PHYS 146 Elasticity part 1: Stress and Strain

PHYS 146 Elasticity part 1: Stress and Strain

In this module we are going to talk
about the effects forces on objects. Now in physics we often treat objects is perfect rigid bodies which means that the never deform and
they never break no matter how large or where we apply
forces to them now of course in real life we know that this
simply just isn’t true if you’ve ever been unfortunate enough to drop your mobile phone you’ll know that the metal case can dent or scratch and the glass can actually shatter and break and so different
materials glass and metal respond differently to
the same applied forces. So clearly we’re going to need to look at material
properties because different materials behave
differently and so we need to be able to quantify
how a material will behave in a particular set of circumstances. So
let’s start looking at something a lot simpler than a mobile phone and
will take a rubber band here now if I apply a tension to this rubber
band you can see that it gets longer so I
simply pull on each and to apply a tension force and the rubber band stretches so we might
think well perhaps we should use force as one
of our quantities to determine how materials
behave but supposing I take a different rubber
band and so here I got the same material, it’s
rubber but now I’ve got a far thinner rubber band and if I pull on it it also stretches however I’m using far less force to produce the same amount of stretch in
this rubber band than I did in the thicker rubber band. So here if I take the same
length of material and I apply the same force I get far
less stretch and to get same amount of stretch I have
to apply a far larger force so force alone is not going to be a useful quantity when we determine
the deformation of a particular material and so what
we’re going to do is we’re going to define a new quantity
called stress and this will take account of the cross-sectional area of the material
so in this case I have a large cross-sectional area and
so I need to provide a large force and in this case I have a small cross-sectional area and so I need to
provide only a small force to get the same amount of stretch so stress is going to be a force per
unit area and what we’re interested in is the
forces inside the material because it’s the force
inside the material that determines how the material will deform so the question is now how do we calculate force per unit area
inside the material and what exactly does that mean and so we will have to switch to the computer to see so
here is our rubber band and we’ve applied to it two external
forces this is one the external forces and its
in equilibrium so what we’re going to do to calculate the
stresses we want to know the internal forces inside the rubber band it’s not the external forces that are
important for stress it’s the internal forces so to find these out
we’re going to cut it in two along plane here and then we’re
going to insert here a thin plane I’m going to put thin in inverted commas
because what we actually mean by this is an
infinitely thin plane so it’s an infinitely thin plane that
we’re going to insert here and we can cut it here it doesn’t have to be
in the mid point, we could cut it anywhere along here and wherever we cut it is where we will
find the stress at that point so if we cut along the
middle here will find that the the stress in the middle. In
this case what we will find is that the stresses in fact constant throughout
this rubber band I’m but in general you cut it where you want to find the
stress so we cut it in two we insert a thin plane and we consider the forces that act on
that plane so this is what we’ve done here’s a
picture now the rubber band with our thin plane in
blue here and these are the forces acting on the plane so we have one here and we have one here so how do we get those forces why
what we did is we look at this part of the rubber
band and so here I have this half of the
rubber band and I have my external force acting on – I’ll show that with a solid arrow
here so this is my external force here – now the rubber band is in equilibrium and so
what this tells me is there must be an equal and opposite
force here and I’ve shown that with an unfilled-in arrow and this is my internal force so this
is the force acting on this side to the rubber band and this is needed in order to be
equal and opposite the external force and keep the rubber
band in equilibrium. Well if I have here my thin plane then what Newton’s third law says is that
there must be a reaction force that acts on this plane, I’ll call that ‘R’ here and that’s a reaction too this
force here and Newton’s third law says the reaction is equal and opposite to the action and so
that means that my reaction force here by Newton’s Third law must be also equal to this F
perpendicular and so this is the force acting on the
plane and so it’s shown here – this force here is my reaction force, this F
perpendicular, shown in blue to show that it’s acting on the plane and I can do exactly
the same thing on this side here I can resolve the
forces on this half of the rubber band and that
gives me the perpendicular force that acts on the other side of the plane and so here we
have the two forces that act on the plane now in general the forces on the plane will allways cancel and in fact if you draw a diagram
where they don’t cancel you got it wrong and the reason for this is because the plane has no mass and so therefore cannot have any net force acting on it
otherwise it would have and infinite acceleration by Newton’s second law so it’s the same argument that keeps the
tension in a massless string constant throughout the string the same argument shows here is that
we’re having this theoretical thin plane and so therefore the forces
acting on either side of it must add to zero in other words there must be no net force acting on the plane so the forces acting on the plane are the internal forces inside this rubber band. So now we can see the forces acting on the
plane and we now want to look at our stress So our stress is based on these internal
forces that act on the plane and it’s just equal to the one of the forces that act on the plane.
