Select and answer 5 of the following 8 questions.
All questions carry 20 points. If you answer more than five questions,
your score will be obtained by summing the best five.
The exam is closed-book.
Useful formulae and material properties are provided on the attached sheet.
Given that 96,500 coulombs will deposit one gram-mole of a substance, what uniform current density (mA/m2) will exist in the protected pipe?
In practice, would the outer surface of the length of pipe be satisfactorily protected if one piece of zinc was attached at one position? Explain.
What alternative methods of protecting the pipe can you suggest?
Model Answer to Question 1
We assume that to protect one mole of iron from corrosion, we need to sacrifice a mole of zinc. We first calculate the mass of iron that will be corroded:
Mass = pi * d * L * density * 0.0001 kg
So no. of kg-moles corroded = pi * d * L * density * 0.0001 / 55.6
So we need pi * d * L * density * 0.0001 / 55.6 kg-moles of zinc
which is pi * d * L * density * 0.0001 * 65.4 / 55.6 kg of zinc
or 2.88 kg.
If we assume that the zinc is mounted as a lump at one end of the pipe, that all the current flows along the pipe, and that the pipe can be treated as solid (since we don't know its inner diameter), then the charge flowing along the pipe is
pi * d * L * density * 0.0001 * 1000 * 96500 / 55.6 coulombs/year
So the flow in amps is
pi * d * L * density * 0.1 * 96500 / (55.6 * 365 * 24 * 3600) amps
So the flow in amps per square meter is
pi * d * L * density * 0.1 * 96500 /(55.6 * 365 * 24 * 3600 * 0.25 * pi * d * d)
which simplifies to
L * density * 9650 /(55.6 * 365 * 24 * 3600 * 0.25 * d) = 172 A/m2
| Load (kN) | Length (cm) |
|---|---|
| 0 | 5.000 |
| 20 | 5.0102 |
| 40 | 5.0197 |
| 50 | 5.0257 |
| 60 | 5.0285 |
| 65 | 5.0355 |
| 70 | 5.0505 |
| 80 (max) | 5.1257 |
| 76 (fracture) | 5.3350 |
The original diameter was 1.262 cm, shrinking to 1.135 cm at maximum load and 1 cm exactly at fracture.
Plot the engineering stress-strain diagram for the sample.
Calculate the elastic modulus of the aluminum.
Calculate the 0.2% offset yield strength. Why is this value important?
Calculate the engineering ultimate tensile strength and the true ultimate tensile strength. Engineers usually use the former rather than the latter; why?
Calculate the % elongation and the % reduction in area at fracture. How might this information be used?
Describe the microstructural changes that are occurring in the elastic and plastic ranges.
Suppose that we stopped the test after applying 50 kN, removed the load, then tested the piece again. Would we get the same curve as before? If not, how would it differ?
Answer to Question 2
The stress-strain diagram is shown here:Fig2 Things to be careful of: plot stress, not force; plot strain, not length.
To calculate the elastic modulus, construct a straight line through the elastic portion of the graph and measure its gradient, which turns out to be 90.9 GPa. (Fig3)
To calculate the 0.2% off-set strength, measure a strain of 0.2% along the x-axis, then construct a line through this point parallel to the elastic portion of the graph and see where it intersects the curve. (Fig4) (It intersects at 540 MPa.) This value is important because it's the greatest load the material can sustain before undergoing permanent deformation.
The engineering ultimate tensile strength is the greatest imposed force divided by the original area of the specimen, which is 80,000/(pi * 0.006312) = 639.5 MPa. The true ultimate tensile strength is the greatest value of (imposed force / actual area). If we evaluate (imposed load / actual area) at the point of greatest load, we get a value of 80,000/(pi * 0.005672) = 790.7 MPa; however, this is not the ultimate tensile strength. If we evaluate the same statistic at the final point of the graph, we get 76,000/(pi * 0.0052) = 967.7 MPa, which is the correct value. (See Fig5)
Engineers usually use the former of these two figures since it's easier to measure; obtaining the true stress requires measuring the cross-sectional area of the specimen as each data point is taken, which is tedious.
