g'day this video is going to provide you
with the skills to understand bending
moment diagrams and the understanding of
these diagrams is critical for any
structural engineer
this video will cover the basics to
allow you to master these diagrams
we'll also cover how they are related to
each other
and what these diagrams actually mean
for the stresses with inside a beam
my name is brendan you're structural
engineer based in Australia and I
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now let's get into it before we get into
drawing these diagrams
there's a couple of basics we need to
cover so we can have a fundamental
understanding of how beams are supported
and loaded so what forces can be on
a beam
there are three primary forces that can
be on a beam that's a concentrated
force
a uniformly distributed load or UDL for
short
or a concentrated moment now that we've
covered loads
will go over the supports that a beam
can have and there's three
typical supports that a beam can have
that's either a pin support
a roller support or a fixed support
and each of them have the unique
capabilities for supporting the
structure
a pin support stops translation
both vertically and horizontally in the
structure but it allows for rotation
a roller support now that only prevents
translation vertically but it allows
translation horizontally and rotation, as
well
so the fixed support that prevents
translation vertically horizontally and
rotation
so each one of those if it prevents that
translation you'll see a corresponding
reaction at that
location now that we've covered both the
supports
and the loads can be on a beam what do
bending moment and shear force diagrams
actually represent they're a graphical
representation of
the stresses seen along the beam so the
shear force diagram is a graphical
representation of the shear stress in
that beam
a shear stress is a stress that's
perpendicular to the axis of the beam
and doesn't really change through its
depth whereas
a bending moment induces a normal force
in the beam
a normal force changes with the depth of
the beam
and can also be either in compression or
tension
so when we have a bending moment that's
causing the beam to sag at the bottom
it's causing a tension at the bottom of
the beam
and causing a compression at the top
because of the strands at the bottom of
the beam
have having to extend further than they
are at the top of the beam that's in
compression
now why these forces actually induced
in the beam they're induced in the beam so
we maintain equilibrium
we know we need to maintain equilibrium
for the structure to be stable
otherwise it'd be moving now for a 2D
frame there's three primary
equations we have for equilibrium and
that is
that the net forces in the horizontal
direction must sum up to zero
all the net forces in the vertical
direction must also sum up to zero
and all the moments must also sum up to
zero and this is for any point along
that beam
from these three simple equations we're
able to equate
what the shear force is what the bending
moments are
and also equate the reactions at the
supports
now before we get into drawing these
diagrams we also need to go through the
sign convention we're going to use
so any load that is vertically going
down that will be a positive
force and likewise we'll use the same
sign convention for shear
so any shear stress that is going down
that will
also be a positive whereas for bending
moments
any bending moment that causes the
bottom of the beam to be in tension so
if the beam is sagging
that would be a positive bending moment
where as any bending moment that's causing
the top of the beam to be in tension so
it's
hogging like over continuous support
that will be a negative bending moment
now let's get started now we start by
drawing a free body diagram
a free body diagram is a graphical
representation of the supports and loads
that are inside that beam
for simplicity in this video we're going
to go through a simply supported beam so
pin on one side and a roll on the other
more complicated structures like
continuous beams these require to
consider boundary conditions
beyond just the three simple equilibrium
equations
now they've gone through both what
supports and loads can be on a beam
and what free body diagrams are let's
get around to
drawing bending moment and shear force
diagrams as we said we need to start by
drawing our free body diagram so let's
take our simply supported beam
with a pin on the left on the roller on
the right and a point load in the middle
what we'll do when you're running
through these examples is have one off
to the left
that will be done by variables so you
can apply that to any problem you have
and one on the right which is a work
example of how we calculate it
what supports do we have well we have
restraints both vertically and
horizontally for the pin
and only a vertical restraint for the
roller and when we look at the load as
the load is vertical
there is no net horizontal force
therefore the horizontal force
on the pin must equal zero now how about
the vertical loads well
we can use our equilibrium equation for
moments to work out what the vertical
reactions are
let's take the moment around Rb it just
means that we only have one unknown at
this point which is the reaction at
right as we know the magnitude of our
point load p
so what moments are around that point
well we have the moment p there's a
distance off right in our worked example
that happens to be five meters but
if we're working with a different
variable
that's not centered it will be l minus x
now that we have
the moment induced by that point load we
just need to divide that by L
to equate it back to the reaction force
so therefore
Ra is equal to P times L minus X
divided by L
now how about Rb we can use that same
moment equation and equate the forces
around
Ra however with the other equilibrium
equation for vertical forces
that they must also equate to zero from
that
we know that P minus Ra must equal to
Rb to satisfy this situation
so now that we have both the reactions
at both points let's get into deriving
the shear force diagram what we'll do is
take a single cut
through the middle of the beam and look
at all the forces to the left of that
point
as we're taking a section cut at that
concentrated point load
it is not seen inside this section cut
so the only force we have to the left of
that point
is the reaction force Ra and because of
that we know that the shear force at
that point must be the same so we get a
straight line
as there's no other forces and parting
upon it until we get there in our worked
example
as you can see here we calculated a
reaction force of 15 and then before we
go draw a straight line
right across till we get to our point
load now that we've calculated
the shear force up to that point load
let's move past there and work out what
the shear force is beyond there
as we can now see that reaction P we now
have
the reaction force Ra and we have P in
this situation
so therefore our A minus P must be the
shear force at this location
now that will again be a straight line
right across until we get back to the
reaction on the far side
and after we get there we just sanity
check to ensure that all that sums up to
zero
so the shear force on that side must
equal the direction forces are being
provided or equated correctly now
how about the bending moment let's take
that take that same station cut through
