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Welcome to thirty first lecture of video course on Tribology. Today’s topic or today’s
lectures topic is selection of rolling element bearings. We have discussed in previous lecture
about rolling element bearings, some description was given about rolling element bearings,
but selection is important. We need to select proper bearing for proper application or for
our application.
So, we are going to discuss that selection criterias in present lecture. Few points,
which we already discussed, will be repeated for proper guidance. We started one example
in previous lecture, and I am just repeating the same slide over here to say that assume
there is a some radial load and axial load on a bearing, bearing load is coming perpendicular
to the axis as well as along the axis, the magnitude of that load is also defined to
say 7.5 kilo Newton is along the radial direction or perpendicular axis and 4.5 kilo Newton
is along the axis. Shaft diameter is given and you are seeing the shaft is rotating;
that means, inner ring is going to rotate. What we need to do select a suitable single
row deep groove ball bearing. Now, in this case we are not going to select
the bearing among the diameter of the bearing only from deep groove ball bearing. One category
of bearing that is a deep groove ball bearing more like we are going to choose different
dimension or we are going to select from the different dimensions what will be the proper
bearing for our application. So, in this case the shaft diameters given
to the bearing board diameter is defined in this case and gets, based on the bearing both
diameter if we select the bearing that is a 1 4 series that is we know to find out the
board diameter we need to multiply last two digit with the 5 number or number 5 that is
a 14 into 5 is a17. Similarly, 1 4 into 570, 1 4 into 570 all
the bearings have a board diameter of 70 mm. We are going to choose bearing among these
bearings. Now in this question particularly we do not have any shielding requirement or
sealing requirement that is why we have removed the option 2 r s 1, 2 r s z, 2 r s 1, 2 r
s z. all these options will removed remaining options are 6 1 8 1 4, 6 1 9 1 4, 1 6 0 1
4. We described previously these are the special
type of the bearings, generally manufacture by one particularly company and then if it
is already mass produced technical develop or manufacturing process have been developed,
then we, the particular company will open it to the open market to the other people
also. Similarly, bearings are also there the 6 0
1 from 6 3 1 4 and we need to select one of the this bearing. To select bearing we require
rating particularly based on axial load to the static load carrying capacity that is
F a is axial load, C 0 is a static load carrying capacity and a static load carrying capacity
is defined over here, 13.2 25 31 same thing in this case 45 45 68.
And as a inner ring is rotating it says the shaft is rotating; that means, inner ring
is going to rotate. Inner ring is rotating that rotation factor will be equal to 1. Coming
to the single row side there is questions raising we need to select a single row deep
groove ball bearing. So, we need to find out what is a this ratio
F a by F r. In our case this ratio F a is a 4.5 divide by 7.5 and we need to compare
with this parameter. This e parameter generally defined on a geometry of the bearing, if this
ratio F a by F r is greater than e then we will choose x parameter and y parameter and
try to find out equivalent load. In previous lecture we did that similar thing
we say the F a by F r is a 0.6 in a most of the cases is a more than what is given as
a F e parameter; that means, we have to select a parameter x and y from this, then we require
F a by C 0 as individual bearing we have particular or typical value of C 0. Naturally, if I am
choosing suppose a I want to find out bearing from this category I will just take up the
two examples 6 0 1 4, 6 3 1 4. Now, we need to choose a bearing or we need
to find out which bearing is going to perform better even though we have other options also
just for example, we picked up only 2 bearing for time bearing the 6 0 1 4, 6 3 1 4 and
C 0 is defined. First case it is a 31 kilo Newton and the second case is a 68 kilo Newton
that is why it is given over here the 4.5 kilo Newton divided by 31 kilo Newton, 4.5
kilo Newton divided by 68 kilo Newton and when we find this ratio this ratio is not
coming it is not exactly matching any of this value. So, we need to do some interpolation.
This is what in the first case the this value is turning out to be 0.145 which is not equal
to 0.11 it is not equal to 0.17 that is why we required a some sort of interpolation between
these 2 to find out exact value of this. Based on that interpolation we can also compare
with the e value Interpol find the, to correct e value from the interpolation and based on
that we can compare and we can say what will be the parameter x and what will be the y.
We know x is not changing at all. So, interpolation does not mean much to this, x will be exactly
0.56 while y will be different it will be somewhere value between 1.45 to 1.31. Up in
correct manner we say that value will range or will be greater than 1.31, but lesser than
1.45. And what we find from our catalog, this value
is a y value is a 1.37. Similarly, we can find out this ratio F a by C 0 that is a 4.5
divided by 68 that is coming out to be 0.0662 again we are not able to get exact value this
exact value over here we have value the 0.056 and 0.084 we need to do interpolation to get
this value and with that interpolation we will be able to find out the correct value
of y, which will be ranging between or will be greater than 1.55 while lesser than1.71.
