WEBVTT

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Shear and moment diagrams are usually
handed to you as a list of rules.

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Jump here by the size
of the load. Slope there.

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Today we derive every one
of them from a single idea:

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a piece of a beam in equilibrium
is still in equilibrium after you cut it.

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Here is where we are going.

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Underneath a loaded beam you draw
two pictures: the shear diagram,

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and the bending moment diagram.

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By the end you will sketch both
of these for a beam you have not seen,

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and know where the biggest moment
sits before computing anything.

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But not by pattern matching.

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Every value in them falls
out of one free body diagram.

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So, the beam.

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Six metres between the supports, a pin
at A, a roller at B, and one point load

00:00:48.581 --> 00:00:51.983
of twelve kilonewtons,
four metres from the left.

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Before we cut anything
we need the reactions.

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Take moments about A.

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The reaction at B acts six metres out,
the load pulls down four metres out,

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and those two have to balance.

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That gives eight kilonewtons at B.

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Vertical equilibrium then
leaves four kilonewtons at A.

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Both supports push upward.

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Now the question that matters.

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A beam does not fail at its supports.

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It fails somewhere inside,

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where the material is carrying
an internal force and an internal moment.

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What are they, and how
do they change as you walk along?

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Here is the one move
the whole subject rests on.

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Cut the beam at a distance x from A,
and throw away everything to the right.

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What is left is a piece with the
reaction at A still pushing up on it.

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That piece cannot
be in equilibrium on its own.

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Four kilonewtons up and nothing down.

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So the part we threw away must have
been holding it, at the cut face,

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with a force and with a moment.

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Call them V and M.

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Those are the internal shear and the
internal bending moment at position x.

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Draw them the standard way every time:

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V pointing down on the cut face,
and M bending the beam into a smile.

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Then a negative answer simply
means it acts the other way.

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Vertical equilibrium of the piece.

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Four up, V down, nothing else.

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So V is four kilonewtons.

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Moments about the cut.

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The reaction acts a distance x away, and M
is the only other thing on the piece.

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So M is four x.

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Look at what x did.

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Nothing at all to the shear:

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four kilonewtons wherever you cut,
as long as you stay left of the load.

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But the moment grows in proportion
to x. Watch the piece get longer.

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That is what we mean by shear
and moment as functions of position.

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One cut, two equilibrium equations,

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and we have both of them for the whole
left hand stretch of this beam.

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But only that stretch.

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Move the cut past the load,
and the free body is a different picture:

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the twelve kilonewton load
now sits on the piece we kept.

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Vertical equilibrium now has three forces.

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Four up, twelve down,
and V drawn downward as before.

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So V is four minus twelve,
which is minus eight kilonewtons.

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And notice:

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the shear did not slide
down to minus eight.

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It jumped, and it jumped
by exactly twelve, the size of the load.

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Hold on to that.

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Moments about the cut again.

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The reaction still acts x away, and
the load now acts on our piece as well,

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a lever arm of x minus four,
turning it the other way.

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That tidies to forty eight minus eight x.

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Check it at the far end.

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Put x equal to six

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and the moment comes out zero, which
is exactly what a roller has to give you.

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A roller cannot resist any moment at all.

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So here is the answer.

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Two expressions for the shear
and two for the moment,

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one pair on each side of the load.

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Nothing there was remembered.

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A cut, a free body, and two
equilibrium equations, done twice.

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Now let us draw them.

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We have four expressions and a beam.

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Drawing the shear and moment diagrams
is nothing more than plotting them,

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underneath the beam,
on the same horizontal scale.

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So: position along the beam across
the bottom, internal shear up the side.

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On the left stretch
the shear is a constant four.

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The diagram starts from zero off the end
of the beam, jumps straight up to four

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where the pin pushes in, and then
runs flat all the way to the load.

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On the right stretch it is minus eight,
so a horizontal line eight below the axis,

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running from the load across to B.

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And those two do not meet.

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At the load they differ by twelve,

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so the diagram falls vertically, right
there, by twelve. The jump is the load.

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Then the roller pushes up eight
and closes the diagram back to zero.

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That is a free check on your arithmetic.

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A shear diagram that does not return

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to zero at the far end
means a reaction is wrong.

