CHAPTER 2 ATOMI ATOMIC C STRUCTURE STRUCTURE AND INTERAT INTERATOMI OMIC C BOND ING
2.3 2.3
(a) In order to determine the number of grams grams in in one amu of material, material, appropriate manipulati manipulation on of the amu/atom , g/mol, g/mol, and atom/mol r elationships is is all that is necessary necessary,, as #g/amu
=
=
1 mol 6.023 × 1023 atoms
1 g/mol 1 amu/atom
1.66 × 10−24 g/amu
2.14 2.14 (c) This port ion of the problem asks that we deter mine for a K + -Cl − ion pair the interato mic spacing spacing ( ro ) and the bonding energy ( E o ). From Equation (2.11) for E N A
=
1.436
B
=
5.86 × 10−6
n
=
9
Thus, Thus, using the solutions from Prob lem 2.13 2.13
ro
=
=
A
1/(1 −n)
nB
1/(1 −9)
1.436
(9)(5.86 × 10−6 )
=
0.279 nm
an d Eo
1.436
= −
1.436 (9)(5.86 × 10−6 )
= −4.57
1/(1 −9)
5.86 × 10−6
+
1.436 (9)(5.86 × 10−6 )
9/(1 −9)
eV
2.19 2.19 The per cent ionic character character is a function function o f the electrone gativities gativities of the ions X A an d to Equation (2.10). The electronegativities of the elements are found in Figure 2.7. 2.7. For TiO 2 , X Ti = 1.5 and X O = 3.5, and therefore, % IC
=
1 − e ( −0.25)(3.5−1.5)
1
2
×
100
=
63.2%
XB
according
CHAPTER 3 STRUCTURE STRUCTURE S OF METALS METALS A ND CERAMICS CERAMICS
3.3 3.3
For this problem, we are asked to calcul calculate ate the volume volume of a unit cell cell of aluminum. aluminum. Aluminum has an FCC crystal structure (Table 3.1). The FCC un it cell cell volume volume may be compu ted from Equation (3.4) as VC
3.7 3.7
= 16R 3
√
2
= (16)(0.143 × 10−9 m) 3
√
= 6.62 × 10−29 m 3
2
This This probl problem em cal calls ls for a demonstrati demonstration on that the A PF for HCP is 0.74. Again, the A PF is just the total sphere-unit cell volume ratio. For HCP, there are the equivalent of six spheres per unit cell, and thus VS
=6
4R 3 3
= 8R 3
Now, Now, the un it cell cell volume volume is just just the produ ct of the ba se area t imes the cell height, height, c. This base area is just three times the area of the parallelepiped ACDE shown below.
D
C
a = 2R 30 60
A
E B a = 2R
a = 2R
The area of ACDE is just the length of CD times the height BC. BC . But CD is just a or 2R , an d BC
= 2R cos(30◦ ) =
2R
√
3
2
Thus, Thus, the base ar ea is just
AREA and since c
= (3)(CD)(BC) = (3)(2R)
2R
√ 3
2
= 6R 2
√
3
= 1.633a 1.633a = 2R (1.633)
VC
= (AREA)(c) = 6R 2c
√
3
= (6R 2 2
√
3)(2)(1.633)R
√ = 12 3(1 .633)R 3
Thus,
A PF
8R 3
VS
= V = √
12 3(1 .633)R 3
C
= 0.74
3.12 3.12.. (a) This This portion of the problem asks that we we compute compute the volume volume of the unit cell cell for Zr. This This volume volume may be computed using Equation (3.5) as as VC
Zr = nA N A
Now, for HCP, n
= 6 atoms/unit cell, and for Zr, A Zr = 91.2 91.2 g/mo l. Thu s,
VC
(6 at oms/unit cell)(91.2 g/mol)
=
(6 .51 g/cm 3 )( 6.023
× 1023 atoms/mol) = 1.396 × 10−22 cm 3 /un it cell = 1.396 × 10−28 m 3 /un it cell
(b) We are now to compute the value valuess of a an d c, given that c/a Prob lem 3.7, 3.7, since a 2R , then, for for H CP
=
VC but, since since c
= 1.593. From the solution to
√
=
3 3a 2 c 2
= 1.593a 593a VC
√
=
3 3(1 .593)a 3 2
= 1.396 × 10−22 cm 3 /un it cell
Now, solving solving for a
a
=
1/3 × 10−22 cm 3 ) √
(2)(1.396
(3)( 3)(1.593)
= 3.23 × 10−8 cm = 0.323 nm And finally c
= 1.593a = (1 .593)(0.323 nm) = 0.515 nm
3.17 3.17 In this problem we are given given that iodine has an orthorhombic unit unit cell cell for whic which h the a, b, and c lattice parameters are 0.479, 0.725, and 0.978 nm, respectively. (a) Given that the atomic packing factor and atomic radius are 0.547 and 0.177 nm, respectively we are to determine the number of atoms in each unit cell. From the definition of the A PF
A PF
n
VS
=V = C
3
4
R 3 ab c
3
we may solve for the number of atoms per unit cell, n, a s
n
= (APF)abc 4 3
=
R 3
(0 .547)(4.79)(7.25)(9.78)(10−24 cm 3 ) 4 (1 .77 10−8 cm) 3 3
×
= 8.0 atom s/unit cell (b) In order to compute the density, we just employ Equation (3.5) as
nA I = abcN
A
(8 a toms/unit cell)(126.91 g/mol) = [(4 .79)(7.25)(9.78) × 10−24 cm 3 /un it cell]( 6.023 × 1023 atoms/mol)
= 4.96 g/cm 3 3.22 This question asks that we generate a three-dimensional unit cell for AuCu 3 using the Molecule D efinition File on the CD-ROM. One set of directions that may be used to construct this unit cell and that are entered on the No tepad are as follows: [DisplayProps] Rotatez 30 Rotatey 15
=− =−
[AtomProps] Gold LtRed,0.14 Copper LtYellow,0.13
=
=
[BondProps] SingleSolid LtGray
=
[Atoms] Au 1 1,0,0,Gold Au 2 0,0,0,Gold Au 3 0,1,0,Gold Au 4 1,1,0,Gold Au 5 1,0,1,Gold Au 6 0,0,1,Gold Au 7 0,1,1,Gold Au 8 1,1,1,Gold Cu 1 0.5,0,0.5,Copper Cu 2 0,0.5,0.5,Copper Cu 3 0.5,1,0.5,Copper Cu 4 1,0.5,0.5,Copper Cu 5 0.5,0.5,1,Copper Cu 6 0.5,0.5,0,Copper
= = = = = = = = = = = = = =
4
[Bonds] B1 Au1,Au5,SingleSolid B2 Au5,Au6,SingleSolid B3 Au6,Au2,SingleSolid B4 Au2,Au1,SingleSolid B5 Au4,Au8,SingleSolid B6 Au8,Au7,SingleSolid B7 Au7,Au3,SingleSolid B8 Au3,Au4,SingleSolid B9 Au1,Au4,SingleSolid B10 Au8,Au5,SingleSolid B11 Au2,Au3,SingleSolid B12 Au6,Au7,SingleSolid
= = = = = = = = = = = =
When saving the se instructions, the file name that is chosen should end with a period followed by mdf and the entire file name needs to be enclosed within quotation marks. For example, if one wants to name the file AuCu3, the name by which it should be saved is “AuCu3.mdf ” . In addition, th e file should be saved a s a “Text Document.” 3.27 In this problem we are asked to show that the minimum cation-to-anion radius ratio for a coordination number of six is 0.414. Below is shown one of the faces of the rock salt crystal structure in which a nions and cations just t ouch along t he e dges, and also the face diagonals.
r A
r C
H
G
F
From triangle FGH, GF
= 2r A
and
FH
= G H = rA + rC
Since FGH is a right t riangle ( G H )2
+ (FH)2 = (FG)2
or (r A
+ r C ) 2 + (r A + r C ) 2 = (2r A ) 2
which leads to rA
+ r C = 2r√ A
2
5
Or , solving for rC /rA rC rA
2
= √ − 1 = 0.414 2
3.29 This prob lem calls for us to predict crystal structure s for several ceramic materials on the basis of ionic charge a nd ionic radii. (a) For CsI, from Table 3.4 r Cs + r I−
nm = 00..170 = 0.773 220 nm
Now, from Table 3.3, the coordination number for each cation (Cs+ ) is eight, and, u sing Table 3.5, the predicted crystal structure is cesium chloride. (c) For KI, from Table 3.4 r K+ r I−
nm = 00..138 = 0.627 220 nm
The coordination number is six (Table 3.3), and the predicted crystal structure is sodium chloride (Table 3.5). 3.36 This problem asks that we compute the theoretical density of diamond given that the C ––C distance and bond angle are 0.154 nm and 109.5 ◦ , respectively. The first thing we need do is to determine the unit cell edge length from the given C ––C distance. The drawing below shows the cubic unit cell with those carbon atoms that bond to one another in one-quarter of the unit cell.
a
y
θ
φ x
From this figure, is one-half of the bond angle or
= 109.5◦ /2 = 54.75◦ , which means that
= 90◦ − 54.75◦ = 35.25◦
since the triangle shown is a right triangle. Also, y Furthermore, x a/4, and therefore,
=
x
= 0.154 nm, the carbon-carbon bond distance.