Remember of course that the net force will be zero so we take one of the forces
acting on one side of the plane and we divide it by the cross-sectional area of the plane so A here is the area
the plane and this force that is acting on the plane we
just take it and we divide it by the area. Now of course in terms of units the forces are measured in newtons in the SI
system, area is measured in square metres and so
therefore stress is measured in newtons per square metre now the other SI unit you can use is the Pascal, abbreviated Pa. That’s a valid SI
unit which has the same dimensions as newtons per square metre. I will not use that. I don’t like using it because it hides the underlying units of the quantity you are
measuring so it’s good to remember that stress is
a force per unit area and that’s a lot more
obvious if you write the units as newtons per square metre so I won’t use Pascal’s now you might say well why bother going through all this
rigmarole love looking at the forces acting on part with the rubber band and then
reaction forces that act on the plane? Wasn’t that a lot of complication? Couldn’t
we just have taken the external forces because of course this is equal in magnitude to the external
force but it’s not the external force it the force acting on this plane, but
couldn’t we just take the external forces and divided by the cross-sectional
area of the rubber band? Well in this case you could have done
that and you would have been absolutely right. However in general you’ve got to be careful when doing that.
Supposing instead of having the rubber band
horizontal I hang it vertically. I’m assuming some tension T now holding up the
rubber band and we will assume that this rubber band
has some mass. So now if I take my plane at the top the rubber band here then here the forces acting on this
plane are going to be quite large because if I
look at this part of the rubber band here the upward force on this part of the
rubber band must support almost all the mass That will be almost equal to the total
weight of the rubber band because it has to
support the entire weight of the rubber band, or almost the entire weight of the rubber band but if I take a cross-section through
here then the force acting on this plane to support this lower half for the
rubber band, or lower little tidbit of the rubber band here, so
the force acting here will be a lot less than ‘mg’ and so here the stress in this rubber band will change from the
top for the rubber band where it will be logged because black part the rubber band to
support me almost the entire way and will become a lot less lower down
here where there’s very little rubber band so in these cases you have to do the
full you have to be I’m careful and do the
fall analysis and look at the forces acting
on these planes to make sure that you’ve got it correct K now the type to stress we looked at
with the rubber band is called normal stress and that’s
because the forces with normal to the internal play we look to act and
normally just a fancy way of saying perpendicular to you forces were right
angles to the plane another two types have normal stress is
what we sow with rob a bank where we have a tensile stress so here the band is on the
tension and so we say it’s a tensile stress but we could also apply the force in the
reverse motion and so here for example if I take this
book I can apply compressive force like crime Porsche the
pool shock and you can see that the I will
play fourth normal to the plane if the plane
to these pages inside the book here the force is normal to the play but it’s
now a compressive forces trying to make the book data so are those the only two types of stressed that the rock well let’s have a
look the simple sheet of paper now we’ll just see what I was trying to
compress the book that people can stand up to a
compressive stress pretty well the book didn’t squeeze down
very much at all and much about for you squeeze the area between the pages so we can also look at tensile stress
and indeed if I along with you can see but it stands up to tensile stress
pretty well right single sheet of paper so if that’s all there is for stress we
should be able to handle any sort of stressed pretty well but look what happens if I do something
a little different it fell apart really quite easily and this trying to stress well I’m
applying different forces in opposite direction
so if this plane of the paper is all stressed plane now I apply force
up once I and down at the other and people fails almost immediately not Piper stress his call shear stress so let’s look at my apt in a bit more
detail okay so here we have block that is under this new type to stress called shear stress and as you can see we have
this external force applied here which is labeled of
parallel and there is a opposite force equal the force applied at the other end I’m
in this is putting the block im sheer stress so once more we’ve divided the block in to using theme playing and this plane
has been laughed a little bit so it’s easier to illustrate the forces on it and the question we want to know is what
are the forces marking on this plane so as we did with normal stress we have to consider
the forces acting on this prior to the block
Samaraweera just taking this problem here I’m gonna look at the forces acting on
it while on this site here you can see that we her their parallel an offer letter again just like they did
before because this is an external force and then here we her wealthy in playing this terms the second now in order for
this blog to be in equilibrium the force acting here must be equal and opposite to the
external force and so therefore this must be bash parallel as well so the contact force
between the screenplay I’m and the block must be better parallel as
well and so what this means is that the
fourth acting on the plane itself right by Newton’s third mall here must be air parallel as well so this is Newton’s third law and it has to be equal so exact parallel
