E344 2013 Summer Solution Set 3

E344 2013 Summer Solution Set 3 1. 13.21 (a) What are the three main components of a whiteware ceramic such as porcelain?
(b) What role does each component play in the forming and firing procedures?
(a) The three components of a whiteware ceramic are clay, quartz, and a flux.
(b) With regard to the role that each component plays:
Quartz acts as a filler material.
Clay facilitates the forming operation since, when mixed with water, the mass may be made to
become either hydroplastic or form a slip. Also, since clays melt over a range of temperatures, the shape of the piece
being fired will be maintained.
The flux facilitates the formation of a glass having a relatively low melting temperature.
2. 13.23 Cite one reason why drying shrinkage is greater for slip cast or hydroplastic products that have smaller
clay particles.
The reason that drying shrinkage is greater for products having smaller clay particles is because there is
more particle surface area, and, consequently, more water will surround a given volume of particles. The drying
shrinkage will thus be greater as this water is removed, and as the interparticle separation decreases.
3. 14.9 For a linear polymer molecule, the total chain length L depends on the bond length between chain atoms d,
the total number of bonds in the molecule N, and the angle between adjacent backbone chain atoms θ, as follows:
⎛ θ ⎞
L = Nd sin ⎜ ⎟
⎝ 2 ⎠
Furthermore, the average end-to-end distance for a series of polymer molecules r in Figure 14.6 is equal to
r = d
A linear polytetrafluoroethylene has a number-average molecular weight of 500,000 g/mol; compute average values
of L and r for this material.
This problem first of all asks for us to calculate, using Equation 14.11, the average total chain length, L, for
a linear polytetrafluoroethylene polymer having a number-average molecular weight of 500,000 g/mol.
It is
necessary to calculate the degree of polymerization, DP, using Equation 14.6. For polytetrafluoroethylene, from
Table 14.3, each repeat unit has two carbons and four flourines. Thus,
m = 2(AC) + 4(AF)
= (2)(12.01 g/mol) + (4)(19.00 g/mol) = 100.02 g/mol
500,000 g/mol
DP = n = = 5000
100.02 g/mol
which is the number of repeat units along an average chain. Since there are two carbon atoms per repeat unit, there
are two C—C chain bonds per repeat unit, which means that the total number of chain bonds in the molecule, N, is
€ bonds. Furthermore, assume that for single carbon-carbon bonds, d = 0.154 nm and θ =
just (2)(5000) = 10,000
109° (Section 14.4); therefore, from Equation 14.11
⎛θ ⎞
L = Nd sin ⎜ ⎟
⎝ 2 ⎠
⎡ ⎛ 109° ⎞⎤
= (10,000)(0.154 nm) ⎢sin ⎜
⎟⎥ = 1254 nm
⎣ ⎝ 2 ⎠⎦
It is now possible to calculate the average chain end-to-end distance, r, using Equation 14.12 as
r = d
N = (0.154 nm) 10,000 = 15.4 nm
4. 14.11 Sketch portions of a linear polystyrene molecule that are (a) syndiotactic, (b) atactic, and (c) isotactic.
Use two-dimensional schematics per footnote 8 of this chapter.
We are asked to sketch portions of a linear polystyrene molecule for different configurations (using twodimensional schematic sketches).
(a) Syndiotactic polystyrene
(b) Atactic polystyrene
(c) Isotactic polystyrene
5. Draw a typical conformation of an average atactic polystyrene molecule for average molecule
weights of: (a) 50,000 g/mole; and (b) 1500 g/mole. Note that different conformations can be
formed because the main chain of polystyrene can rotate without breaking the C-C sigma bonds.
6. 14.22 Explain briefly why the tendency of a polymer to crystallize decreases with increasing molecular weight.
The tendency of a polymer to crystallize decreases with increasing molecular weight because as the chains
become longer it is more difficult for all regions along adjacent chains to align so as to produce the ordered atomic
7. 15.15 Briefly explain how each of the following influences the tensile or yield strength of a semicrystalline
polymer and why:
(a) Molecular weight
(b) Degree of crystallinity
(c) Deformation by drawing
(d) Annealing of an undeformed material
(a) The tensile strength of a semicrystalline polymer increases with increasing molecular weight. This
effect is explained by increased chain entanglements at higher molecular weights.