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Watch the shear as the cut walks along.

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Nothing happens to it
until it reaches the load,

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and then in one step it
is on the other side of the axis.

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Now the moment, on its own
axis underneath the shear.

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On the left stretch it was four x, a
straight line climbing out of the origin.

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It arrives at sixteen
kilonewton metres under the load.

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On the right, forty eight minus
eight x brings it back down,

00:05:46.507 --> 00:05:49.654
in another straight line, to zero at B.

00:05:50.357 --> 00:05:51.655
For a point load,

00:05:51.739 --> 00:05:56.999
then, the moment diagram is a triangle,
and its peak sits directly under the load.

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Put the two side by side
and two things stare at you.

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Where the shear is constant,
the moment is straight.

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And where the shear is bigger,
the moment climbs faster.

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Look at the numbers.

00:06:11.723 --> 00:06:15.938
On the left the shear is four
and the moment gains four units per metre.

00:06:16.135 --> 00:06:20.524
On the right the shear is minus eight
and the moment loses eight per metre.

00:06:20.721 --> 00:06:23.589
The slope of the moment
diagram is the shear.

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And here is the one that earns its keep.

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The moment peaks exactly
where the shear crosses the axis,

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which on this beam happens inside
the vertical jump at the load.

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We have not proved any of that.

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We have noticed it, on one beam.

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So let us take a completely different
kind of load and see whether it survives.

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Second beam.

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The same six metre span
and the same supports, but instead

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of one point load it carries four
kilonewtons on every metre of its length.

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A uniformly distributed load.

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The total is four times six,

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twenty four kilonewtons,
and by symmetry each support takes half.

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Twelve up at each end.

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Cut it at x and keep the left
piece, exactly as before.

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Here is the only new idea in this example.

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The load sitting on our piece is four
kilonewtons per metre times x metres,

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and it acts
through the middle of that stretch.

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Vertical equilibrium.

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Twelve up, four x down, V down.

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So the shear is twelve minus four x.

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It is not constant any more.

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It falls off at four per metre,
which is exactly the load intensity.

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Moments about the cut.

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The reaction gives twelve x.

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The load on our piece gives four x,
times its lever arm of x over two,

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turning the other way.

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So the moment is twelve
x minus two x squared.

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A parabola.

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Watch the free body grow.

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The resultant of the load grows with it,

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and its arrow slides out
to stay at the middle of the piece.

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Plot the shear.

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Twelve minus four x is a straight ramp,

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starting at plus twelve at the pin and
finishing at minus twelve at the roller,

00:08:10.914 --> 00:08:13.631
crossing the axis
at the middle of the span.

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And the moment is a parabola,
zero at both supports.

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By symmetry its top is at three metres:
twelve times three, minus two times nine,

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eighteen kilonewton metres.

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Both of the things we noticed on the first
beam survived, and they got sharper.

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The shear was flat when there was no load;

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now it slopes, at exactly
minus the load intensity.

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The shear is straight, and the moment
has gone up one degree to a parabola.

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And the maximum moment is where
the shear passes through zero.

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Not under the heaviest
load, not at a support.

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Where the shear crosses the axis.

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There is a third thing hiding
here, and it is the useful one.

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Shade the area under the shear diagram,
from the left end out to some position x.

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Now watch the shaded area
and the moment together.

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At three metres the shaded triangle
is half of three times twelve,

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which is eighteen.

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And the moment there is eighteen.

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Keep going.

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Past the middle the shear is negative,
so the new shading counts against you,

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and by the far end the positive
and negative areas have cancelled exactly.

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The moment is back to zero.

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So the area under the shear,

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between two points, is the change
in the moment between those two points.

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We have now seen it twice.

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Time to prove it once.

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Both beams told us the same three
things, so let us prove them once,

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for any beam and any load.

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Take a slice of beam of length d x, out
of a beam carrying a distributed load w.

00:09:50.643 --> 00:09:55.647
On its left face the rest of the beam
pushes with a shear V and a moment M.

00:09:55.751 --> 00:10:00.659
On its right face, a distance d x
along, both have had a chance to change:

00:10:00.743 --> 00:10:04.250
V plus d V, and M plus d M.