And what we get after interpolation that is a y is equal to 1.65. Now using this parameter
0.56 and 1.37 I can find out equivalent load. How to find out equivalent load ? This x needs
to be multiplied with a radial load; that means, the 7.5 kilo Newton into 0.56 plus
4.5 kilo Newton of axial load into y that is a 1.37, this summation will be equivalent
load which is turning out to be 1 0 3 6 5. Same thing for second bearing which is a 6
3 1 4. We have x is equal to 0.56 same multiply with a 7500 Newton, second will be a 1.65
into 4500 Newton. So, it will be summation the 0.56 into 7500 plus 1.65 into 4500 Newton
or summation turn out to be 11625. In previous lecture, I mentioned if I want
to select based on the applied load bearing, the first bearing will be a proper choice
because in this case equivalent load is lesser than second bearing. First case is a 1 0 3
6 5 which is a lesser than 1 1 6 2 5, but this is not all.
We need to find out how to choose a proper bearing based on other criteria’s also in
this case it is the only the load or we say equivalent load and equivalent load states
that this is a better bearing because equivalent load is a lesser and we know very well that
lesser load will be preferable for the bearing life.
So, the question comes for to estimate the bearing life or if the life is given or estimated
life is a comes in the question then how to proceed further for that purpose this is the
next slide says that consider the shaft rotational speed.
Generally bearing life is given in rotations it is the dynamic load, it is a fatigue load.
So, how many rotation that bearing can survive. Naturally we require rotation per minute reading.
As well as how much life is expected, in this case this question has been added with a 1000
RPM as a rotating speed of the shaft and the expected bearing life is a 3000 hours. We
require bearing to survive without any pit on the surface for 3000 hours or minimum of
3000 hours, with a 90 percent reliability. Of course, if the reliability increases naturally
we know whether we have to count a parameter e 1 which was discussed in previous to previous
lecture that need to be accounted. So, when we add these things, these parameters
we need to consider the life equation which was a derived in previous lectures and that
is equivalent to this equation we say the this c is a dynamic load carrying capacity;
that means, bearing can survive for 10 is to 6 cycles for this kind of the load. And
‘a’ decides, is decided based on whether the geometry is a ball geometry or the roller
geometry for ball bearing is equal to 3, while for roller bearing ‘a’ is equal to 10
by 3; that means, 3.33 that is a slightly more than ball bearing because we know roller
bearing can sustain slightly more load compared to the ball bearing.
Same relation for other loads, we know the p 1 is a lesser than C , L 1 will be greater
than 10 is to 6. Similarly, for other cases p 2, L 2, p 3, L 3. We can utilize this relation.
Now, using previous slides we can find out what is equivalent to p. We can substitute
the value of p and L 1 is already given to us, what is the expected life. If we convert
in the rotation, to convert in a rotation we have using this relation. What is that
? It says that we are counting this speed, we know this is 1000 rotation per minute.
If I multiply with a 60 then in that case it will be 60000 rotation per hour and we
require bearing life in hour, that is why this equation has come. Bearing life in hours
because we require 3000 hours, can be rearranged in such a way the C by p power to a.
That means here we are more important about the ratio of the C by p, it is not only the
value of p, but what is the value of that bearing or what is the value of C for that
bearing is important. This ratio is important for us for both the bearing speed will remain
same because same we have to mount on the same shaft the shaft is speed is given that
is defined for us. So, we need to select a bearing, there is
C by p ratio in this case. Of course, other cases also will come when we consider the
friction or we say that we require low friction by high load carrying capacity, naturally
bearing choice will be different than what we are doing it, just now we are concentrating
on the load and we are concentrating on the bearing life.
If bearing life is 3000 hours are expected naturally C by p is going to be a dominating
parameter in this case. So, for 2 bearings what we given in, we picked up when in previous
slide that is 6 0 1 4 and 6 3 1 4 we know very well 6 3 is a high bearing load carrying
capacity series. It will be dimension will be more than 6 0 1 4 naturally if there is
space constraint, say the consideration come and there is a restriction or constraint in
on the space then we have to also choose accordingly. Then we say that we do not have a more space
available diametric space they have to keep in within this envelope dimensions. Naturally,
we have to choose that bearing or from that angle of bearing, however in this question
nothing is been defined, we are free to choose bearing there is no restriction on dimensions.
Now, for 6 0 1 4 the equivalent load equivalent radial load turn out to be 1 0 3 6 5; however,
for a 6 3 1 4 bearing it was a 1 1 6 2 5 and based on this criteria equivalent criteria
we say that this bearing is a better option compared to this bearing.
