MECH0007 Fundamentals of Materials
Summary of BCC questions & answer hints
Dr Adam Wojcik, 2020-21 session
Below, you will find a summary of all the BCC questions that were posed in the sync
sessions that we held, and some additional ones to make up 14 - from which you can
choose any 4 to tackle, provide answers to, and submit. I’ve also added in relevant clues
that I might have mentioned in the sync sessions, just in case these were missed or not
clear from the recordings.
There is an exemplar answer given for one question, at the end of this document – so you
can get an idea of the sort of length we are expecting. This is therefore excluded from the
14 you can choose from!
Answers can be up to 1 side of A4 long ideally, including any images or illustrations that
you wish to add in. There is no set format – just try to answer the questions as if they were
encountered in a written exam – a few short paragraphs that provide some background to
the question, detail your answer, and perhaps some illustrations, images or diagrams to
back things up.
Remember, “a picture can save a thousand words”, particularly for an engineer!
You may insert data or images which you obtain from a variety of sources including the
Internet, digital books, or the ppt presentations that were created for this module. It is good
practice to acknowledge where you got such illustrations, by referencing, but you are being
graded on content – not on presentation.
Although the answers to these questions can all be found in the Detailed Notes, the BCC
questions are designed to make you think a little deeper than just a straightforward reading
of those notes. Using other resources, such as some of the books in your bibliography, or
online, will be beneficial in many different ways.
Read the clues given here carefully – whilst they are not the complete answer, they are
generally a lot more comprehensive than just “clues”. If you are particularly stuck on a
question, either find another to work on, or email Dr Wojcik for more help.
Grading is based on correctness and comprehensiveness of the answer, and will be either
A, B, or C (worth 100, 60 or 40 marks respectively, or 0 if no submission is made). One
mark will then be awarded, calculated as the average of the four.
After the deadline (26th April 2021) a booklet of worked solutions will be provided to give
you some formative feedback.
Submission is via the portal under the Assessment tab on the MECH0007 Moodle pages.
Please submit your four answers as a single document, preferably a PDF.
Make sure you put your name on the first page, or your student number.
Q1. How many materials go to make up a typical domestic lightbulb?
A1. See the exempla answer at the end of this document. The answer required a few
paragraphs of text and a diagram. If you need more space for an answer, you can use
more, but do try to keep it to one page per question maximum.
Q2. What properties are required for the material used for a gas turbine blade?
A2. To answer this fully, you need to think about the conditions in the environment in
which the material is being used – what is it like in a jet engine? In the turbine, gas
temperatures can exceed 1200 DegC, so any metal in there needs to have a high melting
point, for example – this cuts out materials such as aluminium alloys. However other
metals that have much higher melting points, such as steel are still unsuitable – why? In a
word you need to consider “oxidation” too. Additionally, there is a phenomenon known as
creep, which all materials suffer from where, at temperatures approaching an appreciable
fraction of their melting point, they begin to suffer time dependent deformation, when under
load. Aside from this, other more traditional selection criteria such as strength (but at high
temperature) and density, will also play a part.
Q3. Given that bonds in both ionic and metallically bonded solids are non-directional and
hence allow atom layers to easily move over each another, why are metals so much more
ductile than ionic solids (e.g., common salt)?
A3. The clue to this one is to understand how the atoms (or more correctly ions) are
organized in both metals and ionic solids. In a metal, the metallic bond means all the ions
are of the same type – negative. Sliding (the process of slip at the core of plastic
deformation) one plane of these ions over the other is relatively easy – and does not result
in a change to the crystal structure or failure of any bonds (given they are all
omnidirectional electrostatic bonds) – so the material doesn’t actually facture. This is
similar for the ionic bond, except that slip is more difficult because of the disruption it
causes to the electrostatic fields – think about the signs of the ions. Slip can happen, in
theory, but the stresses required are so close enough to those actually needed to break the
bonds, that fracture tends to happen first.
Q4. Can you mathematically infer that the toughness of a material is related to the area
under the stress-strain curve?