= a4 = ysin
6
Or a
= 4 y sin = (4)(0.154 nm)( sin 35.25◦ ) = 0.356 nm = 3.56 × 10−8 cm
The u nit cell volume, V C , is just a3 , that is VC
= a 3 = (3 .56 × 10−8 cm) 3 = 4.51 × 10−23 cm 3
We must now utilize a modified Equation (3.6) since there is only one atom type. There are eight equivalent atoms per unit cell (i.e., one equivalent corner, three equivalent faces, and four interior atoms), and therefore
= Vn ANC C
A
cell)(12.01 g/g-atom) = (4 .51 × 10−(823atcmoms/unit 3 /un it cell) (6 .023 × 1023 atoms/g-atom)
= 3.54 g/cm 3
The measured density is 3.51 g/cm 3 . 3.39 (a) We are asked to compute the den sity of CsCl. Modifying the result of Prob lem 3.4, we get a
+ 2r Cl− = 2(0 .170 nm)√ + 2(0 .181 nm) = 2r Cs+√ 3
3
= 0.405 nm = 4.05 × 10−8 cm From Equation (3.6)
=
n (A Cs
For th e CsCl crystal structure, n
+ A Cl ) = n (A Cs + A Cl ) a3NA
VC NA
= 1 formula unit/unit cell, and thus
unit/unit cell)(132.91 g/mol + 35.45 g/mol) = (4 .(105formula × 10−8 cm) 3 /un it cell( 6.023 × 1023 formula u nits/mol)
= 4.20 g/cm 3
(b) This value of the density is greater than the measured density. The reason for this discrepancy is that th e ionic radii in Table 3.4, used for this computation, were for a coord ination numb er of six, when, in fact, the coordination number of both Cs+ and Cl− is eight. Under these circumstances, the actual ionic radii and unit cell volume ( V C ) will be slightly greater than calculated values; consequently, the measured density is smaller tha n t he calculated den sity. 3.45 We are asked in this problem to comput e the atomic packing factor for the CsCl crystal structure. This requires that we take the ratio of the sphere volume within the unit cell and the total unit cell volume. From Figure 3.6 there isth e equivalent of one Cs and one Cl ion per unit cell; the ionic radii of these two ions are 0.170 nm and 0.181 nm, respectively (Table 3.4). Thus, the sphere volume, V S ,
7
is just VS
= 43 ( )[(0.170 nm) 3 + (0 .181 nm) 3] = 0.0454 n m 3
For CsCl the u nit cell edge length, a, in terms of th e ato mic radii is just a
+ 2r Cl− = 2(0 .170 nm)√ + 2(0 .181 nm) = 2r Cs+√ 3
3
= 0.405 nm Since V C
= a3 VC
= (0 .405 nm) 3 = 0.0664 n m 3
And, finally the a tomic packing factor is just A PF
VS
0.0454 nm 3
= V = 0.0664 nm 3 = 0.684 C
3.50 (a) We are asked for the indices of the two directions sketched in the figure. For direction 1, the projection on the x-axis is zero (since it lies in the y-z plane), while pro jections on t he y- and z-axes ar e b/2 and c, respectively. This is an [012] direction as indicated in the summary below.
Projections Projections in terms of a, b, and c R eduction to integers E nclosure
x
y
z
0a 0 0
b /2 1/2 1 [012]
c 1 2
3.51 This problem asks for us to sketch several directions within a cubic unit cell. The [110], [121], and [012] directions are indicated below. _ [012]
z _ _ [121]
_ [110]
y x 3.53 This prob lem asks that we deter mine indices for several directions that have been drawn within a cubic unit cell. Direction B is a [232] direction, the determination of which is summarized as follows.
8
We first of all position the origin of the coordinate system at the tail of the direction vector; then in term s of this new coordinate system x 2a
Projections Projections in terms of a, b, and c R eduction to integers
3 2 3 2
E nclosure
y
−b −1 −3
z 2c 3 2 3 2
[232]
Direction D is a [136] direction, the determination of which is summarized as follows. We first of all position the origin of the coordinate system at the tail of the direction vector; then in terms of this new coordinate system
Projections Projections in terms of a, b, and c R eduction to integers
x
y
a
b
6 1
2 1
6 1
2 3
E nclosure
[136]
z
−c −1 −6
3.56 This prob lem asks that we deter mine the Miller indices for planes that have been drawn within a unit cell. For plane B we will move the origin of the unit cell one unit cell distance to the right along the y axis, and one unit cell distance parallel to the x axis; thus, this is a (112) plane, as summarized below. x
y
Intercepts
−a
−b
Intercepts in term s of a, b, and c
−1 −1
−1 −1
Reciprocals of intercepts E nclosure
z c 2 1 2 2
(112)
3.58 For plane B we will leave the origin a t t he unit cell as shown; this is a (221) plane, as summarized below.
Intercepts Intercepts in term s of a, b, and c R eciprocals of intercepts E nclosure
x
y
a
b
2 1
2 1
2 2
2 2 (221)
9
z c 1 1
3.59 The (1101) plane in a hexagonal unit cell is shown below. z
a 2
a
3
a _ (1101)
1
3.60 This problem asks that we specify the Miller indices for planes that have been dr awn within hexagonal unit cells. (a) For this plane we will leave the origin of the coor dinate system as shown; thus, this is a (1100) plane, as summar ized below.
Intercepts Intercepts in terms of a’s and c R ecipr ocals of inter cepts E nclosure
a1
a2
a3
z
a 1 1
−a −1 −1
∞a ∞
∞c ∞
0 (1100)
0
3.61 This problem asks for us to sketch several planes within a cubic unit cell. The (011) and (102) planes are indicated below. z
_ _ (011)
y
x
_ (102)
10
3.63 This problem asks that we represent specific crystallographic planes for various ceramic crystal structures. (a) A (100) plane for the rock salt crystal structure would appear as
+ Na Cl
3.64 For the unit cell shown in Problem 3.21 we are asked to determine, from three given sets of crystallographic planes, which are equivalent. (a) The unit cell in Prob lem 3.21 is body-center ed te tragonal. On ly the ( 100) (front face) and (010) (left side face) p lanes are equivalent since th e dimensions of these p lanes within th e un it cell (and therefore the distances between adjacent atoms) are the same (namely 0.40 nm 0.30 nm ), which are different than the (001) (top face) plane (namely 0.30 nm 0.30 nm).
×
×
3.66 This question is concerned with the zinc blende crystal structure in ter ms of close-packed planes of anions. (a) The stacking sequence of close-packed planes of anions for the zinc blende crystal structure will be the same as FCC (and not HCP) because the anion packing is FCC (Table 3.5). (b) The cations will fill tetrahedral positions since the coordination number for cations is four (Table 3.5). (c) Only one-half of the tetr ahedral positions will be occupied because there are two tetrahedral sites per anion, and yet only one cation per anion. 3.70* In this problem we ar e to compute the linear densities of several crystallographic planes for the face-centered cubic crystal structure. For FCC t he linear density of the [100] direction is computed as follows: The linear density, LD , is defined by the ratio
= LLc
LD
l
where Ll is the line length within the unit cell along the [100] direction, and Lc is line length passing through intersection circles. Now, Ll is just the unit cell edge length, a which, for FCC is related to the atomic radius R according t o a 2R 2 [Equation (3.1)]. Also for this situation, Lc 2R an d therefore
=
LD
√
=
=
2R 2R
√ = 0.71
11
2
3.73* In this problem we are to compute the planar densities of several crystallographic planes for the body-centered cubic crystal structure. Planar density, PD , is defined as PD
= AA c
p
where A p is the total plane area within the unit cell and A c is the circle plane a rea within th is same plane. For (110), that portion of a plane that passes through a BCC unit cell forms a rectangle as shown below.
4R
R
3
4R 2 3
√ √ In terms of the atomic radius R, the length of the rectangle base is whereas the height is 4R a = √ . Therefore, the area of this rectangle, which is just A p is 4R 2 , 3
3
Ap
=
√ √ 2 4R 16R 2 2 √ √ = 3
4R
3
3
Now for the number equivalent atoms within this plane. One-fourth of each corner atom and the entirety of the center atom belong to the unit cell. Therefore, there is an equivalent of 2 atoms within th e unit cell. Hen ce Ac
= 2( R 2)
an d PD
=
2R 2
√ = 0.83
16R 2 2 3
3.80* Using the data for aluminum in Table 3.1, we are asked to compute the interplanar spacings for the (110) and (221) sets of planes. From the table, aluminum has an FCC crystal structure and an atomic r adius of 0.1431 nm. U sing Equation (3.1) the lattice par ameter, a, may be computed as a
= 2R
√
2
= (2)(0.1431 nm)( 12
√
2)
= 0.4047 nm
Now, the d110 interplanar spacing may be determined using Equation (3.11) as d 110
=
(1) 2
a
+ (1) 2 + (0)
0.4047 nm = √ = 0.2862 nm 2 2
3.84* From the diffraction pattern for -iron shown in Figure 3.37, we are asked to compute the interplanar spacing for each set of planes that has been indexed; we are also to determine the lattice parameter of Fe for each peak. In order to compute the interplanar spacing and the lattice parameter we must employ E quations ( 3.11) an d (3.10), respectively. For the first peak which occurs at 45.0◦ d 110
n (1)(0.1542 nm) 45.0◦ = 0.2015 nm = 2sin = (2) sin
2
A nd a
= d hkl (h ) 2 + (k )2 + (l) 2 = d 110 √ = (0 .2015 nm ) 2 = 0.2850 nm
(1) 2
+ (1) 2 + (0) 2
Similar computations are made for the other peaks which results are tabulated below: Peak Index
2
d hkl (nm)
a (nm)
200 211
65.1 82.8
0.1433 0.1166
0.2866 0.2856
13
CHAPTER 4 POLYMER STRUCTUR ES
4.4
We are asked to comput e the numbe r-average degree of polymerization for polypropylene, given that the n umber-average molecular weight is 1,000,000 g/mol. The mer molecular weight of polypropylene is just m
If we let
nn
= 3(A C ) + 6(A H ) = (3)(12.01 g/mol) + (6)(1.008 g/mol) = 42.08 g/mol
represent the number-average degree of polymerization, then from Equation (4.4a) nn
4.6
=
Mn m
106 g/mol
= 42.08 g/mol = 23,700
(a) From the tabulated data, we are asked to compute This is carried out below.