and opposite certain acting upwards as opposed to the
downwards force attracting on the block itself and so therefore this force here it’s
just their parallel likewise we can do the
same thing on this block here and we’ll find this false on the other side of the
plane now noticed this is faded slightly so it’s on the other side of the plane is also their parallel and so this is how we calculate the internal forces acting on of them play and as you can
see across these forces are equal and opposite the equal in
magnitude the perfect parallel 1x this way the other its the in the opposite direction so they equal
and opposite as we required before so now the question is what’s the stress well we know the stress is force per
unit area the only difference between this shear
stress so and normal stress is that the forces on the plane mara parallel to the plane
the light in the plane itself the not
perpendicular to the plane and so therefore the shear stress is
just the force which is this parallel force and that is divided by the
cross-sectional area of the object at the plate right
surfaces the plane that fills the object the object of
cross-sectional area a here then we just take the parallel
force acting in the plane and divide it by the cross-sectional area the plane and
so again or Unix are going to be Newton’s per
square meter or if you like Pascal’s but as I said
before right don’t like you Pascal’s I like to
use newman’s per square meter so this is just the definition of shear stress
the only difference is is it now we have a force the lorries in
the plane rather than perpendicular to the plane no this does raise a question which is how do I pick the plane
supposing I have some less obvious object like a cue which doesn’t have necessarily
particularly long axis I’m for which I could take the playing parallel a perpendicular to the
long axis or in fact I might have forces being added in multiple directions so I might have a force half acting like
this but I might also have a force p acting up and down white back of this force
might not exactly man’s I may have a larger force here at one
corner and a small force here this corner at
the Q so in this case how on earth can I
possibly pick the right play well of course in general for problems like these the isn’t right
playing in the problems will be leaving this
course there will always be an obvious plane or you’ll be told which plan to pick but for general
problems like this were to be on the level of discourse their is no correct answer and so what
you do is you take three planes because you’ve
got a three-dimensional object you take one plane like this you take another plane perpendicular to it I’m you take a third plane thats
mutually perpendicular to you the other two planes because you
got three dimensions you could pick 3 perpendicular planes and this is
easier to see perhaps if I droid as the XY I’m sad axes surfaces axe this is why re: and this is a sad I can take the ex white playing I can take the backs said plane and I can take why that plane and 52 these planes I’m in a multiplier have
for you to these planes I’m gonna have a shear stress and I’m gonna have a normal stress and so what I’ve got is this is going to
give me 6 quantities and I write these down in the matrix and so I’m gonna end up with three by three matrix and it’s a metric so I have the normal stresses go along this
diagonal here and then these things are these passive
calm inside the I’m shear stresses so this is the same
shear stress like we’re here this year stress component is the same
report here the shear stress components is the same
and I put here and you end up with what’s called a tenser quantity and so does dress is in fact mine either a scaler nor vector it’s something we call the
tensor however the cancer treatment stress is well beyond the level of discourse or
you need to know is that it is attends the quantities
night the scaler nor vector we’re not going to deal with it is a
tensor we’re gonna do with it in simple situations where you can calculate shear stress and the normal stress in a
single plane and that plane will try to beat given
will be extremely obvious but in general stress
is a lot more complicated but we are gonna do with a complication here now we’ve already defined a
quantity calls rest which is a measure of if you like
the fourth acting on a rigid object and now what we need to do is we need to
look at how the optics 24 so we need to go to
quantify the defamation vortex so when I apply my
stress to this rob a bank you can see the
material structures and deforms answer the question is now
how do we quantify this defamation well the simple way to do this that
might come to you first glance is well let’s just measure
the extension all the material so if I started with
this size and I structure to hear I can measure
the change in length material I may be out by itself is a useful quantity but let’s zoom in and have a look a bit
more closely at this program to see maybe why that’s not such a good
idea okay and what you can see here is that I’ve
marked one centimeter lengths have the rubber band and you can see
that when there’s no tension applied the remand these are all one centimeter
apart map what what happens when I apply stress each of these marks gets further apart and so this point here I’ve got the more about one and a half centimeters roughly apart so each one centimeter length the rubber band has stretched by half a
centimeter so if I now measure the distance between
the first and the last one instead of three centimeters it’s now four and a half centimeters or if I
measure the extension of the first two centimeters a rubber band it’s now been
extended by one centimeter see you can see that the absolute
extension is not for use for quantity instead what we want to do is to find a quantity
called strain which will be the extension her unit
length and to see how to do that will look at