A4. If you assume that toughness defined as the work done in fracturing a material, then
you need to define work in terms of the stress-strain curve. If work is defined as force x
distance, then it is a simple enough proof to show that this is related to the area under the
stress-strain curve. Care must be taken however in defining what area – as the work done
in most materials is a combination of elastic deformation and plastic deformation. The
elastic work is usually returned to the system when fracture occurs.
Q5. Why is it so difficult to make amorphous (glassy) metals?
A5. To answer this, you need to recall what crystallinity actually means – long-range
ordering of atoms, ions or molecules (depending on what type of material we are talking
about). Thermopolymers, for example, contain long chains of carbon atoms which need
arranging and stacking neatly to create crystallinity. This might be easy for the first few
molecules or parts of molecules, but becomes progressively more difficult as crystallization
proceeds and also as the molecular mass of the polymer increases (so some high
molecular mass polymers are almost always amorphous). Similarly, glassy materials are
those where attaining a regular arrangement is made somehow difficult. In window glass,
silicon dioxide is prevented from crystallizing by the addition of other oxides called network
modifiers (such as CaO). Metallic ions have no such problems arranging themselves – for
very good reason – they are super small. So how do you prevent something that is so
small from organizing itself as it freezes?
Q6. Why is it necessary to undercool a liquid before it can start to solidify?
A6. This is a question you can answer either literally or mathematically. The literal
answer centres on the fact that it is necessary to form stable nuclei before crystallization
can proceed to completion and that nuclei are deemed stable when they have reached a
specific size – or critical radius. The critical radius is different for different materials and
defines a nucleus with a given number of atoms (or molecules). The way in which this
critical size is reached is simply by random means – if enough atoms come together by
random movements (at the freezing point) to form a nucleus of critical size, it will remain
stable and survive to grow bigger. If not, it will decompose (in effect melting back into a
liquid again). During the time we are waiting for these random processes to form nuclei of
the right size, the solidifying liquid may continue to cool, by transferring heat to its colder
surroundings – so it remains liquid below its freezing point and is thus “undercooled”. A
better explanation considers why a critical nucleus size is actually needed – and for that
you need to consider thermodynamics – specifically the energy associated with the volume
of the nucleus and its surface.
Q7. Why do dendrites form in a metal?
A7. This is again a question that can be answered quickly or more fully. The quick
answer is that as a solid is forming from a liquid, in a mould or container, there will usually
be a distinct interface between the solid and liquid parts– and this interface will move
parallel to the direction of heat flow, but in the opposite direction. We call this interface, a
“growth front”. Theoretically even if there are no mould walls, and growth is occurring from
a nucleus in the middle of the liquid, there will still be a growth front (but perhaps a radial
one). One would expect this front to be uniform, but statistical variations in the speed of
growth mean that perturbations will occur and some parts of the front will grow faster than
others – this leads to protuberances (solid parts that stick out) being formed. These
become the arms of the dendrite. The mystery is why these protuberances become stable
– why do they not just melt back? Presumably, if the liquid is cooling down, the
protuberance will be entering into hotter liquid – so it should just melt back and a uniform
growth front occur as a consequence. If you can understand why the protuberance
becomes stable, you can fully explain dendritic growth.
Q8. Why are intermediate compounds generally very hard & exhibit high melting points?
A8. The clue is in the name – compounds. Compounds are formed from elements that
have already reacted. Why do things react? The answer is again a thermodynamic one –
as everything seeks a lower energy state. In other words, the process of forming the
compound leads to a happier energy state for the “system” – which means that excess
energy has been “lost”. To return to the original unreacted elements, means that that
energy needs to be put back in. For example, consider the production of aluminium by the
smelting of aluminium oxide (bauxite). This requires huge amounts of energy, simply
because this is reversing a chemical reaction where a lot of energy was released. This
mean that bauxite is far more thermodynamically stable than aluminium, and we would
therefore expect that it is far more inert, unreactive and possesses a higher melting point
(all of which is true). In terms of the high hardness, the link with energy is more difficult to
see, but comes from the type of bonding that occurs when the solid material is formed.