the concentrated point load look at the
force to the left of it
as we can see here the only force we
have again is the reaction force
so therefore the bending moment at this
point is Ra times X
which which in our case as it happens to
be the center of the bame is P L
on two or 75 kilonewtons now we could have
taken that session cut at any point
along that beam
it will just be that reaction force
times the distance we are are away
now after we move past P we could do the
same summing the moments around that
point where we have the concentrated
point load and the reaction
interacting with the bending moment
diagram however
as there is no forces to the right of it
we can take that same section cut and
look to the right and see that the only
force we have beyond there is the
reaction Rb
so therefore we go back to that same
equation we had before but we change
instead of Ra
we change it for Rb and again we get a
straight line all the way
to support as you can see here the
concentrated point load creates a
triangular shape and from this we can
also see
that where the shear force diagram
crosses zero is where our peak moment
occurs
now let's get into uniformly distributed
loads so we draw our same free body
diagram
with a pin on the left a roll on the
right and a uniformly distributed load
a distance off our Ra reaction
as you can see like like the
concentrated point load there is no
horizontal reactions and there's only
one horizontal restraint
so therefore to satisfy the horizontal
equilibrium equation
this must equal zero now how about the
reactions well we can take the same
equilibrium equation for a moment by
summing the forces around Rb
with one additional step we need to
equate that uniformly distributed load
to a point load and center it on the
center of the uniformly distributed load
so as you can see here the formula looks
a little bit more complicated it's
really broken down to the same
principles where you
times the length of the
uniformly distributed load is equal to
the magnitude of that load
and the distance it is of Rb is equal to
L minus X
minus A on two as that is the center
point of the uniformly distributed load
so from that we now have the reactions
at each point let's take a station cut
at the start of the uniformly
distributed load and look at the loads
to the left of that point
as we can see the only load to the left
of that point is the reaction
therefore the only force imparting upon
is that reaction force
until we get to the start of the uniformaly
distributed load we have a straight line
equal to the reaction force
once we move into the uniformly
distributed load we start having forces
and parting upon it depending on how
long we've moved across
so the shear force will reduce at a
gradient equal to the
magnitude of the uniformly distributed
load which in our case was three kilonewtons
this gradient will remain constant until
we get to the other side of the
uniformly distributed load at which
point will go back to a straight line
as there is no more additional forces
imparting opon it now for a sanity
check we've gone back across and we know
that the bottom of that uniform
distributed load must be equal to the
reaction Rb
which it does so we said therefore we
satisfy that situation all net vertical
forces must sum up to zero
how about the bending moment we take
that same section cut at the start of
the uniformly distributed load
this equation leads back to the simple
equation we're using for point loads
it's also a straight line up until this
point so therefore
the at the start of that uniformly
distributed load is just the reaction
force to the left of our reaction Ra
times the distance we are away once we
start moving into the uniformly
distributed load we need to start
counterbalancing that bending moment by
the amount of load that's imparted from
the uniformly distributed load in the
section card
we could do this by equating each of
those little points to a point load
equating the moment it induces at that
point
however there is an easier way and
that's understanding the relationship
between bending moments and shear forces
so what is that relationship it has a
relationship in calculus we know that
the integral
of the shear force is equal to the
bending moment diagram and likewise the
differential of the bending moment
diagram is equal to the shear force
the relationship also extends back to
the loads that are applied on the
structure where the inverse differential
the shear force is equal to the loads
and reactions
so if we know one of these diagrams we
should be able to work out the others
from this what this also tells us is
that we're able to work at
peak bending moment is where it crossed
the center line
of the share force diagram as this is the
calculus relationship
let's go back to our example so
calculating the peak bending moment is
as simple as equating the area under the
curve
up to the point across zero as we
already know the bending moment
at the start of the uniformly
distributed load we only need to
calculate the triangular area under the
curve up to the where it crosses zero
from this we have
everything we need to know to calculate
this as we know that up until the start
of the uniformly distributed the shear
force is equal to Ra
and we know that the gradient of that
line is U therefore the distance
is from the start of the uniformly
distributed load to where it crosses the
zero point is just Ra on U
now that we have the distance and the
magnitude of the shear force at that
point
it is as simple as equating these two together
so it's just Ra times the length to the
zero point divided by two
equals the area under the curve now if
you add that to Mx which the
value that we calculated at the start of
the distributed load this will be the
peak moment we could just sum
all the areas under the curve and get
the same answer so when we're going
through our bending moments and shear
force diagrams it's just a quick sanity
check to ensure that we've drawn them
correctly
all the shear forces in the bending
moment sum to zero as the peak bending
moment occur
where we cross the zero point
now i'll just quickly break down what happens if
we have a uniform distributor in point
load
but all the same principles apply
simplify this question you can split it
off into its two separate components
one we do a calculation for the
uniformly distributed reload as you can
see off to the left
and one where we do the calculation for
the point load which is off to the right
after you derive each of these diagrams
we can super position on top of each
other to get the final result
as you can see here let's try and check
that our bending moment of shift
diagrams are correct
do both of them sum up to zero as we can
see here they have both equated to zero
as we finish across from support to
support where is thepeak bending moment
occur
where we cross the zero point as we can
see here again the peak bending amount
of fifty one point four two kilonewtons
does occur
where across the zero point and where
the uniformly distributed load is
is the only section that is not a
straight line so as we can see we've
satisfied
ourselves for these requirements
hopefully this helps you master these
diagrams and if you want me to cover
anything else like continuous beams for
example
please comment below and now here's a
link to some other videos about career
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