Skip to main content Skip to navigation


<?xml version="1.0"?>

<!DOCTYPE TEI.2 SYSTEM "base.dtd">





<publicationStmt><distributor>BASE and Oxford Text Archive</distributor>


<availability><p>The British Academic Spoken English (BASE) corpus was developed at the

Universities of Warwick and Reading, under the directorship of Hilary Nesi

(Centre for English Language Teacher Education, Warwick) and Paul Thompson

(Department of Applied Linguistics, Reading), with funding from BALEAP,

EURALEX, the British Academy and the Arts and Humanities Research Board. The

original recordings are held at the Universities of Warwick and Reading, and

at the Oxford Text Archive and may be consulted by bona fide researchers

upon written application to any of the holding bodies.

The BASE corpus is freely available to researchers who agree to the

following conditions:</p>

<p>1. The recordings and transcriptions should not be modified in any


<p>2. The recordings and transcriptions should be used for research purposes

only; they should not be reproduced in teaching materials</p>

<p>3. The recordings and transcriptions should not be reproduced in full for

a wider audience/readership, although researchers are free to quote short

passages of text (up to 200 running words from any given speech event)</p>

<p>4. The corpus developers should be informed of all presentations or

publications arising from analysis of the corpus</p><p>

Researchers should acknowledge their use of the corpus using the following

form of words:

The recordings and transcriptions used in this study come from the British

Academic Spoken English (BASE) corpus, which was developed at the

Universities of Warwick and Reading under the directorship of Hilary Nesi

(Warwick) and Paul Thompson (Reading). Corpus development was assisted by

funding from the Universities of Warwick and Reading, BALEAP, EURALEX, the

British Academy and the Arts and Humanities Research Board. </p></availability>




<recording dur="00:50:36" n="6751">


<respStmt><name>BASE team</name>



<langUsage><language id="en">English</language>



<person id="nm0808" role="main speaker" n="n" sex="m"><p>nm0808, main speaker, non-student, male</p></person>

<person id="om0809" role="observer" n="o" sex="m"><p>om0809, observer, observer, male</p></person>

<person id="sm0810" role="participant" n="s" sex="m"><p>sm0810, participant, student, male</p></person>

<person id="sm0811" role="participant" n="s" sex="m"><p>sm0811, participant, student, male</p></person>

<person id="sm0812" role="participant" n="s" sex="m"><p>sm0812, participant, student, male</p></person>

<person id="sm0813" role="participant" n="s" sex="m"><p>sm0813, participant, student, male</p></person>

<person id="sm0814" role="participant" n="s" sex="m"><p>sm0814, participant, student, male</p></person>

<person id="sm0815" role="participant" n="s" sex="m"><p>sm0815, participant, student, male</p></person>

<person id="sm0816" role="participant" n="s" sex="m"><p>sm0816, participant, student, male</p></person>

<person id="sm0817" role="participant" n="s" sex="m"><p>sm0817, participant, student, male</p></person>

<person id="sm0818" role="participant" n="s" sex="m"><p>sm0818, participant, student, male</p></person>

<person id="sm0819" role="participant" n="s" sex="m"><p>sm0819, participant, student, male</p></person>

<person id="sm0820" role="participant" n="s" sex="m"><p>sm0820, participant, student, male</p></person>

<person id="sm0821" role="participant" n="s" sex="m"><p>sm0821, participant, student, male</p></person>

<person id="sm0822" role="participant" n="s" sex="m"><p>sm0822, participant, student, male</p></person>

<person id="su0823" role="participant" n="s" sex="u"><p>su0823, participant, student, unknown sex</p></person>

<person id="sm0824" role="participant" n="s" sex="m"><p>sm0824, participant, student, male</p></person>

<person id="sm0825" role="participant" n="s" sex="m"><p>sm0825, participant, student, male</p></person>

<person id="su0826" role="participant" n="s" sex="u"><p>su0826, participant, student, unknown sex</p></person>

<personGrp id="ss" role="audience" size="s"><p>ss, audience, small group </p></personGrp>

<personGrp id="sl" role="all" size="s"><p>sl, all, small group</p></personGrp>

<personGrp role="speakers" size="21"><p>number of speakers: 21</p></personGrp>





<item n="speechevent">Lecture</item>

<item n="acaddept">Engineering</item>

<item n="acaddiv">ps</item>

<item n="partlevel">UG2</item>

<item n="module">Polymers</item>




<u who="nm0808"> what <pause dur="0.2"/> you wanted a copy of the <pause dur="0.7"/> these are some of the notes anyway </u><u who="om0809" trans="latching"> <gap reason="inaudible due to background noise" extent="1 sec"/> </u><u who="nm0808" trans="latching"> <vocal desc="laughter" iterated="y" dur="1"/><pause dur="5.9"/> can you take <pause dur="0.6"/> can you take one and pass it round please </u><u who="sm0810" trans="overlap"> <gap reason="inaudible" extent="1 sec"/> </u><u who="nm0808" trans="overlap"> <trunc>the</trunc> these are the notes on the <pause dur="1.3"/> <gap reason="inaudible" extent="1 word"/><pause dur="0.7"/> # <pause dur="0.3"/> stuff that we did last <pause dur="0.3"/> week </u><gap reason="break in recording" extent="uncertain"/> <u who="nm0808" trans="pause"> quiet <pause dur="1.4"/> questions on what we did last week <pause dur="4.6"/> okay so we <pause dur="1.3"/> move on <pause dur="1.9"/> with <pause dur="1.7"/> the reinforced materials <pause dur="1.4"/> and as we said last week <pause dur="1.2"/> what we need to establish <pause dur="1.0"/> are the relationship between the <pause dur="0.7"/> intrinsic properties of the two constituents <pause dur="0.2"/> the fibres <pause dur="0.5"/> on the one hand <pause dur="0.9"/> and the matrix the polymer matrix <pause dur="0.2"/> on the other <pause dur="1.1"/> and we want to get the rules <pause dur="1.1"/> which will give us <pause dur="0.7"/> the elastic <pause dur="0.3"/> and the strength properties <pause dur="0.5"/> of the composite <pause dur="0.2"/> as a function <pause dur="0.7"/> of the fibre content <pause dur="1.1"/> fibre length <pause dur="0.9"/> and fibre orientation <pause dur="1.6"/> as we said last week we start with a simple <pause dur="1.1"/> possible system which is <pause dur="1.2"/> continuous fibres <pause dur="0.5"/> embedded in the matrix <pause dur="1.2"/> and the fibres are running <pause dur="1.1"/> continuously from one end <pause dur="0.3"/> of the <pause dur="0.3"/> ideal strip <pause dur="0.7"/> to the other <pause dur="1.3"/> we know the properties of the matrix <pause dur="1.0"/> in terms of <pause dur="1.2"/><kinesic desc="writes on board" iterated="y" dur="6"/> that's Young's modulus <pause dur="1.0"/>

Poisson's ratio <pause dur="1.0"/> sheer modulus <pause dur="1.1"/> similiarly <pause dur="0.4"/> we do know the properties of the fibres <pause dur="1.0"/><kinesic desc="writes on board" iterated="y" dur="5"/> again Young's modulus <pause dur="0.9"/> Poisson's ratio <pause dur="1.1"/> and sheer modulus <pause dur="1.8"/> and so long as we know how many fibres we've actually put <pause dur="0.2"/> in the system <pause dur="0.9"/> okay <pause dur="2.8"/><kinesic desc="writes on board" iterated="y" dur="6"/> so we've got fibre content <pause dur="1.9"/> okay we can try to see how we <pause dur="0.3"/> work on that <pause dur="1.0"/> so the first <pause dur="1.9"/> property that we need to <pause dur="0.8"/><kinesic desc="writes on board" iterated="y" dur="5"/> determine <pause dur="1.1"/> is <pause dur="0.2"/> the modulus <pause dur="2.5"/> in the direction of the fibres <pause dur="0.2"/> if we pull that system <pause dur="1.3"/> in this fashion <pause dur="1.7"/> and all this approach <pause dur="0.2"/> here is called <pause dur="0.3"/> rule of mixtures <pause dur="0.5"/> and you will see why <pause dur="0.3"/> the end <pause dur="0.2"/> result <pause dur="1.0"/> is that <pause dur="0.4"/> is a very simple expression <pause dur="0.8"/> which says that the modulus in the longitudinal direction <pause dur="0.9"/> sometimes we call that <pause dur="1.6"/> so this is <kinesic desc="writes on board" iterated="y" dur="7"/> E-L <pause dur="0.9"/> or sometimes called E-one <pause dur="0.7"/> or sometimes in some books it's called E-one-one <pause dur="0.3"/> okay they're all equivalent <pause dur="2.1"/> all it says is that simply that the modulus of the composite <pause dur="0.2"/> in the direction of the fibres <pause dur="0.5"/> is simply <pause dur="0.8"/> proportional <pause dur="0.2"/> to

the amount <pause dur="0.2"/> of fibre <pause dur="0.5"/> B-F <pause dur="0.8"/> and to the modulus of the fibres <pause dur="0.2"/> E-F <pause dur="1.0"/> and now we'll see how we get here and we define the terms <pause dur="0.2"/> inside here <pause dur="2.8"/> the starting point <pause dur="0.6"/> for <pause dur="0.6"/><kinesic desc="writes on board" iterated="y" dur="8"/> deriving that expression <pause dur="3.4"/> is <pause dur="2.9"/> to make the assumption first of all that the fibres are <pause dur="0.6"/> perfectly bonded <pause dur="0.2"/> to the matrix <pause dur="0.7"/> so that the <trunc>de</trunc> <pause dur="0.2"/> two are obliged to be formed <pause dur="0.2"/> together <pause dur="0.7"/> as you pull <pause dur="0.3"/> on the system <pause dur="0.6"/> the fibres <pause dur="0.3"/> and <pause dur="0.5"/> the <pause dur="0.2"/> matrix <pause dur="0.7"/> will experience the same <pause dur="0.2"/> strain <pause dur="0.3"/> if they're bonded together <pause dur="0.6"/> they've got different moduli <pause dur="0.6"/> but <pause dur="0.2"/> they will be strained to the same amount <pause dur="1.5"/> so the assumption is that <pause dur="0.3"/> E of the <trunc>fi</trunc> epsilon of the fibres <pause dur="0.4"/> epsilon-F <pause dur="0.6"/><kinesic desc="writes on board" iterated="y" dur="5"/> is the same as <pause dur="0.4"/> epsilon-M <pause dur="0.7"/> which is also <pause dur="1.0"/> the <pause dur="0.2"/> strain <pause dur="0.2"/> applied to the composite <pause dur="0.2"/> as a whole <pause dur="0.3"/> as i mentioned last week that holds only <pause dur="0.3"/> until you start getting failure <pause dur="0.9"/> mechanisms <pause dur="0.3"/> beginning to operate <pause dur="2.0"/> okay so if we make this assumption <pause dur="1.8"/> how do we move on from here <pause dur="1.0"/> well to all intents and purposes what you have <pause dur="0.2"/> is a system <pause dur="0.8"/> of two springs <pause dur="0.2"/>