The percentage elongation at failure is 6.7%, and the reduction in area is 100(1 - 1/1.2622) = 37.2%. This information might be used to calculate the degree of work-hardening in the piece.
In the elastic range, the lengths of the interatomic bonds are increasing. When we enter the plastic range, dislocations start to move through the crystal lattice, allowing crystal planes to slip over one another.
If we re-tested the piece after loading to 50 kN, we would get the same curve as before, since the 50 kN point is only about halfway up the elastic region of the graph. We would not expect any permanent changes in the specimen until entered the plastic region.
Given that the glass transition temperature for amorphous polyethylene is -60 C and the melting temperature for crystalline polyethylene is 105 C, compare and explain the shapes of the stress-strain curves for low and high density polyethylene at 110 C, 23 C and -195 C.
Answer to Question 3
We note that one ethylene mer weighs (2*12 + 4*1 = 28 amu). So 1 gram-mole of ethylene is 28 grams. So 1 kg of ethylene is 1000/28 = 35.7 gm-mols.
When ethylene polymerises, we break open the double bond, which requires 719 kJ per gram mole, and form two single bonds -- one between the two carbon atoms, and half of each of the two single bonds linking this mer to those on the left and right. Forming the single bonds releases 2 * 368 = 736 kJ/mole. So each mole of polyethylene formed releases 736 - 719 = 17 kJ.
If the reaction occurs adiabatically, then 35.7 * 17 = 607 kJ are released. So the temperature rise is Delta T, where
E = mc Delta T (and c is the specific heat of polyethylene, given in `Useful Facts')
So Delta T = 607,000/(1 * 2300) = 264 K.
At - 195 C, both amorphous and crystalline polyethylene are glassy, and will show brittle behaviour. (The crystalline variety will be stronger and denser.) At 23 C, amorphous polyethylene will be viscoelastic, elongating to a considerable degree, while crystalline polyethylene will be leathery. At 110 C, both varieties will be liquid, and will distort without limit in response to an imposed stress. (See Fig6)
Dilithium is radioactive. During any one-hour period, one dilithium atom in every thousand will decay, releasing 12.6 MeV of thermal energy. If a sample of kryptonite is kept in a perfectly insulated, mirror-lined container, how much will its temperature rise during the first hour?
Answer to Question 4
This makes use of the equation
E = h nu
where h is Planck's constant and nu is the frequency of light. We use the chart of the electromagnetic spectrum provided to determine that green light has a wavelength of about 0.54 * 10-6 m; then we use
nu = c / wavelength
to determine that nu is 5.55 * 1014 Hz. Substituting this into E = h nu, we get
E = 6.63 * 5.55 * 10-20 J
= 6.63 * 5.55 / 16.02 eV
= 2.296 eV
So this is the distance that the dilithium impurities are above the energy level of the top of the valence band.
For the second part of the question, suppose we have a kilogram of kryptonite. This is 1000/450 gm-moles of kryptonite, which contains
2 * (1000/450) * 6 * 1023 atoms
(The factor of 2 is added since each molecule of cavorite contains two atoms.) One atom in 10,000 is dilithium, and one dilithium atom in every thousand decays during an hour. So in an hour,
2 * (1000/450) * 6 * 1016 decays occur and
2 * (1000/450) * 6 * 1016 * 12.6 MeV of energy are released.
This corresponds to
2 * (1000/450) * 6 * 1016 * 12.6 * 1.602 * 10-13 Joules
which is 0.538 MJ.
