Callister_Materials_Solutionsr

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Unit Cells 
Metallic Crystal Structures 
 3.2 If the atomic radius of lead is 0.175 nm, calculate the volume of its unit cell in cubic meters. 
 
 Solution 
 Lead has an FCC crystal structure and an atomic radius of 0.1750 nm (Table 3.1). The FCC unit cell 
volume may be computed from Equation 3.6 as 
 
 
 
 
 3.3 Show for the body-centered cubic crystal structure that the unit cell edge length a and the atomic 
radius R are related through . 
 
 Solution 
 This problem calls for a demonstration of the relationship for BCC. Consider the BCC unit cell 
shown below 
 
 
From the triangle NOP 
 
 
 
And then for triangle NPQ, 
 
 
 
But = 4R, R being the atomic radius. Also, = a. Therefore, 
 
 
 
And solving for a: 
 
 
 
 
 3.4 For the HCP crystal structure, show that the ideal c/a ratio is 1.633. 
 
 Solution 
 We are asked to show that the ideal c/a ratio for HCP is 1.633. A sketch of one-third of an HCP unit cell is 
shown below. 
 
Consider the tetrahedron labeled as JKLM, which is reconstructed as follows: 
 
 
The atom at point M is midway between the top and bottom faces of the unit cell--that is = c/2. And, since 
atoms at points J, K, and M, all touch one another, 
 
 
 
where R is the atomic radius. Furthermore, from triangle JHM, 
 
 
or 
 
 
Now, we can determine the length by consideration of triangle JKL, which is an equilateral triangle, 
 
 
From this triangle it is the case that the angle subtended between the lines and is 30°, and 
 
 
which reduces to the following: 
 
 
Substitution of this value for into the above expression yields 
 
 
And when we solve for c/a 
 
 
 
 
 3.5 Show that the atomic packing factor for BCC is 0.68. 
 
 Solution 
 The atomic packing factor is defined as the ratio of sphere volume (VS ) to the total unit cell volume (VC ), 
or 
 
 
Because there are two spheres associated with each unit cell for BCC 
 
 
 
Also, the unit cell has cubic symmetry, that is VC = a3. But a depends on R according to Equation 3.4, and 
 
 
Thus, 
 
 
 
 
 3.6 Show that the atomic packing factor for HCP is 0.74. 
 
 Solution 
 The APF is just the total sphere volume-unit cell volume ratio—i.e., (VS /VC ) . For HCP, there are the 
equivalent of six spheres per unit cell, and thus 
 
 
The unit cell volume (VC) for the HCP unit cell was determined in Example Problem 3.3 and given in Equation 3.7b 
as 
 
 
And because c = 1.633a = 2R(1.633) 
 
 
 
Thus, 
 
 
 
Density Computations 
 3.7 Molybdenum (Mo) has a BCC crystal structure, an atomic radius of 0.1363 nm, and an atomic weight 
of 95.94 g/mol. Compute and compare its theoretical density with the experimental value found inside the front 
cover of the book. 
 
 Solution 
 This problem calls for a computation of the density of molybdenum. According to Equation 3.8 
 
 
 
 
For BCC, n = 2 atoms/unit cell. Furthermore, because VC = a3, and (Equation 3.4), then 
 
 
 
Thus, realizing that AMo = 95.94 g/mol, using the above version of Equation 3.8, we compute the theoretical density 
for Mo as follows: 
 
 
 
 
 
= 10.22 g/cm3 
 
The value given inside the front cover is 10.22 g/cm3. 
 
 
 3.8 Strontium (Sr) has an FCC crystal structure, an atomic radius of 0.215 nm and an atomic weight of 
87.62 g/mol. Calculate the theoretical density for Sr. 
 
 Solution 
 According to Equation 3.8 
 
 
 
 
For FCC, n = 4 atoms/unit cell. Furthermore, because VC = a3, and (Equation 3.1), then 
 
 
 
Thus, realizing that ASr = 87.62 g/mol, using the above version of Equation 3.8, we compute the theoretical density 
of Sr as follows: 
 
 
 
 
 
= 2.59 g/cm3 
 
The experimental density for Sr is 2.54 g/cm3. 
 
 3.9 Calculate the radius of a palladium (Pd) atom, given that Pd has an FCC crystal structure, a density of 
12.0 g/cm3, and an atomic weight of 106.4 g/mol. 
 
 Solution 
 We are asked to determine the radius of a palladium atom, given that Pd has an FCC crystal structure. For 
FCC, n = 4 atoms/unit cell, and (Equation 3.6). Now, the density of Pd may be expressed using a 
form of Equation 3.8 as follows: 
 
 
 
 
 
Solving for R from the above expression yields 
 
 
 
 
Incorporation into this expression values for APd , n, and ρ leads to the following value of R: 
 
 
 
= 1.38 × 10-8 cm = 0.138 nm 
 
 
 3.10 Calculate the radius of a tantalum (Ta) atom, given that Ta has a BCC crystal structure, a density of 
16.6 g/cm3, and an atomic weight of 180.9 g/mol. 
 