Mn ,
the number-average molecular weight.
Molecular wt R ange
Mean M i
xi
xi M i
8,000–16,000 16,000–24,000 24,000–32,000 32,000–40,000 40,000–48,000 48,000–56,000
12,000 20,000 28,000 36,000 44,000 52,000
0.05 0.16 0.24 0.28 0.20 0.07
600 3200 6720 10,080 8800 3640
Mn
= xi M i = 33,040 g/mol
(c) Now we are asked to compute nn (the number-average degree of p olymerization), using the Equation (4.4a). For po lypropylene, m
= 3(A C ) + 6(A H ) = (3)(12.01 g/mol) + (6)(1.008 g/mol) = 42.08 g/mol
A nd nn
g/mol = Mmn = 33040 = 785 42.08 g/mol
4.11 This problem first of all asks for us to calculate, using Equat ion (4.11), 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 numbe r-average de gree of po lymerization, nn , using Equation (4.4a). For PTFE, from Table 4.3, each mer unit has two carbo ns and four fluorines. Thus, m
= 2(A C ) + 4(A F ) = (2)(12.01 g/mol) + (4)(19.00 g/mol) = 100.02 g/mol
an d nn
g/mol = Mmn = 500000 = 5000 100.02 g/mol
14
which is the number of mer units along an average chain. Since there are two carbon atoms per mer unit, there are two C ––C chain bonds per mer, which means that the total number of chain bonds in the molecule, N , is just (2)(5000) 10,000 bonds. Furt hermo re, assume that for single carbo n-carbon bonds, d 0.154 nm and 109◦ (Section 4.4); therefore, from Equation (4.11)
=
=
=
L
= Nd sin
2
= (10,000)(0.154 nm)
◦ 109
sin
2
= 1254 nm
It is now possible to calculate the average chain end-to-end distance, tion (4.12) as r
=d
√
N
= (0 .154 nm)
√
10000
r,
using Equa-
= 15.4 n m
4.19 For a poly(styrene-butadiene) alternating copolymer with a number-average molecular weight of 1,350,000 g/mol, we are asked to determine the average number of styrene and butadiene mer units per molecule. Since it is an alternating copolymer, the number of both types of mer units will be the same. Therefore, consider them as a single mer unit, and determine the number-average degree of polymerization. For the styrene mer, there are eight carbon atoms and eight hydrogen atoms, while the butadiene mer consists of four carbon atoms and six hydrogen atoms. Therefore, the styrenebutadiene combined mer weight is just m
= 12(A C ) + 14(A H ) = (12)(12.01 g/mol) + (14)(1.008 g/mol) = 158.23 g/mol
From Equation (4.4a), the numbe r-average de gree of po lymerization is just nn
g/mol = Mmn = 1350000 = 8530 158.23 g/mol
Thus, there is an average of 8530 of both mer types per m olecule. 4.28 G iven that polyethylene has an orthor hombic unit cell with two equivalent mer units, we are asked to compute the density of totally crystalline polyethylene. In orde r t o solve this problem it is necessary to employ Equation (3.5), in which n repre sents the numb er of mer u nits within the unit cell (n 2), an d A is the mer molecular weight, which for polyethylene is just
=
A
= 2(A C ) + 4(A H ) = (2)(12.01 g/mol) + (4)(1.008 g/mol) = 28.05 g/mol
Also, V C is the unit cell volume, which is just the product of the three unit cell edge lengths in Figure 4.10. Thus,
= VnAN C
A
ers/uc)(28.05 g/mol) = (7 .41 × 10−8 cm)(4.94 × 10(2−8mcm)(2 .55 × 10−8 cm)/uc(6.023 × 1023 mers/mol)
= 0.998 g/cm 3
15
CHAPTER 5 IMPERFECTIONS IN SOLID S
5.1
In order to compute the fraction of atom sites that are vacant in lead at 600 K, we must employ Equation (5.1). As stated in the problem, Q V 0.55 eV/atom. Thus,
=
NV N
QV
0.55 eV/atom
= exp − kT = exp − (8 .62 × 10−5 eV/atom-K)(600 K)
= 2.41 × 10−5 5.4
This problem calls for a det ermination of the number of atoms per cubic meter of aluminum. In order to solve this problem, one must employ Equation (5.2),
N
= NAA A l Al
The density of Al (from the table inside of the front cover) is 2.71 g/cm 3 , while its ato mic weight is 26.98 g/mol. Thus, 23
N
= (6 .023 × 10
atoms/mol)(2.71 g/cm 3 )
26.98 g/mol
= 6.05 × 1022 atoms/cm 3 = 6.05 × 1028 atoms/m3 5.9
In the d rawing below is shown the atoms on the ( 100) face of an FCC unit cell; the interstitial site is at the center of the edge.
R
R 2r
a
The diameter o f an atom that will just fit into this site ( 2r) is just the difference between that unit cell edge length ( a) and the radii of the two host atoms that are located on either side of the site ( R );
16
that is 2r
= a − 2R
However, for FCC a is related to R according t o Equation (3.1) as a r gives a
− 2R = 2R r= 2
= 2R
√
2; therefore, solving for
√
2
− 2R = 0.41R 2
5.10 (a)For Li+ substituting for Ca 2+ inCaO, oxygen vacancieswould be created.For each Li+ substituting for Ca 2+ , one positive charge is removed; in order to maintain charge neutrality, a single negative charge may be removed. Negative charges are eliminated by creating oxygen vacancies, and for every two Li+ ions added , a single oxygen vacancy is formed. 5.15 This problem asks that we dete rmine th e composition, in ato m per cent, of an alloy that contains 98 g tin and 65 g of lead. The concentration of an element in an alloy, in atom percent, may be computed using Equation (5.5). With this problem, it first becomes necessary to compute the number of moles of both Sn and Pb, for which Equation (5.4) is employed. Thus, the numbe r o f moles of Sn is just m Sn
98 g = A = 118.69 = 0.826 mol g/mol
n m Sn
Sn
Likewise, for Pb n m Pb
g = 207.65 = 0.314 mol 2 g/mol
Now, use of Equation (5.5) yields C Sn
=n
n m Sn m Sn
+ n m × 100 Pb
0.826 mol = 0.826 mol + 0.314 mol × 100 = 72.5 at% Also, C Pb
0.314 mol = 0.826 mol + 0.314 mol × 100 = 27.5 at%
5.27 This problem asks us to determine the weight percent of Nb that must be added to V such that the resultant alloy will contain 1 .55 1022 Nb atoms per cubic centimeter. To solve this problem, employment of Equation (5.18) is necessary, using the following values:
×
= N Nb = 1.55 × 1022 atoms/cm 3 3 1 = Nb = 8.57 g/cm 3 2 = V = 6.10 g/cm A 1 = A Nb = 92.91 g/mol A 2 = A V = 50.94 g/mol N1
17
Thus C Nb
=
100 1
+ NN AAV − V Nb
=
Nb
Nb
100 23
6.10 g/cm3 × 20 atoms/mole)(6 .10 g/cm 3 ) 1+ − 8.57 g/cm 3 (1 .55 × 1022 atoms/cm 3 )(92.91 g/mol) (6 .023
= 35.2 wt% 5.30 In this problem we are given a general equation which may be used to d eterm ine the Bur gers vector and are asked to give Burgers vector representations for speci fic crystal structures, and then to compute Burgers vector magnitudes. (a) The Burgers vector will point in that direction having the highest linear density. From Problem 3.70 the linear density for the [110] direction in FCC is 1.0, the maximum possible; ther efore for FCC b
= a2 [110]
(b) For Al which has an FCC crystal structure, R [Equation (3.1)]; therefore b
= a2 =
h2
= 0.1431 nm (Table 3.1) an d a = 2R
√
2
= 0.4047 nm
+ k 2 + l2
0.4047 n m 2
(1) 2
+ (1) 2 + (0) 2 = 0.2862 nm
5.37 (a) We are asked for the number of grains per square inch ( N ) at a magnification of 100X, and for an A STM grain size of 4. From Equation (5.16), n 4, and
=
N
= 2(n −1) = 2(4−1) = 23 = 8
18
CHAPTER 6 DIFFUSION
6.8
This prob lem calls for computat ion of the diffusion coefficient for a steady-state d iffusion situation. Let us first convert the carbon concentrations from wt% to kg C/m3 using Equation (5.9a); th e densities of carbon and iron (from inside the front cover of the book) are 2.25 and 7.87 g/cm 3 . For 0.012 wt% C
C C
= =
CC CC C
C Fe
+
Fe
×
103
0.012 0.012
99.988
+ 7.87 g/cm 3
2.25 g/cm 3
= 0.944 kg C/m3
×
103
Similarly, for 0.0075 wt% C
C C
=
0.0075 0.0075 99.9925
+ 7.87 g/cm 3
2.25 g/cm 3
= 0.590 kg C/m3
×
103
Now, using a form of Equation (6.3)
D
=− J
xA
− xB CA − CB
= −(1 .40 × 10−8 kg/m2-s)
−10−3 m 0.944 kg/m 3 − 0.590 kg/m 3
= 3.95 × 10−11 m 2 /s 6.13 This problem asks us to compute the nitrogen concentration ( Cx ) at the 1 mm position after a 10 h diffusion time, when diffusion is non steady-state. Fro m Equation (6.5)
Cx
− C o = C x − 0 = 1 − er f √ x Cs − C o 0.1 − 0 2 Dt
= 1 − er f
10−3 m (2)
(2 .5
× 10−11 m 2 /s)( 10 h )( 3600 s/h )
= 1 − erf(0.527)
19
Using data in Table 6.1 and linear interpolation z
erf(z)
0.500 0.527 0.550
0.5205 y 0.5633
0.527
− 0.500 = y − 0.5205 0.550 − 0.500 0.5633 − 0.5205 from which y
= erf(0.527) = 0.5436
Thus, Cx
− 0 = 1.0 − 0.5436 0.1 − 0 This expression gives Cx
= 0.046 wt% N
6.15 This problem calls for an estimate of the time necessary to achieve a carbon concentration of 0.45 wt% at a point 5 mm from the surface. From Equat ion (6.6b), x2 Dt
