Understanding Materials Science Concepts
Concept Check 5.2
Question: What point defects are possible for MgO as an impurity in Al2O3? How many Mg2+ ions must be added to form each of these defects?
Answer: For every Mg2+ ion that substitutes for Al3+ in Al2O3, a single positive charge is removed. Thus, in order to maintain charge neutrality, either a positive charge must be added or a negative charge must be removed.
Positive charges are added by forming Al3+ interstitials, and one Al3+ interstitial would be formed for every three Mg2+ ions added.
Negative charges may be removed by forming O2- vacancies, and one oxygen vacancy would be formed for every two Mg2+ ions added.
Concept Check 5.3
Question: The surface energy of a single crystal depends on crystallographic orientation. Does this surface energy increase or decrease with an increase in planar density? Why?
Answer: The surface energy of a single crystal depends on the planar density (i.e., degree of atomic packing) of the exposed surface plane because of the number of unsatisfied bonds. As the planar density increases, the number of nearest atoms in the plane increases, which results in an increase in the number of satisfied atomic bonds in the plane, and a decrease in the number of unsatisfied bonds. Since the number of unsatisfied bonds diminishes, so also does the surface energy decrease. (That is, surface energy decreases with an increase in planar density.)
Concept Check 5.4
Question: Does the grain size number (n of Equation 5.19) increase or decrease with decreasing grain size? Why?
Answer: Taking logarithms of Equation 5.19 and then rearranging such that the grain size number n is the dependent variable leads to the expression
Thus, n increases with increasing N. But as N (the average number of grains per square inch at a magnification of 100 times) increases the grain size decreases. In other words, the value of n increases with decreasing grain size.
Chapter 6
Concept Check 6.2
Question: Consider the self-diffusion of two hypothetical metals A and B. On a schematic graph of ln D versus 1/T, plot (and label) lines for both metals given that D0(A) > D0(B) and also that Qd(A) > Qd(B).
Answer: The schematic ln D versus 1/T plot with lines for metals A and B is shown below.
As explained in the previous section, the intercept with the vertical axis is equal to ln D0. As shown in this plot, the intercept for metal A is greater than for metal B inasmuch as D0(A) > D0(B) [alternatively ln D0(A) > ln D0(B)]. In addition, the slope of the line is equal to –Qd/R. The two lines in the plot have been constructed such that negative slope for metal A is greater than for metal B, inasmuch as Qd(A) > Qd(B)
Chapter 7
Mechanical Properties
Concept Check 7.1
Questions: Of those metals listed in Table 7.3:
(a) Which will experience the greatest percent reduction in area? Why?
(b) Which is the strongest? Why?
(c) Which is the stiffest? Why?
Table 7.3 Tensile stress-strain data for several hypothetical metals to be used with Concept Checks 7.1 and 7.6
Yield Strength Tensile Strength Strain at Fracture Elastic Modulus
A 310 340 0.23 265 210
B 100 120 0.40 105 150
C 415 550 0.15 500 310
D 700 850 0.14 720 210
E Fractures before yielding 650 350
Answers:
(a) Material B will experience the greatest percent area reduction since it has the highest strain at fracture, and, therefore is most ductile.
(b) Material D is the strongest because it has the highest yield and tensile strengths.
(c) Material E is the stiffest because it has the highest elastic modulus.
Concept Check 7.2
Question: Make a schematic plot showing the tensile engineering stress–strain behavior for a typical metal alloy to the point of fracture. Now superimpose on this plot a schematic compressive engineering stress-strain curve for the same alloy. Explain any differences between the two curves.
Answer: The schematic stress-strain graph on which is plotted the two curves is shown below.
The initial linear (elastic) portions of both curves will be the same. Otherwise, there are three differences between the two curves, which are as follows:
(1) Beyond the elastic region, the tension curve lies below the compression one. The reason for this is that, during compression, the cross-sectional area of the specimen is increasing—that is, for two specimens that have the same initial cross-sectional area (A0), at some specific strain value the instantaneous cross-sectional area in compression will be greater than in tension. Consequently, the applied force necessary to continue deformation will be greater for compression than for tension; and, since stress is defined according to Equation 7.1 as
the applied force is greater for compression, so also will the stress be greater (since A0 is the same for both cases).
(2) The compression curve will not display a maximum inasmuch as the specimen tested in compression will not experience necking—the cross-sectional area over which deformation is occurring is continually increasing for compression.
(3) The strain at which failure occurs will be greater for compression. Again, this behavior is explained by the lack of necking for the specimen tested in compression.
Concept Check 7.3
Question: When citing the ductility as percent elongation for semicrystalline polymers, it is not necessary to specify the specimen gauge length, as is the case with metals. Why is this so?
Answer: The reason that it is not necessary to specify specimen gauge length when citing percent elongation for semicrystalline polymers is because, for semicrystalline polymers that experience necking, the neck normally propagates along the entire gauge length prior to fracture; thus, there is no localized necking as with metals and the magnitude of the percent elongation is independent of gauge length.
\
Concept Check 7.5
Question: An amorphous polystyrene that is deformed at 120C will exhibit which of the behaviors shown in Figure 7.26?
Answer: Amorphous polystyrene at 120C behaves as a rubbery material (Figure 7.29, curve C); therefore, the strain-time behavior would be as Figure 7.26(c).
Concept Check 7.6
Question: Of those metals listed in Table 7.3, which is the hardest? Why?
Answer: Material D is the hardest because it has the highest tensile strength.