nm0808: what you wanted a copy of the these are some of the notes anyway 
om0809: 
nm0808: [laughter] can you take can you take one and pass it round please 
sm0810: 
nm0808: the-, these are the notes on the er stuff that we did last week 
nm0808: quiet questions on what we did last week okay so we move on with the 
reinforced materials and as we said last week what we need to establish are the 
relationship between the intrinsic properties of the two constituents the 
fibres on the one hand and the matrix the polymer matrix on the other and we 
want to get the rules which will give us the elastic and the strength 
properties of the composite as a function of the fibre content fibre length and 
fibre orientation as we said last week we start with a simple possible system 
which is continuous fibres embedded in the matrix and the fibres are running 
continuously from one end of the ideal strip to the other we know the 
properties of the matrix in terms of that's Young's modulus 
Poisson's ratio sheer modulus similiarly we do know the properties of the 
fibres again Young's modulus Poisson's ratio and sheer modulus and so long as 
we know how many fibres we've actually put in the system okay so we've got 
fibre content okay we can try to see how we work on that so the first property 
that we need to determine is the modulus in the direction of the fibres if we 
pull that system in this fashion and all this approach here is called rule of 
mixtures and you will see why the end result is that is a very simple 
expression which says that the modulus in the longitudinal direction sometimes 
we call that so this is E-L or sometimes called E-one or sometimes in some 
books it's called E-one-one okay they're all equivalent all it says is that 
simply that the modulus of the composite in the direction of the fibres is 
simply proportional to 
the amount of fibre B-F and to the modulus of the fibres E-F and now we'll see 
how we get here and we define the terms inside here the starting point for 
deriving that expression is to make the assumption first of all that the fibres 
are perfectly bonded to the matrix so that the de-, two are obliged to be 
formed together as you pull on the system the fibres and the matrix will 
experience the same strain if they're bonded together they've got different 
moduli but they will be strained to the same amount so the assumption is that E 
of the fi-, epsilon of the fibres epsilon-F is the same as epsilon-M which is 
also the strain applied to the composite as a whole as i mentioned last week 
that holds only until you start getting failure mechanisms beginning to operate 
okay so if we make this assumption how do we move on from here well to all 
intents and purposes what you have is a system of two springs 
acting in parallel you've got on the one hand the soft spring which is the 
matrix and in parallel with that you've got a stiffer string which is the fibre 
and these two are obliged to stretch by the same amount when they're pulled 
okay so that one the way one can think about the system at the microscopic 
level in terms of equations it's very simple all we do is to start with the 
standard way in which we do a lot of these things I-E force equilibrium so if 
we think about force equilibrium what do we have inside here can i rub off that 
okay 
sm0811: mm 
nm0808: so what do we know well we know that if we put a load at this end of 
the material so we apply a load P to the composite as a whole we just take that 
thing and we pull with some force okay that load inside the material's 
going to be shared some of it will be carried by the fibres P-F and some of it 
will be carried by the matrix P-M okay all i'm saying here is the external 
force is the sum of the two internal forces so that's the starting point and 
you will see that from now on is relatively simple is a question of definition 
if i want to move from force to stress what do i do is to divide by the cross-
sectional area so if i want the stress in the composite that will be now the 
load in the composite divided by the area of composite okay and if i do the 
same thing for the other two the load in the fibre the proportion of the load 
carried by the fibres will be the stress in the fibre times the cross-sectional 
area of fibre okay force equals stress times area so here P-F becomes sigma-F 
times P-F and P-M here becomes 
sigma-M times A-M where A-F is the total area or cross-sectional area of fibre 
A-M is the cross-sectional area of matrix if you slice that thing this way in 
fact we might do that if we slice a standard composite what we're going to see 
is something like this the fibres are packed together not necessarily in a very 
regular array but you've got cross-sectional area of fibres here and then 
you've got cross-sectional area of matrix here so A-F is the total cross-
sectional area of fibres A-M is the total cross-sectional area of matrix okay 
so far problems okay so all we do now is to say okay well if er if i divide 
okay by A-C throughout do i have to do that to get the whole thing to be 
correct here okay i've got A-F now ov-, over A-C and i'm going to define that 
