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Now I am going to discuss mathematical model
for material removal in ultrasonic machining
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now these models there are various models
that have been proposed by various researchers
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for example Miller proposed mathematical model
for understanding the mechanics of cutting
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in ultrasonic machining in the year 1957 M
C Shaw proposed a model for the same purpose
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in the year 1956
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Then Kazantsev and Rosenberg proposed a model
for understanding the mechanism of material
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removal in ultrasonic machining in the year
1965 and there are various other models which
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have been proposed and reported in the literature
survey However I am going to discuss only
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one that was proposed by M C Shaw
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Now let us understand the mechanism of material
removal as per the analysis reported by Professor
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M C Shaw he made certain assumptions during
the development of the model as we see he
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assumed that abrasive grains are spherical
in shape and they are having constant diameter
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that is small D and equal to 2 capital R
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However in reality it is not so the shape
is not spherical and the size of the abrasive
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particles also not uniform it varies within
a certain range Second assumption which is
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important to note is volume of the material
chipped off from work piece is equal to half
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the volume of a sphere of diameter equal to
2 small R here please note that this 2 small
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R is not equal to the diameter of the sphere
2 R is something different as we can see
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So let us see the second assumption again
volume of the material chipped off from the
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work piece is equal to half the volume of
a sphere of diameter equal to 2 R note that
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here 2 R is not equal to D because D is the
diameter of the abrasive particle or the grit
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while 2 R is the assumed diameter of the fractured
volume considered as sphere
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As you can see here that this is the grit
and when it is penetrating upto this depth
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it is assumed that the volume of this is equal
to the semi sphere having the radius equal
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to 2 R as you can see here here is the tool
which is vibrating ultrasonically and here
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is the grit having diameter D and this is
the 2 R is the diameter of the hemisphere
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of the fractured volume
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That figure indicates clearly that capital
R is equal to D by 2 where small D is the
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diameter of the abrasive grit and at the bottom
you can see small R is the radius of the hemisphere
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of the fractured volume of the material and
then other things H is the penetration to
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which the abrasive particle is penetrating
inside the work piece surface and capital
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R minus H is the rest of the arm of this figure
shown over here it will be utilized for the
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calculation in the following slide
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So material removal by throwing mechanism
and hammering mechanism two of the mechanism
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that have been considered I will show you
what do they mean in fact in the following
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slide as we can easily understand here suppose
this is the tool and here is the work piece
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and say this is the abrasive particle in the
slurry when this tool is vibrating ultrasonically
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this will hit this abrasive particle as a
result of this this particle will move towards
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the work piece and penetrate inside the work
piece as in this particular case or in this
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particular case
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So this is known as throwing model because
here particle is thrown by the tool towards
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the work piece In the second case this is
the work piece this is the tool and if the
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gap between the bottom face of the tool and
top face of the work piece is smaller than
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the size of the abrasive particle then this
tool will be hitting the abrasive particle
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and part of the abrasive particle will penetrate
inside the tool part of it will penetrate
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inside the work piece and this is known as