(b) Increasing the degree of crystallinity of a semicrystalline polymer leads to an enhancement of the
tensile strength. Again, this is due to enhanced interchain bonding and forces; in response to applied stresses,
interchain motions are thus inhibited.
(c) Deformation by drawing increases the tensile strength of a semicrystalline polymer. This effect is due
to the highly oriented chain structure that is produced by drawing, which gives rise to higher interchain secondary
bonding forces. Moreover, drawing orients the polymer moleculaes, so the loading is increasingly along the main
chains rather than between the chains. Hence, the load is increwasingly carried by stronger intra-molecular primary
bonds rather than weaker inter-molecular secondary bonds.
(d) Annealing an undeformed semicrystalline polymer produces an increase in its tensile strength because
there would be an increase in crystallinity.
8. 15.32 Of those polymers listed in Table 15.2, which polymer(s) would be best suited for use as ice cube trays?
In order for a polymer to be suited for use as an ice cube tray it must have a glass-transition temperature
below 0°C. Of those polymers listed in Table 15.2 only low-density and high-density polyethylene, PTFE, and
polypropylene satisfy this criterion.
9. 15.39 Cite the primary differences between addition and condensation polymerization techniques.
For addition polymerization, the reactant species have the same chemical composition as the monomer
species in the molecular chain. This is not the case for condensation polymerization, wherein there is a chemical
reaction between two or more monomer species, producing the repeating unit. There is often a low molecular
weight by-product for condensation polymerization; such is not found for addition polymerization.
A. 7 Carbons, 2 Oxygens, 4 Hydrogens: 7 x 12 + 2 x 16 + 4 x 1 = 120 g/mole
B. nHBA = 280,00 g/mole/120 g/mole = 2333
nPE = 280,000 g/mole / 28 g/mole = 10,000 i.e. a molecule of PE would contain many more monomer units
compared to a molecule of poly(HBA).
C. We can expect that the mechanical properties for these two specimens would be different, because the higher
molecular weight sample would suffer many more entanglements than the lower molecular weight sample.
11. 6.4 A cylindrical specimen of a titanium alloy having an elastic modulus of 107 GPa (15.5 × 106 psi) and an
original diameter of 3.8 mm (0.15 in.) will experience only elastic deformation when a tensile load of 2000 N (450
lbf) is applied. Compute the maximum length of the specimen before deformation if the maximum allowable
elongation is 0.42 mm (0.0165 in.).
We are asked to compute the maximum length of a cylindrical titanium alloy specimen (before
deformation) that is deformed elastically in tension. For a cylindrical specimen
⎛ d0 ⎞ 2
A0 = π ⎜ ⎟
⎝ 2 ⎠
where d0 is the original diameter. Combining Equations 6.1, 6.2, and 6.5 and solving for l0 leads to
⎛ d 0 ⎞ 2
Δl Eπ ⎜ ⎟
Δl E
Δl Eπ d 02
⎝ 2 ⎠
l0 = =
= = σ
(0.42 × 10 −3 m)(107 × 10 9 N /m 2 ) (π)(3.8 × 10 −3 m) 2
= (4)(2000 N)
= 0.255 m = 255 mm (10.0 in.)
12. 6.25 Figure 6.21 shows the tensile engineering stress–strain behavior for a steel alloy.
(a) What is the modulus of elasticity?
(b) What is the proportional limit?
(c) What is the yield strength at a strain offset of 0.002?
(d) What is the tensile strength?
Using the stress-strain plot for a steel alloy (Figure 6.21), we are asked to determine several of its
mechanical characteristics.
(a) The elastic modulus is just the slope of the initial linear portion of the curve; or, from the inset and
using Equation 6.10
σ − σ1 (200 − 0) MPa
E = 2
= 200 × 103 MPa = 200 GPa ( 29 × 106 psi)
The value given in Table 6.1 is 207 GPa.
(b) The proportional limit is the stress level at which linearity of the stress-strain curve ends, which is
approximately 300 MPa (43,500 psi).
(c) The 0.002 strain offset line intersects the stress-strain curve at approximately 400 MPa (58,000 psi).