00:10:04.849 --> 00:10:07.358
Vertical equilibrium of the slice.

00:10:07.531 --> 00:10:11.258
V up on the left, the load w
d x pressing down on the top,

00:10:11.258 --> 00:10:13.786
and V plus d V down on the right.

00:10:13.870 --> 00:10:16.042
The two V's cancel.

00:10:16.954 --> 00:10:20.740
What is left is d V equals minus w d x.

00:10:20.844 --> 00:10:25.000
Divide through by d x, and the
derivative of the shear with respect

00:10:25.000 --> 00:10:28.019
to position is minus the load intensity.

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Read that as a picture.

00:10:30.685 --> 00:10:33.612
Where there is no load, the shear is flat.

00:10:33.925 --> 00:10:38.175
Where the load is uniform, the shear
is a straight ramp sloping downward.

00:10:38.627 --> 00:10:41.519
And where the load is heavy,
the shear falls steeply.

00:10:42.315 --> 00:10:44.963
Now take moments about
the right hand face.

00:10:45.160 --> 00:10:47.873
The two moments oppose
each other and leave d M.

00:10:47.957 --> 00:10:51.832
The shear on the left face
acts a distance d x away.

00:10:51.916 --> 00:10:55.052
And the load has a lever arm of half d x.

00:10:55.755 --> 00:10:58.508
That last term has d x squared in it.

00:10:58.704 --> 00:11:03.244
As the slice shrinks it dies away
faster than everything else, so it goes.

00:11:03.441 --> 00:11:07.122
What survives is d M equals V d x.

00:11:08.035 --> 00:11:10.636
The derivative of the moment is the shear.

00:11:10.833 --> 00:11:12.746
The slope of the moment diagram,

00:11:12.830 --> 00:11:17.347
at any point at all, is the height
of the shear diagram at that same point.

00:11:17.543 --> 00:11:20.180
That is the relation we kept noticing.

00:11:21.484 --> 00:11:25.432
Integrate the two of them and out
comes the other half of the folklore.

00:11:25.535 --> 00:11:29.111
The change in shear between
two sections is minus the area

00:11:29.111 --> 00:11:31.446
under the load diagram between them.

00:11:32.149 --> 00:11:35.258
And the change in moment
between two sections is the area

00:11:35.342 --> 00:11:37.746
under the shear diagram between them.

00:11:38.360 --> 00:11:41.751
Exactly the shaded triangle
we were watching a moment ago.

00:11:42.803 --> 00:11:46.450
Now every rule you were ever
handed is a line in this table.

00:11:46.646 --> 00:11:50.513
No load: the shear is constant
and the moment is straight.

00:11:50.710 --> 00:11:55.587
A uniform load: the shear is straight
and the moment is a parabola.

00:11:55.900 --> 00:11:58.548
Each one is a degree higher than the last.

00:11:59.344 --> 00:12:03.432
A point force is an enormous load
intensity over no length at all.

00:12:03.628 --> 00:12:08.621
Its area is finite, so the shear jumps
by the size of the force, and the moment,

00:12:08.621 --> 00:12:11.640
whose slope is the shear, simply kinks.

00:12:12.344 --> 00:12:16.022
And a concentrated couple contributes
nothing to vertical equilibrium,

00:12:16.106 --> 00:12:20.364
so the shear does not notice it at all.
It sits in the moment equation instead,

00:12:20.448 --> 00:12:23.119
so the moment jumps
by the size of the couple.

00:12:23.315 --> 00:12:26.277
That is the line people
get wrong most often.

00:12:26.980 --> 00:12:29.787
And finally the sentence you actually use.

00:12:29.871 --> 00:12:32.910
The moment is stationary
where its derivative vanishes,

00:12:32.994 --> 00:12:35.015
and its derivative is the shear.

00:12:35.467 --> 00:12:39.133
So the biggest moment sits
where the shear diagram crosses the axis,

00:12:39.217 --> 00:12:41.633
or at a jump that carries it across.

00:12:42.430 --> 00:12:44.633
Two derivatives and two areas.

00:12:44.717 --> 00:12:48.177
Everything else about
these diagrams is a consequence of them.