acting <pause dur="0.5"/> in parallel <pause dur="1.0"/> you've got on the one hand <pause dur="1.3"/> the soft <pause dur="0.3"/> spring <pause dur="0.9"/> which is the matrix <pause dur="1.5"/><kinesic desc="writes on board" iterated="y" dur="1"/> and <pause dur="0.6"/> in parallel with that <pause dur="0.7"/> you've got <pause dur="0.2"/> a stiffer string <pause dur="1.3"/><kinesic desc="writes on board" iterated="y" dur="7"/> which is <pause dur="0.2"/> the fibre <pause dur="0.7"/> and these two <pause dur="0.3"/> are obliged <pause dur="0.7"/> to stretch <pause dur="0.3"/> by the same amount <pause dur="0.2"/> when they're pulled <pause dur="1.2"/> okay <pause dur="2.2"/> so that one <pause dur="0.4"/> the way one can think about <pause dur="0.7"/> the system <pause dur="0.5"/> at the microscopic level <pause dur="2.1"/> in terms of <pause dur="0.2"/> equations it's very simple all we do <pause dur="0.2"/> is to <pause dur="0.8"/> start with the <pause dur="0.2"/> standard way in which we do a lot of these things <pause dur="0.3"/> I-E <pause dur="0.4"/> force equilibrium <pause dur="11.7"/><kinesic desc="writes on board" iterated="y" dur="9"/> so if we think about force equilibrium <pause dur="1.3"/> what do we have <pause dur="0.3"/> inside here <pause dur="2.7"/> can i <pause dur="0.4"/> rub off that <pause dur="2.2"/> okay </u><u who="sm0811" trans="latching"> mm </u><pause dur="3.7"/> <u who="nm0808" trans="pause"> <event desc="wipes board" iterated="y" dur="2"/> so what do we know well we know that <pause dur="0.3"/> if we put a load <pause dur="1.3"/> at this end <pause dur="1.6"/><kinesic desc="writes on board" iterated="y" dur="5"/> of the material <pause dur="1.2"/> so <pause dur="0.6"/> we apply a load P <pause dur="0.8"/> to the composite as a whole we just take that thing and we pull <pause dur="0.3"/> with some force <pause dur="0.8"/> okay that load <pause dur="0.2"/> inside the material's

going to be shared <pause dur="0.6"/> some of it <pause dur="1.0"/> will be carried by the fibres <pause dur="0.6"/> P-F <pause dur="1.0"/> and some of it <pause dur="1.1"/><kinesic desc="writes on board" iterated="y" dur="3"/> will be carried by the matrix <pause dur="0.4"/> P-M <pause dur="1.4"/> okay <pause dur="0.6"/> all i'm saying here is <pause dur="0.9"/> the external force <pause dur="0.2"/> is <pause dur="0.4"/> the sum <pause dur="0.2"/> of the two <pause dur="0.5"/> internal forces <pause dur="1.6"/> so that's the starting point and you will see that from now on is <pause dur="0.4"/> relatively simple is a question of definition <pause dur="0.6"/> if i want to move <pause dur="0.2"/> from force to stress <pause dur="0.5"/> what do i do is to divide <pause dur="0.2"/> by the cross-sectional area <pause dur="0.6"/> so if i want the stress in the composite <pause dur="0.8"/><kinesic desc="writes on board" iterated="n"/> that will be now the load in the composite <pause dur="0.6"/><kinesic desc="writes on board" iterated="y" dur="1"/> divided by the area of composite <pause dur="2.1"/> okay <pause dur="2.9"/> and <pause dur="1.7"/><kinesic desc="writes on board" iterated="y" dur="1"/> if i do the same thing for the other two <pause dur="0.6"/> the load in the fibre <pause dur="0.3"/> the proportion of the load carried by the fibres <pause dur="0.9"/> will be <pause dur="0.5"/> the stress in the fibre <pause dur="0.5"/> times <pause dur="0.2"/> the cross-sectional area of fibre <pause dur="1.0"/> okay <pause dur="0.4"/> force equals <pause dur="0.3"/> stress times area <pause dur="1.1"/> so here P-F <pause dur="0.3"/><kinesic desc="writes on board" iterated="y" dur="3"/> becomes sigma-F <pause dur="1.0"/> times <pause dur="0.2"/> P-F <pause dur="1.6"/> and P-M here <pause dur="3.9"/><kinesic desc="writes on board" iterated="y" dur="4"/> becomes <pause dur="0.2"/>

sigma-M <pause dur="0.9"/> times <pause dur="0.9"/> A-M <pause dur="0.3"/> where <pause dur="0.2"/> A-F <pause dur="1.6"/> is the <pause dur="0.6"/> total area <pause dur="0.4"/> or cross-sectional area of fibre <pause dur="0.4"/> A-M is the <pause dur="0.2"/> cross-sectional area of matrix <pause dur="0.2"/> if you slice that thing this way <pause dur="2.2"/> in fact we might do that <pause dur="1.1"/><kinesic desc="writes on board" iterated="y" dur="13"/> if we slice <pause dur="0.5"/> a standard composite <pause dur="1.0"/> what we're going to see <pause dur="1.0"/> is something like this <pause dur="1.1"/> the fibres are packed <pause dur="0.3"/> together <pause dur="0.8"/> not necessarily in a very regular <pause dur="1.3"/> array <pause dur="0.9"/> but you've got <pause dur="0.9"/> cross-sectional area of <pause dur="0.9"/> fibres here <pause dur="0.6"/> and then you've got cross-sectional area of matrix <pause dur="0.2"/> here <pause dur="0.9"/> so <pause dur="0.2"/> A-F is the total cross-sectional area of fibres <pause dur="0.4"/> A-M is the total cross-sectional area of matrix <pause dur="2.3"/> okay so far <pause dur="1.5"/> problems <pause dur="1.0"/> okay so all we do now is to say okay well if # <pause dur="2.2"/> if i divide <pause dur="1.2"/> okay by A-C throughout do i have to do that <pause dur="2.6"/><kinesic desc="writes on board" iterated="y" dur="5"/> to get <pause dur="0.2"/> the whole thing to <pause dur="0.3"/> be correct <pause dur="0.8"/> here <pause dur="1.1"/> okay <pause dur="2.1"/> i've got <pause dur="0.2"/><kinesic desc="writes on board" iterated="y" dur="5"/> A-F now <trunc>ov</trunc> <pause dur="0.3"/> over A-C <pause dur="1.3"/> and i'm going to define that ratio <pause dur="0.3"/> the cross-sectional area ratio between fibres and matrix <pause dur="0.5"/> as the volume fraction of fibres <pause dur="1.3"/><kinesic desc="writes on board" iterated="y" dur="1"/>

it is <pause dur="0.4"/> strictly speaking the area fraction <pause dur="0.3"/> A-F over <pause dur="0.2"/> A-<pause dur="0.2"/>total <pause dur="0.8"/> but <pause dur="0.3"/> because the fibres are infinitely long in the other direction okay <pause dur="0.7"/> the <pause dur="0.4"/> area the fraction <pause dur="0.2"/> of area <pause dur="0.7"/> occupied by the fibres is the same as the fraction of volume <pause dur="0.6"/> occupied by the fibres <pause dur="1.0"/> and <pause dur="0.4"/> when one deals with composites normally <pause dur="0.4"/> that is the variable <pause dur="0.4"/> that <pause dur="0.2"/> expresses the <pause dur="0.2"/> quantity of fibre inside is the volume fraction <pause dur="0.3"/> rather than the mass fraction <pause dur="1.5"/> so <pause dur="0.7"/> we define A-F over A-C as V-F <pause dur="0.5"/><kinesic desc="writes on board" iterated="y" dur="5"/> and similarly we define A-M <pause dur="0.6"/> over A-C <pause dur="0.5"/> as <pause dur="0.2"/> V-M <pause dur="1.9"/> and obviously because those are <pause dur="0.9"/> volume fractions <pause dur="0.9"/> and not absolute volumes then we've got to have <kinesic desc="writes on board" iterated="y" dur="4"/> V-F <pause dur="0.7"/> plus V-M <pause dur="1.7"/> must be <pause dur="0.2"/> equals to one <pause dur="3.4"/> well the final step <pause dur="0.2"/> in the calculation <pause dur="0.9"/> is to <pause dur="0.8"/> take this expression here <pause dur="0.4"/> remember what we want <pause dur="0.5"/> is to arrive <pause dur="0.9"/> at the expression here <pause dur="0.5"/> on the <pause dur="1.5"/> of for the Young's modulus of the material <pause dur="0.6"/> so what we've got here now <pause dur="4.1"/> we've got sigma-C <pause dur="1.2"/><kinesic desc="writes on board" iterated="y" dur="8"/>

equals this time we can write this one as sigma-F <pause dur="0.2"/> times <pause dur="0.5"/> V-F <pause dur="0.5"/> plus sigma-M <pause dur="0.4"/> times <pause dur="0.2"/> V-M <pause dur="1.9"/> you notice that this is <pause dur="1.4"/> the <trunc>s</trunc> <pause dur="0.2"/> exact same <pause dur="0.6"/> expression as this one <pause dur="0.6"/> where instead of having the moduli <pause dur="0.4"/> we've got the stresses <pause dur="0.5"/> okay <pause dur="0.4"/> in other words <pause dur="0.2"/> the stress <pause dur="0.5"/> inside the composite <pause dur="0.7"/> is shared <pause dur="0.3"/> in proportion <pause dur="1.3"/> between <pause dur="0.3"/> the fibres and the matrix in proportion <pause dur="0.5"/> to the <pause dur="0.2"/> volume fraction <pause dur="1.0"/> and finally <pause dur="0.6"/> because we want to define the modulus <pause dur="0.3"/> what we do is to divide <pause dur="0.2"/> by the strain <pause dur="0.9"/> the modulus <pause dur="0.3"/><kinesic desc="writes on board" iterated="y" dur="4"/> of the composite will be the stress in the composite <pause dur="0.5"/> divided by the strain <pause dur="0.5"/> in the composite <pause dur="1.5"/> from <pause dur="0.3"/> the definition of Young's modulus <pause dur="0.2"/><vocal desc="clears throat" iterated="n"/><pause dur="1.2"/> and if we divide by the strain <kinesic desc="writes on board" iterated="y" dur="15"/> throughout here <pause dur="0.7"/> then we get sigma-F <pause dur="0.9"/> over epsilon-C times V-F <pause dur="1.1"/>

