WEBVTT
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hello so in this lecture we will review
some of the basic concepts of solid mechanics
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that you would have learned in your first
year of a undergraduate engineering course
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so even though this course is on fluid
mechanics as you can see here however when
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we talk about flow in the cardiovascular system
the pipes or the tubes or the channels the
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arteries the veins in which the flow happen
they are not the rigid tube as we encountered
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in our day to day day to day life or in
the industrial applications they are rather
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a flexible tubes so in this flexible tube
the flow happens and the stresses that are
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applied on the channel wall because of that
the tube is deformed or the channel wall have
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deformation and because of those deformation
the shape of the channel is changed consequently
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the flow behaviour the velocities the pressure
inside the channel will change so it is important
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to understand the stress stand relationship
in the solid walls so we will briefly review
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the basic concepts which are relevant to cardiovascular
fluid mechanics in this lecture ok
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so just briefly a let us look at solids
which are elastic so the elastic solids or
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elastic materials are those when a force is
applied they deform but after the force is
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removed the material comes back to its original
configuration original position then such
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materials are called elastic materials so
for example if we have a elastic bar suspended
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from a surface let us say this bar is of
length l and it is being pulled by a force
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say f then as a result of this the the
bar has
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the length of the bar initially was l naught
now it has become l the change in the length
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of the bar is delta and a is cross section
of the bar so as a result of this there will
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be the stress which will be acting in the
normal direction so we can say the normal
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stress and this bar is sigma is equal to f
over a where a is the area of cross section
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and strain epsilon is equal to delta over
l naught so the elastic modulus
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e is defined as sigma over epsilon ok for
a elastic material you might remember that
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the relationship between stress and a strain
sigma and epsilon is a linear relationship
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and so from this we can say that the slope
is equal to the elastic modulus e the assumption
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that we had here that the material is homogeneous
so homogeneous material means that e is same
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throughout the bar or throughout the material
that we are considering isotropic material
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mean that the behaviour or the elastic behaviour
is same in all directions that is if the stress
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strain relationship is same if we apply is
stress in the x direction or we apply a stress
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in the y direction or the z direction so most
of the engineering material behaviour can
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be classified in these two different material
behaviour ductile behaviour as we can see
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here this is for the ductile material and
this graph is for the brittle material so
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in the brittle material the relationship is
almost linear and then breaks off wherever
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it goes through the first to heal distress
and then a at at a certain value of
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the highest value the nuking a cursor and
then material ruptures and the nuking the
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area changes significantly ok so these
materials behave as an elastic material up
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to certain limit which is called deal stress
and after that their behaviour changes ok
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so when we define a strain we consider
epsilon is equal to delta over l or l naught
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and the that delta s the we consider del[ta]-
epsilon is equal to delta over l or l minus
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l naught over l naught however if we consider
the strain truly the engi[neer]- so this is
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called engineering strain which we have defined
just now in the previous slide now true strain