00:12:49.309 --> 00:12:53.176
Third beam, and it has both
of the things people find awkward.

00:12:53.372 --> 00:12:57.294
It runs eight metres,
but the supports are at zero and at six,

00:12:57.378 --> 00:13:00.002
so the last two metres
hang past the roller.

00:13:00.799 --> 00:13:05.187
The loads are twelve kilonewtons down
at two metres, four kilonewtons down

00:13:05.187 --> 00:13:09.190
at the free tip, and a couple of eight
kilonewton metres applied at five metres,

00:13:09.274 --> 00:13:11.190
twisting counterclockwise.

00:13:11.987 --> 00:13:14.573
Reactions first. Moments about A.

00:13:14.657 --> 00:13:19.577
The twelve acts at two, the four acts
at eight, the reaction at B acts at six,

00:13:19.661 --> 00:13:22.483
and the couple goes straight
into the sum as eight.

00:13:22.680 --> 00:13:24.851
It has no lever arm at all.

00:13:24.955 --> 00:13:27.545
A couple is the same about every point.

00:13:28.458 --> 00:13:32.472
That comes to forty eight,
so the reaction at B is eight kilonewtons,

00:13:32.556 --> 00:13:36.759
and vertical equilibrium then
leaves eight kilonewtons at A as well.

00:13:36.863 --> 00:13:38.953
Both supports push upward.

00:13:39.866 --> 00:13:44.627
Now build both diagrams without cutting
anything, using only what we proved.

00:13:45.079 --> 00:13:47.378
Start at zero off the left end.

00:13:47.575 --> 00:13:52.402
At A the reaction pushes up eight,
so the shear jumps to plus eight,

00:13:52.486 --> 00:13:56.330
and with no load between there
and the twelve, it runs flat.

00:13:57.382 --> 00:14:00.769
At two metres the twelve
kilonewton load drops it by twelve,

00:14:00.853 --> 00:14:03.060
from plus eight to minus four.

00:14:03.372 --> 00:14:05.184
Flat again after that.

00:14:05.497 --> 00:14:08.656
There is no distributed
load anywhere on this beam.

00:14:10.056 --> 00:14:14.665
Here is where the couple acts, at five
metres, and the shear passes straight

00:14:14.665 --> 00:14:16.849
through it without so much as a flinch.

00:14:16.952 --> 00:14:21.435
There is no vertical force in a couple, so
there is nothing for the shear to notice.

00:14:22.138 --> 00:14:27.677
At B the reaction adds eight, taking
it from minus four up to plus four.

00:14:27.873 --> 00:14:29.972
It runs flat along the overhang,

00:14:30.056 --> 00:14:33.888
and the four kilonewton load
at the tip brings it home to zero.

00:14:34.340 --> 00:14:36.314
The diagram closes.

00:14:37.227 --> 00:14:41.744
Now the moment, and every piece
of it is an area under that shear.

00:14:41.940 --> 00:14:47.758
From A out to two metres, a rectangle
eight high and two wide: sixteen.

00:14:47.954 --> 00:14:52.054
So the moment climbs in a straight
line to sixteen kilonewton metres.

00:14:53.106 --> 00:14:58.226
From two to five the shear is minus four,
so the moment falls at four per metre.

00:14:58.841 --> 00:15:03.730
Three metres of that is minus twelve,
taking it from sixteen down to four.

00:15:04.155 --> 00:15:06.524
And at five metres, the couple.

00:15:06.837 --> 00:15:09.423
The shear ignored it. The moment cannot.

00:15:09.507 --> 00:15:15.266
It drops vertically by eight, the size of
the couple, from plus four to minus four.

00:15:15.970 --> 00:15:19.276
Below the axis now,
and the shear is still minus four,

00:15:19.360 --> 00:15:24.539
so the moment goes on falling at four per
metre for the last metre to the roller.

00:15:25.153 --> 00:15:26.547
Minus eight.

00:15:27.344 --> 00:15:30.793
Over the overhang the shear
is plus four, so the moment climbs

00:15:30.793 --> 00:15:35.623
at four per metre for two metres
and lands on zero exactly at the free tip.

00:15:35.727 --> 00:15:37.005
It has to.