plus sigma-M <pause dur="1.0"/> over epsilon-C <pause dur="0.2"/> times <pause dur="1.0"/> V-M <pause dur="2.3"/> okay <pause dur="2.8"/> i'm just using <pause dur="0.2"/> pretty <pause dur="0.3"/> standard definitions and nothing else <pause dur="0.5"/> and now <pause dur="1.9"/> now comes a point <pause dur="1.1"/> when i'm going to make use of that assumption here <pause dur="1.2"/> because if <pause dur="0.5"/> as we have assumed <pause dur="0.6"/> the strain in the composite <pause dur="0.2"/> is the same as the strain in the matrix <pause dur="0.6"/> is the same as the strain in the fibres <pause dur="0.6"/> then <pause dur="0.2"/> here epsilon-C <pause dur="0.2"/> epsilon-C epsilon-C <pause dur="0.5"/> can be replaced <kinesic desc="writes on board" iterated="y" dur="10"/> by sigma-F <pause dur="0.6"/> over epsilon-F <pause dur="0.3"/> times V-F <pause dur="0.8"/> plus here <pause dur="0.4"/> sigma-M <pause dur="0.2"/> over epsilon-M <pause dur="0.9"/> times <pause dur="0.2"/> V-M <pause dur="1.9"/> okay <pause dur="0.8"/> it's the same number <pause dur="0.5"/> with a different <pause dur="0.3"/> name but it's the same number <pause dur="1.0"/> but that allows me then to introduce <pause dur="1.3"/> the <pause dur="0.2"/> elastic modulus of the fibres because stress of the fibre over strain in the fibre is modulus of the fibre <pause dur="0.6"/> so <kinesic desc="writes on board" iterated="y" dur="4"/> this is <pause dur="0.2"/> E-F <pause dur="1.4"/> times V-F <pause dur="0.3"/> and here <pause dur="0.6"/> saying <pause dur="0.2"/> the ratio of stress of the matrix to the strain of the matrix is the Young's modulus of

the matrix <pause dur="0.8"/><kinesic desc="writes on board" iterated="y" dur="10"/> and this is therefore going to be E-M <pause dur="0.3"/> times <pause dur="0.3"/> V-M <pause dur="1.8"/> and this is E-<pause dur="0.5"/>C <pause dur="0.5"/> or <pause dur="0.7"/> E-<pause dur="0.4"/>L <pause dur="3.4"/> okay <pause dur="1.8"/> and that's that <pause dur="1.8"/> so it's very simple <pause dur="1.0"/> in this particular example <pause dur="0.2"/> to derive <pause dur="0.7"/> the equation which <pause dur="0.3"/> makes a prediction <pause dur="0.7"/> about the Young's modulus of the composite in the fibre direction <pause dur="0.4"/> when the fibres are all parallel <pause dur="0.2"/> and continuous <pause dur="1.7"/> and <pause dur="0.4"/> as i said <pause dur="0.2"/> all it says is that the Young's modulus <pause dur="0.5"/> is proportional <pause dur="0.7"/> to the quantity of fibres that you add in the system <pause dur="0.3"/> which is <pause dur="0.4"/> not unexpected <pause dur="3.1"/> like all predictions of any kind <pause dur="0.7"/> one needs to <pause dur="0.4"/> make some assessment as to <pause dur="0.7"/> how good they are <pause dur="0.3"/> in <pause dur="0.4"/> representing reality <pause dur="1.0"/> and people have <pause dur="0.6"/> # <pause dur="0.2"/> used that equation to make predictions <pause dur="0.5"/> about <pause dur="0.2"/> elastic properties of composites <pause dur="0.4"/> and <pause dur="0.4"/> that prediction <pause dur="0.2"/> works very well <pause dur="1.0"/> so that one is good <pause dur="1.1"/> in the sense that the <pause dur="0.6"/> predictions you make are very very close <pause dur="0.7"/> to the <pause dur="0.3"/> # measured values <pause dur="2.0"/> okay <pause dur="1.8"/> clear <pause dur="2.3"/> okay well we can proceed the same way and i'm not going to do it <pause dur="0.7"/> but i'll leave for you

to <pause dur="1.0"/> play around with the <pause dur="0.9"/> # equations which need to <pause dur="0.7"/> be set up <pause dur="0.9"/> for example <pause dur="0.7"/> the other modulus that we need <pause dur="0.9"/> you've got all that in your notes # <pause dur="0.2"/> anyway <pause dur="0.7"/><kinesic desc="writes on board" iterated="y" dur="21"/> the other modulus <trunc>w</trunc> that we need <pause dur="0.2"/> is the one which is <pause dur="0.5"/> normal <pause dur="0.9"/> to the direction of the fibres which we called <pause dur="0.2"/> E-<pause dur="0.6"/>two <pause dur="0.4"/> or sometimes E-T <pause dur="0.5"/> for transverse or sometimes E-two-two <pause dur="0.8"/> okay again this is the fibre <pause dur="1.2"/> and you load the system this way <pause dur="2.5"/> now if you had to start <pause dur="2.3"/> the calculation which <pause dur="0.4"/> eventually leads you to that <pause dur="0.8"/> final expression <pause dur="0.7"/> what assumption are you going to make here <pause dur="0.2"/> to be able to <pause dur="0.3"/> get going <pause dur="2.2"/> without looking at your notes <pause dur="4.4"/> the previous case we assumed that the two springs were deforming in parallel <pause dur="0.4"/> and that the deformation in both of them was the same <pause dur="0.8"/> here we are pulling in this direction now <pause dur="0.7"/> okay <pause dur="0.9"/> can we make the same assumption <pause dur="0.5"/> that the load is the same <pause dur="0.4"/> in the two <trunc>sys</trunc> that the <pause dur="0.8"/> the strain is the same in the two systems </u><u who="sm0825" trans="overlap"> no </u><pause dur="0.6"/> <u who="nm0808" trans="pause"> we can't <pause dur="0.6"/> it's obvious this this is much more

compliant than the fibre <pause dur="0.6"/> okay so when we stretch this thing <pause dur="0.2"/> in that direction <pause dur="0.6"/> the matrix will deform more than the fibre will deform <pause dur="1.2"/> so the assumption that we used <pause dur="0.4"/> for the previous equation <pause dur="0.3"/> is not holding here <pause dur="0.5"/> but what <pause dur="0.9"/> is the assumption which holds here </u><pause dur="0.7"/> <u who="sm0812" trans="pause"> series </u><pause dur="1.0"/> <u who="nm0808" trans="pause"> yeah we've got two springs in series that's correct <pause dur="0.5"/> so this time we've got a soft spring <pause dur="1.7"/><kinesic desc="writes on board" iterated="y" dur="9"/> which is the matrix <pause dur="1.1"/> and then we've got a stiff spring <pause dur="0.8"/> which is the fibre and this time we're pulling on this like that <pause dur="1.5"/> so <vocal desc="clears throat" iterated="n"/><pause dur="0.5"/> if they are in series if we've got two springs in series <pause dur="0.7"/> okay and we pull <pause dur="0.4"/> okay what assumption can we make </u><pause dur="2.0"/> <u who="sm0813" trans="pause"> the same stress </u><pause dur="0.6"/> <u who="nm0808" trans="pause"> the same stress <pause dur="0.4"/> that's right <pause dur="1.1"/> okay in the previous case <pause dur="0.4"/> it was the same strain <pause dur="0.5"/> which was imposed on the system <pause dur="0.5"/> here we've got the same stress <pause dur="0.2"/> in the sense that if we pull that <pause dur="0.5"/> if we take <pause dur="0.5"/> the force <pause dur="0.4"/> the cross-sectional area of matrix and the cross-sectional area of fibre are the same <pause dur="0.9"/>

okay <pause dur="0.6"/> so <pause dur="0.5"/> this <pause dur="0.6"/><kinesic desc="writes on board" iterated="y" dur="7"/> thing here <pause dur="0.4"/> the assumption the starting assumption here <pause dur="0.4"/> is this time that sigma-F <pause dur="0.5"/> equals sigma-M <pause dur="0.4"/> equals sigma-C <pause dur="1.0"/> and then if you do exactly what i did for the previous case <pause dur="0.5"/> by introducing the deformations <pause dur="0.6"/> now <pause dur="0.3"/> in the matrix <pause dur="0.2"/> and <pause dur="0.2"/> in the <pause dur="2.0"/> in the fibres <pause dur="0.3"/> the which means it's actually in that system <pause dur="0.6"/> the total deformation <pause dur="0.3"/> is the sum of the deformation of one spring plus the deformation of the other spring <pause dur="0.6"/> okay because that's what springs in series are <pause dur="0.7"/> so the starting point is effectively is that <kinesic desc="writes on board" iterated="y" dur="2"/> the delta-<pause dur="0.2"/>L of the composite <pause dur="0.6"/> the increase in length <pause dur="0.3"/> in the direction two <pause dur="0.7"/> of the composite <pause dur="0.6"/> is going to be equal to the <kinesic desc="writes on board" iterated="y" dur="6"/> delta-<pause dur="0.5"/>L <pause dur="0.9"/> of the fibre <pause dur="1.1"/> plus the delta-L <pause dur="0.4"/> of the matrix <pause dur="2.1"/> and then by defining strain <pause dur="0.2"/> as a function of the increase in length by introducing the Young's modulus <pause dur="0.4"/> you get <pause dur="0.2"/> to <pause dur="0.5"/> eventually <pause dur="0.5"/> to that equation here <pause dur="1.1"/> and as i said you can do that as an exercise <pause dur="0.6"/> it's <pause dur="0.2"/> good for you <pause dur="2.1"/> and this equation here <pause dur="0.4"/>