= constant
But since the temperature is constant, so also is D constant, and x2 t
= constant
or x21 t1
=
x22 t2
Thus, (2 .5 mm) 2 10 h
= (5 .0 tmm)
2
2
from which t2
= 40 h
6.21 (a) Using Equation (6.9a), we set up two simultaneous equations with Q d an d D o as unknowns. Solving for Q d in terms of temperatures T1 an d T2 (1273 K and 1473 K) and D 1 an d D 2 (9 .4 10−16 and 2.4 10−14 m 2 /s), we get
×
×
Qd
ln D 2 = −R ln1/DT1 − − 1/T 1
2
.4 × 10−16 ) − ln(2.4 × 10−14 )] = − (8 .31 J/mol-K)[ln(9 1/(1273 K) − 1/(1473 K)
= 252,400 J/mol
20
Now, solving for D o from Equation (6.8) Do
= D 1 exp
Qd
RT 1
= (9 .4 × 10−16 m 2 /s) e xp
252400 J/mol (8 .31 J/mol-K)(1273 K)
= 2.2 × 10−5 m 2 /s (b) Using these values of D o an d Q d , D at 1373 K is just D = (2 .2 × 10−5 m 2 /s) e xp
−
252400 J/mo l (8 .31 J/mol-K)(1373 K)
= 5.4 × 10−15 m 2 /s 6.29 For this problem, a d iffusion couple is prepa red using two hypothe tical A and B metals. A fter a 30-h heat treatment at 1000 K, the concentration of A in B is 3.2 wt% at the 15.5-mm position. After another heat treatment at 800 K for 30 h, we are to determine at what position the composition will be 3.2 wt% A . In order to make this determination, we must employ Equation (6.6b) with t constant. That is x2
= constant
D Or
x2800 D 800
x21000
=D
1000
It is necessary to compute both D 800 an d D 1000 using Equation (6.8), as fo llows: D 800
= (1 .8 × 10−5 m 2 /s) e xp
−
152000 J/mo l (8 .31 J/mol-K)(800 K)
= 2.12 × 10−15 m 2 /s 152000 J/mo l D 1000 = (1 .8 × 10−5 m 2 /s) e xp − (8 .31 J/mol-K)(1000 K)
= 2.05 × 10−13 m 2 /s Now, solving for x800 yields
x800
= x1000
D 800 D 1000
= (15 .5 mm)
× 10−15 m 2 /s 2.05 × 10−13 m 2 /s 2.12
= 1.6 m m
21
CHAPTER 7 MECHA NICAL PROPERTIES
7.4
We are asked to compute the maximum length of a cylindrical titanium alloy specimen that is deformed elastically in tension. For a cylindrical specimen
Ao
=
do
2
2
where do is the original diameter. Combining Equations (7.1), (7.2), an d (7.5) and solving for lo leads to lo
=
=
=
7.9
E d 2o l 4F (107 × 109 N/m 2 )( )( 3.8 × 10−3 m) 2 (0 .42 × 10−3 m) (4)(2000 N) 0.25 m
250 mm (10 in .)
=
This problem asks that we calculate the elongation l of a specimen of steel the stress-strain behavior of which is shown in Figure 7.33. First it becomes necessary to compute the stress when a load of 23,500 N is applied as
=
F Ao
F
=
2
do
23500 N
=
2
10 × 10−3 m 2
2
=
300 MPa (44,400 psi)
Referring to Figure 7.33, at this stress level we are in the elastic region on the stress-strain curve, which corresponds to a strain of 0.0013. Now, utilization of Equation (7.2) yields l
= ε lo =
(0 .0013)(75 mm)
=
0.10 mm (0.004 in .)
7.14 (a) We are asked, in this portion of the problem, to determine the elongation of a cylindrical specimen of aluminum. U sing Equations (7.1), (7.2), an d (7.5) F
d 2o
=
E
l
lo
4
Or l
=
4Fl o
=
d 2o E (4)(48,800 N)(200 × 10−3 m)
( )(19 × 10−3 m) 2 (69
×
109 N/m 2 )
=
0.50 mm (0.02 in .)
(b) We are now called upon to determine the change in diameter, d. Using Equation (7.8)
=−
εx
= −
εz
22
d /d o l/lo
From Table 7.1, for Al,
=
d
0.33. Now, solving for d yields
= −
ld o
= −
lo
= −1.6 ×
(0 .33)(0.50 mm)(19 mm) 200 mm
10−2 mm ( −6.2 × 10−4 in .)
The d iameter will decrease. 7.16 This problem asksthat we compute Poisson’s ratio for the metal alloy. From Equations(7.5) an d (7.1) F /A o
εz
=
E
=
F
=
E
Since the tran sverse strain
εx
do 2
4F
=
2
E
d 2o E
is just
εx
d
=
do
and Poisson’s ratio is defined by Equation (7.8) then
= −
εx
= −
εz
= −
(8
×
d /d o
4F
d 2o E
=−
d o d E 4F
10−3 m) ( −5 × 10−6 m) ( )(140 × 109 N/m 2 ) (4)(15,700 N)
=
0.280
7.21 (a) This portion of the pro blem asks that we compute the elongation of the brass specimen. The first calculation ne cessary is that of the applied stress using Equation (7.1), as
=
F Ao
F
=
2
do 2
5000 N
=
6 × 10−3 m 2
2
=
177 MPa (25,000 psi)
From the stress-strain plot in Figure 7.12, this stress corresponds to a strain of about 2 .0 × 10−3 . From the definition of strain, Equation (7.2), l
= ε lo =
(2 .0 × 10−3 )(50 mm)
=
0.10 mm (4 × 10−3 in .)
(b) In order to determine the reduction in diameter d, it is necessary to use Equation (7.8) an d the definition of lateral strain ( i.e., εx = d/do ) as follows: d
=
d o εx
= −d o εz = −(6
= −3.6 ×
mm)(0.30)(2.0 × 10−3 )
10−3 mm ( −1.4 × 10−4 in .)
7.27 This problem asks us to determine the d eformation character istics of a steel specimen, the stressstrain beh avior of which is shown in Figure 7.33. (a) In order to ascertain whether the deformation is elastic or plastic, we must first compute the stress, then locate it on the stress-strain curve, and, finally, note whether this point is on the elastic
23
or plastic region. Thus,
=
F Ao
44500 N
=
10 × 10−3 m 2
2
565 MPa (80,000 psi)
=
The 565 MPa point is past the linear portion of the curve, and, therefore, the deformation will be both elastic and plastic. (b) This portion of the problem asks us to compute the increase in specimen length. From the stress-strain curve, the strain at 565 MPa is approximately 0.008. Thus, from Equation (7.2) l
= ε lo =
(0 .008)(500 mm)
=
4 mm (0.16 in .)
7.29 This problem calls for us to make a stress-strain plot for aluminum, given its tensile load-length data, and then to determine some of its mechanical characteristics. (a) The data are plotted below on two plots: the first corresponds to the entire stress-strain curve, while for the second, the curve extends just beyond the elastic region of deformation.
300
) a P M (
200
s s e r t 100 S
0 0.000
0.002
0.004
0.006
Strain
24
0.008
0.010
0.012
(b) The elastic modulus is the slope in t he linear elastic region as E
=
ε
=
200 MPa
−
0 MPa
0.0032 − 0
=
62.5 × 103 MP a
=
62.5 GPa (9.1 × 106 psi)
(c) For the yield strength, th e 0.002 strain o ffset line is drawn dashed. It intersects the stress-strain curve at appro ximately 285 MPa (41,000 psi). (d) The tensile strength is approximately 370 MPa (54,000 psi), corresponding to the maximum stress on the complete stress-strain plot. (e) The ductility, in percent elongation, is just the plastic strain at fracture, multiplied by onehundred. The total fracture strain at fracture is 0.165; subtracting out the elastic strain (which is about 0.005) leaves a plastic strain of 0.160. Thus, the du ctility is about 16% E L. (f) From Equation (7.14), the modulus o f resilience is just 2 y
Ur
=
2E
which, using data computed in the problem yields a value of Ur
=
(285 MPa) 2 (2)(62.5
× 103
6.5 × 105 J/m 3 (93 .8 in .-lb f /in .3 )
=
MPa)
7.32 This prob lem asks us to calculate the modu li of resilience for the m aterials having the stress-strain behaviors shown in Figures 7.12 and 7.33. According to Equation (7.14), the modu lus of resilience U r is a function o f the yield strength an d th e mo dulus of elasticity as 2 y
Ur
=
2E
The values for y an d E for the brass in Figure 7.12 are 250 MPa (36,000 psi) and 93.9 GPa (13 .6 × 106 psi), respectively. Thus Ur
=
(250 MPa) 2 (2)(93.9 × 103 MPa)
=
3.32 × 105 J/m 3 (47 .6 in .-lb f /in .3 )
7.41 For this problem, we are given two values of εT an d T , from which we are asked to calculate the true stress which produ ces a true plastic strain of 0.25. Employing Equation (7.19), we may set up two simultaneous equations with two unknowns (the unknowns being K an d n), as log(50,000 psi)
=
logK + n log(0.10)
log(60,000 psi)
=
logK + n log(0.20)
From these t wo expressions, n
=
log(50,000)
εT
=
log(60,000)
log(0.1) − log(0.2)
log K Thus, for
−
=
4.96 or K
=
=
0.263
91,623 psi
0.25 T
=
K( εT ) 2
=
(91 ,623 psi)(0.25) 0.263
25
=
63,700 psi (440 MPa)
7.45 7.45 This This problem call callss for for us to utilize utilize the appropriate data from Problem 7.29 in in order to determine the values values of n an d K for this material. From Equ ation (7.32) (7.32) the t he slope and intercept of a log T versus log εT plot will yield n and log K, respectively respectively.. H owever, Equation (7.19) is only valid in the r egion of plastic plastic deformation t o the point of ne cking; cking; thus, thus, only the 7th, 8th, 9th, 9th, and 10th 10th da ta points may b e ut ilized. ilized. The log-log log-log plot with t hese dat a po ints is giv given en b elow. elow.