hammering model
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Other modes of material removal like cavitation
and chemical action of the slurry on the work
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piece are neglected in this particular model
Radius of the crater formed due to fracture
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are as I have shown in the figure in the earlier
slide here crater is assumed to be hemi-spherical
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as shown over here with the dash dash line
Radius R in terms of abrasive particle radius
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and depth of penetration is small H as I have
shown earlier they can be expressed like R
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square is equal to capital R square minus
within bracket R minus H whole square and
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this is very clear from the previous slide
where the figure is shown
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So R square and R square will cancel we are
left with 2 R H minus small H square and this
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comes out to be approximately equal to 2 R
H here H is the depth of penetration and as
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you can see here this and this is very small
and square of this depth of penetration H
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square will become still smaller hence it
can be neglected now let us find out what
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is the volume of material removed per grit
per cycle and let it be expressed by VG
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Now this VG will be equal to half of the volume
of the sphere now here you can see we have
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assumed that this is the semi sphere so actually
this is the sphere so half of the volume of
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the sphere the volume of the sphere is given
by 4 by 3 pi R cube and half of it becomes
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half multiplied by 4 by 3 pi R cube and this
comes out equal to 2 by 3 pi within bracket
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2 R H raise to power 3 by 2 and here as you
can see in the previous line we have shown
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that R is equal to approximately 2 R H so
if we take this one then we get this equal
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to K1 H into D whole raise to power 3 by 2
millimeter cube per grit per cycle this is
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the volume where K1 is the constant which
can be evaluated by the experimental results
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H and D are to be given in millimeter Here
H is the depth of penetration as have been
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shown over there
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Now let N be the number of impacts per cycle
where N is an inverse function of cross-sectional
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area of the grit and whose diameter is D Now
let us see here two figures on the left side
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there is a abrasive particle of the smaller
size having diameter D and on the right side
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you has the abrasive particles of the larger
size
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Now as you can see larger the size of the
particle larger number of the abrasive particles
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are covering the same area So you can write
N is equal to K2 by or K2 upon D square where
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K2 is a constant of proportionality
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Now is every grit active? This is the question
which arises as you can see here in the figure
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suppose this is the tool and there are various
abrasive particles whichever particle is hit
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by the tool the question is whether every
particle is reaching to the work piece and
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hitting the work piece or some of them are
just dying out their energy
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Energy die out while travelling from the tool
towards the work piece so what it is done
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let K be the probability that a particular
grit under the tool is active or it is hitting
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the work piece and removing the material then
the volume VC is the material removed per
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cycle can be given by K multiplied by VG into
number of impacts which becomes equal to K
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multiplied by K1 into K2 H D raise to power
3 by 2 divided by D square that comes out
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to be equal to K into K1 into K2 under the
root H cube by D
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Now here we have substituted the value of
VG that was found in the previous slide so
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volumetric material removal rate MRRV is equal
to K K1 K2 multiplied by F under the root
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H cube by D and this unit is cubic millimeter
per second Now here we can see it is multiplied
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by F where F is the frequency of vibration
of the tool and here you can see you can determine
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K K1 K2 from the experimental results F is
already known that is frequency of vibration