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if we represent it by epsilon then the true
strain will be sum of the incremental strains
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let us say a small change in the length divided
by the instantaneous length so if we do
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that continuously then we will have this as
integral d l over l and the limit from l naught
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the initial length to l the final length so
that will be equal to l n l over l naught
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where l is equal to l naught into one plus
epsilon so this is l n one plus epsilon
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so this gives a relationship to us that epsilon
one is equal to l n one plus epsilon where
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epsilon is the engineering strain and epsilon
one is the true strain and we can say that
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for small deformations that is epsilon is
small epsilon one is almost equal to epsilon
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the other point is that the materials which
do not follow a linear relationship between
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sigma and epsilon for them there is not
a constant value of elas[tic]- elastic
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modulus any number of biological model
beha[ve]- materials do not behave elastically
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do not have elastic behaviour so for such
a materials one can define incremental elastic
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modulus so the at one particular point
the e incremental or the incremental elastic
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modulus is defined as d sigma over d epsilon
so which is the slope at that particular point
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where the elastic modulus is being sought
ok
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now poissons ratio so when a rod that first
example that we considered when a rod it subjected
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to a stress a normal stress then the rod not
only it elongates but to conservate volume
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the area of the rod also changes so there
is because of the normal stress let us
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say sigma x f over is equal to f over a there
is not only the strain in the axial direction
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but there is strain in the transverse direction
also so the poissons ratio is defined as transverse
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strain divided by the normal strain and because
it has the sign with it so the poissons
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ratio is mod of or the magnitude of transverse
ratio and or the transverse can also be said
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as lateral strain divided by the or normal
not in place of normal axial probably is a
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better word so the transverse or lateral strain
divided by the axial strain so for example
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nu is equal to epsilon
y divided by epsilon x and if we consider
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the sign because epsilon y is going to be
negative similarly this will be also equal
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to minus epsilon z over epsilon x for an isotropic
material
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so one can write that epsilon x is equal to
in case of a axial stress sigma x over
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e and one can also find epsilon y is equal
to minus nu sigma y over e and epsilon z is
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equal to minus nu sigma z over e ok so this
is poissons ratio
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now we will talk about shearing stress and
shearing strain so if we consider a planar
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surface in the x y coordinate system let us
consider a planar surface
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so in this it is subjected to a shear stress
which is tangential stress tau y x tau x y
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tau x y so as we have discussed already in
the fluid mechanics review that tau y x has
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to subscript so y is the direction of the
shear stress whereas y is the the plane
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at which shear stress is being applied so
as a result of this shear stress the surface
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deforms by an angle and so the two angles
the earlier angle as we saw that as pi by
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two the two angles are reduced by say gamma
x y and the two angles so this and these angles
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are reduced by gamma x y and the two angles
are increased by the same value so pi by two
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plus gamma x y so under elastic limits
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the relationship between the shear stress
is tau x y is equal to g gamma x y where where
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g is called shear modulus ok so that is the
relationship between shear stress and shear
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strain and like normal stress and normal strain