2.60
2.58
2.56
) a P M ( s s e r t s e u r t
2.54
2.52
2.50
g o l 2.48
2.46 -2.2
-2.0
-1.8
-1.6
-1.4
-1.2
log true strain The slope yields a value of 0.136 for n, whereas th e inter cept gives a value of 2.74 2.7497 97 for log K, and thus K = 562 562 MPa . 7.50 7.50 For this problem, the load is given given at which which a circular circular specimen specimen of aluminum oxide fractures when subjected to a three-point bending test; we are then are asked to determine the load at which a specimen specimen of the same ma terial having a square cross-s cross-section ection fractures. It is first necessary to compute the flexural strength of the alumina using Equation (7.20b), and then, u sing sing this value, we we ma y calculate calculate th e value of Ff in Equation (7.20a). From Equation (7.20b) fs
=
F f L
=
R3
(950 N)(50 × 10−3 m) −3
( )( 3.5 × 10
m) 3
=
352 × 106 N/m 2
=
Now, solving solving for Ff from Equation (7.20a), realizing realizing t hat b F f = =
352 352 MPa (50,000 psi)
=
d
=
12 mm, yields
2fs d 3 3L (2)(352 × 106 N/m 2 )(12 × 10−3 m) 3 (3)(40 × 10−3 m)
=
10,100 N ( 2165 2165 lb f )
7.54 7.54* * (a) This part of the prob lem asks us to determine the flexural strength of nonporous MgO assuming that the value of n in Equation (7.22) is 3.75. Taking natural logarithms of both sides of
26
Equation (7.22) yields ln fs
=
ln o
−
nP
In Table 7.2 it it is noted that for P = 0.05, fs = 105 MPa. For the nonporous material P ln o = ln fs . Solving for ln o from the above equation gives and using these data gives ln o
=
ln fs
=
ln(105 ln(105 MPa )
+
=
0 and,
nP +
(3 .75)(0.05)
=
4.841
or o
=
e 4.841
=
127 127 MPa (18,100 psi)
(b) Now we are asked to compute the volume percent porosity to yield a fs of 62 MPa (9000 psi). Taking the natural logarithm of Equation of Equation (7.22) and solving for P leads to P
=
= =
ln o
−
ln fs
n ln(127 ln(127 MPa )
−
ln(62 MPa)
3.75 0.19 or 19 vol%
7.65 7.65 This problem calls for estimations of Brinell and R ockwell ockwell hardnesses. hardnesses. (a) For the brass specimen, specimen, the stress-strain stress-strain be havior for which is shown in Figure 7.12, 7.12, the t he tensile tensile strength is 450 MPa (65,000 psi). From Figure 7.31, the hardness for brass corresponding to this tensile strength is about 125 HB or 70 HRB. 7.70 7.70 The working stresses stresses for the t wo alloys alloys,, the stress-strain stress-strain behaviors of which which are shown in Figures 7.12 an d 7.33, are calculated calculated by dividing dividing the yield yield stren gth by a factor of safety, safety, which which we will will take to be 2. For the brass alloy (Figure 7.12), since y = 250 MPa (36,000 psi), the working stress is 125 MPa (18,000 psi), whereas for the steel alloy ( Figure 7.33), y = 570 MPa (82,000 psi), and, therefore, w = 285 MPa (41,000 (41,000 psi).
27
CHAPTER 8 DE FORMATI FORMATION ON A ND STRENGTHENING STRENGTHENING MECHAN ISMS ISMS
8.7
I n t h e m an an n er er o f Figure Figure 8.6b, 8.6b, we are to sketch the atomic packing for a BCC 110 type plane, and with with arrows indicate two d ifferent ifferent 111 type directions. Such is shown below.
{ }
Schmid factor for an FCC crystal oriented with its [100] direction 8.10* We are asked to compute the Schmid parallel to the loading axis. With this scheme, slip may occur on the (111) plane and in the [110] direction as noted in the figure below. z
[111]
φ y
λ
_ [110] [100]
x
The a ngle between the [100 [100]] and [110] [110] directions, directions, , is 45◦ . For the (111) plane, √ the angle
between its normal ( which which is the [111] [111] direction) direction) and the [100] [100] direction, direction, , is tan −1 ( a a 2 ) therefore cos cos
= 54.74◦;
= cos(45◦) cos(54.74◦) = 0.408
8.20 8.20 We are asked to determine the grain grain diameter for an iron which which will will give give a yield yield strength of 205 205 MPa (30,000 psi). The best way to solve this problem is to first establish two simultaneous expressions of Equation (8.5), solve for o an d ky , and finally determine the value of d when y 205 205 MPa. The data pertaining to this problem may be tabulated as follows:
=
28
d −1/2 (mm) −1/2
d (mm)
y 135 MPa 260 MPa
5 8
× 10−2 × 10−3
4.47 11.18
The two equations thus become
= o + (4 .47)k y 260 MPa = o + (11 .18)k y which yield the values, o = 51.7 MPa and ky = 18.63 MPa(mm) 1 2 . At a yield strength of 205 MPa 135 MPa
/
= 51.7 MPa + [18.63 MPa(mm) 1 2]d −1 2 or d−1 2 = 8.23 (mm) −1 2 , which gives d = 1.48 × 10−2 mm . /
205 MPa
/
/
/
8.25 This problem stipulates that two previously undeformed cylindrical specimens of an alloy are to be strain hard ened by reducing their cross-sectional area s. For one specimen, the initial and de formed radii are 16 mm an d 11 mm, r espectively. The second specimen with an initial rad ius of 12 mm is to have the same deformed hardness as the first specimen. We are asked to compute the radius of the second specimen after deformation. In order for these two cylindrical specimens to have the same deformed hardness, they must be deformed to the same percent cold work. For the first specimen % CW
2 2 = A o A− A d × 100 = r or−2rd × 100 o
=
o
(16 mm) 2
− (11 mm) 2 × 100 = 52.7%CW
(16 mm) 2
For the second specimen, the deformed radius is computed using the above equation and solving for rd as rd
CW = r o 1 − %100
= (12mm) 1 − 52.7%CW = 8.25mm 100 8.27 This problem calls for us to calculate the p recold-worked radius of a cylindrical specimen of copper that has a cold-worked ductility of 25%EL. From Figure 8.19(c), copper that has a ductility of 25% E L will have experienced a deformation of about 11% CW. For a cylindricalspecimen, Equation (8.6) becomes
% CW
Since rd
=
r 2o
r 2d r 2o
−
× 100
= 10 mm (0.40 in.), solving for ro yields ro
=
rd
1
−
% CW 100
=
10mm
1
11.0
− 100
29
= 10.6 m m ( 0.424 in .)
8.35 In this problem, we are asked for the length of time required for the average grain size of a brass material to increase a specified amount using Figure 8.25. (a) At 500◦ C, the time n ecessary for the average grain diameter to increase from 0.01 to 0.1 mm is appro ximately 3500 min. (b) At 600◦ C the time req uired for this same grain size increase is appro ximately 150 min. 8.45* This problem gives us the tensile strengths and associated number-average molecular weights for two polymethyl methacrylate materials and then asks that we estimate the tensile strength for Mn 30,000 g/mol. Eq uation ( 8.9) provides the dependence of the tensile strength on Mn . Thus, using the data provided in the problem, we may set up two simultaneous equations from which it is possible to solve for the two constants TS∞ an d A . These equ ations are as follows:
=
107 MPa
= TS ∞ − 40000Ag/mol
170 MPa
= TS ∞ − 60000Ag/mol
Thus, the values of the two constants are TS∞ 296 MPa a nd A 7.56 106 MPa-g/mol. Substituting these values into an equation for which Mn 30,000 g/mol leads to
=
TS
=
=
×
= TS ∞ − 30000Ag/mol = 296 MPa − = 44 MPa
7.56
× 106 MPa-g/mol 30000 g/mol
8.54 This prob lem asks that we compute t he fraction of possible crosslink sites in 10 kg of polybutadiene when 4.8 kg of S is added, assuming that, on the average, 4.5 sulfur atoms participate in each crosslink bond. G iven the butadiene mer unit in Table 4.5, we may calculate its molecular weight as follows: A(butadiene)
= 4(A C ) + 6(A H ) = (4)(12.01 g/mol) + 6(1.008 g/mol) = 54.09 g/mol 10000 g
which means that in 10 kg of butadiene there are 54.09 g/mol 184.9 mol. For the vulcanization of polybutadiene, there ar e two possible crosslink sites per mer—on e for each of the two carbon atoms that are doubly bonded. Furthermore, each of these crosslinks forms a bridge between two mers. Therefore, we can say that there is the equivalent of one crosslink per mer. Therefore, let us now calculate the number of moles of sulfur ( nsulfur) that react with the butad iene, by taking the mole ra tio of sulfur t o but adiene, and th en dividing this ratio by 4.5 atoms per crosslink; this yields the fraction of p ossible sites th at are crosslinked. Thus n sulfur
=
g = 32.4800 = 149.7 mol 06 g/mol
A nd 149.7 mol fraction sites crosslinked
30
=
184.9 mol 4.5
= 0.180
8.D1 This prob lem calls for us to d eterm ine whether o r not it is possible to cold work steel so as to give a minimum B rinell hardness of 225 and a d uctility of at least 12% EL . According to Figure 7.31, a Brinell hardness of 225 corresponds to a tensile strength of 800 MPa (116,000 psi). Furthermore, from Figure 8.19(b), in order to achieve a tensile strength of 800 MPa, deformation of at least 13% CW is necessary. Finally, if we cold work t he steel to 13% CW, then the ductility is reduced to only 14% EL from Figure 8.19(c).Therefore,it is possible to meet both of these criteria by plastically deforming the steel. 8.D6 This problem stipulates that a cylindrical rod of copper or iginally 16.0 mm in diameter is to be cold worked by drawing; a cold-worked yield strength in excess of 250 MPa and a ductility of at least 12%EL are required, whereas the final diameter must be 11.3 mm. We ar e to explain how this is to be accomplished. Let us first calculate the percent cold work and attendant yield strength and ductility if the dr awing is carried out without interrup tion. From Equation (8.6) % CW
=
do
2
dd
2
=
2
− × − × 2
100
2
do 2
16 mm
2
2
2
11.3 mm 2
2
16 mm
100
= 50%CW
2
At 50% CW, the copper will have a yield strength on the order of 330 MPa (48,000 psi), Figure 8.19(a), which is adequ ate; however, the ductility will be about 4% EL , Figure 8.19(c), which is insuf ficient. Instead of perform ing the dr awing in a single operat ion, let us initially draw some fraction of the total deformation, then anneal to recrystallize, and, finally, cold work the mater ial a second time in order to achieve the final diame ter, yield strength, an d d uctility. Reference to Figure 8.19(a) indicates tha t 21% CW is necessary to give a yield strength of 250 MPa. Similarly, a m aximum of 23% CW is possible for 12% E L [Figure 8.19(c)]. The average of th ese two values is 22% CW, which we will use in t he calculations. If t he final diameter after the first dr awing is do , then 22%CW
A nd, solving for do yields do
=
d o
2
2
− ×
2
d o 2
= 12.8 mm (0.50 in.).