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and D is the diameter of the grit which is
also known but we do not know the depth of
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penetration H that is given over here
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So we will be using two different models the
question arises how to know the depth of penetration
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there are the two models proposed by the Professor
M C Shaw one is the throwing model another
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is the hammering model which I have already
explained to you this is the throwing model
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and this is the hammering model or this is
the throwing model and the hammering model
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As you can see clearly grit is thrown by the
tool and it is penetrating inside the work
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piece F TH is the force acting when the tool
is hitting the abrasive grit in the throwing
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model
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A grain is thrown on to the work piece by
the tool as we have already seen in the slide
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and this is very clear from the slide on the
figure on the right side and we can see T
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is the time that is equal to 0 that means
when time is zero the tool is just touching
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the abrasive particle Now the motion of vibration
can be expressed or the displacement can be
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expressed as Y is equal to A by 2 sin 2 pi
FT where we already know that A by 2 is the
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amplitude of vibration F is the frequency
of vibration and T is the time at which we
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are trying to find out the displacement
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Now velocity of tool can be evaluated by DY
over DT that is equal to Y dot and if you
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differentiate the earlier equation it comes
out to be pi AF cos 2 pi FT The maximum velocity
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that a tool can achieve is given by Y dot
maximum equal to pi AF and this will be maximum
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only when value of the cos 2 pi FT is equal
to 1 so for cos 2 pi FT is equal to 1 we achieve
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Y dot maximum is equal to pi AF
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Now here there is very important assumption
in determining the maximum velocity of the
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grit particle or the tool specially when we
are assuming that the grit also leaves the
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tool at this particular maximum velocity at
which the tool is vibrating This tells the
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tool vibration where amplitude of vibration
and the time both are shown
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Now from this we can find out the kinetic
energy content by the abrasive particle we
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already know kinetic energy is given by half
MV square while M is the mass of the abrasive
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particle V is the velocity of the abrasive
particle with which it is moving towards the
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work piece or it is hitting the work piece
and as you can see here the mass is given
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by pi by 6 D cube rho A where rho A is the
density of the abrasive particle and small
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D is the diameter of the abrasive particle
and pi square A square F square is the V square
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or the maximum velocity of the abrasive particle
with which it is hitting the work piece
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Now full kinetic energy is absorbed by the
work piece before the particle comes to rest
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it is assumed it will result in penetration
equal to H THW suffix THW or subscript THW
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indicates TH stands for throwing model and
W stands for work piece into the work piece
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then the average work done is given by half
FH THW where H we can determine from the figure
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as you can see here it is assumed that the
variation in the force is triangular in nature
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Assuming that the force vibration during force
variation during penetration is triangular
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in nature Average contact stress sigma W on
the work surface is equal to brittle fracture
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hardness of the work piece material with this
assumption sigma W is given by F divided by
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pi H THW D now here pi DH THW just a minute
H THW
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Now this portion pi D is the periphery of
the spherical hemi-spherical penetration that
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we are assuming and this gives the
H THW so from this we can find out whatever
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the area over which force F is acting from
there we can determine the sigma W So we get
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F is equal to pi H this will be pi H THW DH
substituting sigma W is equal to H it is known