okay the assumption as we said equals stress <pause dur="0.5"/> and one important consideration which <pause dur="0.7"/> matters <pause dur="0.3"/> when you want to refine <pause dur="0.7"/> the thing <pause dur="0.2"/> and why <trunc>i</trunc> which explains also why this thing <pause dur="0.2"/> doesn't really work that well <pause dur="0.2"/> like the other one <pause dur="0.5"/> is that <pause dur="0.6"/> in this calculation one neglects the fact that a Poisson's ratios <pause dur="0.5"/> all the fibres in the matrix are <pause dur="0.7"/> different <pause dur="0.6"/> when you are pulling in that direction <pause dur="0.5"/> if the Poisson's ratio are different as you might expect the lateral contraction <pause dur="0.6"/> of the fibre and the lateral contraction of the matrix will be different <pause dur="0.4"/> there's going to be a mismatch at the interface <pause dur="0.6"/> okay so that creates problems <pause dur="0.9"/> but anyway this is the simple <pause dur="0.4"/> or the simplest <pause dur="0.6"/> approximation one can get <pause dur="0.6"/> and again the question is <pause dur="0.2"/> how does it work <pause dur="0.5"/> well unfortunately this one doesn't work <pause dur="0.2"/> well at all <pause dur="0.8"/> okay in other words <pause dur="0.2"/> the predictions <pause dur="0.9"/> of the transverse modulus that this equation <pause dur="0.4"/> makes <pause dur="0.3"/> are not very good <pause dur="1.7"/> and that means that <pause dur="0.4"/> they can be used only <pause dur="0.7"/> as a very <pause dur="0.2"/> first approximation but if you really need <pause dur="0.2"/>

information <pause dur="0.3"/> or more detailed information <pause dur="0.6"/> then you will have to do and get some measurements <pause dur="0.4"/> or <pause dur="0.4"/> use some of the more sophisticated <pause dur="0.4"/> # models <pause dur="0.5"/> which <pause dur="0.2"/> introduce these effects here <pause dur="2.4"/> if we plot <pause dur="1.0"/> these two equations <pause dur="0.3"/> as a function of the variable involved <pause dur="0.8"/> which is <pause dur="0.3"/> essentially the <pause dur="0.2"/> volume fraction <pause dur="1.7"/> of <pause dur="0.7"/> # <pause dur="1.2"/> fibres okay <pause dur="0.7"/> we can start here <pause dur="0.5"/><kinesic desc="writes on board" iterated="y" dur="2"/> this is the modulus <pause dur="0.9"/> of the matrix by itself <pause dur="1.4"/> and up here <pause dur="1.3"/><kinesic desc="writes on board" iterated="y" dur="5"/> we've got <pause dur="0.2"/> the modulus of the fibres <pause dur="1.0"/> by themselves <pause dur="0.8"/> this is not quite to scale because remember that <pause dur="0.2"/> the order of magnitude difference is much bigger than i <pause dur="0.2"/> can draw on the board <pause dur="0.5"/> okay E-M is of the order of one G-P-A <pause dur="0.7"/> E-F is of the order of seventy <pause dur="0.2"/> G-P-A <pause dur="1.8"/> for glass <pause dur="0.5"/> okay <pause dur="0.4"/> but nevertheless <pause dur="0.4"/> the first equation <pause dur="0.2"/> the one we saw <pause dur="1.8"/> it's a linear relationship <pause dur="0.7"/> between <pause dur="0.5"/> the <pause dur="0.7"/> # <pause dur="0.6"/> fibre modulus <pause dur="0.5"/> volume fraction and composite modulus <pause dur="1.0"/> so as we increase the volume <kinesic desc="writes on board" iterated="y" dur="5"/> fraction <pause dur="0.2"/> of fibres <pause dur="0.5"/> this thing is going to go up <pause dur="1.5"/> like this <pause dur="0.6"/> so the modulus increases linearly <pause dur="1.7"/> with <pause dur="0.4"/>

the amount of fibre <pause dur="0.2"/> that you put in <pause dur="1.0"/> okay <pause dur="2.2"/> and if you plot <pause dur="0.3"/> on the same graph <pause dur="0.2"/> the other one <pause dur="1.4"/> the one which gives the modulus <pause dur="1.0"/> parallel <pause dur="0.3"/> to the fibre direction <pause dur="1.9"/> this one here <pause dur="2.2"/> okay you find something different <pause dur="1.3"/> but obviously <pause dur="0.3"/> it's got to start at the same point <pause dur="0.3"/> down here <pause dur="2.0"/><kinesic desc="writes on board" iterated="y" dur="2"/> and it's got to end up at the <pause dur="0.3"/> same point <pause dur="0.3"/> up here <pause dur="0.5"/> because if we got only fibres we have to measure the modulus of the fibres <pause dur="0.5"/> and the way it goes from that <pause dur="0.2"/> point to that point <pause dur="0.6"/> is <pause dur="0.2"/> somewhat like this <pause dur="5.2"/><kinesic desc="writes on board" iterated="y" dur="3"/>

okay <pause dur="3.1"/> so this is <pause dur="2.0"/><kinesic desc="writes on board" iterated="y" dur="3"/> E-<pause dur="1.2"/>one <pause dur="0.3"/> or E-L <pause dur="1.8"/> and this one <pause dur="0.2"/> is <pause dur="0.4"/> E-two <pause dur="0.8"/><kinesic desc="writes on board" iterated="y" dur="2"/> or <pause dur="0.6"/> E-<pause dur="0.6"/>sub-T <pause dur="2.6"/> so <pause dur="0.3"/> the modulus increases far more slowly <pause dur="0.8"/> with <pause dur="0.2"/> the volume fraction of fibres in the second case <pause dur="0.5"/> okay and then obviously it's got to start shooting up <pause dur="0.8"/> to catch up <pause dur="0.3"/> with the value of the modulus of the fibre <pause dur="0.7"/> when we get up there <pause dur="1.9"/> you realize of course that <pause dur="1.7"/> it's impossible to get <pause dur="0.3"/> volume fractions <pause dur="0.7"/> of what <vocal desc="laugh" iterated="n"/><pause dur="0.6"/> okay volume fraction of one means we've <trunc>ju</trunc> just got fibres and nothing else <pause dur="0.4"/> volume fraction of <pause dur="1.2"/> # <pause dur="1.0"/> one doesn't mean a composite means just pure fibres <pause dur="0.9"/> and the question is realistically <pause dur="0.9"/> how far can we go <pause dur="0.2"/> along these two curves <pause dur="0.7"/> okay <pause dur="0.7"/> well in practice okay <pause dur="0.2"/> you see the problem <pause dur="3.2"/><kinesic desc="writes on board" iterated="y" dur="3"/> okay this is the cross-section of your composite <pause dur="0.7"/> and in an ideal world <pause dur="0.2"/> okay you can pack <pause dur="0.3"/> the fibres <pause dur="0.6"/><kinesic desc="writes on board" iterated="y" dur="4"/> either <pause dur="1.6"/> in a nice <pause dur="0.7"/> cubic <pause dur="1.0"/> array <pause dur="0.3"/> like this <pause dur="1.3"/> or you can pack the fibres <pause dur="1.1"/> in <pause dur="1.4"/><kinesic desc="writes on board" iterated="y" dur="7"/> a hexagonal array like this <pause dur="3.9"/>

okay <pause dur="0.5"/> it's pretty obvious but so just looking at the pictures <pause dur="0.5"/> that if we can pack them in a hexagonal array <pause dur="0.2"/> we are going to pack more fibres <pause dur="0.5"/> in the same volume because the fibres are nesting <pause dur="1.6"/> with each other <pause dur="0.2"/> much better <pause dur="0.5"/> okay <pause dur="0.5"/> and <pause dur="0.3"/> if you work out and you've got again the calculation <pause dur="0.7"/> in your notes in detail <pause dur="0.5"/> if you <pause dur="0.3"/> simple geometry <pause dur="0.5"/> okay if <kinesic desc="writes on board" iterated="y" dur="3"/> you get a hexagonal array the packing the maximum packing <pause dur="0.8"/> is of the order of ninety per cent <pause dur="0.3"/> or thereabouts <pause dur="2.5"/> and if you pack <pause dur="1.0"/> in a cubic array <pause dur="0.4"/> you can get something of the order of <pause dur="0.3"/> seventy per cent <pause dur="0.4"/> or thereabouts <pause dur="1.4"/> but remember that those are ideal situations which in practice is impossible <pause dur="0.7"/> remember that the fibres are six to ten microns in diameter <pause dur="0.6"/> okay <pause dur="0.2"/> and short of puttting them one by one <pause dur="0.6"/> okay in in in a very nice neat <pause dur="0.4"/> kind of arrangement it's virtually impossible by any process in technology <pause dur="0.8"/> to arrive at volume fractions <pause dur="0.3"/> anywhere near these values <pause dur="0.6"/> so where are the limits <pause dur="0.6"/> well the limit in practice <pause dur="0.2"/> it's at about <pause dur="4.7"/><kinesic desc="writes on board" iterated="y" dur="4"/>

sixty per cent V-F <pause dur="1.1"/> these are for the high <pause dur="0.8"/> # <pause dur="0.5"/> performance <pause dur="0.3"/> composites that we're going to see <pause dur="0.7"/> next term <pause dur="0.2"/> in the composites course <pause dur="0.7"/> okay <pause dur="0.5"/> and <pause dur="0.7"/> for most reinforced plastics on the other hand <pause dur="0.4"/> those that we can process by <pause dur="0.5"/> injection moulding <pause dur="0.2"/> extrusion <pause dur="0.4"/> vacuum forming <pause dur="0.6"/> and the other methods <pause dur="0.4"/> we are already far below that <pause dur="0.8"/> okay so these are the high performance fibres <pause dur="2.2"/><kinesic desc="writes on board" iterated="y" dur="8"/> sorry high <pause dur="0.2"/> performance composites <pause dur="2.9"/> which <pause dur="0.2"/> as i said <pause dur="0.2"/> is the subject of next term lecture <pause dur="1.2"/> whereas <pause dur="0.3"/> here we are more or less <pause dur="2.0"/> in the band which goes from about say <pause dur="0.7"/> ten per cent <kinesic desc="writes on board" iterated="y" dur="3"/> five per cent say <pause dur="0.5"/> to about twenty-five per cent <pause dur="1.9"/> so the useful region <pause dur="0.7"/> of reinforcement <pause dur="0.6"/> for <pause dur="0.2"/> the # reinforced plastics <pause dur="0.5"/> okay is <pause dur="0.4"/> about five to <trunc>twenty</trunc> <event desc="drops pencil case" n="su826" iterated="n"/> <pause dur="0.7"/> twenty-five per cent <pause dur="0.5"/> it's okay <pause dur="0.8"/> pick it up if you like or leave it there <pause dur="1.3"/> okay <pause dur="2.4"/> now unfortunately as i said last week <pause dur="1.0"/> having the modulus <pause dur="1.0"/> in the <pause dur="2.7"/><kinesic desc="writes on board" iterated="y" dur="8"/>