 Solution 
 It is possible to compute the radius of a Ta atom using a rearranged form of Equation 3.8. For BCC, n = 2 
atoms/unit cell. Furthermore, because VC = a3 and (Equation 3.4) an expression for the BCC unit cell 
volume is as follows: 
 
 
 
For Ta, Equation 3.8 takes the form 
 
 
 
 
Solving for R in this equation yields 
 
 
Upon incorporation of values of n, ATa, and ρ into this equation leads to the atomic radius of Ta as follows: 
 
 
 
= 1.43 × 10−8 cm = 0.143 nm 
 
 
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	CHAPTER 10
	PHASE TRANSFORMATIONS
	PROBLEM SOLUTIONS
	Differentiation of this expression with respect to a is as
	If we set this expression equal to zero as
	Substitution of this expression for a in the above expression for G yields an equation for G* as
	For computation of the activation free energy, Equation 10.7 is employed. Thus
	Inasmuch as 4 atoms are associated with each FCC unit cell, the total number of atoms per critical nucleus is just
	Now for 200 K supercooling, it is first necessary to recalculate the value G* of using Equation 10.7, where, again, the T42Tm42T – T term is replaced by the number of degrees of supercooling, denoted as T, which in this case is 200 K. Thus
	And, from Equation 10.8, the value of n* is
	Now, for 300 K supercooling the value of G* is equal to
	We now want to manipulate Equation 10.17 such that t is the dependent variable. The above Equation 10.17a may be written in the form:
	And solving this expression for t leads to
	Now, using this equation and the value of k determined above, the time to 90% transformation completion is equal to
	Now, solving for t from this expression leads to
	Now, the rate is computed using Equation 10.18 as
	At this point we want to compute t41T0.541T, the value of t for y = 0.5, which means that it is necessary to establish a form of Equation 10.17 in which t is the dependent variable. Rearrangement of Equation 10.17b above leads to
	And solving this expression for t leads to
	And solving for t leads to
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	Using a modified form of Equation 16.2, the minimum thermal conductivity k40Tmin40T is
	And for the specific gravity
	Therefore, we compute the specific stiffness as follows:
	Or, when x = 1.25 mm (0.05 in.) and  = 0.90, then
	(a) The longitudinal tensile strength is determined using Equation 16.17 as follows:
	(b) The longitudinal elastic modulus is computed using Equation 16.10a as
	While transverse modulus expression is as follows:
	Solving these two expressions simultaneously for E41Tm41T and E41Tf41T leads to the following values:
	For elastic deformation, combining Equations 6.1 and 6.5 yields the following expression:
	But, we may write the following expression as a way of defining Vm:
	Vm = Am/Ac
	which, upon rearrangement gives
	Now, the volume fraction of fibers is equal to Vm = Am/Ac; upon substitution of this expression for V40Tm40T into Equation S16.11f leads to the following:
	Now, using values for F41Tf41T and F41Tm41T from the problem statement
	And substitution of the values for E41Tf41T and E41Tm41T given in the problem statement into the first equation above leads to
	And, solving for V41Tf41T yields, V41Tf41T = 0.258.
	(b) We are now asked for the tensile strength of this composite. The computation requires the use of Equation 16.17—viz.,
	Which means that F41Tf41T= 44.7F41Tm41T.
	Now, the stresses of each of the fiber and matrix phases are determined using Equation 6.1 as follows:
	Or, solving for V40Tf
	Also
	And, an expression for 41Tm41T analogous to the one for 41Tf41T given in Equation S16.14a above is as follows:
	From which we solve for F41Tm41T as follows:
	Inasmuch as l >> l41Tc41T (that is, 8.0 mm >> 0.18 mm), then use of Equation 16.17 is appropriate. Therefore,
	Inasmuch as l < lc (0.50 mm < 0.80 mm), then use of Equation 16.19 is required. Therefore,
	Solving this expression for V41Tf41T leads to V41Tf41T = 0.397.
	Solving this expression for K yields
	For Vf = 0.3
	And, for Vf = 0.4
	For the carbon-fiber reinforced epoxy
	And, for the aramid-fiber reinforced epoxy
	For the normalized 1040 plain carbon steel, the ratio is
	For the 7075-T6 aluminum alloy
	For the C26000 brass (cold worked)
	For the AZ31B (extruded) magnesium alloy
	For the annealed Ti-5Al-2.5Sn titanium alloy
	For the carbon-fiber reinforced epoxy
	And, for the aramid-fiber reinforced epoxy
	For the normalized 1040 plain-carbon steel
	For the 7075-T6 aluminum alloy
	For the cold worked C26000 brass
	For the extruded AZ31B magnesium alloy
	For the Ti-5Al-2.5Sn titanium alloy
	And
	Since, in general,  = /E (Equation 6.5), making substitutions of the form of Equation 6.5 into the previous expression yields the following:
	And, finally, taking the reciprocal of this equation leads to
	Because this value is less than the specified minimum (i.e., 55 GPa), glass is not an acceptable candidate.
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