Now, we are counting C because we are interested in ratio of C by p. C for 6 0 1 4 is a 39700
Newton. It is more like a 3 times or may be 3.8 times of this while coming to this side
dynamic load carrying capacity that is C is equal to 1 1 1 0 0 0 or say 111 kilo Newton
which is almost a 3 times of this. So, here slight increase in the p, but here
the major increase or more increase in the C; that means, we are going to get a better
value of C by p for this kind of bearing. Naturally this bearing is going to survive
for much longer time compared to this bearing because if this ratio is a higher then there
is a power series also, it is a cubic series naturally if this ratio is a more than 1 ‘a’
will always be in favor of the situation for us.
So, if I substitute all these value. If I substitute C over here, p over here, speed
is already defined as a 1000 RPM we will be able to find out the relation or we will be
able to find out whether this bearing can survive for 3000 hours or not or this bearing
can survive 3000 hours or not.
If we do that what we are going to get 937 hours for the first kind of bearing there
is a 6014 bearing, 937 hours which is lesser than 3000 hours; that means, this bearing
cannot be selected for our application, there defining is for the 3000 hours; however, if
the bearing life estimated bearing life are 600 hours, 700 hours, 800 hours, 900 hours
this bearing is a better option, because it have a lesser envelope dimensions and naturally
it will be lesser costlier compared to this. Then coming to the this bearing we have a
much longer life as a 1 4 5 0 9 bearing. Naturally this is a much, we say the high speed coming
over here or we say operating hours almost a 5 times compared to what we require we can
go ahead with a some other bearing like a 6 2 1 4 bearing which will show lesser dimension
as well as other lesser life which will be more than 3000 hours.
So, based on this example we can say that equivalent load as well as the operating hours
should be considered when we go for the bearing selection that is important for us to choose
a proper bearing. In this case what we have picked up the two bearing 1 is not able to
sustain or survive for the life which we require, other bearing is showing much longer life
compared to what we require. Naturally an intermediate bearing or in between
bearing which will be coming in the catalog should be selected or should be checked first
and then selected bearing based on that. So, that is giving some idea how to choose a proper
bearing for our application. In addition there are some statements which generally we quote
or which we specify or some notes generally be specified and the thumb rule applied load
equivalent radial load should be lesser than 10 percent of C, that is a recommendation
it should be lesser than 0.1 times C .If the load is more than that bearing life will not
be much. So, thumb rule says p need to be lesser than
10 percent of C and here if you can see this is a 39700 and this is a 10365, ratio will
be coming if something like that which is much larger than what we require. In this
case applied load is very high from that point of view this bearing is not selected and cannot
be selected. So, it is another way if we know what is a p and what is a C directly based
on that ration we can say or this bearing will not give a good life. So, we should not
select it we should not move to the towards the direction we should reject this bearing
at the first instant), we do not have to do this kind of calculations itself.
Coming to other side you say the bearing need to sustain some load; however, there will
be some sort of vibration because the we know the rolling elements will be having some separation
from inner ring and outer ring there will be some clearance, if there is no load there
will be lot of vibration lot of noise that is why there will be and there is a need of
some load on that. And ofcourse this there is a form rule that
p need to be exactly like this because we have seen in a number of books and we find
that every where this value is different. Thumb rule says that it should be greater
than 3 percent of C it should not happen that you are applying very low load on a bearing
and bearing is a making lot of noise in that because of noise and noise as well as a vibration
there is a possibility of early failure of the bearing. So, it should be avoided.
However as I mentioned this is not a firm rule 0.03 because we have seen the SKF catalog,
they recommend 0.02 C to be imposed for the roller bearing. The roller bearing minimum
load need to be 2 percent of C. However slightly lesser is been applied for the ball bearing,
here the ball bearing they recommend 1 percent of the C the load need to be more than 1 percent
of C and for roller bearing it need to be more than 2 percent of the C. Of course, if
they are too many parameters involved too much geometry involved in that doing a detail
analysis will give the right value, but as a thumb rule or the starting purpose we can
say p as far as possible from economic point of view need to be more than 3 percent of
the C. We can stick to that rule. There is another consideration or a you say
that this bearing life an estimated without correction factors. What we say the life adjusting
factors, we assume here bearing life for the ninety percent reliability. If we required
other than ninety percent reliability that factor should come over here. Similarly, we
are assuming there is a ideal lubrication whatever the lubrication wherever the lubrication
is required, that is getting supplied or that is been fulfilled completely when there is
a possibility of the starvation there is a possibility of excess of lubrication that
has not been counted over here. So, there will be a some factor counted over
here. Another thing is that we are assuming the bearing material is perfect, there is
no crack formation, there is no manufacturing related problems and then the kind of the
material which we are using is perfect. If material is not perfect then that factor also
should be accounted; however, the most of the industries they go out to the better material
and the material parameter or material factor is generally equal to 1, we need to adjust
this bearing lifefor the reliability as well as for lubrication parameter. For lubrication
related aspect or lubrication parameter we will be discussing and we will start the lecture
on the lubrication of rolling element bearing, for time being we are assuming this relation.