fibre direction <pause dur="1.8"/> one <pause dur="1.4"/> and having the modulus in the transverse direction <pause dur="0.2"/> two <pause dur="0.8"/> is not enough <pause dur="1.0"/> to get all the information we need <pause dur="0.6"/> to do any calculation <pause dur="0.6"/> relating to design stress analyses or whatever <pause dur="0.2"/> of <pause dur="0.8"/> # <pause dur="0.5"/> reinforced <pause dur="0.3"/> plastic <pause dur="0.8"/> materials <pause dur="1.2"/> we need <pause dur="0.2"/> on top and above these two <pause dur="0.5"/> the sheer modulus <pause dur="0.4"/> which <pause dur="0.4"/> comes into play <pause dur="1.0"/><kinesic desc="writes on board" iterated="y" dur="2"/> if we try to deform <pause dur="0.4"/> the system <pause dur="0.4"/> that way sliding <pause dur="0.2"/> essentially <pause dur="0.7"/> the matrix with respect to the fibre <pause dur="0.6"/> and then we need the Poisson's ratio <pause dur="1.1"/> # <pause dur="0.9"/> I-E the contraction which occurs in one direction when we pull the material in the other direction <pause dur="1.7"/> and <pause dur="1.4"/> so we need <pause dur="0.5"/> G-one-two <pause dur="1.1"/><kinesic desc="writes on board" iterated="y" dur="2"/> and we need a new one-two <pause dur="0.8"/> i remember the definition <pause dur="0.4"/> one-two means <pause dur="0.6"/> pulling in the one direction and measuring the protraction in the <pause dur="0.3"/> two direction <pause dur="0.7"/> and we need a new two-one <pause dur="1.6"/><kinesic desc="writes on board" iterated="y" dur="2"/> which is this time pulling <pause dur="0.2"/> in the two direction and looking at the contraction <pause dur="0.2"/> occurring in the one direction <pause dur="0.8"/> and as we said last week <pause dur="0.2"/> these two are

not the same <pause dur="0.8"/> so in theory we've got five numbers one <pause dur="0.6"/><kinesic desc="writes on board" iterated="y" dur="3"/> two <pause dur="0.4"/> three <pause dur="0.5"/> four and five <pause dur="1.6"/> that we need <pause dur="0.2"/> to characterize the material <pause dur="0.5"/> fully <pause dur="0.5"/> in terms of its elastic behaviour <pause dur="1.7"/> okay well one can do # <pause dur="0.5"/> calculations <pause dur="0.8"/> based on the same <pause dur="0.4"/> principles <pause dur="0.9"/> as the one we've seen and again i'm not going to go into the details <pause dur="0.5"/> okay <pause dur="1.0"/><kinesic desc="changes transparency" iterated="y" dur="5"/> but this is how one would set up <pause dur="1.2"/> the calculation for the sheer modulus G-one-two <pause dur="0.5"/> or G-L-T <pause dur="0.3"/> as it's called sometimes <pause dur="3.7"/><kinesic desc="writes on board" iterated="y" dur="2"/> okay if you look at the distortion <pause dur="0.9"/> in each one of the components and again playing around with the basic definitions of sheer strain <pause dur="0.6"/> bringing in the modulus <pause dur="0.2"/> and the load sharing between the fibres and the matrix <pause dur="0.6"/> and this is again a system of <pause dur="0.6"/> springs in series essentially <pause dur="0.6"/> okay <pause dur="0.2"/> you get an equation <pause dur="0.2"/> which predicts the sheer modulus <pause dur="1.1"/> and <pause dur="0.6"/> the equation <pause dur="0.2"/> looks <pause dur="0.3"/> almost <pause dur="0.3"/> in fact it is the same identical equation <pause dur="0.5"/> than we have <pause dur="0.6"/> obtained <pause dur="0.7"/> for the <pause dur="0.3"/> transverse modulus E-two <pause dur="1.7"/> and <pause dur="0.5"/> in the same way as the <pause dur="0.3"/> other one did not

work very well <pause dur="0.3"/> this one also is not very accurate so we've got the same problem <pause dur="0.6"/> that we can predict <pause dur="0.2"/> this number here <pause dur="0.8"/> quite well <pause dur="1.2"/><kinesic desc="writes on board" iterated="y" dur="1"/> this one is not very good <pause dur="3.4"/><kinesic desc="writes on board" iterated="y" dur="4"/> this one is not very good either <pause dur="0.8"/> so again either measurement <pause dur="0.5"/> or <pause dur="0.3"/> more sophisticated modelling <pause dur="0.4"/> and theories <pause dur="1.9"/> and the last <pause dur="0.4"/> one <pause dur="0.8"/> is the Poisson's ratio that we need as well <pause dur="2.9"/> and nu-one-two is what <pause dur="0.5"/> on the <pause dur="1.4"/><kinesic desc="writes on board" iterated="y" dur="2"/> transparency's <pause dur="0.8"/> called nu-L-T <pause dur="1.4"/> this one on the other hand <pause dur="0.6"/> follows the same rule <pause dur="0.3"/> as the longitudinal modulus <pause dur="1.0"/> E-L <pause dur="0.3"/> or <pause dur="0.2"/> E-one <pause dur="1.0"/><kinesic desc="writes on board" iterated="y" dur="3"/> so this one <pause dur="0.2"/> is okay <pause dur="1.8"/> this one <trunc>i</trunc> <pause dur="0.2"/> also is a good prediction <pause dur="0.5"/> of the <pause dur="1.4"/> # measured values <pause dur="0.2"/> of the material <pause dur="2.1"/> and fortunately <pause dur="0.7"/> fortunately we don't have to worry <pause dur="0.9"/> about <pause dur="0.2"/> the last one <pause dur="2.6"/> nu-two-one <pause dur="0.7"/> is not <pause dur="0.2"/> independent <pause dur="0.7"/> of the other <pause dur="0.5"/> properties <pause dur="1.0"/> and there is this relationship <pause dur="0.7"/> which exists <pause dur="0.5"/> between the two Poisson's ratio <pause dur="0.9"/>

one-two two-one on the one hand <pause dur="0.7"/> and the corresponding moduli in the same directions <pause dur="0.4"/> E-one or E-L <pause dur="0.5"/> and E-two or E-T <pause dur="1.2"/> so <pause dur="0.3"/> if we know <pause dur="0.9"/> this <pause dur="0.4"/> and we know this and we know this <pause dur="0.4"/> then using that equation which is on the board here <pause dur="0.5"/> we can <pause dur="0.3"/> calculate this <pause dur="1.6"/> nevertheless at the end <pause dur="0.2"/> of all that <pause dur="0.9"/> the important message <pause dur="0.2"/><kinesic desc="writes on board" iterated="y" dur="8"/> is that we've got four elastic <pause dur="4.5"/> properties <pause dur="0.2"/> which we need <pause dur="0.5"/> to <pause dur="0.4"/> predict <pause dur="0.2"/> or measure <pause dur="1.2"/> or a combination of both <pause dur="0.9"/> and <pause dur="0.5"/> the important thing of course is the fact that <pause dur="3.1"/><kinesic desc="writes on board" iterated="y" dur="3"/> they are all <pause dur="0.8"/> independent of each other <pause dur="1.1"/> okay <pause dur="0.5"/> the fifth one is the only one which can be recovered <pause dur="0.6"/> from the <pause dur="0.3"/> other four <pause dur="1.9"/> so we got <pause dur="0.2"/> a more complex system <pause dur="0.2"/> than isotropic materials <pause dur="0.2"/> where we need only two <pause dur="0.2"/> numbers <pause dur="0.2"/> E-<pause dur="0.3"/>G or E-nu <pause dur="0.2"/> or G-nu <pause dur="0.8"/> and that means that even the simplest type of calculation gets more cumbersome <pause dur="0.5"/> when you're dealing with <pause dur="0.2"/> an isotropic fibre enforced materials <pause dur="0.4"/> well that's life we can't help it <pause dur="0.6"/> # the

only justification for that <pause dur="0.3"/> increase in complexity <pause dur="0.7"/> is the fact that <pause dur="1.5"/> we are going to get some benefits <pause dur="0.5"/> in terms of <pause dur="0.5"/> higher moduli <pause dur="0.2"/> higher strengths <pause dur="0.8"/> # fatigue life <pause dur="0.3"/> # <pause dur="0.6"/> reduction in <pause dur="0.2"/> creep <pause dur="0.2"/> or stress relaxation behaviour <pause dur="0.4"/> all these kind of things <pause dur="2.7"/> questions <pause dur="1.9"/> all clear <pause dur="1.3"/> please do have a go <pause dur="0.2"/> at <pause dur="0.2"/> trying <pause dur="0.2"/> to <pause dur="0.4"/> do the simple <pause dur="0.8"/> mechanics <pause dur="0.2"/> of <pause dur="0.5"/> deriving <pause dur="0.7"/> these equations here because it's a good exercise <pause dur="0.5"/> and it forces you to think <pause dur="0.9"/> # <pause dur="0.6"/> in the way that i would like you to think because we'll need that <pause dur="0.9"/> as we move on <pause dur="0.3"/> to the <pause dur="0.3"/> more <pause dur="0.2"/> complex things <pause dur="1.5"/> any more questions about that <pause dur="0.3"/> all happy with that <pause dur="1.8"/> okay well just to give you some <pause dur="0.3"/> numbers okay if you remember <pause dur="1.3"/> what <pause dur="0.2"/> i said here that the volumes of the matrix down here <pause dur="0.4"/> is typically of the order <pause dur="0.3"/> say of one G-P-A <pause dur="1.5"/><kinesic desc="writes on board" iterated="y" dur="2"/> the modulus of the fibre up here is typically of the order <pause dur="0.2"/> of <pause dur="0.2"/><kinesic desc="writes on board" iterated="y" dur="3"/> seventy <pause dur="1.4"/> G-P-A <pause dur="1.6"/> and you can see that <pause dur="0.6"/> even if you have a volume fraction of fibres of say sixty per cent <pause dur="0.7"/> okay you still got <pause dur="0.4"/> a modulus which is <pause dur="0.2"/>