31
11.3 2
2
100
CHAPTER 9 FAILURE
9.7
We are asked for the critical crack tip radius for an Al 2 O 3 material. From Equation (9.1b)
m
Fracture will occur when thus
m
a
1/2
t
reaches the fracture stren gth of the mater ial, which is given as E /10;
E 10 O r, solving for
= 2o
= 2o
1/2
a
t
t
2
t
From Table 7.1, E
o = 400a 2 E
= 393 GPa, and thus, t
−3
× 10 mm)(275 MPa) = (400)(2(393 × 103 MPa) 2
2
= 3.9 × 10−7 mm = 0.39 nm 9.8
We may determine the critical stress required for the propagation of a surface crack in soda-lime glass using Equat ion (9.3);takingthe value of69GPa (Table 7.1) as the modulusof elasticity, we get
c
= =
9.12*
2E s a
(2)(69
× 109 N/m 2)( 0.30 N/m) = 16.2 × 106 N/m 2 = 16.2 MPa ( )( 5 × 10−5 m)
This prob lem deals with a tensile specimen, a drawing of which is provided. (a) In this portion of the problem it is necessary to compute the stress at point P when the applied stress is 100 MPa (14,500 psi). In order to determine the stress concentration it is necessary to consult Figure 9.8c. From the geometry of the specimen, w/h (25 mm) /(20 mm) 1.25; furthermore, the r/h ratio is (3 mm) /(20 mm) 0.15. Using th e w/h 1.25 curve in Figure 9.8c, th e m Kt value at r/h 0.15 is 1.7. An d since Kt , then
=
m
9.15*
= =
=
=
=
o
= K t o = (1 .7)(100 MPa ) = 170 MPa (24,650 psi)
Thisproblem callsfor us to determine the value of B , the minimum compon ent thicknessfor which the condition of plane strain is valid using Equation (9.12) for th e m etal alloys listed in Table 9.1.
32
For t he 2024-T3 aluminum alloy
B
2
= K lc
= 2.5
(2 .5)
y
√
44 MPa m 345 MPa
2
=
0.041 m
= 41 mm (1.60 in .)
For the 4340 alloy steel temper ed at 260◦ C B 9.19
√
50 MPa m 1640 MPa
2
=
0.0023 m
= 2.3 mm (0.09 in .)
For this problem, we are given values of Klc , , a n d Y for a large plate and are asked to determine the m inimum length of a surface crack that will lead t o fracture. A ll we nee d do is to solve for ac using Equation (9.14); therefore
ac 9.26
= (2 .5)
1
=
2
= K lc
Y
√
1
55 MPa m
(1)(200 MPa)
2
=
0.024 m
= 24 mm (0.95 in .)
This problem first provides a tabulation of Charpy impact data for a ductile cast iron. (a) The plot of impact energy versus temperature is shown below. 140 120 100
J , y g r e n E t c a p m I
80 60 40 20 0 -200
-150
-100
-50
0
Temperature, C °
(b) This port ion of the prob lem asks us to deter mine the ductile-to-brittle transition temp eratu re as that temperatur e corresponding to the average of the maximum and minimum impact energies. From these data, this average is Average
= 124 J2+ 6 J = 65 J
As indicated on the plot by the one set of dashed lines,t he ductile-to-brittle tran sition temperat ure according to this criterion is about 105◦ C.
−
(c) Also as noted on the plot by the other set of dashed lines, the ductile-to-brittle transition temperature for an impact energy of 80 J is about 95◦ C.
−
33
9.31
We are asked to determine the fatigue life for a cylindrical red brass rod given its diameter (8.0 mm) and t he m aximum tensile and compressive loads ( 7500 N and 7500 N, r espectively). The first thing th at is necessary is to calculate values of max an d min using Equation (7.1). Thus
+
max
= FAmax =
F max
o
= min
( )
do
2
2
8.0 × 10−3 m 2
= 2
150
× 106 N/m 2 = 150 MPa (22,500 psi)
F min
=
=
7500 N
−
2
do 2
7500 N
( )
−× 8.0
10−3 m 2
=− 2
150
× 106 N/m 2 = −150 MPa ( −22,500 psi)
Now it becomes necessary to compute the stress amplitude using Equation (9.23) as
a
= max −2 min = 150 MPa − 2( −150 MPa ) = 150 MPa (22,500 psi)
From Figure 9.46 f o r t he r ed br ass, the n umber of cycles to failure at this stress amplitude is about 1 105 cycles.
×
9.33
T his p r ob le m first provides a tabulation of fatigue data (i.e., stress amplitude and cycles to failure) for a brass alloy. (a) These fatigue data are plotted below.
300
a P M , e d u t i l p 200 m a s s e r t S 100 5
6
7
8
Log cycles to failure
34
9
10
(b) As indicated by one set of dashed lines on the plot, the fatigue strength at 5 [log(5 105 ) 5.7] is about 250 MPa.
× 105 cycles
(c) As noted by the other set of dashed lines, the fatigue life for 200 MPa is about 2 (i.e., the log o f the lifetime is abou t 6.3).
× 106 cycles
×
9.34
=
We are asked to compute the maximum torsional stress amplitude possible at each of se veral fatigue lifetimes for the brass alloy, the fatigue behavior of which is given in Problem 9.33. For each lifetime, first compute the number of cycles, and then read the corresponding fatigue strength from the above plot. (a) Fatigue lifetime (1 yr)(365 days/yr)(24 h/day)(60 min/h)(1500 cycles/min) cles. The stress amplitude corresponding to this lifetime is about 130 MPa.
=
(c) Fatigue lifetime (24 h) (60 min/h) (1200 cycles/min) corresponding to this lifetime is about 195 MPa.
=
9.48
= 7.9 × 108 cy-
= 2.2 × 106 cycles. The stress amplitude
This problem asks that we determine the total elongation of a low carbon-nickel alloy that is exposed t o a tensile stress of 40 MPa (5800 psi) at 538 ◦ C for 5000h; the instantaneous and primary creep elongations ar e 1.5 mm ( 0.06 in.). From the 538◦ C line in Figure 9.43, the steady-state creep rate, ε˙s , is about 0.15% /1000 h (or 1.5 10−4 % /h) at 40 MPa. The steady-state creep strain, εs , therefore, is just the product of ˙s and time as ε
×
εs
= ˙s × (time) = (1 .5 × 10−4% /h)(5000 h) = 0.75% = 7.5 × 10−3 ε
Strain and elongation are related as in Equat ion (7.2); solving for the steady-state elongation, ls , leads to ls
= lo s = (750 mm)(7.5 × 10−3) = 5.6 mm (0.23 in .) ε
Finally, the total e longation is just th e sum o f this ls and the total of both instantaneous and primary creep elongations [i.e.,1.5 mm (0.06in.)]. Therefore, the total elongation is 7.1 mm (0.29in.). 9.52*
The slope of the line from a log ε˙s versus log that is n
plot yields the value of n in Equation (9.33);
˙s = log log ε
We are asked to determine the values of n for the creep data at the three temperatures in Figure 9.43. This is accomplished by taking ratios of the differences between two log ε˙s and log values. Thus for 427◦ C n
log ε˙s
− log(10−2 ) = log (85 MPa) − log( 55 MPa) = 5.3
log ε˙s
− log(10−2 ) = log (59 MPa) − log( 23 MPa) = 4.9
= log
log(10−1 )
and for 538◦ C n
= log
log(1.0)
35
9.55*
This problem gives ε˙s values at two different temperatures and 70 MPa (10,000 psi), and the stress exponent n 7.0, and asks that we determ ine the steady-state creep rate at a stress of 50 MPa (7250 psi) and 1250 K. Taking the natural logarithm of Equation (9.34) yields
=
ln ε˙s
Qc = ln K 2 + n ln − RT
With the given data there are two unknowns in this equation —namely K2 an d Q c . Using the data provided in the problem we can set up two independent equations as follows: ln[1.0
Qc × 10−5 (h ) −1] = ln K 2 + (7 .0) ln(70 MPa ) − (8 .31 J/mol-K)(977 K)
ln[2.5
Qc × 10−3 (h ) −1] = ln K 2 + (7 .0) ln(70 MPa ) − (8 .31 J/mol-K)(1089 K)
Now, solving simultaneo uslyfor K2 an d Q c leadsto K2 2.55 105 (h ) −1 an d Q c 436,000 J/mol. Thus it is now possible to solve for ε˙s at 50 MPa and 1250 K using Equation (9.34) as
=
˙
εs
= K 2
n
exp
×
=
− Qc
RT
5
˙s = [2.55 × 10 (h ) −1 ](50MPa) 7.0 exp ε
−
436000 J/mol (8 .31 J/mol-K)(1250 K)
= 0.118 (h) −1 9.D1*
This prob lem asks us to calculate the minimum Klc necessary to ensure t hat failure will not occur for a flat plate given an expression from which Y ( a/W) may be determined, the internal crack length, 2a (20 mm), the plate width, W (90 mm), and the value of (375 MPa). First we must compute the value of Y ( a/W) using Equation (9.10), as follows:
Y( a /W)
= =
W a
ta n
a
W
1/2
90 mm ( )(10 mm)
ta n
( )(10 mm) 90 mm
1/2
= 1.021
Now, using Equation (9.11) it is possible to de termine Klc ; thus K lc
9.D7*
√ = Y( a /W) a = (1 .021)(375 MPa)
( )(10
× 10−3 m) = 67.9 MPa
√
√
m (62.3 ksi in .)