Again we use the specific heat equation,
Delta T = E/mc = 0.538 * 106 / 1500 = 359 C
Answer to Question 5
Fatigue is the most common failure mode, accounting for about 90% of all failures. Metals, polymers and ceramics (except glasses) are all subject to fatigue. It results from an oscillating stress -- either because the part is in an application, such as a connecting rod, where it is alternately transmitting a push and a pull, or because it's under a nominally constant load but subject to vibration. The mechanism of failure is that a crack begins to grow while the part is in tension; the tip of the crack would normally widen, reducing the stress concentration, but the crack tip is then sharpened during the compression phase, so the stress concentration is maintained and the crack grows further during the next tension phase. This mechanism produces a characteristic pattern of striations on the fracture surface. These striations are typically microscopic, and represent the advance of the crack tip during a single cycle of tension and compression. The failed surface may also display beachmarks; these are larger than striations, usually visible to the naked eye, and correspond to longer interuptions in the crack-forming process.
Fatigue failure occurs more readily at low temperatures. There is a trade-off between the size of the imposed load and the number of cycles to failure. Some materials will eventually fail even under minimal load, but for others, such as steel, there is a fatigue limit such that fatigue failure will not occur for imposed stresses below this limit.
Creep is the migration of atoms away from the highly stressed regions in a part under sustained static stress. It is found in all materials, but in metals is only important at temperatures above about 40% of the melting temperature. Boilers, high-pressure steam pipes and turbine blades are all prone to failure by creep.
Ordinary glassware often breaks when immersed in hot water, whereas Pyrex glass doesn't. Deduce the approximate critical value of the thermal shock resistance parameter for washing-up applications. The fracture strengths of soda glass and pyrex are both approximately equal to 70 MPa.
Answer to Question 6
We assume that there is a sharp boundary between the portion of the tile at 315 C and the portion at 20 C. (In reality, there will be a steep negative-exponential curve between the surface at 315 and the lower regions of the tile at 20 C; but this curve will be very steep, since the tile is a poor conductor, so assuming a sharp boundary is fairly close to the truth, and errs on the side of caution.) We further assume that the stresses in orthogonal directions are independent, rather than being added to give a combined stress. Lastly, we assume that we need only consider tensile and compressive stresses. (There will also be a shear stress, as the heated parts of the tile try to move sideways with respect to the cooler parts; but we don't yet know how to deal with shear stresses.)
Since the tile is a ceramic, it is more likely to fail in tension than in compression. So we consider the cooler part of the tile to be put into tension by the expansion of the hotter part.
The hotter part expands by alpha l (315 - 20), so if the cooler part is stretched by the same amount, it will experience a strain of
strain = alpha (315 - 20)
And hence a stress of
Stress = E alpha (315 - 20)
which evaluates to 137.4 MPa.
For the second part of the question, we evaluate the thermal stress resistance parameter for the two glasses. The TSRP is defined as
TSRP = sigmaf k /(E alphal)
plugging in numbers from the `Useful Facts', this evaluates to 244 W/m for soda glass, and 529 W/m for pyrex. We deduce that the critical value of TSRP for washing-up-related activities is somewhere in the range 244-529 W/m.
Comment on this assertion, illustrating your answer with relevant examples from the course or from other sources.
Answer to Question 7
This is probably not a good question to attempt in an exam, since it's an essay-type question, which is harder to get 100% right than a calculation. Still, if you are going to attempt it, here are some of the points you might make:
An overall observation is that not all that much has changed in the syllabus since 1904. The mechanical properties of materials are much the same; being able to calculate bond strengths from Schrodinger's equation is not a big improvement, since the equation can only be solved for a few simple molecules, and other factors, such as the presence of dislocations, make its predictions quite inaccurate for the macroscopic strength of materials.
Almost all of our discussion of crystals and crystal structure has treated atoms as hard indivisible spheres, linked by springs, a picture that would have been familiar to Democritus.
So the only places I can think of where quantum theory has had much impact on the contents of this course is in the optical properties of materials, where it provides an explanation for the fact that gases and vapours have line spectra in adsorption and emission -- this fact was known long before the advent of quantum theory, but was not understood. Similarly, quantum theory provides an explanation for why some materials are transparent, and for the operation of the laser.