We will be carrying this relation for a next calculations also.
We discussed something about the deep groove ball bearing and that is, it gives a good
position like accuracy and it is cheap one this is more like a low cost factor, readily
available in the market. However there are some other bearings also with having a some
contact angle and we are trying to show how other bearings are better than deep groove
ball bearing when there is a axial load. These deep groove ball, these kind of the bearing
which can sustain some axial load having some contact angle. Contact angle can be 20 degree,
30 degree, 40 degree, 50 degree depends on the which bearing you are going to select
that is that is why this bearing are known as angular contact ball bearing. This is the
first picture shows a this angular contact and if I draw a line perpendicular to the
axis of this ring, we find that this is not merging with a load axis or the contact axis.
There will be some angle between this. This is a single row similarly we have a double
row also, 2 row bearings with naturally we know very well load carrying capacity of the
2 rows will be higher than the single row with they have number of rolling elements
and wherever load will be shared. Naturally load carrying capacity of this bearing will
be higher or on higher side compared to this bearing.
There is another option available, we say the instead of going for the this kind of
bearing we want what we want we want more contact. Particularly in a for axial direction
why not we go ahead with a 4 contact , 4 point contact bearings which can sustain some axial
force as well as a radial force and also it is advisable from the size point of view.
We do not have as high size as a these kind of bearings.
So, we can think about 4 contact ball bearing or 4 point contact ball bearing, double row
angular contact bearing, single row contact bearing depends what is a requirement. If
the load requirement says that we do not have an axial force naturally we will not choose
any of this bearings because they are going to show a high coefficient of friction. From
friction point of view this bearings are less advisable or lesser advisable compared to
deep groove ball bearing compared to roller bearings.
Now, there is a some sort of a specification if we see some catalog and in this case we
am referring a SKF catalog that for contact angle they do not write a contact angle to
40 degree, they do not write a 25 degree, they do not write 30 degree. What they give
they give us a some sort of suffix at the end of the bearing series and then for 40
degree they give suffix equal to B, So, if we find any time bearing with a B specification
in a suffix we say contact angle is a 40 degree, similarly if the contact angle is the lesser
that is a 25 degree then it will be known as AC series or say a suffix will be with
AC and if contact angle is 30 the suffix will be A.
So, if we are scanning through the catalogs, we can say what will be the contact angle
because for our calculation purpose we require a contact angle. So, that we can figure out
with what will be a contact life of the bearing.
Just I discussed in a previous slide about double row contact bearing or we say the angular
contact bearing and double row contact bearing, that is the this is the 3 arrangement have
been shown here this is a tandem. We say that bearing is supporting or we say enhancing
the load carrying capacity in one direction in this case.
Now, they are in a same arrangement. There is a other possibility of back to back arrangement
or face to face arrangement or some time it is known as o bearing and this is known as
x bearing. If you see the previous slide this is what we are saying that double row o bearing,
x bearing, tandem bearing. And this all three bearing are shown in this
slide, say there is a tandem arrangement, there is a back to back arrangement, there
is a face to face arrangement every configuration have has its own advantages. What we say that
back to back bearing is going to provide. Back to back bearing is going to provide more
and more rigidity more and more attachment or fixation of that bearing.
And that is why they the primary factor is that we require rigid mounting. We do not
want the shaft if it is deflected or displaced in any way, then we should go ahead with back
to back bearing. While in this situation particularly inner ring need to be clamped, but the outer ring is rotating
naturally from the load point of view this is not very highly recommend from positional
of view this is can be recommended. Coming to the face to face bearing in this
case what we are going to get a some sort of a flexibility in outer ring; that means,
we can allow some sort of misalignment in this kind of bearing there is a possibility
of slight misalignment what will happen in this case thee roller or this ball will get
shifted to one side it will be losing load carrying capacity for that, but it can sustain
some misalignment because of the this kind of the groove.
And in this case outer ring will be clamped. So, that this will be preferable choice for
a general purposes and if we require specially positional accuracy then we can go ahead with
this kind of bearing arrangement. This is a giving introduction or we say some sort
of a description about when the two bearings are arranged are in a tandem arrangement;
that means, we are supporting the load in one direction or in the back to back arrangement
or the face to face arrangement.
If I compare all these bearing what we say the 4 point contact ball bearing, double row
angular contact bearing, single row angular contact bearing and single row angular contact
bearing for the paired mounting for the tandem mounting whatever we show that pervious slide
the paired mounting and tandem mounting. And for same diameter of the shaft we say assuming
that these bearing is going to get mounted on 25 mm diameter, all the bearings are 0
5 C that is why I have shown here d 25, 25, 25, 25.
Now, outer dimension that is a envelop outer dimension of we say that final dimension of
a outer ring is also same is 52, 52, 52, 52 mm. Coming to the length side say 4 point
contact bearing is a 15 mm. Coming to the double row it is 20.6 mm. Single row angular
contact bearing which will be naturally lesser than this dimension that is the 15 mm.