00:15:37.201 --> 00:15:39.536
There is nothing
beyond the tip to bend it.

00:15:40.332 --> 00:15:42.260
Check that by hand if you like.

00:15:42.573 --> 00:15:45.581
Cut the overhang anywhere
and keep the right piece:

00:15:45.777 --> 00:15:50.093
one four kilonewton load,
a lever arm of eight minus x,

00:15:50.177 --> 00:15:52.407
bending the beam the wrong way up.

00:15:52.604 --> 00:15:55.705
Negative, and shrinking
to nothing at the tip.

00:15:56.501 --> 00:15:59.137
Two things to carry away from this beam.

00:15:59.334 --> 00:16:02.481
A couple jumps the moment
and leaves the shear alone.

00:16:02.584 --> 00:16:05.484
And an overhang drives
the moment below the axis,

00:16:05.568 --> 00:16:09.760
which means tension on the top of the
beam, which is where the steel has to go.

00:16:10.844 --> 00:16:15.929
Last beam, and you have not seen it.
Eight metres, simply supported, with six

00:16:15.929 --> 00:16:20.945
kilonewtons per metre spread over the left
half only and nothing on the right half.

00:16:21.649 --> 00:16:24.494
Here is the whole method, in five lines.

00:16:24.691 --> 00:16:27.718
Reactions, then jumps and slopes,

00:16:27.802 --> 00:16:32.482
then areas, then the maximum,
then the check that it closes.

00:16:33.279 --> 00:16:36.983
One: reactions,
and these you do have to compute.

00:16:37.087 --> 00:16:40.709
The load totals twenty four kilonewtons
acting through the middle of the left

00:16:40.709 --> 00:16:47.049
half, which is much nearer A. So A takes
eighteen of it, and B takes only six.

00:16:47.362 --> 00:16:50.358
Everything after this comes
out of the two derivatives.

00:16:51.154 --> 00:16:52.142
Two:

00:16:52.246 --> 00:16:53.396
the shear.

00:16:53.592 --> 00:16:57.316
It starts at zero, jumps
to plus eighteen at A, and then

00:16:57.400 --> 00:17:01.825
under the distributed load it slopes
down at six per metre for four metres.

00:17:02.022 --> 00:17:06.365
That is a fall of twenty four,
from plus eighteen to minus six.

00:17:07.277 --> 00:17:11.016
Past the load there is nothing
pressing down, so the shear runs flat

00:17:11.016 --> 00:17:16.520
at minus six all the way to B, where the
six kilonewton reaction closes it to zero.

00:17:17.223 --> 00:17:18.115
Three:

00:17:18.199 --> 00:17:19.326
the moment.

00:17:19.754 --> 00:17:21.903
Zero at a simple support.

00:17:21.995 --> 00:17:24.340
Its slope is the shear, which starts big

00:17:24.340 --> 00:17:29.090
and positive and is falling, so the
moment leaves A steeply and bends over.

00:17:29.275 --> 00:17:31.702
A parabola, concave down.

00:17:32.603 --> 00:17:33.602
Four:

00:17:33.694 --> 00:17:36.517
it peaks where the shear crosses the axis.

00:17:36.818 --> 00:17:41.497
Eighteen divided by six is three,
so the maximum sits three metres in.

00:17:41.694 --> 00:17:45.027
Not four, and certainly not at midspan.

00:17:46.079 --> 00:17:48.688
After that the shear
is a constant minus six,

00:17:48.772 --> 00:17:53.731
so the moment comes down as a straight
line, and it has to arrive at zero at B.

00:17:53.927 --> 00:17:57.423
Five: it does, and the sketch is finished.

00:17:58.475 --> 00:18:01.250
None of that needed
the bending moment function.

00:18:01.447 --> 00:18:06.019
If you do want the number, it is the area
under the shear out to three metres:

00:18:06.103 --> 00:18:10.306
half of three times eighteen,
twenty seven kilonewton metres.

00:18:11.218 --> 00:18:15.213
So: cut, equilibrium, and two derivatives.

00:18:15.665 --> 00:18:19.416
The jumps, the slopes
and the areas are not rules to remember.

00:18:19.612 --> 00:18:23.038
They are what those two derivatives
look like once you draw them.