ratio the cross-sectional area ratio between fibres and matrix as the volume 
fraction of fibres 
it is strictly speaking the area fraction A-F over A- total but because the 
fibres are infinitely long in the other direction okay the area the fraction of 
area occupied by the fibres is the same as the fraction of volume occupied by 
the fibres and when one deals with composites normally that is the variable 
that expresses the quantity of fibre inside is the volume fraction rather than 
the mass fraction so we define A-F over A-C as V-F and similarly we define A-M 
over A-C as V-M and obviously because those are volume fractions and not 
absolute volumes then we've got to have V-F plus V-M must be equals to one well 
the final step in the calculation is to take this expression here remember what 
we want is to arrive at the expression here on the of for the Young's modulus 
of the material so what we've got here now we've got sigma-C 
equals this time we can write this one as sigma-F times V-F plus sigma-M times 
V-M you notice that this is the s-, exact same expression as this one where 
instead of having the moduli we've got the stresses okay in other words the 
stress inside the composite is shared in proportion between the fibres and the 
matrix in proportion to the volume fraction and finally because we want to 
define the modulus what we do is to divide by the strain the modulus of the 
composite will be the stress in the composite divided by the strain in the 
composite from the definition of Young's modulus and if we divide by the strain 
throughout here then we get sigma-F over epsilon-C times V-F 
plus sigma-M over epsilon-C times V-M okay i'm just using pretty standard 
definitions and nothing else and now now comes a point when i'm going to make 
use of that assumption here because if as we have assumed the strain in the 
composite is the same as the strain in the matrix is the same as the strain in 
the fibres then here epsilon-C epsilon-C epsilon-C can be replaced by sigma-F 
over epsilon-F times V-F plus here sigma-M over epsilon-M times V-M okay it's 
the same number with a different name but it's the same number but that allows 
me then to introduce the elastic modulus of the fibres because stress of the 
fibre over strain in the fibre is modulus of the fibre so this is E-F times V-F 
and here saying the ratio of stress of the matrix to the strain of the matrix 
is the Young's modulus of 
the matrix and this is therefore going to be E-M times V-M and this is E- C or 
E- L okay and that's that so it's very simple in this particular example to 
derive the equation which makes a prediction about the Young's modulus of the 
composite in the fibre direction when the fibres are all parallel and 
continuous and as i said all it says is that the Young's modulus is 
proportional to the quantity of fibres that you add in the system which is not 
unexpected like all predictions of any kind one needs to make some assessment 
as to how good they are in representing reality and people have er used that 
equation to make predictions about elastic properties of composites and that 
prediction works very well so that one is good in the sense that the 
predictions you make are very very close to the er measured values okay clear 
okay well we can proceed the same way and i'm not going to do it but i'll leave 
for you 
to play around with the er equations which need to be set up for example the 
other modulus that we need you've got all that in your notes er anyway the 
other modulus w-, that we need is the one which is normal to the direction of 
the fibres which we called E- two or sometimes E-T for transverse or sometimes 
E-two-two okay again this is the fibre and you load the system this way now if 
you had to start the calculation which eventually leads you to that final 
expression what assumption are you going to make here to be able to get going 
without looking at your notes the previous case we assumed that the two springs 
were deforming in parallel and that the deformation in both of them was the 
same here we are pulling in this direction now okay can we make the same 
assumption that the load is the same in the two sys-, that the the strain is 
the same in the two systems 
sm0825: no 
nm0808: we can't it's obvious this this is much more 
compliant than the fibre okay so when we stretch this thing in that direction 
the matrix will deform more than the fibre will deform so the assumption that 
we used for the previous equation is not holding here but what is the 
assumption which holds here 
sm0812: series 
nm0808: yeah we've got two springs in series that's correct so this time we've 
got a soft spring which is the matrix and then we've got a stiff spring which 
is the fibre and this time we're pulling on this like that so if they are in 
series if we've got two springs in series okay and we pull okay what assumption 
can we make 
sm0813: the same stress 
nm0808: the same stress that's right okay in the previous case it was the same 
strain which was imposed on the system here we've got the same stress in the 
sense that if we pull that if we take the force the cross-sectional area of 