In the thermal properties of materials, quantum physics provides the explanation for why specific heats of materials fall to zero as the temperature drops to zero -- classical theory predicts that specific heats should be independent of temperature.
Quantum physics provides the explanation for the photo-electric effect when we come to the electrical properties of materials; it also explains the effect of doping semi-conductors, and provides a basis for designing diodes and transistors.
On the other hand, a classical-physics version of the course could simply take certain facts as given, without attempting to understand them. At the moment, for example, we don't have a good theory of how high-temperature superconductors work, but we are still able to design apparatus that makes use of them.
Answer to Question 8
Referring to the phase diagram, if we start with a 50:50 composition (line 1 in the diagram), we can begin by melting it, then allowing it to cool slowly. From the tie-line rule, we can see that the first solid to form will be about 80% nickel. We allow more solid to form, down to about 65% nickel, then separate the liquid from the solid and re-melt the solid (line 2 in the diagram). We again cool and allow alpha-phase to precipitate out; the first solid to form is now about 90% nickel. We allow more solid to form, then separate the liquid and solid portions and re-melt the solid (line 3 in the diagram). By iteration of this process, any desired purity may be achieved. The process is most conveniently accomplished by having a series of hot and cold zones, and moving a tilted conveyor between the zones, so that the liquid fraction will move in one direction and the solid fraction in the opposite direction.
This process is called zone refining.
Segregation is a related idea: if a part is formed by cooling a mixture of two or more metals, the first solids to form will have a different composition from those that form later on, unless we're at the eutectic point. So the properties of the final solid will be inhomogenous; in particular, parts of the solid may melt at a temperature below the nominal solidus temperature for an alloy of the given composition, leading to failure of the part. This can be prevented by annealing the part at just below melting temperature for an extended period, allowing diffusion to smooth out the inhomogeneities in composition.
Answer to Question 9
This is not entirely a trick question -- the fact that mercury is a liquid can be predicted from relativistic quantum mechanics. However, the proof for this is a journal-level physics paper, and, though I have a copy of the paper in question, I have not so far managed to follow it beyond halfway down the first page. If you should come across this question in the final, therefore, it would probably be best to skip it.
Avogadro's number: 6 * 1023 atoms/gram-mole
Boltzmann's constant: 8.12 * 10-5 eV/atom.K
Planck's constant: 6.63 * 10-34J-s
1 eV = 1.602 * 10-19 J
Some bond energies: C--C: 368 kJ/gram-mole; C=C: 719 kJ/gram-mole
Electrostatic attraction:
F = q1q2 /(4 pi epsilon0 x2)
where
1 / (4 pi epsilon0) = 9 * 109 farads/metre
TSR = sigmaf k/ (E alphal)
Some useful material properties:
| Material | Density (kg/m3) | Atomic (or Molecular) Wt. | Valency |
|---|---|---|---|
| Zinc | 7100 | 65.4 | 2 |
| Iron | 7800 | 55.9 | 2 |
| Carbon | 1800 | 12 | 4 |
| Hydrogen | 0.0899 | 1 | 1 |
| Dilithium | 21000 | 300 | 2 |
| Cavorite | 13500 | 450 | - |
| Material | Density (kg/m3) | Sp. Ht. (J/kg.C) | k (W/m.K) | E (MPa) | alphal |
|---|---|---|---|---|---|
| Polyethylene | 1200 | 2300 | 0.38 | 200-1000 | 120 * 10-6 |
| Kryptonite | 13500 | 1500 | 75 | 3100 | 0.8 * 10-6 |
| Soda-glass | 3200 | 830 | 2.2 | 70000 | 9 * 10-6 |
| Pyrex | 3200 | 840 | 1.7 | 75000 | 3 * 10-6 |
