Macaulay's Method (Double Integration Method) Video Lecture - Civil Engineering (CE)

Hello friends here in this video we will
see what is Macaulay's method and how it
is to be used so let us get started
Macaulay's method it is also called as
double integration method this is the
other name of Macaulay's method it is
also called as the double integration
method next in this method a section is
taken extreme far away from the left
hand support so in my college method we
have to take a section at extreme far
away from the left hand support of the
beam then in this is systemic is
systematic calculation of bending moment
is done at the section next
in my college method we use the double
integration concept to get slope and
deflection respectively
so in my college method we use the
double integration concept to get slope
and deflection respectively next your
let me explain that with the help of a
diagram here I am considering a simply
supported beam having length L then on
this simply supported beam there are
three loads I am just taking this
example that there are three loads w1 w2
and w3 all three are point loads as we
can see here next the distances of the
load are w1 is located at distance a1
from the left support w2 is located at
distance a 2 from the left sense of wood
and w3 is located at distance a 3 from
the left hand support now suppose for
such a beam which I have drawn here we
need to find out the slope and
deflection at various points by using
makalah's method then the steps would be
so I am going to write the steps in
makalah's method
and the steps are first is calculation
of support reactions that is first we
are going to find the support reactions
at a it is RA at B it is RB so first we
will get the value of support reactions
that is RA and RB next after getting the
support reaction we will take a section
X X at a distance small X which is small
X from left hand support so here left
hand support is point a so from point a
we will consider the section X extreme
far away from the left portion so
suppose I am taking the section here
that is extreme far away from a and just
before B so this is the section xx
located at a distance small X now when
we are using the makalah's method the
first thing would be taking moments
about section xx so here I will go I am
going to take the moment at this section
xx and for that whatever are the
clockwise moment I will treat them as
positive and anti-clockwise moment I
take them as negative so taking the
moments moment at xx will be equal to
clockwise moment RA into X that is
positive this is the first section next
w1 will produce anti-clockwise moment
now the distance between xx section and
w1 is X minus a because X minus a 1 will
give us the distance between W 1 and
section xx so that is minus because it
produces anti-clockwise moment W 1 into
X minus a 1 the second section next W 2
is also producing anti-clockwise moment
and the distance between xx section and
W 2 is X minus a 2 so it is minus W 2
into X minus a 2 then after this W 3 is
producing anti-clockwise moment so again
negative minus W 3 into the distance
between W 3 and XX section is X minus a
3 X minus a 3 so now once we get this
bending moment equation this bending
moment moment at xx can be made equal to
e aí d square y by DX square because we
can write the bending moment at any
section by using this formula we can
replace moment at xx by E I D square Y
by DX square and this term I will copy
it as it is
now once I have reached here this year I
have completed the third step now in the
fourth step integrating the bending
moment equation once we get so now
integrating the bending moment equation
we have here the bending moment equation
I will call this as equation 1 now
integrating this on the left hand side I
have a I it is a constant I will keep it
as it is a I is a constant next
integration of d square Y by DX square
that is dy by DX is equal to now RA will
remain as it is because it is a constant
value integration of X that will be X
square by 2 plus c1 that is the constant
of integration this will be our first
term next minus w-1 will remain as it is
constant next thing is this bracket will
be integrated as a whole so X minus a 1
that will become X minus a 1 whole
square divided by 2 the second term then
minus w2 will remain as it is and X
minus a 2 will become X minus a 2 whole
square divided by 2 at last - w3 X minus
a 3 square whole square divided by 2 so
here I have integrated once I will keep
this as
equation number second next the fifth
step is integrating equation two again
we get now I am in case of my college
method once we get dy by DX this dy by
DX represents slope that is the angle by
which the beam has been bent so once we
get dy by DX equation which is from this
we will be getting the slope equation
now we have to integrate the slope
equation to get deflection so I am
integrating this equation number second
so here we have e aí dy by DX
integration is y ra will remain as it is
this two will be written here X square
integration becomes X cube by 3 plus C 1
into X plus another constant C 2 because
I am integrating it twice next minus W 1
by 2 X minus a 1 whole square so
integration becomes whole cube divided
by 3 the next term minus W by 2 X minus
a 2 whole square becomes whole cube
divided by 3 minus W 3 by 2 X minus a 3
whole square integration becomes whole
cube divided by 3 so therefore e I into
y is equal to R a 2 into 3 that becomes
6 X cube plus C 1 X plus C 2 minus W 1
upon 6 X minus a 1 cube similarly W 2 by
6 X minus a 2 whole cube
- w3 by 6x minus a 3 whole cube so now
here also we have an equation I will
call it as equation number third next if
we look at equations again an equation
third in both the equations we are
having constraints in equation two we
are having c1 in equation 3 we have c1
c2 both so we need to find this constant
so the next step would be step number
five in step number six we are going to
find the constants by applying boundary
conditions so by applying the boundary
conditions we will have to find the
constants and if I look into the diagram
the boundary condition is when the
distance X is zero here deflection Y is
zero so the first boundary condition
would be at X is equal to zero Y is zero
and the second boundary condition is at
X is equal to capital L at X is equal to
L y is again zero so deflection is zero
so after reaching here the next step
would be we would be putting that is in
equation seven applying boundary
conditions in deflection equation one at
a time to get constants
c1 and c2 so once we know the boundary
conditions we can apply them or put the
values in the deflection equation which
is equation number three to get the
values of c1 c2 and once we know c1 c2
that is step number eight in makalah's
method once c1 and c2 are known then put
their values in equations two and three
respectively that is once we know c1 c2
values first we are going to put the c1
value in equation number second and that
will give us the slope equation then c1
c2 values we will put in equation three
that gives us Y that is deflection
equation so one c1 and c2 unknown we can
put their values and the in equation two
entries respectively
hence we will get the slope and
deflection equation respectively and
from them we can calculate slope and
deflection at any point in a given beam
so once we get after putting the
constants c1 and c2 in equation 2 and 3
respectively from Equation 2 we will get
slope equation that is dy by DX and once
the constants are known we know the
value of RA just we have to shift the
distances and once we shift the values
of X that is different values of X we
will get the different values of slope
similarly we can get different values of
deflection from Equation 3 so here and
this
video we have seen how to use my college
method what is meant by my college
method and by using the double
integration method we can get slope and
deflection for a given beam