matrix and the cross-sectional area of fibre are the same 
okay so this thing here the assumption the starting assumption here is this 
time that sigma-F equals sigma-M equals sigma-C and then if you do exactly what 
i did for the previous case by introducing the deformations now in the matrix 
and in the in the fibres the which means it's actually in that system the total 
deformation is the sum of the deformation of one spring plus the deformation of 
the other spring okay because that's what springs in series are so the starting 
point is effectively is that the delta- L of the composite the increase in 
length in the direction two of the composite is going to be equal to the delta- 
L of the fibre plus the delta-L of the matrix and then by defining strain as a 
function of the increase in length by introducing the Young's modulus you get 
to eventually to that equation here and as i said you can do that as an 
exercise it's good for you and this equation here 
okay the assumption as we said equals stress and one important consideration 
which matters when you want to refine the thing and why i-, which explains also 
why this thing doesn't really work that well like the other one is that in this 
calculation one neglects the fact that a Poisson's ratios all the fibres in the 
matrix are different when you are pulling in that direction if the Poisson's 
ratio are different as you might expect the lateral contraction of the fibre 
and the lateral contraction of the matrix will be different there's going to be 
a mismatch at the interface okay so that creates problems but anyway this is 
the simple or the simplest approximation one can get and again the question is 
how does it work well unfortunately this one doesn't work well at all okay in 
other words the predictions of the transverse modulus that this equation makes 
are not very good and that means that they can be used only as a very first 
approximation but if you really need 
information or more detailed information then you will have to do and get some 
measurements or use some of the more sophisticated er models which introduce 
these effects here if we plot these two equations as a function of the variable 
involved which is essentially the volume fraction of er fibres okay we can 
start here this is the modulus of the matrix by itself and up here we've got 
the modulus of the fibres by themselves this is not quite to scale because 
remember that the order of magnitude difference is much bigger than i can draw 
on the board okay E-M is of the order of one G-P-A E-F is of the order of 
seventy G-P-A for glass okay but nevertheless the first equation the one we saw 
it's a linear relationship between the er fibre modulus volume fraction and 
composite modulus so as we increase the volume fraction of fibres this thing is 
going to go up like this so the modulus increases linearly with 
the amount of fibre that you put in okay and if you plot on the same graph the 
other one the one which gives the modulus parallel to the fibre direction this 
one here okay you find something different but obviously it's got to start at 
the same point down here and it's got to end up at the same point up here 
because if we got only fibres we have to measure the modulus of the fibres and 
the way it goes from that point to that point is somewhat like this 
okay so this is E- one or E-L and this one is E-two or E- sub-T so the modulus 
increases far more slowly with the volume fraction of fibres in the second case 
okay and then obviously it's got to start shooting up to catch up with the 
value of the modulus of the fibre when we get up there you realize of course 
that it's impossible to get volume fractions of what [laugh] okay volume 
fraction of one means we've ju-, just got fibres and nothing else volume 
fraction of er one doesn't mean a composite means just pure fibres and the 
question is realistically how far can we go along these two curves okay well in 
practice okay you see the problem okay this is the cross-section of your 
composite and in an ideal world okay you can pack the fibres either in a nice 
cubic array like this or you can pack the fibres in a hexagonal array like this 
okay it's pretty obvious but so just looking at the pictures that if we can 
pack them in a hexagonal array we are going to pack more fibres in the same 
volume because the fibres are nesting with each other much better okay and if 
you work out and you've got again the calculation in your notes in detail if 
you simple geometry okay if you get a hexagonal array the packing the maximum 
packing is of the order of ninety per cent or thereabouts and if you pack in a 
cubic array you can get something of the order of seventy per cent or 
thereabouts but remember that those are ideal situations which in practice is 
impossible remember that the fibres are six to ten microns in diameter okay and 
short of puttting them one by one okay in in in a very nice neat kind of 
arrangement it's virtually impossible by any process in technology to arrive at 
volume fractions anywhere near these values so where are the limits well the 
limit in practice it's at about 
