Beam on Elastic Foundation Video Lecture - Civil Engineering (CE)

and welcome back in this video I'm going
to introduce the elastic curve equation
and then talk about how we can use it to
solve for displacement and slope of
beams at any point I'm going to do it in
general terms in this video and in the
next couple videos will do exactly
examples so to get started let's
remember what we were talking about with
pure bending we had this expression
where we have 1 over Rho is equal to M
over e I and this basically just comes
from if we had this typical number let's
say it's a beam and we're applying only
pure bending to it so it will look
something like this will or have an
externally applied moment on each end
well if we draw the shear force diagram
and bending moment diagram we're going
to see that we get that constant
internal moment now if we consider
another type of structure let's say we
have a cantilever beam like this
with a fixed rigid connection to some
some wall so we'll just shade it in like
that all right and let's say that we are
applying just a single point load at the
end well we're going to find that we get
a non zero shear force diagram and then
a bending moment diagram is going to
reveal that our bending moment is
actually changing along the length of
the beam so in this case with the point
load on a cantilever beam the expression
for our moment as x increases basically
as we go left from right we're going to
say that the moment in terms of X is
equal to negative P times X right so
when when x equals L what's over getting
that negative P tile there basically at
that support so this is pretty much the
case where where any time where we're
having transverse loads basically you
notice perpendicular to the member that
we're looking at so any type of point
loads or distributed loads and that sort
of thing we're going to be having to
take an account for that that the radius
of curvature is is well this is actually
that the moment is not constant along
the length and so the radius of
curvature expression has to be slightly
adjusted so we'll say that radius of
curvature is just equal to the
expression for the moment in terms of X
/ II I all right and it also turns out
that 1 over Rho 1 over the radius of
curvature also happens to be equal to
the second derivative of Y so we'd say
the Y Rd squared Y over DX squared and
basically when we're talking about the
second derivative of Y we'll essentially
define some of the things that are that
we're saying in here because this that
I'm going to circle here in green this
is the basis for that all of these
problems were resolved in for slope and
deflection so to get started
EW he R is the flexural rigidity this
thing here is just the second derivative
of Y and Y is basically the vertical
displacement of the loaded structure
from if undeformed position and we're
also going to need the intermediate step
between y and its second derivative so
we'll be looking for just be Y DX just
the first derivative of Y and as you
know that is basically just the slope
and we can write that as tan theta or
using the small angle approximation we
can just call that theta so basically
what we want to do is we'll be able to
given and given the type of structure
and loading that we have we'll be able
to find out what our moment in terms of
X is like in this case and then all we
need to do is we just need to integrate
this expression twice to be able to find
out what Y is and that will reveal us we
can set y equal to the rest of the
expression that we get and that will
give us basically the position of the
deformed structure at any any value of x
and then we can also we can just we can
just take the derivative of that and
find out what this slope is because once
this bends down a little bit they'll
have different amounts of slopes that
are changing
basically the radius of curvature will
be changing as we go along you know
progress as we go down the length of the
member so quickly just looking at what
that double integration looks like we're
going to basically have we have M of X
and we can just bring up the EEI onto
the other side so we'll have e
and then that dy d squared Y over d x
squared so basically what we want to do
is we want to integrate this each side
from zero to X and this is a constant E
I so let us throw that in zero to X this
will return us we are through on little
DX things there to be all proper so
we're going to just leave this left hand
side the same as it is so we'll have
from zero to X we have M X DX the
integral of the second derivative will
just be the first derivative so we have
e aí x d y DX and then well we'll have
an integration constant left over from
all of this stuff so we'll just call
that c1 alright so there we did we just
did the first integral um we can also
rewrite this knowing that dy DX is is
going to be equal to theta the slope so
we'll just I'll just rewrite that here
so we'll just get rid of dy DX we can
write this in as theta right and we're
saying that at PR dy DX is equal to
theta and this expression becomes super
important for us because we're going to
have I'll show you in a second we're
going to be using boundary conditions to
figure out what the constant is and then
the rest of this will just be in terms
of X and y and we'll be able to solve
for for this type of thing bear with me
for one second the next thing that we
want to do here is we want to keep
