Three Moment Equation Video Lecture - Civil Engineering (CE)

alright guys welcome back in this video
it's going to introduce three moment
equation and talk about how we can use
it to solve for statically indeterminate
beams to get the reactions and the shear
force diagram of bending moment diagrams
so over here we have three moment
equation there's more than one way to
write it but this is the way that I like
to write it makes sense to me basically
what it is is it compares three moments
across two spans of a beam so if this is
a span one and this is span two and then
we're going to have three points we're
gonna have Point a which is on the left
of span one point B which is on the
right a span one and the left of mat two
or basically the point right in between
and see the point on the far right hand
side is bad to know some people write
this a little bit differently some
people write I'll see an R for like left
center right some people when they're
talking about using subscripts one and
two like I would have it all ones to be
the length of section one some people
might refer to that even as like l8b or
something so just depending on just
watch out for the subscripts when you're
when you're looking at this and
depending on who's written the equation
it might look slightly different
depending on yeah what that person
prefers the other thing to notice in
here is if you have a constant e I the
whole way across then this e is gonna
dropped out the whole way if E is like
double on one side or something then
just pay attention to that make sure to
include a two at the bottom basically
but otherwise AI flexural rigidity is in
every denominator and so usually that
drops out in these equations as well
alright so what we want to do basically
is of we want to be able to draw the
bending moment diagram for a problem
like this but what we can do is instead
of just drawing the entire many mo
diagram right away we can use the
principle of superposition to kind of
draw it a little bit by parts so the
first part when we're doing it by a
superposition would be to just draw each
span as if it was simply supported
that's carrying the applied loads so
those would look like this where for the
first simply supported beam with a point
load right in the middle we'd get this
triangular shape that's a height of PL
over 4 and then for a simply supported
beam with a distributed load we'd get
this parabola parabolic shape with a
height Bava WL squared over 8 now you
can find these values and tables or you
can just
do it with a you know virtual cut and
everybody diagram and you'll figure out
that these are the the heights there of
these been among the diagrams for the
simply supported situations now when we
do draw these as simply supported you'll
notice that the ends of simply supported
beams the the moments are always zero
now that's cool when we compare it to
the actual structure because ma the
internal moment at a will actually be
zero and the internal moment at C will
be 0 because they are at the ends of
spans that aren't you know subjected to
a applied load or at a rigid connection
or something like that and that's going
to be helpful for us when we use the pre
moment equation because you'll notice
there is M a and B and M C in here well
in a problem like this we know that M a
and M C are going to be 0 and so really
that's going to simplify this a lot but
they might not always be 0 for example
if you have an overhanging beam
something like this maybe with a point
load at the end ma would no longer be 0
but really ma would be really easy to
find because you could just take a
virtual cut there and the magnitude of
this point load times the distance and
you'd be able to find MA but so MA
typically the the edge moments like ma
and C will be known and it were pretty
easy to find just by by using statics
before you actually proceed with the
rest of the problem so anyways moving on
with the principle of superposition so
we start out with the bending moment
diagrams as if they were simply
supported and just carrying the moments
basically caused by the applied loads so
then what we want to add in is the
moments just caused by the supports
because we need them we need to include
the end moments here if they are
existing because the actual moments on
the actual bending moment diagram are
maybe not zero at all of these points
and so to account for any nonzero
moments at a or C we draw them on like
this but in this case we'll get to it in
a second but we do know that MA and M C
are going to be 0 anyways and then we
also draw the internal moment at point B
here due to the reaction of e ok so we
have the internal moments caused by the
applied loads and the internal moments
of the supports and if we were to add
those together using the principle of
superposition we would end up getting
something that looks like this for this
problem no M a and M C in this problem
are zero so basically the superposition
of this graph in this graph well it's
just basically this graph because this
one is just a horizontal line that
equals zero because it would be zero
here and zero there so basically we have
this triangle and this parabola is
coming from right here and then we have
this long triangle like that it goes the
whole way along the bottom now if you
add these two together basically what we
can even do is I like to put a little
bar on this side and label this as the
bending moment diagram so this is like
the actual bending moment diagram if we
just add this line at the top to this
line at the bottom basically at every
point going along the span and if we did
that it's going to look something like
this
where this would be the actual bending
moment diagram for this system and you
can look at this like where we're adding
zero plus this negative value we're
getting this negative value down here at
this peak right here we have this
positive value plus this negative value
nets out to this positive value and then
somewhere in here like if we add you
know the peak of the parabola plus half
of this triangle about the height here
we're getting the height of the parabola
mid span here now the interesting thing
to note that is this is a good way to do
these problems graphically if you're
dealing with point loads because adding
triangles to other triangles straight
lines it's really easy to do graphically
and you can generate the bending moment
diagram pretty rapidly
but in these regions where we're adding
parabolas to straight lines or triangles
it gets a lot Messier and this problem
becomes actually a little bit trickier
and this sort of becomes not the best
way to solve the problem but it is
possible but you'll just notice like
it's complicated because the actual
maximum of this parabolic shape is not
going to be at the mid-span of this
section it's gonna be over to the right
a little bit and it's just kind of a
headache I'll do an example I think
where we actually solve it this way but
you and I'll compare it to an example of
solving it the other way just so you
know but this is this is one way to
basically get the bending moment diagram
is to fully like graphically do this
with addition the other way that we can
go about this is if we come back up to
three moment equation we basically just
plug and chug the whole way along only
things in here that we don't know yet is
this a 1 a 1 is just the area this is a
1 this is the area of the simply
supported bending moment diagram what
that's carrying the point loads and then
this is a to the simply supported area
of the area of the minimum diagram of
the simply supported diagram that's
carrying the loads in this section and
then X bar 1 basically the centroid of
this of this shape is that the distance
from that Center right to basically the
left hand side so this would be X bar 1
and then the distance from this centroid
over to the right hand side basically
was going to the outsides that were that
we're looking in it for the two span
section this is going to be X bar two so
basically you just you find this area
you find the distance of the centroid if
it's not if this shape is not
symmetrical so I decide to me I might
have to do a little bit of composite
areas or something to define the
location of that centroid but basically
you just plug and chug in here in this
example we know that ma is 0 MC is 0 so
the only real unknown in this whole
problem and this whole expression would
be an B you would calculate for an B and
then basically you would get once you
know and B you would know the internal
moments at a B and C and if you've
watched this slope deflection method
problems then from here on it becomes
exactly the same method so once you once
you determine M a and B and M C then
what we do is we draw each span with a
virtual cut just before each support and
once we label on M a and B and M C which
is the actual moments right at those
cuts then we can label on the a the
shear a the shear on the left hand side
of be the share on the right hand side
of beat and then the share at C then the
other thing you have to do also is you
draw on the applied loads and now you
can just use these as Freebody diagrams
to just solve for the shears each end
because you can just do sum of the
moments about at one point and then do
this sum of vertical forces
and then from there you'll know the
shear and the moment just on the end of
each span just to the side of each of
those reactions and then from there you
can use that to plot the shear force
diagram and then the bending moment
diagram and then off the shear force
diagram you can calculate or you can
just pretty much pick off what the what
their reactions are so this method seems
to be a lot more straightforward and and
usually easier and less prone to error
then trying to graphically draw the
bending moment diagram by summing up the
two functions but like I said I want to
show some examples of both but yeah
that's kind of the general method about
how the three moment equation works it's
kind of weird to like to talk about it
without doing an example because it's a
little bit abstract but I join me in the
next video and we'll go through an
example with actual numbers and you'll
see that it's actually not so bad