sixty per cent V-F these are for the high er performance composites that we're 
going to see next term in the composites course okay and for most reinforced 
plastics on the other hand those that we can process by injection moulding 
extrusion vacuum forming and the other methods we are already far below that 
okay so these are the high performance fibres sorry high performance composites 
which as i said is the subject of next term lecture whereas here we are more or 
less in the band which goes from about say ten per cent five per cent say to 
about twenty-five per cent so the useful region of reinforcement for the er 
reinforced plastics okay is about five to twenty-, twenty-five per cent it's 
okay pick it up if you like or leave it there okay now unfortunately as i said 
last week having the modulus in the 
fibre direction one and having the modulus in the transverse direction two is 
not enough to get all the information we need to do any calculation relating to 
design stress analyses or whatever of er reinforced plastic materials we need 
on top and above these two the sheer modulus which comes into play if we try to 
deform the system that way sliding essentially the matrix with respect to the 
fibre and then we need the Poisson's ratio er I-E the contraction which occurs 
in one direction when we pull the material in the other direction and so we 
need G-one-two and we need a new one-two i remember the definition one-two 
means pulling in the one direction and measuring the protraction in the two 
direction and we need a new two-one which is this time pulling in the two 
direction and looking at the contraction occurring in the one direction and as 
we said last week these two are 
not the same so in theory we've got five numbers one two three four and five 
that we need to characterize the material fully in terms of its elastic 
behaviour okay well one can do er calculations based on the same principles as 
the one we've seen and again i'm not going to go into the details okay but this 
is how one would set up the calculation for the sheer modulus G-one-two or G-L-
T as it's called sometimes okay if you look at the distortion in each one of 
the components and again playing around with the basic definitions of sheer 
strain bringing in the modulus and the load sharing between the fibres and the 
matrix and this is again a system of springs in series essentially okay you get 
an equation which predicts the sheer modulus and the equation looks almost in 
fact it is the same identical equation than we have obtained for the transverse 
modulus E-two and in the same way as the other one did not 
work very well this one also is not very accurate so we've got the same problem 
that we can predict this number here quite well this one is not very good this 
one is not very good either so again either measurement or more sophisticated 
modelling and theories and the last one is the Poisson's ratio that we need as 
well and nu-one-two is what on the transparency's called nu-L-T this one on the 
other hand follows the same rule as the longitudinal modulus E-L or E-one so 
this one is okay this one i-, also is a good prediction of the er measured 
values of the material and fortunately fortunately we don't have to worry about 
the last one nu-two-one is not independent of the other properties and there is 
this relationship which exists between the two Poisson's ratio 
one-two two-one on the one hand and the corresponding moduli in the same 
directions E-one or E-L and E-two or E-T so if we know this and we know this 
and we know this then using that equation which is on the board here we can 
calculate this nevertheless at the end of all that the important message is 
that we've got four elastic properties which we need to predict or measure or a 
combination of both and the important thing of course is the fact that they are 
all independent of each other okay the fifth one is the only one which can be 
recovered from the other four so we got a more complex system than isotropic 
materials where we need only two numbers E- G or E-nu or G-nu and that means 
that even the simplest type of calculation gets more cumbersome when you're 
dealing with an isotropic fibre enforced materials well that's life we can't 
help it er the 
only justification for that increase in complexity is the fact that we are 
going to get some benefits in terms of higher moduli higher strengths er 
fatigue life er reduction in creep or stress relaxation behaviour all these 
kind of things questions all clear please do have a go at trying to do the 
simple mechanics of deriving these equations here because it's a good exercise 
and it forces you to think er in the way that i would like you to think because 
we'll need that as we move on to the more complex things any more questions 
about that all happy with that okay well just to give you some numbers okay if 
you remember what i said here that the volumes of the matrix down here is 
typically of the order say of one G-P-A the modulus of the fibre up here is 
typically of the order of seventy G-P-A and you can see that even if you have a 