integrating want to integrate this again
once more but we just want to make note
that that is a super important
expression so we're going to integrate
the side again so we have from 0 to X MX
DX and we want to integrate that again
doing a second integration and then we
want to integrate this stuff so II I
times the integral of well here we can
write it out e I times the integral of
DX
d one plus the integral from 0 of DX 0
to x times this is a constant so we can
actually bring the constant out of that
integration and this will this be DX and
when we go and solve for this basically
it's going to look like this we'll have
0 to X the integral 0 to X of DX times
the integral of 0 to X M X DX is going
to be equal to that constant e I times
well the integral of the derivative is
just going to leave us with Y and then
the C 1 X so that will just be C 1 X and
then we're going to have a second
integration constant C 2 so this is also
going to be the key equation for us when
we're solving for when we want to find
out what the expression for Y is so
basically we rearrange this one to give
us an expression for Y and then we
rearrange this one to give us an
expression for theta and that will so
the y1 will give us basically the
position of the deform structure at any
point in this one will give us the slope
of the different slopes along the length
of that beam at any point but the
problem is we do have some unknowns in
here we have these integration constants
so basically this is an incomplete
solution so basically because this is a
differential equation we're going to be
using our boundary conditions to be able
to solve what these constants are and
then we'll have no unknowns in these
expressions and I'll just give you an
example here of what some boundary
conditions would be for typical problems
like this so in the case of a simply
supported beam with a constant
distributed distributed load on it we
know that yeah in the middle once we
apply this load it's going to be flexing
down and we'll get some slopes in here
and we'll actually get some vertical
displacement of that beam but we know
right at the reactions because these are
these are you know fixed to the Earth's
right they're not moving same over here
and you get the point these are all
fixed some reference standard that we're
taking our deflections off of so that
one's fixed to the earth and and like
that so we'll just change the color here
maybe we know that in this case so at x
equals zero right if X is going this way
at x equals zero the displacement at a
you know I should probably also label in
call that a B so at x equals zero the
displacement at Y or the deflection at Y
is going to be zero because this isn't
getting pushed into the ground right
extinct but here also at x equals L
we're saying that the displacement Y so
the displacement at B sorry Y B is going
to also be equal to zero because not
going into the ground it's staying
exactly where it is in these two sets of
boundary conditions fix this so we put
in like x equals zero and we'd also set
the Y's to be zero that would actually
give us one of the integration constants
and then we do the same thing we'd plug
in all the X is equal to L and all the Y
is equal to zero and that would give us
the other integration constant and then
we'd be able to work back plug that
stuff back into here because we would
know c1 and that we be able to find like
the slope at each you know at any point
and then we wouldn't have these unknowns
anywhere so it's a little hard to kind
of visualize what that is what I would
haven't done one with a real problem yet
but I just want to show you this so
hopefully by the next videos you kind of
see what's going on here in the case of
like the example from this video for a
cantilever with the point load the
boundary conditions that we can work
with are we know that at x equals L the
displacement of the beam at B is going
to be 0 and the other thing that we know
is that at x equals L the slope is going
to be 0 the fixed connection so whatever
angle that is to begin with it doesn't
we just assume that it doesn't actually
get deformed right at the connection so
in this case when we're using your
boundary conditions we would have to
plug in into this expression we would
set all the X is equal to L and all the
Y is equal to 0 and then we solve for
whatever we can and then we plug in
because we have a theta here we
to use this expression so whatever our
expression is with exes we'd set the X
is equal to L and then the state is
equal to zero and then we get that other
c1 and then we'd be able to solve those
and and give actual expressions with no
unknown integration constants so again
you'll see these in the next couple
videos and these last two are a little
bit more complicated each span is that
has a continuous moment is going to give
us two integration constants and as we
cross Point C we're actually going to be
getting a discontinuous moment
expression so we have to do two sets
will actually end up getting four
integration constants using four
boundary conditions and those so these
ones are a little bit harder and we will
do these also in future videos so
hopefully that was a decent introduction
to the concept and join me in the next
video and we'll actually put some real
numbers to one of these problems and
we'll it'll be easier to understand
what's going on once you see it not in
as general of terms