are relationship or a graph can be plotted
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between the shear stress and shear strain
and the slope of it is the shear modulus
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now based on this a now we can write a generalised
hookes law so if we have a cubic material
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or a general material which have a cubic shape
and subjected to a three kind of a stresses
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or three stresses in three different directions
sigma x sigma y and sigma z and a different
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shear stresses then we can write using the
principles of of super position that is the
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material subjected to different sigma to so
we can write sigma x is equal to or sorry
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epsilon x is equal to sigma x over e minus
nu sigma y over e minus nu sigma z over e
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so the first deformation in the x direction
is coming because of the stress in the x direction
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whereas the other two components are because
of the two stresses in the y and z directions
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respectively similarly one can write sigma
y is equal to sigma sorry epsilon y is equal
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to sigma y over e minus nu sigma x over e
minus nu sigma z over e similarly epsilon
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z is equal to sigma z over e minus nu sigma
x over e minus nu sigma y over e one can also
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write the relationships between tau x y
is equal to g gamma x y tau y z is equal to
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g gamma y z tau z x is equal to g gamma z
x so one can remember in this case that the
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material is isotropic so the e and g they
are same in all the directions ok
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now for the polar coordinates because
in the cardiovascular system what we are going
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to encounter are the arteries or the cylindrical
tubes and these cylindrical tubes it is easier
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to work in polar coordinates so in polar coordinates
we have r theta and z so like cylindrical
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like cartesian coordinate the relationship
in the polar coordinate can be written as
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epsilon r epsilon theta and epsilon z they
are equal to sigma r over e minus nu sigma
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theta over e minus nu sigma z over e similarly
sigma theta over e minus nu sigma r over e
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minus nu sigma z over e equal to sigma z over
e minus nu sigma theta over e minus nu sigma
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r over e similarly one can write the relationships
for the shear stresses in the r theta coordinates
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now one can also re cast these equations
so that one can write the stresses in terms
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of the three strains epsilon x epsilon y and
epsilon z so that is algebra and one need
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to reconstitute or a reframe these equations
so as to obtain shear stress or or not
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the shear stress of but the extra stress normal
stress in terms of the three strains so that
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is left for you as an exercise now
we will look at bulk modulus so bulk modulus
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so when we consider let us say a cubic volume
and this volume has unit dimension the dimension
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is in direction is one so the v is one now
after the stresses in the three directions
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epsilon x epsilon y and epsilon z the strains
are sorry the stresses are sigma x sigma y
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and sigma z so the resultant strains are epsilon
x epsilon y and epsilon z
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so the new volume will be one plus epsilon
x one plus epsilon y into one plus epsilon
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z and if one neglect the higher order terms
then this will be equal to one plus epsilon
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x plus epsilon y plus epsilon z so the change
in volume delta v is equal to because the
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initial volume is one so the change in volume
is epsilon x plus epsilon y and epsilon z
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and because the initial volume is one so delta
v over v is equal to epsilon x plus epsilon
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y and v is the initial volume so v naught
epsilon x plus epsilon y plus epsilon z now
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we can substitute the values of epsilon x
epsilon y and epsilon z so we will have a
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this is equal to one minus two nu when nu
is the poissons ratio into sigma x plus sigma
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y plus sigma z so the volume is strain
which is the ratio of change in volume to
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the original volume or the initial volume