volume fraction of fibres of say sixty per cent okay you still got a modulus 
which is 
significantly higher of course than the one of the pure polymer matrix so if we 
imagine a volume fraction of sixty per cent well our prediction for E-one or E-
L is simply er point-six times seventy plus point-four times one okay so this 
is forty-two here plus we can forget about that point-four okay so at best okay 
with all the fibres parallel and the maximum amount of fibre that we can 
practically introduce in a polymer okay with all the fibre in one direction the 
modulus the maximum modulus we're going to get is about forty-two G-P-A well 
that's a obviously a great deal more than the one G-P-A we started from er with 
the pure polymer but again compare that to steel which is at two-hundred-and-
ten G-P-A or aluminium which is at seventy G-P-A so we begin to catch up in 
modulus terms we begin to catch up a little bit with aluminium but we're still 
a 
long way from catching up with the modulus of steel there are other fibres 
which are used when you really need a very high modulus that you may want to 
use to com-, er to er compete with steel and we're talking about carbon fibres 
Kevlar fibres and things of this kind which again we're going to see next term 
so we leave them at that for the moment more relevant to this course is what 
happens down here because this is the typical range of volume fractions that 
we're going to use for injection moulded er polymers okay so if we take a 
volume fraction say of twenty per cent okay well again our prediction for E-one 
will be of the order of point-two times seventy plus point-eight times one okay 
and you can see that we're going to get something of the order of fourteen G-P-
A which is respectable remember 
that you started from one okay which is a very low value indeed so getting from 
one to something like ten fifteen okay it's not a bad shot at all the problem 
is that this implies as we said that all the fibres are parallel okay and 
highly er oriented in one direction in practice okay it's not easy to achieve 
that parallel orientation of the fibres when you are using processes like 
injection moulding because the flow of the material which is carrying the 
fibres with it okay is not necessarily going to give you a nice er oriented 
system so you're going to lose some of this inevitably okay this is the maximum 
you can get if you like at that volume fraction but you're going to lose some 
of that because the fibres are not going to be packing exactly parallel to each 
other and you're going to lose also a little bit more because the fibres are 
not going to be continuous and this is what we're going to see later on er in 
the course so we've got to sacrifice 
something if you like when we go to short fibres we lose some of their 
potential and when we go to non-parallel fibres we lose something else but that 
is for tomorrow can i remind you that we've got the tutorial tomorrow okay 
whatever is the tutorial time slot i think it's eleven to twelve in G-twenty-
six okay okay so if you have no more questions okay we move on to the next 
thing we all happy okay well the next thing is okay well what happens to the 
strength we talked about the modulus okay this is one of the properties that we 
need the other one property that we need has to do with the strength of the 
material and here get thing's get a great deal more complicated i'm afraid but 
if you start thinking in terms of strains and er either equal strains or equal 
stresses or the like things get into shape quite happily so let's talk about 
strength now the first comment to make is that the only strength that i'm 
going to be interested in is the one in the direction of the fibre because in 
the direction normal to the fibre okay this is my little sample again it's 
pretty obvious that if i pull this in that direction the only strength i'm 
going to get is the strength of the matrix i started from okay the fibres are 
not carrying a great deal of stress in that direction we can verify that from 
the models we've just talked about so the strength in that direction is 
essentially equal to is roughly the same as the strength of the matrix so i'm 
not going to worry too much it's low it's low it's whatever it is there's 
precious little i can do about it 
whatever however in this direction here which is the direction in which i'm 
going to take maximum advantage and benefit out of the fibres then i want to 
know what kind of predictions i can make about the strength of the material in 
that direction so when we talk about strength is strength parallel to the fibre 
direction okay and remember that when we are loading the system in the 
direction of the fibres the way in which we derived the elastic modulus was to 
assume that the composite strain was the same as the fibre strain was the same 
as the matrix strain okay that was the initial assumption okay and that initial 
assumption led then to the expression for the Young's modulus in the 
longitudinal direction well if we're talking about strength we're talking about 
failure we're talking about something which as we carry on pulling is going to 