We are asked in this prob lem to estimate the maximum tensile stress that will yield a fatigue life of 2.5 107 cycles, given values of ao , ac , m, A , and Y . Since Y is indepe ndent of crack length we may utilize Equation (9.31) which, upon integration, takes the form
×
N f
ac
1
= A m/2 ( ) m Y m
36
ao
a −m /2 da
And for m
= 3.5 N f
ac
1
= A 1.75 ( ) 3.5Y 3.5 1.33
a −1.75 da
ao
= − A 1.75( ) 3.5 Y 3.5
1
1
a 0c .75
− a 0.75 o
Now, solving for from this expression yields
= =
1.33
1
N f A 1.75 Y 3.5 a 0o.75
1
− a 0.75 c
1/3.5
1.33 (2 .5
× 107)( 2 × 10−14 )( ) 1.75(1 .4) 3.5
1
(1 .5
1
× 10−4) 0.75 − (4 .5 × 10−3) 0.75
1/3.5
= 178 MPa This 178 MPa will be the maximum tensile stress since we can show that the minimum stress is a compressive one —when min is negative, is taken to be max . If we take max 178 MPa, and since m is stipulated in the problem to have a value of 25 MPa, then from Equation (9.21)
=
min
Therefore
min
= 2m − max = 2(25 MPa) − 178 MPa = −128 MPa
is negative and we are justified in taking max to be 178 MPa.
9.D16* We are asked in this problem to calculate the stress levels at which the rupture lifetime will be 5 years and 20 years when an 18-8 Mo stainless steel component is subjected to a temperature of 500◦ C (773 K). It first becomes necessary, using the specified temperature and times, to calculate the values of the Larson-Miller parameter at each temperature. The values of tr corresponding to 5 and 20 years are 4 .38 104 h a n d 1.75 105 h, respectively. Hence, for a lifetime of 5 years
×
T(20 And for tr
×
+ logtr ) = 773[20 + log(4.38 × 104 )] = 19.05 × 103
= 20 years T(20
+ logtr ) = 773[20 + log(1.75 × 105 )] = 19.51 × 103
Using the curve shown in Figure 9.47, the stress values corresponding to the five- and twenty-year lifetimes are approximately 260 MPa (37,500 psi) and 225 MPa (32,600 psi), respectively.
37
CHAPTER 10 PHASE D IAGRAMS
10.5
This problem asks that we cite the phase or phases present for several alloys at specified temperatures. (a) For an alloy composed of 90 wt% Zn-10 wt% Cu and at 400 ◦ C, from Figure 10.17, phases are present, and Cε
=
87 wt% Zn-13 wt% Cu
C
=
97 wt% Z n-3 wt% Cu
ε
an d
(c) For an alloy composed of 55 wt% Ag-45 wt% Cu and at 900 ◦ C, from Figure 10.6, only the liquid ph ase is present; its compo sition is 55 wt% A g-45 wt% Cu. 10.7
This problem asks that we determine the phase mass fractions for the alloys and temperatures in Problem 10.5. (a) For an alloy composed of 90 wt% Zn-10 wt% Cu and at 400 ◦ C, an d Co
=
90wt% Zn
Cε
=
87wt% Zn
C
=
97wt% Zn
ε
an d phases are present,
Therefore, using modified forms of Equation (10.2b) we get Wε
=
W
=
C − Co C − Cε Co
−
Cε
C − Cε
=
=
97 − 90 97 − 87 90 − 87 97 − 87
=
0.70
=
0.30
(c) For an alloy composed of 55 wt% Ag-45 wt% Cu and at 900 ◦ C, since only the liquid p hase is present, then WL = 1.0. 10.9
This problem asks that we determine the phase volume fractions for the alloys and temperatures in Problem 10.5a, b, and c. This is accomplished by using the technique illustrated in Example Prob lem 10.3, and t he r esults of Problem 10.7. (a) This is a Cu-Zn alloy at 400 ◦ C, wherein Cε
=
87 wt% Zn-13 wt% Cu
C
=
97 wt% Zn-3 wt% Cu
Wε
=
0.70
W
=
0.30
Cu
=
8.77 g/cm 3
Zn
=
6.83 g/cm 3
38
Using these data it is first necessary to compute the densities of the Equation (5.10a). Thus ε
=
+
Cu
100
=
87 6.83
=
an d phases using
100 C Zn ( ε) C Cu( ε) Z n
ε
g/cm 3
13
+
=
7.03 g/cm 3
=
6.88 g/cm 3
8.77 g/cm 3
100 C Zn ( ) C Cu( ) +
Zn
Cu
100
=
97 6.83 g/cm 3
3
+
8.77 g/cm 3
Now we may determine the V ε an d V values using Equation 10.6. Thus, Wε Vε
=
ε
Wε ε
+
W
0.70 7.03 g/cm 3 0.70 0.30
=
7.03 g/cm 3
+
=
0.70
=
0.30
6.88 g/cm 3
W V
=
Wε ε
+
W
0.30 6.88 g/cm 3 0.70 0.30
=
7.03 g/cm 3 10.12
+
6.88 g/cm 3
(a) We are asked to deter mine how much sugar will dissolve in 1500 g of water a t 90◦ C. From the solubility limit curve in Figure 10.1, at 90◦ C the maximum concentration of sugar in the syrup is about 77 wt% . It is now possible to calculate th e ma ss of sugar using Equation (5.3) as C sugar (wt%)
=
77wt%
=
Solving for msugar yields msugar
=
m sugar m sugar
+
m water
m sugar m sugar
+
1500 g
×
100
×
100
5022 g.
(b) Again using this same plot, at 20 ◦ C the solubility limit (or the concentration of the saturated solution) is about 64 wt% sugar.
39
(c) The mass of sugar in this saturated solution at 20 ◦ C (msugar ) may also be calculated using Equation (5.3) as follows: 64 wt%
m sugar
=
m sugar
+
×
1500 g
100
which yields a value for msugar of 2667 g. Subtracting the latter from the former of these sugar concentrations yields the amount of sugar that pr ecipitated out of the solution upon cooling msugar ; that is m sugar 10.21
=
m sugar
−
m sugar
=
5022 g − 2667 g
=
2355 g
Up on cooling a 50 wt% Pb-50 wt% Mg alloy from 700◦ C and utilizing Figure 10.18: (a) The first solid phase forms at the temperature at which a vertical line at this composition intersects the L-( + L) phase boundary—i.e., about 550◦ C; (b) The composition of this solid phase corresponds to the intersection with the -( + L) phase boundar y, of a tie line constructed across the + L phase region at 550 ◦ C—i.e., 22 wt% Pb-78 wt% Mg; (c) Complete solidification of the alloy occurs at the intersection of this same vertical line at 50 wt% Pb with the eutectic isotherm —i.e., about 465◦ C; (d) The composition of the last liquid phase remaining prior to complete solidification corre sponds to the eutectic composition —i.e., about 66 wt% Pb-34 wt% Mg.
10.24
(a) We are given that the mass fractions of and liquid phases are both 0.5 for a 30 wt% Sn-70 wt% Pb alloy and asked to estimate the temperature of the alloy. Using the appropriate phase diagram, Figure 10.7, by trial and error with a ruler, a tie line within the + L phase region that is divided in h alf for an alloy of this composition exists at abou t 230◦ C. (b) We are now asked to determine the compositions of the two phases. This is accomplished by noting the intersections of this tie line with both the solidus and liquidus lines. From these intersections, C = 15 wt% Sn, and CL = 42 wt% Sn.
