in the last class we were looking at what
is called as a half car model
and then we reduced it to what is called as
quarter car model 
okay we reduced this to what is called quarter
car model now let us understand this there
been lot of equations on it let us understand
what we are trying to do now with the quarter
car model as i told you in last class that
you can have a full car model but to quickly
understand the effect of suspension system
okay and two of the very important frequencies
that go with this system of a suspension sprung
mass and unsprung mass as well as the tire
of course okay
we are looking at this quarter car model in
actuality we have to consider the road which
gives the input how road is represented how
that goes as an input all those things are
important and we are going to see that from
the next class onwards okay next four to 5
classes we will understand the road the statistical
nature of the road how road is represented
how that affects the performance of the vehicle
all those things will do that from next class
so we quickly finish some of the things that
we want to do in the quarter car model and
then we will go to half car model and finish
that part as well hope we will be able complete
it or else we will maybe it will spill over
so what is a quarter car model 
that is the what is called as the m sprung
mass that is unsprung mass right if i change
anything just tell me i hope i used the same
notation if there is any difference in the
notation tell me okay this one of problems
like and so on
right so this is the road input r of t this
we called it as y and that we called it as
z right i think that is why we are left now
let us write the equations for this as i said
this is cut into two halves the front and
the rear and we had put ms to be distributed
mass between the front and the rear from the
total sprung mass which we called as b/l this
is the front and for the rear we said that
it is ms*a/l right
so the system was the same you have to now
rewrite or replace ms by the corresponding
front or rear springs by front and rear and
so on okay in other words in the quarter car
model we are delinking both right sometimes
we will analyse it with complete ms okay as
one vehicle and the suspension and so on as
one whole vehicle as well okay
now let me write down the two governing equations
because why two because i have two degrees
of freedom which is y and z so ms*z double
dot is the first one okay which is we had
already discussed this cs*z dot-y dot – ks*z-y
same fashion again write down mus*y double
dot okay so write down yesterday we did that
write down the forces that are acting so it
will be –c*y-z-k*y-z
there are the two forces that are acting okay
and then the tire force -kt*y-r right these
are the two forces that are acting now we
can do two things as i told you we are going
to look at that is the input of course look
at it as linear time in variant system okay
r can be expressed in terms of the sinusoidal
and exponential function and so on and so
forth right
so what we are going to do to follow this
kind of system approach we can find out the
eigenvalues and eigenvectors and then just
express them express the result as a sum of
eigenvalues and the eigenvectors so that is
what we are going to do i think we did that
already yesterday yeah we did that right and
did we right down z/r that we also wrote down
okay fine i think z/r and y/r both of them
we wrote down sorry yes so we wrote down both
of them and so
z=magnitude=kt root of ks squared+cs squared
omega squared/d square+c square omega square
e square okay and y/r=kt*root of ks-ms omega
square whole square+c square omega square/d
square+c square omega square e square okay
i think this is where we stopped right like
d and e we had written than already right
so the other one this is the displacement
the ratio of the displacements z and y due
to r right the other one is actually what
is called road holding
okay road holding is the force that the road
exerts or the vehicle exerts onto the road
so how do i express the road holding what
is road holding here what is the force kt*r-y
okay so that is the force that supply it please
note that all of them are perturbations about
the static equilibrium okay in sense that
it is the equilibrium due to the weights and
so on it is about that that we are oscillating
so that i substitute it rearrange it and so
on so let me call that how i called it so
let me call that as h*e power i omega t that
is the force so that h/kt*r okay = omega square
root of ms mus omega square –ks*ms+mus whole
square+c square omega square ms okay the techniques
are simple the only thing is that the problem
is you have huge i would say equations are
very long that is because of the number of
terms that are involved so apart from that
they are not very difficult to understand
i suggest that you derive it because of the
lack of time i am not going to do that please
derive it and check how this equations are
obtained okay so most of the questions which
i am going to express now i am not going to
derive it because all of them are just algebra
bring it to the left hand side bring it to
the right-hand side put them together divide
it you know those are the things they are
not very conceptually very difficult to understand
right
“professor - student conversation starts:
in the model why we are not consideration
the dampers of the tire yeah the damping usually
the damper effect is not very high for the
tire absolutely absolutely but those effects
to into rolling assistance those effects are
for rolling assistance okay professor - student
conversation ends”
it is a very difficult topic see sometimes