start breaking down giving up if we plot the stress versus strain curve of the 
composite we just do a pure tensile test in a strip like this one okay what 
we're going to get is the material is going to deform elastically okay the two 
phases the matrix phase and the fibre phase are going to carry the same strain 
okay so everything goes hand in hand and then something happens okay what are 
the events which can happen 
sm0814: the fibres can snap 
nm0808: the fibres can snap first okay so we get the situation where the fibres 
break before the matrix breaks in other words what does that mean it means that 
the failure strain of the fibres is less than the failure strain of the matrix 
okay if you're pulling 
the fibres and the matrix in parallel okay the one which is giving up first is 
the one which has got the lowest failure strain okay so that's one case any 
more 
sm0815: can you have it the other way round 
nm0808: of course we can have another situation which is where there's fibre 
strain maximum is actually greater than the matrix strain any more offers 
sm0816: they're equal 
nm0808: we'll come to back in i-, to your question yeah the the third one is 
the t-, ch-, sort of simple case where epsilon-F- M-maximum is the same as 
epsilon-M- maximum so those are the three cases we need to think about okay and 
going back to your question when you're saying they're falling apart i think 
what you're meaning is that you've got an i-, er an interfacial f-, sh-, 
failure between the fibres and the matrix okay that is something which happens 
more often with the short fibre enforcement remember 
here we are in an ideal world getting hold of the bits at the two ends and we 
pull simultaneously so you can't have a relative sliding when we look at the 
short fibres that can happen of course so for the long fibre enforcement 
continuous infinitely long or parallel okay those are the three situations 
which we have to analyse and the simplest one to look at not surprisingly is 
the last one and that the one who is although it's a very unrealistic one 
because there is really ver-, virtually no system that i'm aware of where 
you've got the situation that the fibre strain is er maximum strain is of the 
same magnitude as the matrix maximum strain out of curiosity do you know what 
is the maximum failure strain of glass fibres what do you think the failure 
strain of glass fibres is likely to be one per cent half a per cent point-two 
per cent point-one per cent ten per cent 
sm0817: very small 
sm0818: two-point 
sm0818: one per cent would be very little 
nm0808: well very little 
is not an engineering [laughter] okay what is typically the strain at which a 
material like a piece of metal yields what is the yield strain of a mild steel 
at what sort of strain levels does a steel or an aluminium yields not fails 
yields okay no idea 
sm0819: is it a constant 
nm0808: er fed up the y-, yie-, a yield strain a yield strain for a metal is of 
the order of point-two to point-three per cent okay that doesn't mean that is 
the ultimate strain this is the strain at which it starts yielding and 
deforming plastically remember that a metal is a polycrystalline system okay a 
crystal can not deform by a great deal okay before you start getting the glide 
and the slip coming in to operation so the yield strain of a metal is of the 
order of point-two point-three per cent and the failure strain of a ductile 
metal can be very large indeed because we know that we can stretch a piece of 
steel plastically quite a lot so 
the failure strain okay can be anything up to say about twenty per cent 
possibly more sorry someone had a question 
sm0820: er yes at point-two to three per cent of what 
nm0808: it's the strain 
sm0820: right 
nm0808: [laughter] of what [laughter] okay [cough] so remember that a [cough] 
glass does not yield glass is a brittle material it will stretch elastically up 
to the point of failure and the question is can you give me some idea now that 
we've put some numbers in okay at what value of the strain that failure in the 
glass fibre's going to occur 
sm0821: nought-point-one-five 
nm0808: point-nought-five okay at least the 
sm0822: nought-point-nought 
nm0808: nou-, 
sm0822: nought-one 
nm0808: nought-point-nought-nought-one well you'll be all surprised that the 
maximum failure strength for glass if glass is perfect if the fibres are not 
damaged can be as high as five per cent [whistle] okay so you can 
get very very large elastic strains now i-, in practice okay you hardly ever 
achieve this value because the fibres inevitably get damaged in handling get 
damaged in processing because they touch each other you get scratches and 
defects but in what one may call ideal er undamaged fibres the failure strains 
are indeed very high and although this is represents a maximum okay epsilon 
failure maximum practical if you like of about two per cent to three per cent 
is not unknown you all seen the bit of glass fibre springs downstairs in the 
labs okay well those are typically working at strains of working [laugh] okay 
which means below the maximum stress they are working at strains of about one-
