Example of Double Integration Method for Beam Deflections Video Lecture - Civil Engineering (CE)

so in this problem we're going to apply
the double integration method to
calculate for a cantilever beam with a
linearly distributed load starting from
0 to 3 Kip per foot 9 foot long the
maximum the maximum slope or yeah
maximum rotation or slope and the
maximum deflection of this cantilever
end and usually just by properties of
cantilever is when you have a loading
like this all in one direction the
maximum deflection and rotation are
usually at the tip of a cantilever okay
so that that's just something to know
because this is fixed the slope here is
0 and it increases gradually as you go
to the end of the beam so the first
thing we're going to do here is and
we're and we're asked to use this
coordinate system using X X defined here
from 0 okay and so the first thing that
we need to do is as always is calculate
reactions calculate reactions okay or
draw a schematic and calculate the
reactions so in this schematic here
really we have a y for the fixed end a
concentrated moment and a here and a X
here okay and really you can draw those
in any direction as long as you
understand that the negative sign that
you get from a global equilibrium
equation means that you you know it's
going in the opposite sense okay but
here so by doing that if I sum forces in
the y equal to 0 or the vertical right
here I have a y minus the resultant of
this you know the triangle which is I'll
put the purple arrow here and that's
going to be one-half the base which is
nine feet times you know I'm talking
about the area of this triangle right
here times three kip per foot which is
what was that number again
13.5 Kipps equal to zero okay and that
tells me that a Y is equal to 13.5 Kip's
and my positive result indicates that my
sense or my direction is correct and I
then to calculate the moment at a at the
fixed support the moment equal to zero
most some moments about a point
I'll some moments about this point right
here okay about the dot this dot ok
equal to zero and I would get here n a
the reaction minus the resultant times
the distance from here to the resultant
and that distance is the centroid of
this triangle which is 2/3 from the zero
side so this is the resultant is 13.5
kips times two thirds times nine feet
equal to zero and then I get that MA is
what is that three 681 kip feet and the
positive indicates that my sense of this
MA is also correct that means it also
means that I know which side compression
and tension is going on to right at
least at the support I know that tension
is at the top compression is at the
bottom and then really nothing is going
to nothing no other boundary conditions
are going to affect that so I know
that's going to be true for the entire
length of the beam anyway so if you get
this wrong you're going to you know you
just you're going to get the whole
problem wrong so it it was really
important that you get these reactions
right okay so now we have that and so in
order to use a double integration method
we want to come up with a moment
function as a function of X here so
we're going to make a cut here right
here so we're going to let's do the
moment function so two is the moment
function
and this moment function I'm going to
draw a schematic of it again here I'm
not going to draw the whole beam out
this time just control a little stick
right here I have at the support here I
have 13.5 Kip's and a moment right here
of 81 kitsy okay internally for my
moment function I'm going to draw a
moment here which I know is going to be
a function of X which is cut over here
at X I also have a shear here but I
don't need that need that shear function
and I also have the distributed load
that's also cut right here whose value I
don't really know right here what that
is because it's varying linearly from 0
to 3 Kip per foot as a function of X so
now let me let me solve for M here using
all these variables and I'm just going
to simply I called I'll call this point
here oh and I'll saw some for moments
about Oh equal to 0 so here positive is
this way and so I'll have em this what
I'm solving for this M let's see oh let
me calculate the resultant of this so
let me use a purple here this purple
this resultant right here is 1/2 W times
X yeah and it has a distance from the
cut of x over 3 yeah okay and here M is
equal so now I have M plus 1/2 W times X
here's that force this is a force that's
the force times the arm times x over 3
minus 13.5 Kip's times X plus 81 Kip
feet equal to 0 and my moment function
now is let's see
Oh minus 1/2 w x squared over 3 plus
13.5 kips times X minus 81 kip x feet ok
so that looks good so far
except now I need to know what W is and
W if I go back up here W is associated
with that point right there W and so I
can look at by similar triangles I can
figure out what W is so here by similar
triangles I would say W over X is equal
to 3 Kip per foot over 9 feet and that
means W is a 3 kit per foot over 9 feet
times X which means W let me write that
one more time how many times have to
write it kit per foot squared 3/9 or one
third kit per foot squared times X and
so now when I put that into here my
moment function finally becomes minus
1/2 times one third kit per foot squared
times X times x squared over three okay
which plus you know this term minus this
term okay which becomes 3 times 3 is 9
so minus X cubed over 18 plus 13 point
oh let me keep some units here to get
per foot squared plus 13.5 Kitt's times
X minus 81 Kip feet and this is my
moment function right there I think
that's the hardest part about this
problem really
it's a moment function
normally it would be finding the
boundary conditions and continuity
conditions in a double integration
method but for this particular problem
it's just really the that the moment
function and and so here now now what we
need to do is apply double integration
so double integrate so we have this
stinking polynomial right so we have
this double integration method double
int right here so now if I integrate
this one time or take the antiderivative
I would get e I times DV DX which is the
same as e I times theta or the slope is
minus X to the fourth over eighteen
times four which is 72 okay
hip per foot square yeah that works
yeah that works that works plus thirteen
point five divided by two which is six
point seven five Kip times x squared
minus 81 Kip feet times X and then oh
plus a constant I'll call that c1 okay
then I want to do e I times V the
displacement function or the equation of
the elastic curve minus X to the fifth
over 72 times five is 360 Kip over feet
squared let me know if any of my algebra
is wrong please or my math 6.75 Kip's
divided by three x cubed minus 81
divided by two
Kip feet x squared plus c1 times X plus
c2
that looks okay and now I want to find I
want to calculus
so any questions so far nothing yeah all
right good
so here for is to calc constants which
means we need boundary conditions and
the boundary conditions for cantilever
are very easy and what are the boundary
conditions so V equals zero at x equals
zero good and theta equals zero at x
equals zero and that tells me if I plug
in x equals zero to this first equation
if I plug in x equals zero I get c1
equals zero Hey and then if I plug in x
equals zero to this one that was easy c2
equals zero
yeah awesome right and then to find the
maximums right I'm going to plug in
numbers I would plug in numbers for I I
that we're given and so to find find V
Max and theta max just substitute x
equals nine feet into the equations and
just make sure you don't mess up your
units and that's it okay