is equal to one minus two nu sigma x plus
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sigma y plus sigma z so one can see here that
if the poissons ratio is point five then delta
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v over v is equal to zero if the material
is subjected to uniform loading and sigma
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x and said sigma y and sigma z are same so
let us say this is a in case of a hydrostatic
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pressure and material subjected to minus p
then delta v over v naught is equal to one
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minus two v v e here one minus two nu over
e minus three p or minus p over k where k
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can be defined as e over three into one minus
two nu and it is also called bulk modulus
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ok
we might also want to remember here that the
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relationship between the shear modulus
g is equal to e over two into one plus nu
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so this is the relationship between the shear
modulus and the elastic modulus ok and you
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have the poisson ratio which relates the two
ok so until now what we have been doing
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is defining a different modulus what we have
done is defined the elastic material and then
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the elastic modulus the elastic modulus for
the material in which the stress stress strain
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relationship is not linear then we have looked
at the generalised hookes law and the
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strains in terms of the three different stresses
and then we have looked at the bulk modulus
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and the shear modulus so now with this
information we would like to apply this
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to a cylindrical tube
so for simplicity let us think or let us assume
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that this cylindrical tube is of thin wall
so our analysis becomes simpler and as a first
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approximation let us assume that the arteries
which we will encounter in the cardiovascular
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system they are elastic tubes and they are
thin walled so that we can apply this analysis
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to them because the the walls of or the
the tubes are can be bend easily so the forces
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that develop in these tubes they are generally
tangential in nature because the the
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vessels they offer very little resistance
to the bending and we consider because
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the the geometry which cylindrical so it is
a axisymmetric geometry as a result no shear
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forces are generated consequently only normal
forces exist in the axial which is the axial
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direction or the in a cylindrical tube
so the axial direction the forces will
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be in this direction and in the angular direction
or circumferential direction
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so the stress in the circumferential direction
is also known as hoop stress hoop stress is
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normal stress in the circumferential direction
so if we consider a half part on the tube
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then the stresses on these portions in this
direction is the circumferential or if we
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take just a small angular portion of the tube
then these stress or this stress is known
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as sigma theta and its called hoop stress
the other stress that will be important is
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longitudinal stress
in the axial direction so you also call it
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axial stress ok now hoop stress we can find
by a force balance on a part of the tubes
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if we consider a let us consider a semi circular
tube the thickness of the tube which say t
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and the radius is r and we consider a depth
of the tube or length of the tube say d
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z the internal pressure or p is what we call
transmural pressure which is the difference
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between inside and outside pressure so the
pressure if the two sides are pressure
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p one and p two then what we consider the
transferral pressure p x along this direction
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normal to it will act the pressure acts normal
to the surface everywhere correct and the
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outside pressure the difference we have
considered so the outside pressure is zero
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here
now if we take this as x direction and this
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as y direction and do the force balance in
let us say x direction then what we will have
32:53.279 --> 33:06.650