point-two one-point-five per cent okay and the failure is about twice as high 
so three per cent is achievable and i think it's important to put that into 
context because the fact that the material is brittle which it 
certainly is okay does not mean that it can not be strong so long as it is in 
fibre form where having no defects is what is making the difference if you take 
a glass pane certainly that one will never get anywhere near to these sort of 
numbers because it's got too many defects surface fractures damage and whatever 
not to be able to do so okay so let's back here now all i'm saying is that when 
we talk about the maximum failure strain of the matrix being the same as the 
maximum failure strain of the fibres which is one particular simple system okay 
it's important to have some idea of what kind of numbers we are talking about 
okay well this is the situation you have and you've got it in your notes again 
so i'm not er going to write all that on the board okay the equation which 
governs the deformation of the material is the one we've just seen before the 
rule of mixtures in other words up to the point okay let me draw the thing a 
little bit like it is there 
okay this is the case i'm considering now same failure strain in the matrix and 
in the fibre and let's assume that that failure strain for the system okay is 
somewhere here so this is epsilon- max when i reach that strain level the 
fibres and the matrix will fail at the same time okay because that's what we 
are assuming however this is how the composite may deform so let's put this 
with a C the fibres of course if one looks at the fibres themselves the fibres 
are stiffer okay and the fibres will deform like this and the matrix is a lot 
softer and the matrix will deform like this 
so i-, if we look at the individual components matrix by itself fibres by 
themselves and then composites okay we get three stress strain curves of this 
kind and this one here the one of the composite of course moves up or down 
depending on the volume fraction the higher the volume fraction it is the 
closer it's going to get to the curve of the fibres the lower the volume 
fraction is the closer it's going to be to the curve for the matrix okay well 
no matter where we are okay no matter where we are between here and here okay 
at any point up this curve this equation here halts okay yeah 
sm0824: is the end modulus of the matrix really a lot lower than that of the 
composite or 
nm0808: yeah well i'll give you a sort of example just from before the modulus 
of the matrix about one per cent er one gigapascal the fibres are seventy 
gigapascal [laugh] okay and in a typical volume fraction of twenty per cent the 
modulus of the composite going to be about twenty twenty-five gigapascal so 
it's a big 
big difference and this is one of the problems of course especially when you go 
to the short fibre enforcement so we can write that okay and this equation here 
is nothing but the equation of that line okay just there as the volume fra-, 
for a given fixed volume fraction okay if i increase the stress in the 
composite okay a stress in the fibre will increase the stress in the matrix 
will increase in proportion and i move up that line when i reach my limit value 
here everything gives up okay the fibres have reached their failure strain the 
matrix has reached its failure strain so the composite can only carry as a 
maximum stress the stress that corresponds to the fibre ultimate stress times 
the volume fraction of fibres plus the matrix ultimate stress times the volume 
fraction of matrix okay and so this is the equation which gives me the 
prediction of the tensile strength of the composite when i have this condition 
here okay it's simply the limit case of the 
rule of mixtures when both components fibres and matrix have reached their 
failure point i can't get a higher stress than that obviously okay it is also 
pretty obvious that if the amount of reinforcement is very low in other words 
if the volume fraction of fibres is very low if i head down here okay then to 
some extent you can neglect that and you end up with a strength which is just a 
little bit higher than the strength of the matrix by itself but obviously that 
would be rather pointless okay we really need to put the fibres in to move the 
strength and the modulus of the matrix significantly beyond its normal level 
when it is unreinforced otherwise 
we are spending a lot of money to er no effect whatsoever questions okay we 
reached the end of the hour so we'll leave it at that tomorrow we are going to 
look at the other two situations both of which exist there are situation where 
the fibres do have a higher failure strain than the matrix that is typical for 
example with thermosetting resins which have got low ve-, very low failure 
strains and other situations sorry where is the opposite where the m-, failure 
strain of the matrix is very large and this is typical of the er crystalline 
thermoplastic all the materials which can deform in a ductile manner over large 
strains okay so don't forget tutorial time tomorrow G-twenty-six from eleven to 
twelve