what we do in engineering is that when it
is very difficult to handle we make an assumption
that it does not exist okay so damping itself
since you asked this question i want to comment
on this damping is actually not a very easy
topic to handle to understand specially in
the first course like this damping can be
classified into what is called as proportional
damping and non-proportional damping okay
and the proportional damping in the sense
that the damping i am sure you had the background
on vibration so you talked about mode ships
mode superposition we are going to see that
again in the next course<so where we had very
nice mattresses okay when i looked at the
mode ships okay and they are orthogonal and
so on right we actually we did what we call
as coordinator transformation and then we
had those nice diagonal terms we uncoupled
the differential equations and so on right
so that this is the beauty of a proportional
damping where c = alpha time m+beta times
k when i write like that my c matrix behaves
nicely and then i am able to get a very similar
expressions because c is now split into m
and k okay i will take some of them to m and
some of the them to k and then i get a nice
matrix okay if the damping is not proportional
in other words if i am not able to write c
= alpha m+beta k which is the case actually
with tires they are nonproportional damping
then the way i have to handle becomes difficult
there is a way of handling it it is quite
an evolved ideas but not here in this course
so people feel that the best way to handle
this is what differences does it make if it
is not much different then i will remove it
in fact you would notice that in the next
step i am going to do i am going to look at
the natural frequencies and even in that case
i am going to remove the damping in order
to calculate the natural frequencies of an
umdamped system okay
does it mean especially you know the more
important damping is this shock absorber does
it mean that the shock absorber does not have
any effect okay when i remove it fortunately
the difference between the natural frequencies
of an undamped system and the damped system
the differences are very very small okay because
though in the whole of mechanical engineering
this system is very highly one of the very
highly damp okay “under damped system”
it is a very highly damp system but still
it is an under damped system okay so the difference
between we are going to see that difference
between the natural frequency of an undamped
system and a damped system will be small so
when i want to calculate the natural frequencies
okay which looks like every mechanical engineer
has such a good hold of it understands it
so is very easy to now calculate the natural
frequency if i do not have an undamped system
but i cannot say the same thing with respect
to the motion or mode shifts for this motion
rather the ratio of z versus y i cannot say
that because once i put this damping here
okay the amount of motion for example in a
typical car if i do not have a damper the
ratio of motion will be 80 times for say for
example for omega 2 but if i put a damper
then the ratio of motion comes down to nearly
10 in another words it is nearly an order
of magnitude difference in the ratio of that
is z/y okay the ratio of the displacement
or ratio of motion will be huge
so i want to calculate simple only natural
frequencies that is fine i can do that but
i do not want to calculate natural frequencies
alone i want look at other things then damping
will have an effect okay and so on and that
is what we are going to see now but in the
case of tire okay it has been a tradition
to remove it so there will be in fact there
had been very interesting papers which are
now question for example tire noise what are
the material parameters which have an effect
on tire noise this is one of the say issues
stiffness has an effect g double prime has
an effect and so on now people are re-looking
at it the people said that initially in the
90s there were papers which said that the
materials do not have an effect but now there
are papers which say that material has an
effect on noise and so on so in simple words
this is a very traditional approach
in fact tire damping here itself our friend
here his research is on the tire damping and
it is not a very easy thing to calculate what
is that you are getting 2% right so it is
a phd topic he is looking at how tire damping
can be calculated how tire damping can be
measured you know in fact he is doing all
that okay it is a topic which is of interest
which i would say is not at this level of
vehicle damage this is an introductory level
okay fine
let me get back so we are now looking at so
that are the two things and now we will look
at a situation where i have two now given
the sinusoidal input i know how to that that
but i have a situation where i have to actually
say tune the suspension how do i optimise
the suspension so in order to optimise the
suspension okay i have to understand how the
system behaves with respect to omega or in
other words excitation
okay so i have know how the system behaves
with respect to excitation why is it because
the road consists of a number of frequencies
so ultimately when i use whatever be the technique
that we will discuss it later to split this
up into a number of frequencies of excitation
then i have to understand that the c value
effects are
the c value effects are not the same at every
frequency this you already know from elementary
vibration classes okay i am going to exploit
that and you will see right now that that
is not going to be very straightforward technique