Solid aluminium is metallically bonded – but what about bauxite?
Q9. Use the phase diagram for water and salt to explain why slush persists on
pavements which have been salted.
A9. Slush is the term we give to a mix of liquid and solid. Use the diagram in your notes
for the salt-water system, or look it up online - and this question should be straightforward
to answer, if you recall what happens in a two-phase region of such a diagram – there, two
phases exist in equilibrium, so if the temperature remains the same, nothing will change in
terms of the relative quantities of those two phases, or their compositions. Try and identify
which part of the phase diagram you will be sitting in, if salt is added to icy pavements.
Imagine what then happens to these if the external temperature remains constant. It is true
that during the winter, and during the day, the temperature of the pavement might begin to
alter – but what then happens to the slush – and why does it still persist?
Q10. How can coring be represented on a phase diagram?
A10. To answer this question, you might need to do some digging on-line. It is
theoretically not possible to “see” coring on a phase diagram, however you could plot a line
of points that represented the ends of the tie line for a cored alloy and join them up to form
a transformation line on the phase diagram. This is easy if you recall what coring is – a
change in composition of a crystallizing solid solution in a metallic alloy, as crystallization
proceeds and the temperature is reduced. If we assume dendritic growth, and a starting
composition of the solid of α1 at temperature T1, then the phase diagram will predict that at
some new lower temperature, T2, (all) the solid will have reached α2. In reality, only the
outside of the dendrite will have reached α2, so the true solid composition is the average of
the two (i.e., somewhere between α1 and α2). This new value is what you can plot on the
phase diagram (note also that the liquid will actually also be a different composition to that
precited by the phase diagram). In this way, a plot of the average solid composition with
temperature, becomes the new transformation line.
Q11. How can we predict the morphology (in terms of the relative layer thickness) of an
eutectic, from a phase diagram?
A11. In a phrase – the Lever Rule. If you apply the lever rule for an alloy of eutectic
composition (this is done because it is the simplest alloy to deal with – as all of it will be
eutectic) – and if you make some assumptions (such as the density of the two phases in
the eutectic are the same), then by applying the lever rule (which gives you weight
fractions), you can predict the volume fraction, and hence the area fraction in 2D (again by
making some assumptions) and then the width fractions. This allows you to draw the
eutectic as a realistically scaled set of layers. The relative width of the layers is an
important part of the morphology of the eutectic (i.e., its “look”).
Q12. Why are the microstructures of peritectic forming alloys usually extensively cored?
A12. This we covered in some detail in a sync class, but is also explained in the Detailed
Notes. The answer lies in the detail of the peritectic transformation – one solid, reacting
with a different liquid (in terms of composition) to generate a totally new solid. If (for
simplicity) you consider an alloy of peritectic composition, then this transformation from one
solid into another, must occur completely. The problem comes when you consider where
the new solid is going to form. It will form wherever the old solid meets the liquid (these
being the two substances that are required to form the new solid) – so on the surface of the
old solid. This produces a coating of new solid, over the old solid, which effectively cuts off
the liquid from the very solid that it needed to react with. Instantly then, a core of old solid,
inside a coating of new solid is generated. To continue with the transformation, the liquid
must gain access to the old solid through the new solid, and can only do so via solid-state
diffusion – which is much slower. The result of this is that although solidification can
continue, the transformation cannot – so a cored microstructure occurs.
Q13. Why can the more compact FCC austenite phase of iron dissolve orders of
magnitude more carbon than the more open BCC ferrite phase?
A13. In a word – shape. Carbon is an interstitial atom, so it will locate itself in-between
the iron atoms in the interstitial spaces in the iron lattice. The shape of these spaces is
what largely controls the solubility of the carbon – rather than the quantity. Whilst the FCC
structure has less “empty space”, what it does have is “better” shaped space. It is quite
easy to use a little bit of simple maths and Pythagoras to work out the maximum radius of a
sphere that can fit into the interstitial spaces in FCC and BCC, and so prove that that FCC
is better at accepting carbon. The sizes of the spheres are still smaller than the diameter of
a carbon atom – so distortion will occur of course, but this distortion is both the origin of the
solubility limit seen in Fe-C alloys, and also of the strengthening effect behind “solid
solution strengthening” per se.