(d) The tensile strength (the maximum on the curve) is approximately 515 MPa (74,700 psi).
13. 7.1 To provide some perspective on the dimensions of atomic defects, consider a metal specimen that has a
dislocation density of 104 mm-2. Suppose that all the dislocations in 1000 mm3 (1 cm3) were somehow removed and
linked end to end. How far (in miles) would this chain extend? Now suppose that the density is increased to 1010
mm-2 by cold working. What would be the chain length of dislocations in 1000 mm3 of material?
The dislocation density is just the total dislocation length per unit volume of material (in this case per cubic
millimeters). Thus, the total length in 1000 mm3 of material having a density of 104 mm-2 is just
( 10 mm )(1000 mm ) = 10 mm = 10 m = 6.2 mi
Similarly, for a dislocation density of 1010 mm-2, the total length is
( 10 mm )(1000 mm ) = 10 mm = 10 m = 6.2 × 10 mi
14. 7.5 (a) Define a slip system.
(b) Do all metals have the same slip system? Why or why not?
(a) A slip system is a crystallographic plane, and, within that plane, a direction along which dislocation
motion (or slip) occurs.
(b) All metals do not have the same slip system. The reason for this is that for most metals, the slip system
will consist of the most densely packed crystallographic plane, and within that plane the most closely packed
direction. This plane and direction will vary from crystal structure to crystal structure.
15. 7.23 (a) From the plot of yield strength versus (grain diameter)–1/2 for a 70 Cu–30 Zn cartridge brass, Figure
7.15, determine values for the constants σ0 and ky in Equation 7.7.
(b) Now predict the yield strength of this alloy when the average grain diameter is 1.0 × 10-3 mm.
(a) Perhaps the easiest way to solve for σ0 and ky in Equation 7.7 is to pick two values each of σy and d-1/2
from Figure 7.15, and then solve two simultaneous equations, which may be created. For example
d-1/2 (mm) -1/2
σy (MPa)
The two equations are thus
= σ0 + 4 k y
= σ0 + 12 k y
Solution of these equations yield the values of
k y = 12.5 MPa(mm)
[1810 psi(mm) ]
σ0 = 25 MPa (3630 psi)
(b) When d = 1.0 × 10-3 mm, d-1/2 = 31.6 mm-1/2, and, using Equation 7.7,
σ y = σ0 + k y d
= (25 MPa) + 12.5 MPa(mm)
](31.6 mm
) = 420 MPa (61,000 psi)
16. 7.32 Experimentally, it has been observed for single crystals of a number of metals that the critical resolved
shear stress τcrss is a function of the dislocation density ρD as
τ crss = τ 0 + A ρ D
where τ0 and A are constants. For copper, the critical resolved shear stress is 2.10 MPa (305 psi) at a dislocation
density of 105 mm-2. If it is known that
€ the value of A for copper is 6.35 × 10 MPa-mm (0.92 psi-mm), compute the
τcrss at a dislocation density of 107 mm-2.
We are asked in this problem to compute the critical resolved shear stress at a dislocation density of 107
mm-2. It is first necessary to compute the value of the constant τ0 (in the equation provided in the problem
statement) from the one set of data as
τ 0 = τ crss − A ρ D
= 2.10 MPa − (6.35 × 10 −3 MPa -­‐ mm) 10 5 mm −2 = 0.092 MPa (13.3 psi)
Now, the critical resolved shear stress may be determined at a dislocation density of 107 mm-2 as
τ crss = τ 0 + A ρD
= (0.092 MPa) + (6.35 × 10 MPa -­‐ mm) 10 mm = 20.2 MPa (2920 psi)
A. From the 0.2% offset method, σy = 160
σT = 185 MPa (the maximum stress on
the stress-strain diagram)
% Elongation to failure = 2.5%
E = (160 – 0)/(0.01 – 0.002) = 20 GPa
(B) Indentation into cortical bone is less
than that into trabecular bone, because according to Figure 5, trabecular bone has a much lower yield strength than
cortical bone.
L = 2inch × (1+ (0.025 − 0.01)) = 2.03 inch , which corresponds to the total elongation minus the elastic
18. B
19. B
20. B
21. B
22. C
23. E
24. C
25. A
26. D
27. E