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that kinetic energy of a particle is equal
to the work done
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So we can equate the work done to the kinetic
energy of the abrasive particle so we can
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see here half MV square is equal to half pi
upon 6 D cube rho A multiplied by pi square
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A square F square is equal to pi H small H
DH THW Now if we simplify this equation we
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get the value of the H THW is equal to pi
AFD multiplied by under root rho A divided
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by 6 capital H now from here one can determine
the depth of penetration in the throwing model
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in the work piece and this value can be substituted
in the equation which we initially
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This value of H THW from equation 3 can be
substituted in the equation 1 derived for
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evaluation of MRR VTH where VTH is volumetric
material removal rate in throwing model by
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the throwing model In the same way H can be
derived if the material removal takes place
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due to hammering action as follows this clearly
indicates how the hammering model is going
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to work and what is happening in the hammering
as you can see here also part of the abrasive
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is penetrated inside the tool and part of
it is penetrated inside the work piece
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Now here it indicates the tool the topmost
position of the tool where from it starts
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vibration mean position of the tool and bottom
most position of the tool and at the bottom
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most position of the tool it is hitting the
abrasive particles or hammering the abrasive
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particle Now it is assumed that the grain
is hammered into work piece as shown in the
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above figure it simply means that the gap
between the bottom face of the tool and top
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face of the work piece is smaller than the
size of the grain that we can see here
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This is the size of the grain and this is
the bottom face of the tool and this is the
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top face of the work piece So tool will hit
the abrasive particle only when this gap becomes
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smaller than the size of the abrasive particle
or the grit size Mean speed of the tool is
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quite low mean static feed force is applied
to the tool and it is expressed as F and this
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is equal to mean force of tool on the grit
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Delta T is the duration of impact FI maximum
is the maximum value of impact force FI which
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is acting on the abrasive particle force will
vary with the depth of penetration by the
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grit exposed contact area between the crane
and the work piece are different and grain
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and tool is also different this indicates
the delta T indicates the period during which
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abrasive is contact with the tool and work
piece both and capital D indicates the cycle
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time
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This average force in this particular situation
can be evaluated with the help of this equation
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F average is equal to 1 upon T integration
0 to T FI T DT where T is the cycle time
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Now there are various positions of the tool
during a cycle you can see here top position
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of the tool which is indicated by C mean position
of the tool is indicated by O when tool touches
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the grit is indicated by A and bottom position
of the tool is indicated by B and work piece
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is shown over there tool are separately shown
there the cycle time is capital T and the
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time required for moving from O to B is T
by 4 and the time during which abrasive and
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tool and work piece are in contact with which
each other is delta T
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The movement from A to B is due to impact
on the abrasive particle as the tool touches
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the abrasive it experiences the resistance
to move Let us assume nature of variation
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of force FI as triangular so you can see here
FI maximum is the maximum force which is experienced
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by the abrasive particle FI is the force and
delta T as I have mentioned earlier is that
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period during which abrasive tool and work
piece are in contact
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Due to this force the abrasive will partly
penetrate in the tool and the depth of penetration
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is H suffix TH and partly in the work piece