Coming to the paired one tandem bearing arrangement that has a 30 mm so, that means from dimensional
point of view from axial length point of view naturally this bearing should be recommended
and this bearing should be recommended, but not this bearing, not this bearing, but this
is only the space consideration. We are concentrating on the space coming to the dynamic load capacity
or c, In this case particularly the dynamic load is a capacity is the 25 kilo Newton while
here it is a just 21 kilo Newton. Coming to this side it is a 15.6 kilo Newton
we say lesser than 16 kilo Newton which is a least in this bearing configuration. In
a single row angular contact bearing occupy the same space the 4 point contact ball bearing,
same outer dimension, same board dimension, but load carrying capacity is inferior compared
to this 25.1 kilo Newton, here the 15.6 kilo Newton naturally this bearing cannot be recommended
in compared to the 4 point contact bearing if we are not considering the friction losess.
Naturally if we consider friction is also having highest important than in this situation
this bearing cannot be recommended, this bearing can be recommended based on the load, this
can bearing can be recommended based on the life estimation say life will be higher in
this case. Coming to the static load capacity, static
load capacity also almost two times compared to this it will not be much plastic deformation
compared to this or we say that plastic deformation in this case will be lesser than this case;
however, coming to the 430 with the fatigue load limit. Lets choose also the fatigue load
limit for the four contact, 4 point contact bearing is higher compared to this. Only the
problem is the running the speed that is 9500 running the speed for the 4 contact bell bearing
reason being it is a higher friction and will generate a high heat if you rotate at a higher
speed. But in this case it is 10000, in case of the
presence of r is 14000, in presence of r single row angular contact bearing showing 15000.
So, if I compare 4 point contact ball bearings with a single row angular contact bearing
other than friction this bearing or 4 point contact bearing is showing a better performance.
So, we should select this kind of bearing; however, other parameters the cost or the
parameters is also like availability in a market bearing is not available in market
or very costly and do not want to invest this much and that much money then we need to think
about at a different way, but from load point of view from my estimation point of view 4
point contact ball bearing is preferable compared to single row.
Coming to the double row here we are able to see in this case the length is increasing
by roughly 25 percent to 30 percent. In this case the same thing is happening in load carrying
capacity and we gain much advantage from the double row angular contact bearing, slight
increase what we say some increment and load carrying capacity similar increment is happening
in the length side. So, this is a double row angular contact bearing is not giving that
much profit as much as expected. Coming to the paired bearing, paired bearing,
we say that we are giving too much length the two bearing are been combined together
and for pairing cases there is a some decrease in decrement in load carrying capacity. Instead
of 15600 now it is, it is not two times in two into 15600 slightly lesser than that was
25.1 kilo Newton which was lesser and was almost same as what we are getting in a 4
point ball bearing. Now, if I compare 4 point contact ball bearing
with a single row angular contact bearings for the paired arrangement with a tandem arrangement,
what we are going to loose, length is almost half, load carrying capacity is almost same.
Either dynamic capacity or static load carrying capacity, fatigue load is also same. So, note
we are not going to much if we are choosing the paired bearings. Here this bearing will
be better option compared to this bearing. So, what I am trying to complete from this
slide is that we need to see all kind of bearings we need to find what is a exact requirement
for us and ensure for our requirement we will be finding a number of bearing and later we
will add the friction aspect, we want to add cost aspect, we want to add availability aspect
based on that we can make a final selection, but if those selections or those criteria’s
are not dominating criteria’s or cost is not a major factor then we can choose a best
bearing load based on the life estimation.
Now, this slide shows a how to choose an angular contact ball bearing, but it was just repetition
of previous one lecture slide. So, that for different contact angles x and y factor will
be different for single row as well as for double row. When we are talking about the
double row both the factors are accounted this ratio is a greater than e or lesser equal
to e; that means, the four cases we need to consider x y factor.
However when that is a lesser than e, x is equal to 1 that need not to be considered
you can say directly F r plus y into F a that can be used whenever there is a this ratio
is lesser than e or equal to e. So, based on that we can find out the equivalent load
we can choose a proper bearing. And am going to show that deep groove ball bearing angular
contact ball bearing comparison, you can see that deep groove ball bearing is generally
a typical choice of a designers may not be a good choice when we think about a signs,
when we think about a proper Tribological knowledge for that purpose I am considering
this example.