Q14. How does tacticity affect the mechanical properties of a polymer?
A14. Tacticity is the organization of side groups on a long chain organic polymer
molecule. Tacticity can be controlled by the polymerisation processes used by the
manufacturers of the polymer raw materials. It is not something that engineers would be
able to change – but it is important to understand how tacticity affects the properties of the
polymer that you’ve chosen for your end-use. Polymers which have a high degree of
ordering to side groups (isotactic) are easier to crystalize, due to the regularity in the
molecules. The molecules can also pack more efficiently and therefore get closer together,
which raises the level of dipole-dipole interactions between molecules. Given that these
interactions are responsible for the bulk properties of thermopolymers, differences in
tacticity can have a large effect on mechanical properties such as strength, toughness and
stiffness, as well as physical properties such as melting point and density. A good answer
here, would add in some diagrams to explain the different degrees of tacticity and their
Q15. Why do the mechanical properties of a plastic bag vary with the direction in which it
A14. To answer this, you need to see your Detailed Notes both at the start of the module
and towards the end, on polymers, and check out the stress-strain response of linear
thermopolymers such as polyethylene (from which such bags are manufactured). The
stress-strain curves will be different for the polymer bag, for specimens cut from different
directions. Along the width of the bag, the polymer will yield very differently to along the
depth of the bag – its ductility width-wise, is far higher than depth-wise. The reasons for
this lie with the manufacturing process. How do they make plastic bags – or more precisely
the polymer tubes, from which plastic bags are cut and sealed? If you imaging what this
process does to the molecules in the polymer sheet, and how this might vary depending on
the direction that we look along – can you then extrapolate this into why the bag behaves
differently in different orientations? Once again, think about interactions, and how
molecules transfer load between each other.
PTO for A1
Exemplar Solution to Beneficial Coffee-time Conundrum (BCC) Q1
MECH0007 Fundamentals of Materials
Question: How many materials go to make up a typical domestic lightbulb?
Answer: Note that this question is designed to illustrate that even apparently simple
everyday objects can be complex assemblages of materials. Often, the end product is also
the result of numerous technological and manufacturing processes too - each of which is
optimised to be cost effective, whilst also delivering a product with the desired
It may come as a surprise to find that the average lightbulb contains at least 14 separate
components, if the cap is included, each of which comes with its own set of production
processes. The diagram below illustrates some of these components. In addition to the
well-known use of tungsten as the filament, there are several different types of glass
employed in the bulb - each one has a specific role in the overall product and is chosen for
specific materials properties. This is the process of "materials selection". By varying the
composition of the glass, for example, the thermal expansion coefficient can be
sequentially altered so that there is a more gradual transition in this property, generated at
the point where the glass envelope joins (and seals) on to the metal wires (these wires
deliver the current to the bulb filament). Such a transition lowers the interfacial stresses in
the joint, hence maximising bulb lifetime.
The wires themselves are not made of tungsten and vary in make-up, depending on
location. The Dumet wire is a copper wire with a core of an iron-nickel alloy with a low
coefficient of thermal expansion. The lead-in wires are usually nickel-plated copper or
nickel, and the filament support wires (which need to support and withstand glowing
tungsten – so well over 1500 DegC) are of molybdenum. These materials are chosen for
their high melting point, and resistance to oxidation. Connections to these wires can be
made mechanically (by crimping) or by resistance welding (depending on the materials to
be joined). If crimping is used, connections must remain ohmic and low in electrical
resistance, so immunity to oxidation is a really important property – else the oxides will act
as effective insulators, initially causing local heating but eventually a loss of connectivity.
To avoid this as much as possible, the glass “envelope” of the bulb is evacuated and back
filled with an inert gas, containing as low an oxygen content as possible.
Dr A Wojcik 2021 学霸联盟