and it is expressed as H suffix WH where W
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stands for work piece H for hammering model
Total indentation H H during hammering is
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given by the sum of the penetration in the
tool and the work piece and it can be expressed
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as H H is equal to H TH plus H WH this is
very clear from this figure H TH is the penetration
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depth inside the tool and H WH is the penetration
depth inside the work piece and force F is
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acting on the grit particle
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Average velocity during the quarter cycle
OB is given by A by 2 divided by T by 4 T
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by 4 is the time taken in the quarter cycle
and A by 2 is the amplitude and this is given
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by 2A by T If delta T be the time required
to travel from A to B that is approximately
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total penetration depth H H then delta T is
equal to H H that is the total depth of penetration
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divided by the velocity that is given by 2A
by T
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So we can simplify this H H is equal to H
TH plus H WH multiplied by T divided by 2A
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now here let F average be the average force
acting during one cycle then F average can
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be written as 1 by 2 F I maximum multiplied
by H TH plus H WH multiplied by T by 2 into
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A and divided by 1 by T
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Here you can see since there is a triangular
force variation that we have assumed so the
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average will be the this is the F maximum
and here is the 0 So average will be F maximum
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divided by 2 that you can see the first term
and H TH plus H WH multiplied by T divided
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by A2 that gives you the time delta T during
which this force is acting and capital T is
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00:26:56,600 --> 00:26:59,950
the cycle time so this will give you the average
force
200
00:26:59,950 --> 00:27:10,140
Or you can write it like this FI maximum is
equal to 4 F average multiplied by A divided
201
00:27:10,140 --> 00:27:22,179
by H TH plus H WH now maximum force per grit
will be obtained when the total force that
202
00:27:22,179 --> 00:27:31,540
is acting on the tool is divided by total
number of particles which are being hit by
203
00:27:31,540 --> 00:27:37,929
the tool so let this be the capital N as the
total number of abrasive particles which are
204
00:27:37,929 --> 00:27:42,399
being hit by the tool
205
00:27:42,399 --> 00:27:46,919
Now what is going to be the maximum stress
developed in the work piece because abrasive
206
00:27:46,919 --> 00:27:52,000
particle is in contact with the tool as well
as the work piece and depth of penetration
207
00:27:52,000 --> 00:27:57,370
in the tool as well as work piece are different
so the area on which the force is acting will
208
00:27:57,370 --> 00:27:58,409
also be different
209
00:27:58,409 --> 00:28:02,669
So the stress is developed in the tool as
well as the work piece will be different let
210
00:28:02,669 --> 00:28:08,580
us find out the maximum stress developed in
the work piece it is given by sigma suffix
211
00:28:08,580 --> 00:28:21,639
W is equal to FI maximum divided by N multiplied
by 1 over pi D H WH the term within the bracket
212
00:28:21,639 --> 00:28:25,350
gives you the area
213
00:28:25,350 --> 00:28:30,620
Same way you can find out the maximum stress
developed in the tool that is sigma T is equal
214
00:28:30,620 --> 00:28:42,290
to FI maximum divided by N within bracket
1 upon pi D H TH now substitute the value
215
00:28:42,290 --> 00:28:51,150
of FI maximum in above equation A then you
get the stress sigma W in the work piece that
216
00:28:51,150 --> 00:29:02,770
is 4 F average A divided by H TH plus H WH
multiplied by 1 over N pi D H WH
217
00:29:02,770 --> 00:29:10,730
Assumption here made is as follows number
of grits acting is inversely proportional
218
00:29:10,730 --> 00:29:17,480
to the square of the diameter for a given
area of tool face I have already shown one
219
00:29:17,480 --> 00:29:27,799
slide regarding this and we have already seen
then N becomes proportional to 1 by D square
220
00:29:27,799 --> 00:29:38,669
and K 2 this becomes equal to K 2 divided
by D square where K 2 is the constant of proportionality
221
00:29:38,669 --> 00:29:47,980
Or you can write down sigma W is equal to
4 F average A D divided by H T plus H W into
222
00:29:47,980 --> 00:29:57,820
K2 whole of it multiplied by 1 over pi H W
now you can further simplify it and you get
223
00:29:57,820 --> 00:30:09,059
it as H W square is equal to 4 F average A
D divided by sigma W pi K 2 within bracket
224
00:30:09,059 --> 00:30:11,020
lambda plus 1
225
00:30:11,020 --> 00:30:21,279
Where lambda is equal to H T over H W and
depth of penetration will be inversely proportional
226
00:30:21,279 --> 00:30:30,679
to the strength of the work piece or the tool
if higher the strength lower will be the depth
227
00:30:30,679 --> 00:30:37,470
of penetration that is why you can see here
H T over H W is equal to sigma W over sigma
228
00:30:37,470 --> 00:30:47,909
T Stress sigma W can be replaced by brinnel
hardness number H both are the same sigma
229
00:30:47,909 --> 00:30:51,549
W is equal to H W that is the work piece brinnel
hardness number
230
00:30:51,549 --> 00:31:00,450
Above equation can now be simplified as H