What is this example you say that lets consider a bearing is subjected to 3000 Newton radial
load with thrust load equal to 2500 Newton you can say the ratio 2.5 divided by 3 it
is a very high ratio more than point 8 is to be carried by the 1 deep groove ball bearing
6 2 1 4 series bearing, for either the already shown over here is last 2 digit is fourteen
1 4 if we multiply with the 5 naturally we are going to get a 70 mm. It is for 70 mm
shaft rotating which is rotating at a 1000 RPM.
So, in this case particularly the inner is rotating naturally v will be equal to 1, determine
equivalent radial load to be used for calculating fatigue life or for estimating the fatigue
life we can find out our first we need to find out equivalent radial load as equal to
p. Compare life of this 6 2 1 4 bearing with that of 7 2 1 4 bearing that is a angular
contact bearing and the here the contact angle is a 30 degree; that means, whatever we choose
from catalog we need to choose for the 30 degree contact angle.
If we scan through the catalog we find out C o a static load carrying capacity 6 to 1
4 is a bearing which we are defining over here is 45 kilo Newton and for the 7 2 1 4
this bear static capacity is slightly on higher side that is 60 kilo Newton.
Now, we say that is almost a 25 percent plus or more than 25 percent high side compared
to the C 0 for the 6 2 1 4 bearing coming to dynamic capacity that is the value of C
for the 6 2 1 4 is 63.7 kilo Newton and for the 7 2 1 4 this bearing load carrying capacity
is 71.5 kilo Newton. So, from this load comparison clearly indicated naturally 7 2 1 4 bearing
is going to be better bearing compared to this bearing, just by scanning catalog we
can find the 7 2 1 4 bearing is going to show true performance compared to 6 2 1 4 bearing.
Now, we need to find out what are additional advantages. This conclusion is based on the
C value 7 2 1 4 bearing C 71.5 kilo Newton which is a greater than 63.7 kilo Newton as
a load carrying capacity or dynamic load carrying capacity of this bearing is higher naturally
7 2 1 4 bearing should perform better. Now, we need to find out second option that is
say what will be the equivalent load because as we earlier mentioned, bearing life will
always be depending on the C by p ratio we are showing here C is higher side, what is
the value of p we need to find out that. And to find out that equivalent load, we require
x parameter and we require y parameter. To find out x parameter y parameter we need to
find out what is F by r. F by r in this case is a 2.5 divided by 3 which is need to be
greater than e, if it is a lesser than e then we need not consider equivalent load or we
say F r will be equivalent load. Now, we have considered two bearings the deep
groove ball bearing and angular contact bearing and static capacity as well as axial load
will be defined. So, we can find out this ratio in this case F by C 0 is a 0.056 for
angular contact 30 degree angle we know and based on that we can find out those parameters
x and y. So, for x parameter for deep groove ball bearing is a 0.56; that means, we have
to consider 56 percent of radial load F r, while y parameter is F panel t. Panel t is
coming 71 percent, if we are not just adding F a, we are panelizing the F a with a 1.71
and that means, enhancing x F a and then adding to the radial load. We are not being the single
combination and in this case this e is equal to 0.26 naturally or F a y F r is greater
than this that we will have to choose x and y parameter to find out what is equivalent
load for 6 2 1 4 bearing, coming to the angular contact ball bearing is 7 2 1 4 bearing we
need to choose 0.39 into F r plus 0.76 into F a.
If we do the this kind of the study its clear this parameter is a lesser here. We know F
r is same for the both the bearing, apply load remains same for the both the bearing,
but x parameter in this case is a less naturally. We are going to get a first component of equivalent
load lesser in this bearing. Coming to the second component if the 0.76 again lesser
than fifty percent of this. Naturally equivalent load which is a required for us is will be
lesser for the angular contact bearing in that case we are getting win-win situation.
First thing is C is on a higher side and p is on lower side, naturally C by p ratio will
be much larger for angular contact bearing for 7 2 1 4 bearing.
Naturally life estimated or estimated life for the bearing for angular contact bearing
will be much larger compared to deep groove ball bearing. So, this analysis is also indicates
we are going to right direction whereas if we replace 6 2 1 4 or say 7 2 1 4 with the
6 2 1 4 will replace the deep groove ball bearing with angular contact bearing will
be better option for us.