before deriving okay let us look at the result
this is not the right way to do it but this
is just to motivate okay now why am i deriving
it that is why i am plotting this here
suppose i now plot i do the derivation of
free vibration system where i find out what
would be the z suppose i plot omega the excitation
frequency versus a normalised z which i would
let me plot it as how do i plot it omega square
zr okay that is normalised plot okay when
it plotted like this then my plot is going
to look like this for c=0 okay at two places
you are going to have difficulties and those
two places are the two natural frequencies
which is omega 1 and omega 2 okay those are
the two places where you going to have trouble
now when i introduce damping then this curve
is going to change obviously all of you know
that this would not be infinity okay and it
will come down right now so for example at
one level it will come down like that and
it will be like this right when i now increase
damping it may come out like this okay and
then pass through the same point may increase
okay pass through the same point and it can
be like that
in other words without damping i have two
peaks and with various values of damping i
have curves where it will be bad it will be
good and so on in other words depending upon
the frequency my z value okay would now vary
r is actually the amplitude of the road excitation
now in other words a good damper in one frequency
you cannot say that it is good or little dampen
out in other frequency okay and so on lowe
damping is good in 1 bad in another
higher damping is good in one bad in another
and so on under these conditions damping optimisation
becomes difficult how do i dampen out okay
number one that is why you have techniques
where the damping varies with the frequency
of excitation and that is in the realm of
controls okay but every car does not have
it it is still lot of things are still in
the what you cal as experimental stage so
given a choice what will you do you then
you do not know i mean what frequencies are
going to be excited may be one of them so
what would be our choice the best choice would
be that i have c as flat as possible okay
is it a correct choice? no but as a first
cut you can look at c as one which is almost
flat which is not that it is high at one place
lot at other place and so on so in other words
optimising the suspension specially the shock
absorber is not a task which is very obvious
because of this kind of graph where it is
behaviour is not going to be the same one
compromise results in an increase at other
place and vice versa
okay in order to understand this is clear?
you are now going to see how we get this graph
that is going to be my derivation right so
with a that background let us quickly run
through this this stuff which you know already
so i am not going to spend a lot of time on
this you have done this already for say single
degree of freedom system right where you know
that there is a crossover point okay where
the damping characteristics are different
on either side of point and so on so same
thing happens here okay hence we would run
through this whole thing
first let us look at we had finished this
part let us first look at undamped quarter
car model and is very simple to arrive at
what we call as the natural frequencies okay
follow the same thing you have the equation
remove c get the determinant expand the determinant
all those things are very straightforward
you can do that so you will get that ultimately
if you have any doubts i will answer it but
i think that is a known thing hence 
so the characteristic equation for the undamped
system go back to that equation and then get
this right
we solved this equation of course for omega
and call that as omega 1 square and omega
2 square okay and this result in 
okay that is expression for omega 1 square
and omega 2 square i am not very happy with
expression it is too long okay let us do some
simple tricks immediately get the two natural
frequencies without much as i said keep on
assuming conscious of the fact that my assumption
will not affect my results okay
this is the standard technique fine but can
someone suggest how else can we do it one
is to do a mathematical for example the first
thing we observe is that the sprung mass is
far far greater than unsprung mass and the
tire stiffness is far far greater than the
ks these two assumptions are not assumptions
they are facts of flare yeah 
this should be ks i will just check that i
think that is ks okay i will just check that
next class i will get this thing
so these two are well known right not let
us get some facts clearly so there are now
2 natural frequencies okay one would say corresponding
to the body and the other corresponding to
that of the unsprung mass okay not that they
actually participate but if you look at the
ratios it is the ratio of excitation okay
we can assume as if that once fixed and the
other is exciting and so on right
typical frequencies we will see in a minute
can be calculated by a very simple assumption
here so what the assumption be what do you
think of this assumption a very simple assumption
here the first is with respect to body okay
since it is an undamped system i am going
to remove this and that this mass is greater
than this okay i can neglect it
so i can actually look at it as if there is
small mass here okay as if it is going to
be like that and two springs in series with
a mass ms so that omega 1 square is written
as k/m okay so this is k spring k tire k tire*k
spring/k tire+k spring*ms okay so this is
very straightforward in fact you can achieve
that by an approximate say for example you