the diagram on the computer okay so the diagram here shows block that starts off with the Lancs l0 getting stretched bite this additional
and due to a force the taxing on it so we
have this perpendicular force here scratches the object out and causes it to get longer so in here the definition of Oscar rain is equal to
the change in length which of course is
justice del Toro quantity here and that’s divided by the original length l0 which is this quantity here and so thats our west rain to this particular case now if you notice that the units for the change in length of course are
language and so this isn’t meters history in SI units on the
original angst in SI units is also gonna be measured in meters and so what this means is that
strain is I mention that she has no units now of course when you taking this ratio
you do have to be careful that you do not have your centimeters
divided by meatless right that would be a bad thing to have
happens you gotta make sure that is sent to meet is divided by
centimetres or meters divided by meters in order to get the time mention
Rt %uh the dimensionless I’m quantity correct otherwise you have
also left over 5 200 or 100 so whatever so for example if this
would be well we did this rubber band and we I am started with and initial links I o0 problem and was three centimeters cuz
reporters 3 one centimeter lengths marked out I’ll
final links I’ll was 4.5 centimeters on so here we want to write all strain may be slightly differently
so strained here is going to be equal to a change in
length which of course is now offline land minus our initial and divided by our
initial and so in this particular example it’s going
to be 1.5 some two meters divided by 3 centimeters another course is just 0.2 song so meet sample we looked at just
before this the strain was that rubber band was 0.5 now this is what we call tensile strain because we’re putting I’m the object apart but you can also harm what’s called a compressive strain on
that’s when you push the Enzian she applied compressive
stressed and then you get it compressive strange now of course a compressive strain you
end up with instead of an extension to the link here you end up with a
reduction in the links so sheer through tensile strain you end up with a I’m Delta I’ll is greater m0 on for a compressive strain here you
will end up with Delta I’ll being less than syrup okay so that’s for tensile compressive and strains a loser in response to
compressive and tensile stresses which are both types of the normal
stressed with the force is perpendicular to the cross-section but we also dealt
which Shia stress so next question is what is the response to a shear stress so to look at the response to shear
stress we’re going to use this textbook so never let it be said that you can’t
learn physics from a textbook I’m perhaps sometimes in more ways than
you realize so here we have the textbook and what I’m going to do is I’m going to
apply a shear strain to the book I’m gonna move the top the book that way I’m you apply four
from the bottom in the book this way or I’m gonna hold this still
I’m try move the top for the book in that
direction and you’ll see what the response is so
here we go so you can see the top have the book slides in the direction the applied force now obviously the
thicker this book is the easier it would be for
me to move it in that direction and so the
displacement will get larger has the thickness of the book increases but you can clearly see that there’s a
displacement due to the a strain the stress and this displacement is known as the
Shia strain but the difference is from the strains we’ve seen before is
that now the thickness of the book is in this
direction and the displacement is in fact direction and so the
displacement we’re going to be interested in is how far this top and moves relative to its original position so you can see
that here it’s a displacement the perpendicular to
the thickness of the book and will switch back to the
computer screen to have a look at that in a bit more detail should now we’ve got here his we’ve got
a block and this block is subject to a shear stress so we have
these two forces that are now and opposite but now
acting in the same plane and so this defamation is a little
bit different we get our original link here and I’ll 0 but we get our dish placement are
extensional change defamation whatever you wanna call it our extension here is perpendicular to
this original links here which is you know 0 so in this case we have what’s called he Shia strain and this is defined in a very similar
manner to on normal strain and it’s now just
the extension divided by the original link
so in many ways it’s the same because this is now perpendicular to the original likes right so there’s a um
this is it ninety degrees to the original links if the object that
we’re concerned about now course again this is measured in
meters and this is measured in meters so again this is a time mention less value and so it has no units and this is what we call the shear
strength so for example if we take that book we said that the
original sickness of the book this is l0 was 5 centimeters on our extension that we got white squeezing it in on
moving sideways walls from 0.5 centimeters than of shear strain to be equal to 0.5 divided by five centimeters and so that is just 0.1 I’m strain and so that’s our definitions for Shia
string sewing Nancy how to measure can quantify both the internal stresses inside an
object and the defamation so those stresses
calls the strain so the next question we have to answer
is how is the stress related to the strain and to do that we have to introduce a
new topic called elastic moduli and that will cover in the next morning

9 comments on “PHYS 146 Elasticity part 1: Stress and Strain

  1. Mirhebib Seydeliyev Post author

    perfect video… now i understand what stress and strain are. great thanks to uuuuu


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