Question 11.6
On the basis of microstructure, briefly explain why gray iron is brittle and weak in tension.
Answer 11.6 Grey iron contains flakes of graphite, and the sharp tips of the flakes make good starting points for cracks, which will propagate when the iron is put under tension.
(Just pointing out that graphite is weak is not a good answer, since nodular iron also contains graphite, but is not brittle or weak in tension.)
Question 11.13
Why must rivets of a 2017 aluminium alloy be refrigerated before they are used?
Answer 11.13
Aluminum will precipitation-harden even at room temperature; this will lead to the rivets becoming harder and more brittle, so they may shatter when used. If they are refrigerated, the mobility of the atoms within the alloy is reduced, so precipitation is postponed. (The rivets will precipitation-harden once they're in place, but that isn't a problem; in fact, it makes the riveted joint stronger.)
Some answers cited the low melting point of aluminium (660 C). This isn't really relevant, since room temperature in Canada never gets anywhere close to that; there's no real risk of an unrefrigerated rivet turning into a puddle of molten aluminium.
Another suggested answer was that keeping the rivet cold increased the cold-work hardening when it was put into place. This might be true, but if this were the major explanation, it would apply to rivets of any metal, not just 2017 aluminum. Also, this explanation only works if the rivet is kept cold until the instant of riveting, which presents practical difficulties: aluminium is a good conductor, and the rivet is loaded into a hand-held metal riveting gun, which will normally be at ambient temperature.
A fourth suggested answer was that keeping the rivet cold allowed it to expand after insertion, thus giving a tighter joint. To see the problem with this answer, consider that the rivet would expand in two directions: radially, towards the sides of the hole; and axially, along its length. The radial expansion would make the rivet a snugger fit in the hole, but the axial expansion would make the rivet longer, so it wouldn't hold the two sheets of metal together as firmly. (Iron rivets for heavy-duty work, such as holding ships together, are often inserted when red-hot, just so they will shrink after insertion and hold the deck plates together more tightly.)
Question 11.26
Cite three sources of internal residual stresses in metal components. What are two possible adverse consequences of these stresses?
Answer 11.26
Work-hardening; the presence of interstitial impurity atoms; the presence of substitutional impurity atoms of a different size from the host metal; a sudden phase transition due to abrupt cooling, as in the quenching of austenite to form martensite; non-uniform cooling of a large, heated solid; these are all potential sources of residual stress. Adverse consequences include the distortion of the finished part (for example, warping); cracking around the stressed region; also, the stressed region has a heightened vulnerability to corrosion. (Though we won't be discussing that for a couple of weeks.)
Question 12.12
Iron titanate, FeTiO3, forms in the ilmenite crystal structure that consists of an HCP arrangements of O2- ions.
Answer 12.12
r(Fe2+)/r(O2-) = 0.55; hence, from Table 12.2,
the coordination number is 6, and the Fe2+ ions occupy octahedral sites.
Question 12.18
Cadmium Sulphide (CdS) has a cubic unit cell, and from X-ray diffraction it is known that the cell edge length is 0.582 nm. If the measured density is 4.82 g/cm3, how many Cd2+ and S2- ions are there per unit cell?
Answer 12.18
Consider a gram-mole of cadmium sulphide. Let n be the number of cadmium ions per unit cell; since the cell must be electrically neutral, n will also be the number of sulphide ions per cell. Its density is given by the formula:
density = n(atomic weight of sulphur + atomic weight of cadmium)/(cell volume * NA)
where NA is Avogadro's number.
Re-arranging and plugging in the numbers, we get:
n = density * (cell volume * NA)/(atomic weight of sulphur + atomic weight of cadmium)
Hence n = (4.82 * (0.582*10-7)3*6.023*1023)/(32.06+112.41)
from which n = 4.
(Note that we convert the cell edge length to centimeters rather than meters, since the density is given in g/cm3; alternatively, we could convert the density to g/m3.)
Question 12.28
Determine the angle between covalent bonds in an SiO4-4 tetrahedron.
Answer 12.28
This is a purely geometric calculation. From the symmetry of the molecule, we deduce that this must be a regular tetrahedron. We may happen to know that the angle between the lines joining the centre of a regular tetrahedron to its vertices is 109 degrees, or we can calculate it as follows:
(See q1228.jpg)
Consider the triangle ABS. Drop a perpendicular from S to meet AB at its midpoint, C. Let the length of side of the enclosing cube be 1.
Then AB = sqrt(2), and AC = 1/sqrt(2).
AS is one-half a major diagonal of the cube, so AS = sqrt(3)/2.
The angle ASC is sin-1 AC/AS = sin-1 sqrt(2/3) = 54.7 degrees
So the angle ASB is 2 * 54.7 = 109.5 degrees.
Question 12.46
The modulus of elasticity for titanium carbide (TiC) having 5 vol% porosity is 310 GPa.
Answer 12.46
This is a case for Equation 12.5 (6th edition):
E=E0(1-1.9P+0.9P2)
Plugging in the numbers, we get that
E0= 341.8 GPa
To answer the second part of the question, we use the same equation, but solve for P. This leads to a quadratic equation, which has two roots. Only one of these roots is in the range [0,1] so we take that value, concluding that the porosity is 17%.