So, now what whatever I mentioned in previous slide is same thing over here F r is a 3000
for the both the bearings. x is a 0.56 for the deep groove ball bearing for angular contact
bearing is 0.39. So, we are multiplied 3000 into 0.39 then naturally this sum is this
multiplication will be lesser than point multiply 0.56 into 3000. So, because of this parameter
is lesser than this when we add, we have to multiply add with this 1.71 into 2500, while
in the second case it will be 0.76 into 2500. Again this product is a lesser naturally summation
will be lesser. And when we find that this for radial load
equivalent on radial load for 6 2 1 4 bearing is a 5955 Newton, well for 7 2 1 4 bearing
this load is a 3070, its almost half of the bearing. naturally we are going to see or
we are going to estimate bearing life much higher or longer or much higher compared to
6 2 1 4 bearing to do that kind of calculation we know about the speed of the equation that
is the 1000 RPM. We already calculated C or we figure out C
from the catalog. C has been estimated or calculated, we can find out bearing life in
hours and that is for the deep groove ball bearing is a 7192 hours not very low life,
but as high as a second bearing that shows a 1,24,420 hours. So, when compared 7 verses
124 cost will not be that much higher, availability will not be that much problematic. If we for
this kind of load carrying capacity or for the axial load on higher is ratio F a by F
r is almost a 0.8 plus than we should choose angular contact bearing compared to deep groove
ball bearing. As I mentioned earlier deep groove ball bearing
is easily available and often designer choose a deep groove ball bearing, but slight more
consideration, slight calculation indicates no that is not that right choice. Right choice
will be angular contact ball bearing which is a better option or the much better option
compared to the deep groove ball bearing right. Of course at it is a simple thing that if
we want to find number of days we need to find out how many operating hours in a day
is a 8 hours, 10 hours, 12 hours, 16 hours based on that we can find out how many days
this kind of bearing is going to survive and generally for machines we calculate a bearing
life in number of days not in number of hours.
Now, there is another we say we have calculated equivalent load, equivalent radial load for
a static load carrying capacity or we say that when the apply load is a static it is
not going to fluctuate in magnitude as well as in direction. Apply load is always pointing
in 1 direction adding cost and magnitude, either radial load and axial load whatever
we have considered combination and based on the combination or we say using those parameter
the x y parameter from catalog we calculate what will be the equivalent load carrying
capacity. There is a possibility route magnitude changes we say that bearing may be operating
at the load p 1, p 2, p 3, p 4 for different hours. That is why we it is given here bearing
operates at a 1000 RPM for apply load equal to 500 Newton. This 500 Newton is been applied
for 100 hours at this for 100 hours applied load to 500 Newton and operating speed was
1000 rotation. Then some change in the system, some difference
in the operational requirement. See then in that case bearing operates at 1200 RPM, slightly
higher RPM compared to this, but lesser load to under 15 Newton and operating hours are
more that the instead of 100 hours, now the operating hours to 250 hours. In this situation
you have to estimate bearing life, that is the big question. Here we do not know what
will be the equivalent load to find out the equivalent load we use this equation. You
see we know what should be the final life 10 is to 6 cycle or operating hours based
on whatever we require, may be we say that for L 1 number of rotation bearing is subjected
to p 1 load. For L 2 number of rotation bearing is subjected to load p 2. For L 3 number of
rotation bearing is subjected to load p 3 and so on.
So, what will be the bearing life? Estimated bearing life will be required L 1 plus L 2
plus L 3 plus L 4 plus L 5 whatever number final number comes repetition of that also.
So, based on that we can find out what will be p equivalent and this is been derived same
equation, we are able to estimate equivalent load based on this relation.
This is what we do simplify this relation what we can do, we can say let us say assume
L is expected life. Would l is expected life they can divided numerator as well as denominator
with the L; that means, here the summation will be L 1 plus L 2 plus L 3 plus L 4 plus
L 5 and all and this is overall life. So, this ratio will turn out to be 1 because all
a summation of L or L’s will be equal to L and that L divided by L will turn out to
be 1, but here what we are going to gain L 1 divided by L let me see the fraction of
the life, 10 percent of the life will subject to load p 1 20 percent life is subjected to
load p2. So, that fraction is important for us, that
is why I say L 1 divided by L is F 1, L 2 divided by L is F 2, L 3 divided by L is F
3 and now if we know F 1, if we know F 2, if we know F 3 and the loads corresponding
load on for those number of rotations then can find out what will be equivalent load.
Once I know the equivalent load we can find out the operating life, here is the total
number of cycles and then we can find out the bearing life in number of hours, number
of days, number of rotations whichever is required.
But here again we are assuming the life bearing is subjected to load for the sometime it is
repeating, it is not continuously changing for L 1 rotation the bearing is subjected
to load p 1 and more like a static load, but magnitude is changing after certain rotation.
There is other possibility load is changing on magnitude from p max to p minimum in the
average rotation, say maximum load and minimum load.
In that case we do not use a summation and p min plus p max divided by 2 because we know
higher load is much more harmful compared to lower load that is why we go ahead with
this kind of ratio say 33 percent of the p min and 66.6 percent of a p max. Again we
are not directly going for the maximum load, we know the bearing is the fatigue is subjected
to fatigue, lower load is always half. So, we want to count that that, that is why we
say the equivalent load for dynamic cases. Dynamic cases in the sense the load is continuously
changing, is changing between the p max and p min in 1 rotation then we should count equivalent
load this 1 this relation. However there is another possibility, a combination.