W square is equal to 4 F A D square divided
231
00:31:00,450 --> 00:31:13,320
by pi K2 C D HW within bracket lambda plus
1 Substitute the value from HW in the equation
232
00:31:13,320 --> 00:31:26,220
equal rate by hammering model comes out to
be equal to K K1 K2 multiplied by the term
233
00:31:26,220 --> 00:31:36,070
given over here 4 F A divided by K2 C D HW
within bracket lambda plus 1 bracket closed
234
00:31:36,070 --> 00:31:43,899
whole raise to power 3 by 4 into D raise to
power 1 by 4 multiplied by F I have already
235
00:31:43,899 --> 00:31:48,269
mentioned F is the frequency of vibration
236
00:31:48,269 --> 00:31:53,909
Now from the theoretical okay let me go back
once again here one point you have to note
237
00:31:53,909 --> 00:32:00,050
that the volumetric material removal rate
is proportional to D raise to power 1 by 4
238
00:32:00,050 --> 00:32:05,809
but practically it is observed something different
rather than D raise to power 1 by 4 I will
239
00:32:05,809 --> 00:32:07,559
explain it later on
240
00:32:07,559 --> 00:32:13,120
From the theoretical calculations it has been
found that material removal rate in throwing
241
00:32:13,120 --> 00:32:20,759
model from throwing mechanism is much much
smaller than MRR VHM that is the volumetric
242
00:32:20,759 --> 00:32:27,070
material removal rate from hammering model
and it is many times so small compared to
243
00:32:27,070 --> 00:32:30,590
hammering model that it can be neglected
244
00:32:30,590 --> 00:32:38,250
The discrepancy is observed between experimental
and theoretical result theoretical results
245
00:32:38,250 --> 00:32:44,409
calculated by Shaw’s model lot of discrepancy
has observed with the experimental results
246
00:32:44,409 --> 00:32:55,559
then later on Professor Shaw tried to explain
them the discrepancy as follows
247
00:32:55,559 --> 00:33:02,380
What he considers that actual shape of the
grain is not spherical and they are not smooth
248
00:33:02,380 --> 00:33:08,580
also now I had mentioned this particular point
earlier also and they have projections of
249
00:33:08,580 --> 00:33:17,769
average diameter and he assumed that it is
equal to D1 and D1 is proportional to D square
250
00:33:17,769 --> 00:33:22,580
or D1 square is proportional to D raise to
power 4
251
00:33:22,580 --> 00:33:30,299
But MRR V is proportional to D raise to power
1 by 4 so if we substitute there the value
252
00:33:30,299 --> 00:33:36,149
of this in the hammering model then we get
D raise to power 4 whole raise to power 1
253
00:33:36,149 --> 00:33:43,289
by 4 you get proportional to D and experimentally
it is observed that volumetric material removal
254
00:33:43,289 --> 00:33:45,320
rate is proportional to D
255
00:33:45,320 --> 00:33:54,090
As F increases MRR V also increases in practice
we want a certain value of F it starts decreasing
256
00:33:54,090 --> 00:34:04,059
because of abrasive grit gets crushed under
heavy load as F increases MRR V also increases
257
00:34:04,059 --> 00:34:10,780
in practice we want a certain value of F that
is the force it starts decreasing because
258
00:34:10,780 --> 00:34:16,980
abrasive grit gets crushed under heavy load
259
00:34:16,980 --> 00:34:28,210
Now let us take one example to understand
how the forces etc can be calculated find
260
00:34:28,210 --> 00:34:35,830
out the approximate time required to machine
a hole of diameter equal to 6 millimeter in
261
00:34:35,830 --> 00:34:44,100
a tungsten carbide plate whose fracture hardness
is equal to 6900 mega pascal and it is of
262
00:34:44,100 --> 00:34:52,360
thickness equal to 1 and half times of hole
diameter the mean abrasive grain size is 0.015
263
00:34:52,360 --> 00:34:58,760
millimeter the feed force is equal to 3.5
newton the amplitude of tool oscillation is
264
00:34:58,760 --> 00:35:06,750
25 micron and the frequency is equal to 25
kilo hertz the tool material used is copper
265
00:35:06,750 --> 00:35:14,560
having fracture hardness equal to 1.5 into
1000 newton per millimeter square The slurry
266
00:35:14,560 --> 00:35:21,930
contains one part abrasive to one part of
water take the values of different constants
267
00:35:21,930 --> 00:35:32,510
as K1 is equal to 0.3 K2 is equal to 1.8 millimeter
square K3 is equal to 0.6 and abrasive density
268
00:35:32,510 --> 00:35:39,390
equal to 2.8 gram per cubic centimeter also
calculate the ratio of the volume removed
269
00:35:39,390 --> 00:35:46,520
by throwing mechanism to the volume removed
by hammering mechanism
270
00:35:46,520 --> 00:35:51,000
This problem has been taken from the book
Advanced Machining Processes published by
271
00:35:51,000 --> 00:35:59,670
Allied Publishers this problem to solve this
particular problem or before solving this
272
00:35:59,670 --> 00:36:05,740
particular problem let us see what are the
available data hole diameter is equal to 6
273
00:36:05,740 --> 00:36:12,820
into 10 raise to power minus 3 meter all the
data that are given have been converted in
274
00:36:12,820 --> 00:36:18,440
meter so that is why you can see here plate
thickness is equal to 1.5 into hole diameter
275
00:36:18,440 --> 00:36:23,160
that comes out to be 9 into 10 raise to power
minus 3 meter
276
00:36:23,160 --> 00:36:29,010
Mean abrasive grain size it should be grit
not the grain mean abrasive grit size is equal
277
00:36:29,010 --> 00:36:36,530
to 1.5 into 10 raise to power minus 5 meter
feed force is equal to 3.5 newton amplitude
278
00:36:36,530 --> 00:36:43,590
of tool oscillation is given by 25 into 10
raise to power minus 6 meter and frequency
279
00:36:43,590 --> 00:36:51,240
of oscillation is 25000 cycles per second
fracture hardness of work piece material that
280
00:36:51,240 --> 00:36:58,050
is sigma W is equal to HW is equal to 6.9
into 10 raise to power 9 newton per meter
281
00:36:58,050 --> 00:36:59,230
square
282
00:36:59,230 --> 00:37:05,200
Fracture hardness of tool material that is
H T is equal to 1.5 into 10 raise to power
283
00:37:05,200 --> 00:37:12,950
9 not 109 this is 10 raise to power 9 newton
per meter square and abrasive grain density