can do a taylor series approximation here
and you will get a very similar result right
so the other omega 2 is arrived at by assuming
that the other end is fixed and you have this
mass mus m unsprung mass 
like that okay this is not in series because
both of them are going to take the loads so
they are in parallel so that omega 2 square
can be written in terms of k/m which is k
in this case is kt+ks/mus right so that gives
me the very simple one minute answer for undamped
natural frequency with all the approximations
that are put in place clear okay yes we will
look at some typical results okay as to just
understand these factors
for example for typical a car which you can
take from one of the text book so let us say
that unsprung mass is 100 kg sprung mass is
1000 ks=70 kt=560 if i calculate now omega
1 square so in terms of cycles per second
hertz so f1 happens to be 125 and f2 happens
to be 1264 and at f1=125 and at f2 okay these
are typical values this omega 1 which is hertz
of course 125 is typical value and usually
is between 1-15 for all the cars is between
1 to 15 for all the vehicles that is what
we call as the body frequency right
and omega 2 which results in frequency of
1264 hertz it a typical value and is called
as wheel hop frequency of the typical value
for the unsprung mass so this is a typical
wheel hop frequency okay so look at that results
yt for the second case look at that 89 or
90 times zt which means that the z is almost
you know negligible displacement and hence
in other words you can assume as if this is
stationary and this guy is jumping but this
is for an undamped system this is what i was
telling if i damp the system okay this would
come down to say -10 or -12 and so on so this
may be not vary much this would vary -10 or
-12
“professor - student conversation starts:
sir why is there is a -sign yeah why is there
is -sign because these are mode shapes so
obviously the way this you know they oscillate
it is oscillating like this okay so that is
how this the mode shapes are for this okay
professor - student conversation ends”
okay fine we had already derived the expression
for the other cases and we had that is the
approximately the omega 1 and omega 2 what
we just now saw okay you can draw this graph
from whatever we know as i said optimization
of the suspension system is not a very straightforward
problem but one of the earliest attempts to
do optimization in the 1950 is by just saying
that look at this graph for various c find
out that c for which the curve let us say
this is a point a all these graphs passes
through point b point c and so on okay
it so happens that looking at the graph people
said that if the slope of the graph the slope
of this curve happens to be flat at a another
words slope is 0 okay or dy/dx=0 at that point
then this curve is flatter almost throughout
the range of omega so this is the first condition
you know very early okay first condition that
they had put
in other words when they duo of omega 2 z
of course r is what is a given as a constant
this=0 okay this is the first cut optimized
c so the result is that c optimized = root
of ms*ks/2*root of kt+2ks/kt
“professor - student conversation starts:
no no these are natural frequencies these
are natural frequencies and that frequencies
for which the mode shapes are calculated okay
these are two mode shapes for that frequency
that frequency of oscillation what is the
motion movement of the unsprung mass and the
sprung mass that is what we are expressing
it here no no we are talking about 0 right
answer=0
this is the free vibration case if it were
to vibrate in that frequency okay then what
would be the relative motion this is relative
motion look at that carefully this it he relative
motion between the sprung mass and the unsprung
mass clear so there is no input here no yeah
see more shapes are usually expressed as a
vector okay you would say z y or z theta and
so on right as a vector for a particular frequency
say omega 1 right i am expressing that as
a ratio z/y normalizing it and expressing
it as a ratio if z happens to be 1y equals
to this and z=1 y is this and so on right
so it just as a ratio i am expressing the
what we call as a mode ships clear professor
- student conversation ends”
so this is an optimized one i am not very
happy with it since we have to move further
away from this okay so in other words i have
to look at the statistical nature of the road
and see whether some optimization can be obtained
from that perspective as i said first cut
this i fine now i am going to shift gears
i am going to quickly run though a case where
i am going to back to my half car model again
make some assumptions and look at from the
quarter car i will go over to the half car
model so what are lessons learned in the quarter
care mode
the lessons learnt is that two of the important
frequencies okay which are not going to change
in fact we have done extensive testing with
many cars in india okay all these cars the
natural frequency when i put an accelerometer
at the action position the body position and
all that all these cars you can very clearly
see this z and y okay have natural frequencies
between 12 to 15 right so that is the natural
frequencies which we have and the natural
frequencies or the wheel hop frequencies is
between 11 to about 135 14
you know this i the range in which we have
got these frequencies so these are two important
frequencies wheel hop and the body frequencies
we will see it is important again in the in
the next class and let us look at what is
called as the half car model i am going to
simply this half car model
okay i am not going to derive it again completely