You say p 1 is a constant load the way we consider in the previous slides, but p 2 is
a rotational load at every degree that load is a changing its direction. So, this is a
not magnitude is changing, but the direction is changing that is why we have divided p
1 constant load always directed to 1 direction while p 2 is continuously changing.
So, in that situation we use this relation. This relation shows say the p 2 1 plus 0.5
in bracket p 1 by p 2 square of the that similarly in the second the p is equal to p 1 one plus
0.5 p 2 by p 1 square of that. Question comes when should I use this relation when then
should I use this relation? This is for the only this situation when p is a constant load,
p 2 is a rotational load. Again these are the equation there is no complete science
behind that it has been based on the experience. Now, wherever the dominating factors comes
if the p 2, if the p 2 is much larger than p 1 then we should use this relation. If p
1 is much larger than p 2 then we should choose this relation what is a overall thing or some
time people can think what will happen if p 1 is equal to p 2, if p 1 is equal to p
2 we can choose any of this relation both the relations because we are going to convert
to the same solution same results you can see p 1 is a p 2 and this will be 1 one plus
0.5. Now, here the p 1 and p 2 is the same relation.
So, whenever p 1 p 2 I can choose any of the relation p 2 is greater than p 1 I will choose
this relation, when p 1 is greater than p 2 I will choose this relation. So, depends
on what is the magnitude of p 1 and p 2 and compare it and based on that we can find out
what will be the equivalent to radial load carrying capacity or equivalent load capacity
required from the system.
Now, once we know p we can estimate bearing life. Let us have an example; see in this
case we have seen 1 ball bearing is running at a 4 piece wise load. What is a piece wise
load? We say that in first case it is subjected to load or the 4 kilo Newton for certain rotation
may be say when is the operating at the 1000 RPM and for the overall time, we say the for
10 percent of the total time is running at the 4 kilo Newton. For that 20 percent of
total time, it is running at 2000 RPM, but lesser load that is a 3 kilo Newton; for 30
percent of the total time this bearing is running at 3000 RPM and applied load 2 kilo
Newton, for 40 percent of total friction of time or time is running at a 4000 RPM and
applied load is 1 kilo Newton. Here we required f 1, f 2, f 3, f 4 and nothing
is been given. So, what we need to do we need to find out first the rotation. How many rotations
at this load? Naturally 0.1 time into 1000, we are calculating based on that we find that
this product is giving 100, here again the 0.2 into this rotation or this RPM is been
the 400. What we are saying that number of rotation have increasing or we do not need
final number of rotations for time being when we are trying to find out the equivalent load
final number of rotation we required when we trying to find out overall life.
In this case we are just calculating fraction and we say that ratio here the 1 into 2, it
is the time which we assume in the minutes or it is a dimensional as such, but if I assume
from that point of view that is the 0.1 into 100 we say total time. I can assume here which
is a can be neglected later; that means, this is not giving absolute value in the rotation,
but it is giving in that proportion. So, the this product is giving value proportion
to rotation multiplied will be the t here and this 100 is one, 400, 900 and 1600 we
sum up and normalize it what we will get a fraction, that fraction F 1 for the first
case is a 0.033 was much lesser; that means, this bearing is not operating at a much longer
time for much longer number of rotation or 4 kilo Newton. That means, happening only
for the some time when we say some rotation or some rotation. Coming to the second rotation
is a 0.133 for 3 kilo Newton the fraction is a 0.3 on a higher side for the 2 kilo Newton.
A major junk a major number of rotations are happening at the 1 kilo Newton.
Now, if we use this relation in this equivalent load what we get p 1 q that is a 4 kilo Newton
is here in the highly penalty on that, but this friction is lesser, since 1 that second
thing is a 3 kilo Newton over here. So, we need to substitute p 2 is equal to 3 kilo
Newton and F 2 is 0.133, for third case what we are saying that this 2 kilo Newton, p 3
is equal to 2 kilo Newton and F 3 is equal to point 3 here in this case the p 4 is equal
to 1 kilo Newton, F 4 is equal to 0.533. Naturally we need to represent this kilo Newton and
Newton we have to say p 1 is 4000, p 2 is a 3000, p 3 is a 2000. p 4 is a 1000 and this
fraction will be multiplied with that when we take a cubic route of this what we get
is equivalent load is 2054 Newton or this equivalent load is a roughly 2 kilo Newton.
And this can be shown over here the major junk of time is happening in 1 and 2 and in
some time between the 3. So, this is a average is coming somewhere here, it is not in between.
It is not 2.5 because the fraction of the time which is spent on the 4 kilo Newton and
a 3 kilo Newton is much lesser than this fraction. So, this is an equivalent load carrying capacity.
They can utilize this to find out what will be the overall life of the bearing based on
this. We will continue on the selection of the bearing in our next lecture. Thank you
for your attention.