284
00:37:12,950 --> 00:37:22,250
rho A is equal to 3.8 into 10 raise to power
3 KG per meter cube So the value of the lambda
285
00:37:22,250 --> 00:37:31,810
can be written as lambda is equal to HW divided
by HT that is equal to 4.6 K1 is equal to
286
00:37:31,810 --> 00:37:39,020
0.3 K2 is equal to 1.8 millimeter square and
that come out to be 1.8 into 10 raise to power
287
00:37:39,020 --> 00:37:46,840
minus 6 meter square and K3 is equal to 0.6
and C is equal to 1 Here C is nothing but
288
00:37:46,840 --> 00:37:48,110
the concentration
289
00:37:48,110 --> 00:37:55,240
The following procedure should be adopted
to solve the problem first we should try to
290
00:37:55,240 --> 00:38:00,380
find out what is time required to machine
the hole we know that the volume of material
291
00:38:00,380 --> 00:38:06,520
removed using ultrasonic machining can be
calculated using the following relationship
292
00:38:06,520 --> 00:38:14,590
that is V is equal to K1 K2 K3 under the root
H raise to power 3 divided by D into F
293
00:38:14,590 --> 00:38:21,000
First step to be followed in the solution
of this problem is calculate the value of
294
00:38:21,000 --> 00:38:27,860
H which is different for throwing model that
is H TH and for hammering model that is the
295
00:38:27,860 --> 00:38:39,500
H WHM Step two after knowing the values of
H TH and H WHM calculate V TH that is the
296
00:38:39,500 --> 00:38:45,500
volumetric material removal by throwing model
and volumetric material removal by hammering
297
00:38:45,500 --> 00:38:50,570
model from the work piece by substituting
these values in above equation 1
298
00:38:50,570 --> 00:39:03,590
Then find the total volume of material removed
per unit time by adding V TH and V WHM in
299
00:39:03,590 --> 00:39:08,520
the third step calculate the total amount
of material to be removed to make the required
300
00:39:08,520 --> 00:39:16,030
hole divide this volume by VS that has been
calculated in the earlier step two to find
301
00:39:16,030 --> 00:39:25,060
the total time required to make the hole In
step four find the ratio of H TH divided by
302
00:39:25,060 --> 00:39:27,230
H WHM
303
00:39:27,230 --> 00:39:39,030
Following the above steps all calculations
are made as followed in step one in equation
304
00:39:39,030 --> 00:39:45,330
1 except H all other parameters are known
let us calculate the value of H TH by throwing
305
00:39:45,330 --> 00:39:52,750
model and the equation is given as follows
now let us substitute the value of pi A F
306
00:39:52,750 --> 00:40:07,370
D rho A and sigma W as follows and we get
H TH equal to 1.78 into 10 raise to power
307
00:40:07,370 --> 00:40:10,900
minus 5 millimeter
308
00:40:10,900 --> 00:40:19,000
So penetration H WHM in the work piece due
to hammering is given by the following equation
309
00:40:19,000 --> 00:40:28,950
that is H WHM is equal to 4 F average A D
divided by pi K2 sigma W multiplied by within
310
00:40:28,950 --> 00:40:36,291
bracket lambda plus 1 bracket closed whole
under root substitute the values which we
311
00:40:36,291 --> 00:40:46,510
have already given in the earlier slides and
we get H WHM equal to 2.192 multiply by 10
312
00:40:46,510 --> 00:40:51,110
raise to power minus 4 millimeter
313
00:40:51,110 --> 00:40:56,820
Now volume removed by throwing mechanism is
given as follows this particular equation
314
00:40:56,820 --> 00:41:00,900
is to be used for the throwing model that
we have already derived
315
00:41:00,900 --> 00:41:07,990
And if we substitute the value of various
parameters we get that the volume of material
316
00:41:07,990 --> 00:41:15,390
removed by throwing model equal to 4.97 into
10 raise to power minus 3 cubic millimeter
317
00:41:15,390 --> 00:41:17,270
per second
318
00:41:17,270 --> 00:41:24,350
Now volume removed by hammering model can
also be found in the same way this is the
319
00:41:24,350 --> 00:41:34,200
equation which we have already derived substitute
the values of various parameters and constants
320
00:41:34,200 --> 00:41:41,880
then we get volume of material removed by
hammering model is equal to 0.2146 cubic millimeter
321
00:41:41,880 --> 00:41:48,040
per second which is much higher than what
we get the volume of material removed by throwing
322
00:41:48,040 --> 00:41:49,040
model
323
00:41:49,040 --> 00:41:58,020
So what is the time required to drill a hole
as I have mentioned earlier that volume of
324
00:41:58,020 --> 00:42:04,660
the hole to be drilled divided by the volumetric
material removal rate by both the models hammering
325
00:42:04,660 --> 00:42:09,310
model as well as throwing model that will
give you the time required for drilling the
326
00:42:09,310 --> 00:42:18,440
hole and this is as you can see values have
been substituted here and you get 19.289 minute
327
00:42:18,440 --> 00:42:30,630
We can also find out the ratio of V TH and
V WHM and by substituting these values of
328
00:42:30,630 --> 00:42:42,760
V TH and V WHM we get the ratio as 0.023 Thus
from this value of 0.023 it is evident that
329
00:42:42,760 --> 00:42:51,060
the material removed by hammering is much
more than by throwing approximately 43 times
330
00:42:51,060 --> 00:42:57,750
Henceforth approximate calculations and for
all practical purposes volume removed by throwing
331
00:42:57,750 --> 00:43:06,620
model can be ignored as compared to the volume
of material removed by hammering model
332
00:43:06,620 --> 00:43:15,090
This way you can make the calculations for
various types of the problems given in the
333
00:43:15,090 --> 00:43:21,680
examination or in the practical lie I have
tried to give a question bank also and these
334
00:43:21,680 --> 00:43:28,840
problems you can solve at your convenience
I have given various problems over as you
335
00:43:28,840 --> 00:43:35,540
can see and for solving these problems this
you can take the help of both these lectures
336
00:43:35,540 --> 00:43:44,400
as well as you can take the help of the book
which I have already mentioned to you Thank
337
00:43:44,400 --> 00:43:51,690
you very much