is p into the area on which it is being
applied so the internal area is this area
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so if we draw this area edge this area
which is this distance is two r and the other
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distance is delta z so you will have p two
r
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d z where r is the radius of the channel this
is equal to t and the hoop stress in this
33:49.049 --> 34:04.559
is sigma theta so sigma theta into the area
of a these two so there are two parts of
34:04.559 --> 34:12.110
the tube here and these parts are the small
area and this is small area this distance
34:12.110 --> 34:25.370
is t and this the length is d z so two
t d z sigma theta d z d z will cancel out
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and what we have is two and two will also
cancel out sigma theta is equal to p r over
34:35.330 --> 34:44.899
t which is the hoop stress
so you find out the relationship for the hoop
34:44.899 --> 34:52.260
stress in terms of the transmural pressure
the radius of the vessel and thickness of
34:52.260 --> 34:58.400
the tube for a thin walled cylindrical tube
for the circumferential stress if the vessel
34:58.400 --> 35:06.130
is close ended so for the close ended vessel
if we can draw a small schematic of the
35:06.130 --> 35:27.220
close ended vessel then we can do a
force equilibrium or force balance
35:27.220 --> 35:40.910
in axial direction then we will have p into
pi r square so what will be the area the in
35:40.910 --> 35:47.069
inner area of the tube that will be a putting
a pressure that will be applying a force
35:47.069 --> 36:06.869
and then this will be balanced by the
axial stress sigma z into two pi r into t
36:06.869 --> 36:15.609
so pi and pi will cancel out and r and r will
cancel out so we will have sigma z is equal
36:15.609 --> 36:26.359
to p r by two t however if the vessel is open
ended then there is no axial stress and
36:26.359 --> 36:39.010
sigma z is equal to zero from there we can
also see that sigma z is equal to sigma theta
36:39.010 --> 36:46.780
by two for a close ended vessel
the epsilon for this tube will be equal to
36:46.780 --> 36:59.760
epsilon the change will be in the radius
of the tubes so the epsilon r will be d r
36:59.760 --> 37:19.839
over r or delta r over r and epsilon theta
will be two pi r plus delta r minus two pi
37:19.839 --> 37:34.331
r over two pi r so two pi two pi will cancel
out and that will also equal to be delta r
37:34.331 --> 37:49.000
over r that will be epsilon theta so we will
have a relationship between epsilon theta
37:49.000 --> 38:08.140
is equal to sigma theta over e which is p
r by t e and that is equal to delta r over
38:08.140 --> 38:19.970
r because this is what epsilon r is so we
can there find a relationship that p r
38:19.970 --> 38:31.560
square over e t is equal to the deformation
of the tube which is subjected to a a internal
38:31.560 --> 38:44.440
pressure of p for a thin walled tube now
the assumption of thin walled tube is good
38:44.440 --> 38:51.839
enough when the ratio of the thickness
and the channel radius is less than point
38:51.839 --> 39:00.930
one or so however for the thicker a walls
the thick wall analysis needs to be performed
39:00.930 --> 39:07.040
and one need to take into account the stress
variation in the walls of the tube for many
39:07.040 --> 39:14.790
cardiovascular application this might be the
case so we will briefly look at the the
39:14.790 --> 39:20.990
formulation that can be developed for the
thick wall tubes and as the arterial walls
39:20.990 --> 39:27.000
are tethered that means their movement in
the axial direction is restricted so the only
39:27.000 --> 39:32.910
plain strain formulation or two dimensional
strain formulation in the r and tethered direction
39:32.910 --> 39:37.670
need to be considered
so for the equilibrium condition let us consider
39:37.670 --> 39:49.829
a part of the tube so we have a tube
which have a thick wall and the inner radius
39:49.829 --> 40:02.849
is r one and the outer radius of the tube
is r two and they consider a small angular
40:02.849 --> 40:20.780
portion which has thickness r and it subtends
a angle of d theta at the centre the the length
40:20.780 --> 40:39.790
of the tube that we consider is d z the internal
pressure is p and we consider only transmural
40:39.790 --> 40:51.040
pressure here so because of that we will have
say we can draw this element here for
40:51.040 --> 41:08.070
clarity and this subtends then angle of d
theta at the centre the stresses in this r
41:08.070 --> 41:22.400
sigma r and sigma r plus d sigma r and the
hoop stress or the angular or stress is
41:22.400 --> 41:50.440
sigma theta now let us balance the force
in the radial direction so if we have force
41:50.440 --> 42:02.280
then if we consider this direction so along
this direction the force will be at this
42:02.280 --> 42:10.930
surface the forces the force stress on this
direction the stress will be there applied
42:10.930 --> 42:29.290
in this direction so sigma r plus d sigma
r into the area of this surface which is r
42:29.290 --> 42:53.109
plus d r d theta into d z minus r sorry
sigma r when this is multiplied by the r r
42:53.109 --> 43:05.030
is the radius and this place and d d r is
this distance so this radius is r and sigma
43:05.030 --> 43:24.190
r r d theta d z
now the force sigma theta will have a component
43:24.190 --> 43:32.589
in the radial direction so if you look at
this angle this is d theta by two so the force
43:32.589 --> 43:46.420
component on this direction is sigma theta
sin d theta by two so we will have another
43:46.420 --> 44:11.790
force sigma theta into the area which is
d r into d z multiplied by sin d theta by
44:11.790 --> 44:18.800
two and there are two components we will