significantly higher of course <pause dur="0.6"/> than the one of the <pause dur="0.7"/> pure <pause dur="0.4"/> polymer matrix <pause dur="1.5"/> so if we imagine a volume fraction <pause dur="0.7"/><kinesic desc="writes on board" iterated="y" dur="13"/> of sixty per cent <pause dur="0.5"/> well <pause dur="0.4"/> our prediction for E-one <pause dur="0.3"/> or E-L <pause dur="0.8"/> is simply <pause dur="0.3"/> # <pause dur="0.2"/> point-six times <pause dur="0.7"/> seventy <pause dur="0.6"/> plus <pause dur="0.5"/> point-four <pause dur="0.4"/> times one <pause dur="1.3"/> okay <pause dur="0.9"/> so this is <pause dur="1.3"/><kinesic desc="writes on board" iterated="y" dur="8"/> forty-two <pause dur="0.9"/> here <pause dur="0.8"/> plus we can forget about that point-four <pause dur="1.1"/> okay <pause dur="0.7"/> so <pause dur="1.7"/> at best <pause dur="0.9"/> okay with all the fibres parallel <pause dur="0.5"/> and the maximum amount of fibre that we can <pause dur="0.3"/> practically introduce in a polymer <pause dur="0.8"/> okay <pause dur="0.6"/> with all the fibre in one direction <pause dur="0.4"/> the modulus the maximum modulus we're going to get <pause dur="0.4"/> is about forty-two <kinesic desc="writes on board" iterated="y" dur="1"/> G-P-A <pause dur="1.2"/> well that's a <pause dur="0.4"/> obviously a great deal more <pause dur="0.5"/> than the one G-P-A we started from <pause dur="0.7"/> # with the pure <pause dur="0.4"/> polymer <pause dur="0.6"/> but <pause dur="0.5"/> again compare that to steel <pause dur="1.5"/><kinesic desc="writes on board" iterated="y" dur="9"/> which is at two-hundred-and-ten <pause dur="1.3"/> G-P-A or aluminium <pause dur="1.4"/> which is at seventy <pause dur="0.3"/> G-P-A <pause dur="1.1"/> so we begin to catch up <pause dur="1.2"/> in modulus terms we begin to catch up a little bit with aluminium <pause dur="0.6"/> but we're still a

long way <pause dur="0.7"/> from <pause dur="1.1"/> catching up with the modulus of steel <pause dur="1.8"/> there are other fibres which are used <pause dur="0.2"/> when you really need a very high modulus <pause dur="0.8"/> that <pause dur="0.8"/> you may want to <pause dur="0.5"/> use to <trunc>com</trunc> <pause dur="0.2"/> # <pause dur="0.3"/> to <pause dur="0.2"/> # compete <pause dur="0.8"/> 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 <pause dur="0.4"/> so we leave them at that for the moment <pause dur="1.6"/> more relevant to this course is what happens down here because this is the typical range <pause dur="0.6"/> of volume fractions <pause dur="0.2"/> that we're going to use <pause dur="0.2"/> for injection moulded <pause dur="0.6"/> # <pause dur="1.7"/> polymers <pause dur="0.5"/> okay so if we take a volume fraction say of <pause dur="0.7"/> twenty per cent <pause dur="1.5"/><kinesic desc="writes on board" iterated="y" dur="1"/> okay <pause dur="0.2"/> well again our prediction for E-one will be of the order of <pause dur="2.0"/><kinesic desc="writes on board" iterated="y" dur="11"/> point-two <pause dur="0.2"/> times seventy <pause dur="1.0"/> plus <pause dur="0.9"/> point-eight times one <pause dur="0.3"/> okay <pause dur="0.5"/> and <pause dur="0.2"/> you can see that we're going to get something of the order of fourteen <pause dur="1.2"/> G-P-A <pause dur="5.1"/><kinesic desc="writes on board" iterated="y" dur="3"/> which is <pause dur="0.2"/> respectable <pause dur="0.6"/> remember

that you started from one <pause dur="0.5"/> okay which is a very low value indeed <pause dur="0.2"/> so getting from one to something like ten fifteen <pause dur="0.5"/> okay it's <pause dur="0.4"/> not <pause dur="0.4"/> a bad <pause dur="0.2"/> shot at all <pause dur="1.7"/> the problem is that <pause dur="0.3"/> this implies <pause dur="0.2"/> as we said that all the fibres are parallel <pause dur="1.2"/> okay and highly <pause dur="0.3"/> # <pause dur="0.5"/> oriented in one direction <pause dur="0.4"/> in practice <pause dur="0.5"/> okay it's not easy to achieve that parallel orientation of the fibres <pause dur="0.4"/> when you are using processes like injection moulding <pause dur="0.4"/> because the flow <pause dur="0.3"/> of the material <pause dur="0.3"/> which is carrying the fibres with it <pause dur="0.4"/> okay is not necessarily going to give you a nice <pause dur="0.6"/> # <pause dur="0.6"/> oriented system <pause dur="0.5"/> so you're going to lose <pause dur="0.3"/> some of this inevitably <pause dur="0.9"/> okay this is the maximum you can get if you like at that volume fraction <pause dur="0.7"/> but <pause dur="0.2"/> you're going to lose some of that because the fibres are not going to be <pause dur="0.3"/> packing <pause dur="0.4"/> exactly parallel to each other <pause dur="1.0"/> and you're going to lose also a little bit more <pause dur="0.4"/> because <pause dur="0.3"/> the fibres are not going to be continuous <pause dur="0.2"/> and this is what we're going to see <pause dur="0.5"/> later on <pause dur="0.7"/> # <pause dur="0.2"/> in the course <pause dur="0.7"/> so we've got to <pause dur="0.3"/> sacrifice

something <pause dur="0.2"/> if you like <pause dur="0.7"/> when we go to short fibres <pause dur="0.6"/> we lose some of their potential <pause dur="0.6"/> and <pause dur="0.3"/> when we go to <pause dur="0.2"/> non-parallel fibres <pause dur="0.3"/> we lose <pause dur="0.2"/> something else <pause dur="1.3"/> but that <pause dur="0.3"/> is <pause dur="0.5"/> for <pause dur="0.2"/> tomorrow <pause dur="0.4"/> can i remind you that we've got <pause dur="0.4"/> the tutorial tomorrow <pause dur="0.6"/> okay <pause dur="0.3"/> whatever is the tutorial time slot i think it's eleven to twelve <pause dur="0.5"/> in G-twenty-six <pause dur="0.4"/> okay <pause dur="1.7"/> okay so if you have no more questions <pause dur="0.2"/> okay <pause dur="0.9"/> we move on to the next thing <pause dur="0.8"/> we all happy <pause dur="1.3"/> okay <pause dur="0.5"/> well the next thing is okay well what happens to the strength <pause dur="0.2"/> we talked about the modulus <pause dur="0.7"/> okay this is one of the properties that we need <pause dur="0.7"/> the other one <pause dur="0.5"/> property that we need <pause dur="0.2"/> has to do with the strength of the material <pause dur="1.3"/> and here get <pause dur="0.4"/> thing's get <pause dur="0.5"/> a great deal more complicated i'm afraid <pause dur="0.6"/> but if you start thinking in terms of <pause dur="0.8"/> strains and <pause dur="0.7"/> # <pause dur="0.8"/> either equal strains or equal stresses or the like <pause dur="0.2"/> things get into <pause dur="0.2"/> shape <pause dur="0.9"/> quite happily <pause dur="2.1"/> so let's talk about strength <pause dur="1.6"/><kinesic desc="writes on board" iterated="y" dur="3"/> now the first comment to make <pause dur="0.2"/> is that the only strength that i'm

going to be <pause dur="0.3"/> interested in <pause dur="0.9"/> is the one <pause dur="0.5"/> in the direction of the fibre <pause dur="0.7"/> because in the direction normal to the fibre <pause dur="1.9"/> okay this is my <pause dur="1.2"/> little <kinesic desc="writes on board" iterated="y" dur="3"/> sample again <pause dur="2.5"/> it's pretty obvious that if i pull this in that direction <pause dur="0.7"/> the only strength i'm <unclear>going to</unclear> get <pause dur="0.2"/> is the strength of the matrix i started from <pause dur="0.7"/> okay <pause dur="1.3"/> the fibres are not carrying a great deal of stress in that direction we can verify that <pause dur="0.7"/> from the models we've just talked about <pause dur="0.5"/> so <pause dur="0.2"/> the strength in that direction is essentially <pause dur="1.5"/> equal to <pause dur="3.2"/><kinesic desc="writes on board" iterated="y" dur="11"/> is roughly the same as the strength of the matrix <pause dur="5.4"/> so i'm not going to <pause dur="0.4"/> worry too much <pause dur="0.6"/> it's low it's low <pause dur="0.4"/> it's whatever it is there's <pause dur="0.2"/> precious little i can do about it <pause dur="2.6"/>

whatever <pause dur="0.9"/> however <pause dur="0.9"/><kinesic desc="writes on board" iterated="y" dur="2"/> in this direction here <pause dur="0.9"/> which is the direction in which i'm going to take maximum <pause dur="0.4"/> advantage <pause dur="0.6"/> and benefit <pause dur="0.2"/> out of the fibres <pause dur="0.5"/> then i want to <pause dur="0.2"/> know <pause dur="0.6"/> what kind of predictions i can make about the strength of the material in that direction <pause dur="0.7"/> so when we talk about strength is strength <pause dur="0.5"/> parallel <pause dur="3.1"/><kinesic desc="writes on board" iterated="y" dur="10"/> to the fibre direction <pause dur="6.0"/> okay <pause dur="1.6"/> and remember <pause dur="0.9"/> that <pause dur="0.2"/> when <pause dur="0.5"/> we <pause dur="0.9"/> are loading the system <pause dur="1.1"/> in the direction of the fibres <pause dur="2.0"/> the way in which we derived <pause dur="0.3"/> the elastic modulus <pause dur="0.5"/> was to assume <pause dur="0.5"/> that the composite strain <pause dur="0.5"/> was the same as the <kinesic desc="writes on board" iterated="y" dur="3"/> fibre strain <pause dur="0.4"/> was the same as the matrix strain <pause dur="2.4"/> okay that was the <pause dur="0.6"/> initial assumption <pause dur="1.3"/> okay <pause dur="1.4"/> and that initial assumption <pause dur="0.4"/> led <pause dur="0.3"/> then to the expression for the Young's modulus <pause dur="0.6"/> in the <pause dur="0.4"/>

