's the Young's Modulus, that's
the proportion, the constant of proportionality
between stress and strain.
And we can write Hooke's Law as stress equals Young's Modulus
times strain, and I like to also put, as you know, equations
in a little red box.
So that's an important one.
And that is, as I said, that is Hooke's Law.