multiply this by two this is equal to zero
44:18.800 --> 44:29.930
now we can straight away cancel out d z
from this and we also assume that sin d theta
44:29.930 --> 44:38.599
by two is equal to d theta by two because
d theta is a small angle and we also will
44:38.599 --> 44:53.359
neglect higher order terms so we will neglect
the multiplication of these sigma r and d
44:53.359 --> 45:01.089
are because that will be a smaller in magnitude
term so we will have sigma r and sigma
45:01.089 --> 45:19.599
r which will basically cancel out so we will
have sigma r d r plus r d sigma r minus sigma
45:19.599 --> 45:25.420
this term is already cancelled out minus
two two will cancel out so we will have sigma
45:25.420 --> 45:47.011
theta d r is equal to zero that will give
us sigma r minus sigma theta over r plus d
45:47.011 --> 45:55.710
sigma r over d r is equal to zero
so this is the first equation that we will
45:55.710 --> 46:05.819
get as a result of the equilibrium condition
so that says the derivative of radial stress
46:05.819 --> 46:11.329
plus the difference between the radial and
hoop stress divided by r the sum of these
46:11.329 --> 46:18.480
two is equal to zero so this is the equilibrium
condition now we substitute the compatibility
46:18.480 --> 46:39.549
conditions in this so if we assume that
u is the radial displacement of the in the
46:39.549 --> 46:50.260
tube instantaneously the local displacement
in the tube so epsilon r is equal to d u by
46:50.260 --> 47:08.710
d r and epsilon theta will be will be or the
angular strain will be equal to two pi
47:08.710 --> 47:25.119
r plus u minus two pi r divided by two pi
r so that will be equal to u by r now from
47:25.119 --> 47:34.700
the previous slide we had the equilibrium
condition d sigma r over d r plus sigma r
47:34.700 --> 47:45.440
minus sigma theta over r is equal to zero
if we substitute those values here and also
47:45.440 --> 47:52.200
substitute the values of epsilon r and epsilon
theta then we will have so first let us find
47:52.200 --> 48:01.710
out this at the substitute sigma r there
then we will have sigma r is equal to e over
48:01.710 --> 48:17.490
one minus nu square into epsilon r is d u
by d r plus nu u by r and epsi[lon]- sigma
48:17.490 --> 48:33.180
theta will be e by one minus nu square epsilon
theta is u by r plus nu d u by d r we substitute
48:33.180 --> 48:41.140
this in here we also need to find out d sigma
over d r so let us do that e by one minus
48:41.140 --> 48:52.990
nu square will be everywhere so that can
be cancelled so we can write d two u by d
48:52.990 --> 49:15.270
r two plus nu by r d u by d r minus nu u by
r square which is differentiation of one over
49:15.270 --> 49:20.549
r plus now we substitute sigma r and sigma
theta here
49:20.549 --> 49:39.630
so we will have sigma r is one over r d
u by d r plus nu u by r square minus sigma
49:39.630 --> 49:57.170
theta r so u by r square minus nu over r d
u by d r is equal to zero now what we will
49:57.170 --> 50:12.190
have is d two u over d r two nu by r d u by
d r is cancelled we have plus one over r d
50:12.190 --> 50:20.660
u by d r that is the only term and a first
order differentiator and then these two terms
50:20.660 --> 50:32.450
will cancel out and we will have minus u by
r square is equal to zero so this is a relationship
50:32.450 --> 50:47.691
for the displacement and say strain
so if we reconstitute or recast it it can
50:47.691 --> 51:03.380
be recast or read it in as d by d r is equal
to one over r d by d r of u r is equal to
51:03.380 --> 51:14.740
zero so one can find out this equation
or one can integrate this equation and get
51:14.740 --> 51:23.280
u is equal to c one r plus c two r in this
form one can get an expression for u for the
51:23.280 --> 51:33.270
displacement and if we substitute back this
in substitute this in back into sigma
51:33.270 --> 51:47.630
r and then we can have boundary conditions
at sigma r is equal to minus p at r is equal
51:47.630 --> 51:56.330
to r one and h zero at r is equal to r two
so one can obtain the both two constants c
51:56.330 --> 52:05.130
one and c two and from that one can obtain
sigma r sigma theta and so on and so forth
52:05.130 --> 52:07.640
ok
so that is the analysis for thick walled
52:07.640 --> 52:15.390
cylindrical tubes and from this analysis one
can obtain relationship for the displacement
52:15.390 --> 52:21.250
or for the deformation in the thick walled
tubes as a result of a transmural pressure
52:21.250 --> 52:28.930
p it this relationship has been used to
measure the properties of the material rather
52:28.930 --> 52:35.480
so one can measure the displacement for a
known transmural pressure and from that displacement
52:35.480 --> 52:41.750
one calculates the properties of the material
for example poissons ratio and the elastic
52:41.750 --> 52:50.380
modulus and so so in summary what we have
looked at is that how the hoop stress or how
52:50.380 --> 52:58.039
the deformation of a thin walled and thick
walled tube can be measured or can be found
52:58.039 --> 53:05.220
out can be calculated for for a given transmural
pressure and for that we also looked at some
53:05.220 --> 53:11.420
of the basics or we have reviewed the basics
of solid mechanics a the relationship between
53:11.420 --> 53:19.710
a stress and a strain for for for normal
stresses and shear stresses and the bulk modulus
53:19.710 --> 53:28.130
and so on and so forth we have looked at the
general hookes law so hope you have
53:28.130 --> 53:35.579
all you can remember now the basics
of the solid mechanics and if you find any
53:35.579 --> 53:44.520
difficulty you can discuss while the course
is on or you can review your solid mechanics
53:44.520 --> 54:06.760
a notes or books that you have studied
in a first year undergraduate courses
54:06.760 --> 54:13.920
thank you