longitudinal direction <pause dur="1.2"/> well if we're talking about strength <pause dur="0.9"/> we're talking about failure <pause dur="0.6"/> we're talking about something <pause dur="0.4"/> which as we carry on pulling is going to start <pause dur="0.9"/> breaking down <pause dur="0.8"/> giving up <pause dur="1.8"/> if we <pause dur="0.2"/> plot <pause dur="2.8"/><kinesic desc="writes on board" iterated="y" dur="4"/> the stress <pause dur="0.3"/> versus strain <pause dur="0.9"/> curve <pause dur="0.3"/> of the composite <pause dur="0.2"/> we just do a <pause dur="0.4"/> pure <pause dur="0.2"/> tensile test <pause dur="0.6"/> in a strip like this one <pause dur="0.7"/> okay what we're going to get is <pause dur="0.3"/> the material is going to deform <pause dur="0.4"/> elastically <pause dur="1.4"/> okay <pause dur="1.9"/> the two <pause dur="1.5"/> phases the matrix phase and the <pause dur="0.6"/> fibre phase are going to carry the same strain <pause dur="0.6"/> okay so everything goes hand in hand <pause dur="2.0"/> and then something happens <pause dur="0.5"/> okay <pause dur="0.2"/> what <pause dur="0.4"/> are the events which can happen </u><pause dur="2.0"/> <u who="sm0814" trans="pause"> the fibres can snap </u><pause dur="0.8"/> <u who="nm0808" trans="pause"> the fibres can snap <pause dur="0.2"/> first <pause dur="0.6"/> okay <pause dur="0.6"/> so we get the situation where the fibres <pause dur="0.2"/> break <pause dur="0.5"/> before <pause dur="0.4"/> the matrix <pause dur="0.3"/> breaks <pause dur="0.2"/> in other words what does that mean <pause dur="0.2"/> it means that the <pause dur="0.5"/> failure <kinesic desc="writes on board" iterated="y" dur="11"/> strain of the fibres <pause dur="1.7"/> is less <pause dur="0.7"/> than the failure strain of the matrix <pause dur="5.2"/> okay <pause dur="1.1"/> if you're pulling <pause dur="1.4"/>

the fibres and the <pause dur="0.2"/> matrix in parallel <pause dur="0.5"/> okay <pause dur="0.5"/> the one which is giving up first is the one which has got the lowest failure strain <pause dur="0.5"/> okay <pause dur="1.1"/> so that's one case <pause dur="1.1"/> any more </u><pause dur="2.9"/> <u who="sm0815" trans="pause"> can you have it the other way round </u><pause dur="0.9"/> <u who="nm0808" trans="pause"> of course <pause dur="0.5"/> we can have another situation which is where there's <kinesic desc="writes on board" iterated="y" dur="6"/> fibre strain <pause dur="0.7"/> maximum is actually greater <pause dur="0.7"/> than <pause dur="0.3"/> the matrix strain <pause dur="3.2"/> any more offers </u><pause dur="1.7"/> <u who="sm0816" trans="pause"> <gap reason="inaudible" extent="3 secs"/> they're equal </u><pause dur="1.6"/> <u who="nm0808" trans="pause"> we'll come to back in <trunc>i</trunc> to your question yeah the the third one is the <trunc>t</trunc> <pause dur="0.3"/> <trunc>ch</trunc> <pause dur="0.7"/> sort of simple case where <pause dur="0.9"/><kinesic desc="writes on board" iterated="y" dur="2"/> epsilon-F-<pause dur="0.2"/>M-maximum is the same as <pause dur="0.4"/> epsilon-M-<pause dur="0.2"/>maximum <pause dur="0.7"/> so those are the three cases we need to think about <pause dur="0.8"/> okay and going back to your question when you're saying they're falling apart <pause dur="0.2"/> i think what you're meaning is that you've got an <trunc>i</trunc> # an interfacial <trunc>f</trunc> <pause dur="0.3"/> <trunc>sh</trunc> failure <pause dur="0.6"/> between the fibres and the matrix okay <pause dur="0.5"/> that is something which happens more often with the short fibre enforcement <pause dur="0.4"/> remember

here we are in an ideal world <pause dur="0.7"/> getting hold of the <pause dur="0.5"/> bits <pause dur="0.7"/> at the two ends and we pull simultaneously <pause dur="0.4"/> so you can't have <pause dur="0.2"/> a relative sliding <pause dur="0.7"/> when we look at the short fibres that can happen of course <pause dur="0.7"/> so for the long fibre enforcement continuous infinitely long <pause dur="0.2"/> or parallel <pause dur="0.5"/> okay <pause dur="0.3"/> those are the three situations <pause dur="0.4"/> which we have to analyse <pause dur="1.4"/> and the simplest one to look at <pause dur="0.4"/> not surprisingly <pause dur="0.4"/> is the last one <pause dur="0.7"/><kinesic desc="writes on board" iterated="y" dur="1"/> and that <pause dur="0.2"/> the one who is although it's a very unrealistic one <pause dur="0.6"/> because <pause dur="0.2"/> there is really <trunc>ver</trunc> virtually no system that i'm aware of <pause dur="1.2"/> where you've got the situation that the fibre strain <pause dur="0.3"/> is <pause dur="0.3"/> # maximum strain <pause dur="0.4"/> is of the same <pause dur="0.2"/> magnitude as the matrix <pause dur="0.3"/> maximum strain <pause dur="0.9"/> out of curiosity <pause dur="0.5"/> do you know what is <pause dur="1.6"/><kinesic desc="writes on board" iterated="y" dur="2"/> the maximum <pause dur="1.2"/> failure strain of glass fibres <pause dur="2.0"/> what do you think the failure strain of glass fibres is likely to be <pause dur="8.2"/> one per cent half a per cent <pause dur="0.3"/> point-two per cent <pause dur="0.2"/> point-one per cent <pause dur="0.6"/> ten per cent </u><pause dur="0.4"/> <u who="sm0817" trans="pause"> very small </u><pause dur="0.3"/> <u who="sm0818" trans="pause"> two-point <pause dur="0.5"/> <gap reason="inaudible" extent="1 word"/> </u><u who="sm0818" trans="overlap"> one per cent would be very little </u><pause dur="0.9"/> <u who="nm0808" trans="pause"> well very little

is not an engineering <vocal desc="laughter" iterated="y" dur="2"/><pause dur="0.4"/> okay what is <pause dur="0.4"/> typically <pause dur="0.3"/> the <pause dur="0.5"/> strain at which a material like a piece of metal <pause dur="0.2"/> yields <pause dur="0.2"/> what is the yield strain <pause dur="0.2"/> of a <pause dur="0.7"/> mild steel <pause dur="0.7"/> at what sort of strain levels <pause dur="0.3"/> does a steel or an aluminium yields <pause dur="2.6"/> not fails yields okay <pause dur="2.7"/> no idea </u><pause dur="2.7"/> <u who="sm0819" trans="pause"> is it a constant </u><pause dur="0.6"/> <u who="nm0808" trans="pause"> # <pause dur="1.5"/> fed up <kinesic desc="writes on board" iterated="y" dur="10"/> the <trunc>y</trunc> <trunc>yie</trunc> a yield strain <pause dur="2.9"/> a yield strain for a metal is of the order of point-two <pause dur="0.5"/> to point-three per cent <pause dur="1.4"/> okay <pause dur="0.3"/> that doesn't mean that is the ultimate strain <pause dur="0.2"/> this is the strain at which it starts yielding <pause dur="0.2"/> and deforming plastically <pause dur="1.0"/> remember that a metal is a polycrystalline system okay <pause dur="0.6"/> a crystal can not deform by a great deal <pause dur="0.2"/> okay before you start getting the glide and the slip <pause dur="0.5"/> coming in to operation <pause dur="1.4"/> 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 <pause dur="0.2"/> we can stretch a piece of steel <pause dur="0.2"/> plastically <pause dur="0.4"/> quite a lot so

the failure strain <pause dur="0.6"/> okay <pause dur="0.6"/><kinesic desc="writes on board" iterated="y" dur="8"/> can be <pause dur="0.8"/> anything <pause dur="0.4"/> up to <pause dur="1.1"/> say <pause dur="2.2"/> about twenty per cent <pause dur="0.7"/> possibly more <pause dur="0.5"/> sorry <pause dur="0.2"/> someone <pause dur="0.2"/> had a question </u><u who="sm0820" trans="overlap"> # <pause dur="0.8"/> yes at point-two <pause dur="0.2"/> to three per cent of what </u><pause dur="1.0"/> <u who="nm0808" trans="pause"> it's the strain </u><pause dur="2.4"/> <u who="sm0820" trans="pause"> right </u><pause dur="0.5"/> <u who="nm0808" trans="pause"> <vocal desc="laughter" iterated="y" dur="1"/> <pause dur="0.2"/> of what <vocal desc="laughter" iterated="y" dur="2"/> <pause dur="0.2"/> okay <vocal desc="cough" iterated="y" dur="1"/><pause dur="1.6"/> so <pause dur="0.4"/> remember that a <vocal desc="cough" iterated="n"/> glass does not yield <pause dur="0.4"/> glass is a brittle material <pause dur="0.5"/> it will stretch elastically up to the point of failure <pause dur="0.6"/> and the question is can you <pause dur="0.2"/> give me some idea now that we've put some numbers in <pause dur="0.5"/> okay at what value of the strain <pause dur="0.5"/> that failure in the glass fibre's going to occur </u><pause dur="1.5"/> <u who="sm0821" trans="pause"> nought-point-one-five </u><pause dur="1.5"/> <u who="nm0808" trans="pause"> point-nought-five <pause dur="0.6"/> okay <pause dur="0.3"/> at least the </u><pause dur="0.3"/> <u who="sm0822" trans="pause"> nought-point-nought </u><pause dur="0.4"/> <u who="nm0808" trans="pause"> <trunc>nou</trunc> </u><u who="sm0822" trans="overlap"> nought-one </u><pause dur="0.3"/> <u who="nm0808" trans="pause"> nought-point-nought-nought-one <pause dur="0.5"/> well you'll be all surprised that the maximum failure strength for glass <pause dur="0.4"/> if glass is perfect if the fibres are not damaged <pause dur="0.4"/> can be as <kinesic desc="writes on board" iterated="y" dur="2"/> high as five per cent <pause dur="1.5"/><vocal desc="whistle" n="su0823" iterated="n"/> okay <pause dur="1.3"/> so you can