because i hardly have 4-5 classes i want to
shift to that statistical nature of the road
when in doubt simplify that is all i am going
to look at the way i am going to look at the
half car model i am going to have that is
all my model okay i am going to remove the
unsprung mass okay this is going to give me
some important inputs remove the unsprung
mass okay put this whole spring in the front
the tire as well as the suspension and call
that as kf and kr right and then just consider
the displacement here as well as the theta
okay
and call that jy in other words what essentially
i have done is to compress some of them and
expressed this in a very straight forward
fashion okay now what i am going to do is
very standard i am going to do how many equations
i am going to get i am going to get two differential
equations okay we will call that as mass m
okay we have removed or we have neglected
the mus so that is ms so we will call that
as ms z double prime+kf*z+a theta a of course
you know what is a that is b+kr*z-b theta=0
okay that is my first expression my second
expression is jy*theta double dot+kf*a*z+a
theta these are the 
two equations i am now looking at natural
frequencies undamped natural frequencies okay
because what happens is that for the proportional
damping this x is not that much affected we
will see that in a minute so i am removing
the damping and we are looking at the undamped
case
how far it is correct lot of questions okay
first cut is fine right now we do not have
time to complete this but what are we trying
to do here we are trying to find out see there
are 2 modes in this let us understand the
physics equations follow there are 2 modes
to it what are the two modes one is what is
called as the bounce mode and the other is
the pitch mode okay a bounce mode and a pitch
mode so those bounce and the pitch modes are
given by these two things okay
now because of the fact look at that equation
because of the fact that the equations are
coupled unless i have a special cut they are
coupled rewrite it you will see that both
these equations are couples equations okay
so what is meant by coupled equation simply
means that z will have an effect on theta
and theta will have an effect and so on there
is only one condition into which i can uncouple
it yes i uncouple it let us know look at before
we go into the details look at the solution
for coupled and uncoupled equations
what do i mean physically by coupled and uncoupled
equations suppose i have a method of uncoupling
the condition is very simple we will see that
in a minute okay so in other words you expand
it i have to make the first equation theta=0
that will be the condition so what will be
condition kf*a=kr*b it becomes an uncoupled
equation that is what you will see everywhere
right here also you will see the same thing
okay i think –b or may this is minus just
check that we will come to that in a minute
so that both the equations will become uncoupled
yes minus there what is meant by uncoupled
equation so when there is a bounce mode if
this is the my vehicle where there is a bounce
mode there were will be pure bounce at a particular
frequency when it gets excited it will be
a pure bounce and when it is a pitch mode
then it will be a pure pitch okay this is
called as the uncoupled vibration they are
not coupled
on the other hand when i have both of them
coupled how it would vibrate i have ms okay
it would vibrate in 2 modes omega 1 and omega
2 okay in one case it will vibrate like this
with respect to a particular point it will
vibrate so in other words there will be displacement
as well as there is a rotation which will
take it to this so it will vibrate like this
okay in another mode it will vibrate like
that
okay both of them are now coupled this is
uncoupled and this is coupled okay point number
1 this is clear? point number 2 what is the
significance why we are doing this yes i understand
the assumption but why we are doing this?
so what is that i want from this i want minimum
vibration okay or minimum disturbance when
i sit in the sprung mass so where i am going
to get the disturbance when the vehicle goes
say for example a front wheel goes over a
bump okay now what happens in this case?
when the front wheel goes over a bump wherever
you are sitting you are going to oscillate
okay because that is the mode so rear wheel
goes over a bump again you are going to oscillate
and where there is a bounce whether you are
sitting in the front or rear you are going
to oscillate so would you prefer this no you
are going to oscillate all the time
okay let us look at this situation suppose
this point happens to be in the front seat
let us say that your front seat which is very
close to the suspension and this happens to
be the rear okay now one of them oscillate
when it oscillates because at the place where
you give the input when you would see that
the front bump would not affect the rear rear
bump would not affect the front and so on
this is a much more optimum from point of
view of the vibration characteristics when
compared to this right so in other words how
best can we get this kind of behaviour or
what are the conditions under which we get
this behaviour when compared to this is going
to the aim of the derivation right the first
thing is very clear i have to avoid coupling
right
the second thing is how do i place what is
this called as the node okay in the front
how do i place this node at the rear this
is my next question okay how i am going to
answer this questions by looking at the mode
shape for omega 1 and omega 2 and that is
what we are going to derive from this we will
do that in the next class