get <pause dur="0.2"/> very very large elastic strains <pause dur="0.3"/> now <trunc>i</trunc> in practice <pause dur="0.6"/> okay you hardly ever achieve this value because <pause dur="0.2"/> the fibres inevitably get damaged in handling <pause dur="0.4"/> get damaged in processing because they touch each other you get scratches and defects <pause dur="0.4"/> but in <pause dur="0.2"/> what one may call ideal <pause dur="0.3"/> # undamaged fibres <pause dur="0.4"/> the failure strains are indeed very high <pause dur="0.7"/> and although this is <pause dur="0.2"/> represents a maximum <pause dur="0.3"/> okay <pause dur="0.5"/> epsilon <pause dur="0.2"/> failure <pause dur="0.2"/> maximum <kinesic desc="writes on board" iterated="y" dur="7"/> practical if you like <pause dur="2.3"/> of about <pause dur="1.2"/> two per cent to three per cent <pause dur="0.7"/> is not unknown <pause dur="1.6"/> you all seen the bit of glass fibre springs downstairs in the labs <pause dur="0.4"/> okay <pause dur="0.4"/> well those are typically working at strains of <pause dur="0.2"/> working <vocal desc="laugh" iterated="n"/><pause dur="0.4"/> okay <pause dur="0.3"/> which means below the maximum stress <pause dur="0.2"/> they are working at strains of about one-point-two one-point-five per cent <pause dur="0.8"/> okay <pause dur="0.4"/> and the failure is about <pause dur="0.2"/> twice as high <pause dur="0.5"/> so <pause dur="0.5"/> three per cent is achievable <pause dur="1.1"/> and i think it's important to put that into context because <pause dur="0.4"/> the fact that the material is brittle <pause dur="0.5"/> which it

certainly is <pause dur="0.5"/> okay does not mean that it can not be strong so long as it is in fibre form <pause dur="0.5"/> where <pause dur="0.5"/> having no defects is what is <pause dur="0.5"/> making the difference <pause dur="0.6"/> if you take a glass pane <pause dur="0.2"/> certainly that one will never get anywhere near <pause dur="0.2"/> to these sort of numbers <pause dur="0.5"/> because it's got too many <pause dur="0.4"/> defects surface fractures damage and whatever not to be able to do so <pause dur="1.2"/> okay so let's back here now <pause dur="0.7"/> all i'm saying is that <pause dur="0.3"/> when we talk about the maximum failure strain of the matrix <pause dur="0.5"/> being the same as the maximum failure strain of the fibres which is <pause dur="0.5"/> one particular simple system <pause dur="0.5"/> okay it's important to have some idea of what kind of numbers <pause dur="1.0"/> we are talking about <pause dur="1.1"/><kinesic desc="changes transparency" iterated="y" dur="2"/> okay well this is the situation you have <pause dur="0.2"/> and you've got it in your notes again so i'm not <pause dur="1.6"/> # <pause dur="2.4"/> going to write all that on the board <pause dur="0.3"/> okay <pause dur="1.5"/> the equation which governs <pause dur="0.2"/> the deformation of the material is the one we've just seen before <pause dur="0.2"/> the rule of mixtures <pause dur="0.4"/> in other words <pause dur="0.5"/> up to the point <pause dur="2.1"/> okay let me draw <pause dur="0.4"/> the thing a little bit like it is there <pause dur="0.9"/>

okay <pause dur="1.1"/> this is the case i'm considering now <pause dur="1.7"/> same failure strain in the matrix and in the fibre <pause dur="0.4"/> and let's assume that that failure strain for the system <pause dur="1.0"/> okay <pause dur="0.8"/> is somewhere here <pause dur="3.4"/><kinesic desc="writes on board" iterated="y" dur="4"/> so this is epsilon-<pause dur="0.5"/>max <pause dur="1.8"/> when i reach that <pause dur="0.2"/> strain level <pause dur="0.5"/> the fibres and the matrix will fail at the same time <pause dur="0.9"/> okay <pause dur="0.4"/> because that's what we are assuming <pause dur="1.2"/> however <pause dur="0.6"/> this is how the composite may deform <pause dur="0.4"/> so let's put this <pause dur="0.6"/><kinesic desc="writes on board" iterated="y" dur="2"/> with a C <pause dur="0.7"/> the fibres of course <pause dur="0.5"/> if one looks at the fibres themselves the fibres are stiffer <pause dur="0.6"/> okay and the fibres <pause dur="1.6"/> will deform <pause dur="1.4"/> like this <pause dur="0.7"/><kinesic desc="writes on board" iterated="y" dur="1"/> and the matrix is a lot softer <pause dur="0.7"/> and the matrix will deform like this <pause dur="1.1"/>

so <trunc>i</trunc> if we look at the individual components <pause dur="0.2"/> matrix by itself fibres by themselves and then composites <pause dur="0.3"/> okay <pause dur="0.3"/> we get three <pause dur="0.2"/> stress strain curves of this kind <pause dur="0.5"/> and <pause dur="0.5"/> this one here the one of the composite <pause dur="0.3"/> of course moves <pause dur="0.8"/> up or down depending on <pause dur="0.3"/> the volume fraction <pause dur="0.3"/> the higher the volume fraction it is <pause dur="0.4"/> the closer it's going to get <pause dur="0.2"/> to the curve of the fibres <pause dur="0.5"/> the lower the volume fraction is <pause dur="0.4"/> the closer it's going to be <pause dur="0.6"/> to the <pause dur="0.7"/> curve for the matrix <pause dur="1.0"/> okay <pause dur="1.6"/> well <pause dur="0.4"/> no matter where we are <pause dur="0.4"/> okay <pause dur="0.7"/> no matter where we are between here and here <pause dur="0.5"/> okay at any point <pause dur="0.2"/> up <pause dur="0.3"/> this curve <pause dur="0.9"/> this equation here halts <pause dur="1.0"/> okay <pause dur="0.8"/> yeah </u><pause dur="0.3"/> <u who="sm0824" trans="pause"> is the end modulus of the matrix <pause dur="0.2"/> really a lot lower than <pause dur="0.3"/> that of the composite <pause dur="0.3"/> or </u><pause dur="0.3"/> <u who="nm0808" trans="pause"> yeah <pause dur="1.0"/> well i'll give you a sort of example just from before <pause dur="0.5"/> the modulus of the matrix about one per cent # one gigapascal <pause dur="0.6"/> the fibres are seventy gigapascal <vocal desc="laugh" iterated="n"/><pause dur="0.6"/> okay and in a typical volume fraction of <pause dur="0.2"/> twenty per cent <pause dur="0.6"/> the modulus of the composite going to be about <pause dur="0.3"/> twenty twenty-five gigapascal so it's a big

big difference <pause dur="1.1"/> and this is one of the problems of course especially when you go to the short fibre enforcement <pause dur="1.6"/> so <pause dur="0.7"/> we can write that <pause dur="0.2"/> okay and this equation here is nothing but the equation of that line <pause dur="1.2"/> okay <pause dur="1.5"/> just there <pause dur="0.6"/> as <pause dur="0.2"/> the volume <trunc>fra</trunc> for a given fixed volume fraction <pause dur="0.6"/> okay if i increase the stress in the composite <pause dur="0.4"/> okay a stress in the fibre will increase the stress in the matrix will increase in proportion <pause dur="0.3"/> and i move up that line <pause dur="0.7"/> when i reach <pause dur="0.4"/> my limit value <pause dur="0.5"/> here <pause dur="0.7"/> everything gives up <pause dur="1.2"/> okay the fibres have reached their <pause dur="0.3"/> failure strain <pause dur="0.7"/> the <pause dur="0.2"/> matrix has reached its failure strain <pause dur="0.6"/> so <pause dur="0.5"/> the composite <pause dur="0.3"/> can only carry <pause dur="0.6"/> as a maximum stress <pause dur="0.5"/> the stress <pause dur="0.2"/> that corresponds to the <pause dur="0.3"/> fibre <pause dur="0.2"/> ultimate stress <pause dur="0.6"/> times the volume fraction of fibres <pause dur="0.5"/> plus <pause dur="0.4"/> the matrix <pause dur="0.2"/> ultimate stress <pause dur="0.9"/> times the volume fraction of <pause dur="0.8"/> matrix <pause dur="0.5"/> okay <pause dur="0.7"/> and so this is the equation <pause dur="0.4"/> which gives me the <pause dur="0.4"/> prediction <pause dur="0.4"/> of the tensile strength of the composite <pause dur="0.5"/> when i have <pause dur="0.2"/> this condition here <pause dur="1.1"/> okay <pause dur="0.3"/> it's simply <pause dur="0.8"/> the limit case of the

rule of mixtures <pause dur="0.4"/> when both <pause dur="0.8"/> components <pause dur="0.2"/> fibres and matrix <pause dur="0.4"/> have reached their failure point <pause dur="0.7"/> i can't get a higher stress than that <pause dur="0.8"/> obviously <pause dur="0.5"/> okay <pause dur="1.5"/> it is also pretty obvious that if <pause dur="0.5"/> the <pause dur="0.2"/> amount of reinforcement is very low <pause dur="0.6"/> in other words if the volume fraction of fibres is very low <pause dur="1.2"/> if i head down <pause dur="0.8"/> here <pause dur="2.0"/> okay <pause dur="0.2"/><vocal desc="clears throat" iterated="n"/><pause dur="1.1"/> then to some extent you can neglect that <pause dur="0.5"/> and you end up with <pause dur="0.5"/> a strength which is just <pause dur="0.2"/> a little bit higher <pause dur="0.5"/> than the strength of the matrix by itself <pause dur="0.8"/> but obviously that would be rather pointless <pause dur="0.6"/> okay <pause dur="0.6"/> we really need to put the fibres in <pause dur="0.2"/> to move the strength and the modulus of the matrix <pause dur="0.7"/> significantly beyond <pause dur="0.2"/> its normal level when it is unreinforced otherwise

we are <pause dur="0.3"/> spending a lot of money <pause dur="0.3"/> to <pause dur="0.9"/> # no effect whatsoever <pause dur="2.8"/> questions <pause dur="2.1"/> okay <pause dur="0.3"/> we reached the end of the hour so we'll leave it at that <pause dur="1.6"/> tomorrow we are going to look at <pause dur="0.8"/> the other two situations <pause dur="1.0"/> both of which exist there are situation where the fibres do have a higher <pause dur="0.3"/> failure strain than the matrix <pause dur="0.3"/> that is typical for example with thermosetting resins <pause dur="0.4"/> which have got low <trunc>ve</trunc> <pause dur="0.2"/> very low failure strains <pause dur="0.5"/> and other situations sorry <pause dur="0.4"/> where is the opposite <pause dur="0.4"/> where the <trunc>m</trunc> <pause dur="0.6"/> failure strain of the matrix is very large and this is typical of <pause dur="0.4"/> the <pause dur="0.5"/> # crystalline thermoplastic <pause dur="0.5"/> all the materials which can deform in a ductile manner over large strains <pause dur="1.1"/> okay so <pause dur="0.9"/> don't forget tutorial time tomorrow G-twenty-six from eleven to <pause dur="0.4"/> twelve