10.28
This problem asksif it ispo ssible to have a Cu-Ag alloy of composition 50 wt% Ag-50 wt% Cu that consists of mass fractions W = 0.60 and W = 0.40. Such a n alloy is not possible, based on the followingar gument. Usingthe appropriate phase diagram, Figure 10.6, and, using Eq uations( 10.1) an d (10.2) let us determine W an d W at just below the eutectic temperature an d also at room temperature. At just below the eutectic, C = 8.0 wt% Ag and C = 91.2 wt% A g; thus, W
=
W
=
C
−
C
−
=
C
1.0 − W
Furthermore, at room temperature, C tions (10.1) an d (10.2) yields W
Co
=
C
−
C
−
91.2 − 50
=
=
91.2 − 8 1.0 − 0.5
=
=
0.50
0.50
0 wt % A g a nd C
Co C
=
100 − 50 100 − 0
=
=
100 wt% Ag; employment ofE qua-
0.50
And, W = 0.50. Thus, the mass fractions of the an d phases, upon cooling through the + phase region will remain approximately constant at about 0.5, and will never have values of W = 0.60 and W = 0.40 as called for in the prob lem.
40
10.35* This problem asks that we d eterm ine the composition of a Pb-Sn alloy at 180◦ C given th at W = 0.57 and We = 0.43. Since there is a primary microconstituent present, then we know that the alloy composition, Co , is between 61.9 and 97.8 wt% Sn (Figure 10.7). Furthermore, this figure also indicates th at C = 97.8 wt% Sn and Ceutectic = 61.9 wt% Sn. Applying the appropriate lever rule expre ssion for W
W
and solving for Co yields Co
=
Co
−
C
−
C eutectic
Co
=
C eutectic
−
61.9
97.8 − 61.9
=
0.57
82.4 wt% Sn.
=
10.47* We are asked to specify the value of F for Gibbs phase rule at point B on the pressure-temperature diagram for H 2 O. Gibbs phase rule in general form is P
+
F
=
C
+
N
For this system, the number of components, C, is 1, whereas N, the number of noncompositional variables, is 2—viz. temperature and pressure. Thus, the phase rule now becomes P
+
F
=
1+2
F
=
3−P
=
3
Or
where P is the number of phases present at equilibrium. At point B on the figure, only a single (vapor) phase is present (i.e., P F
=
3−P
=
3−1
=
1), or
2
=
which means that bot h temper ature and pressure are necessary to de fine the system. 10.54
This problem asks that we compute the carbon concentration of an iron-carbon alloy for which the fraction of total ferrite is 0.94. Application of the lever rule [of the form of Equation (10.12)] yields
W
=
0.94
=
C Fe 3 C
−
C o
C Fe 3 C
−
C
=
6.70 − C o 6.70 − 0.022
and solving for Co C o 10.59
=
0.42 wt% C
This prob lem asks that we deter mine the carbon concentra tion in an iron-carbon alloy, given the mass fractions of proeutectoid ferrite and pearlite. From E quation (10.20)
Wp which yields Co
=
=
0.714
=
0.55 wt% C.
41
C o
−
0.022
0.74
10.64
This problem asks if it is possible to have an iron-carbon a lloy for which W = 0.846 and WFe3 C = 0.049. In order to ma ke t his determination, it is necessary to set up lever ru le expressions for th ese two mass fractions in terms of the alloy composition, then to solve for the alloy composition of each; if both alloy composition values are equal, then such an alloy is possible. The expression for the mass fraction of total ferrite is
W
=
Solving for this Co yields Co
C Fe 3 C
−
C Fe 3 C
−
=
10.70
=
6.70 − C o
=
C
=
6.70 − 0.022
0.846
1.05 wt% C. Now for WFe3 C we ut ilize E quation (10.23) as
W Fe 3 C This expression leads to C1
Co
=
C 1 − 0.76 5.94
0.049
=
1.05 wt% C. And, since Co
=
C1 , this alloy is possible.
This problem asks that we determine the appr oximate Brinell hardn ess of a 99.8 wt% Fe-0.2 wt% C alloy. First, we compute th e mass fractions of pearlite and pr oeute ctoid ferrite using Equations (10.20) an d (10.21), a s Wp W
=
=
C o
−
0.022
0.74 0.76 − C o 0.74
=
0.20 − 0.022 0.74
0.76 − 0.20
=
0.74
=
=
0.24
0.76
Now, we compute the B rinell hardness of the alloy as H B alloy
=
H B W
=
(80)(0.76) + (280)(0.24)
+
H Bp Wp =
128
10.73* We are asked to consider a steel alloy of composition 93.8 wt% Fe, 6.0 wt% Ni, and 0.2 wt% C. (a) From Figure 10.36, the eutectoid temperature for 6 wt% Ni is approximately 650 ◦ C (1200◦ F) . (b) From Figure 10.37, the e utectoid composition is approximately 0.62 wt% C. Since the carbon concentration in th e alloy (0.2 wt% ) is less than th e eut ectoid, the pro eutectoid ph ase is ferrite. (c) Assume that the -( + Fe 3 C) phase boundary is at a negligible carbon concentration. Modifying Equation (10.21) leads to W
=
0.62 − C o 0.62 − 0
=
0.62 − 0.20 0.62
=
Likewise, using a m odified E quation (10.20) Wp
=
C o
−
0
0.62 − 0
=
42
0.20 0.62
=
0.32
0.68
CHAPTER 11 PHA SE TRA NSFORMATIONS
11.4
This problem gives us the value of y (0.40) at some time t (200 min), and also the value of n (2.5) for the recrystallization of an alloy at some temperature, and then asks that we determine the rate of recrystallization at this same tempe ratur e. It is first ne cessary to calculate t he value of k in Equation (11.1) as
k =
−
= −
ln(1 − y) tn ln(1 − 0.4)
=
min) 2.5
(200
7
−
9.0 × 10
At this point we want to compute t0.5 , the value of t for y = 0.5, also using Eq uation (11.1). Thus
t 0.5
=
=
1/n
ln(1 − 0.5) −
k
1/2.5
ln(1 − 0.5) −
9.0 × 10
7
−
=
226.3 min
And , therefore, from Equation (11.2), the rate is just
rate
11.7
1 =
t 0.5
1 =
226.3 min
=
3
−
4.42 × 10
(min)
1
−
This problem asks us to consider the percent recrystallized versus logarithm of time curves for copper shown in Figure 11.2. (a) The rates at the different temperatures are determined using Equation (11.2), which ra tes are tabulated below:
◦
Temperature ( C)
R ate (min)
135 119 113 102 88 43
0.105 4.4 × 10 2.9 × 10 1.25 × 10 4.2 × 10 3.8 × 10
1
−
2
−
2
−
2
−
3
−
5
−
43
(b) These data are plotted below as ln rate versus the reciprocal of absolute tem perature.
-2
) n i m / 1 ( e t a R
-4
-6
-8
n l -10
-12 0.0024
0.0026
0.0028
1/T
0.0030
0.0032
(1/K)
The activation energy, Q, is related to the slope of the line drawn through the data points as Q
Slope(R)
= −
where R is the gas constant. The slope of this line is −1.126 × 104 K, and thus Q
( 1.126 × 104 K)(8.31 J/mol-K)
= − − =
93,600 J/mo l 3
(c) At room temperature (20 C), 1/T = 3.41 × 10 this 1/T value gives ◦
−
ln(rate)
K
1
−
. Extrapolation of the data in the plot to
∼ = −
12.8
or rate
∼ =
e
12.8
−
=
6
−
2.76 × 10
(min)
1
−
But since rate
=
1 t 0.5
then t 0.5
=
∼ =
11.15
1 rate
∼ =
1 2.76 × 10
6
−
3.62 × 105 min
∼ =
(min)
1
−
250 days
Below is shown an isothermal transformation diagram for a eutectoid iron-carbon alloy, with a time-temperature path that will produce (a) 100% coarse pearlite.
44
11.18
Below is shown an isotherma l transformation diagram for a 0.45 wt% C iron-carbon alloy, with a time-temperature path that will produce (b) 50% fine pearlite and 50% bainite.
45
11.20* Below is shown a cont inuous cooling tran sformat ion diagram for a 1.13 wt% C iron-carbon alloy, with a continuous cooling path that will produce (a) fine pearlite and proeutectoid cementite.
11.34
This problem asksfor estimates of Rockwell hardness values for specimens of an iron-carbon alloy of eutectoid composition that have been subjected to some of the heat treatments described in Problem 11.14. (b) The microstructural product of this heat treatment is 100% spheroidite. A ccording to Figure 11.22(a) the ha rdness of a 0.76 wt% C alloy with spheroidite is about 87 HR B. (g) The microstructural product of this heat treatment is 100% fine pearlite. A ccording to Figure 11.22(a), the hardn ess of a 0.76 wt% C alloy consisting of fine pearlite is about 27 HR C.
11.37
For this problem we are asked to describe isother mal heat treatme nts required to yield specimens having several Br inell hard nesses. (a) From Figure 11.22(a), in order for a 0.76 wt% C alloy to have a R ockwell hardness of 93 H RB, the m icrostructure must be coarse pearlite. Thus, utilizing the isotherm al transformation diagram for this alloy, Figure 11.14, we must r apidly cool to a temperature at which coarse pearlite forms (i.e., to abo ut 675 C), allowing the specimen to isothermally and completely transform to coarse pearlite. At this temperature an isothermal heat treatment for at least 200 s is required. ◦
11.D1 This problem inqu ires as to the possibility of producing an iron-carbon alloy of eutectoid composition that h as a minimum hardness of 90 HR B and a minimum ductility of 35% RA . If the alloy is possible, then the continuous cooling heat treatment is to be stipulated. According to Figures 11.22(a) an d (b), the following is a tabulation of Rockwell B hardnesses and percents reduction of area for fine an d coarse pea rlites and spheroidite for a 0.76 wt% C alloy. Microstructure
HRB
% RA
Fine pearlite Coarse pearlite Spheroidite
>100 93 88
22 29 68
Therefore, none of the microstructure s meets both of the se criteria. Both fine and coarse pearlites are hard enough, but lack the required ductility. Spheroidite is suf ficiently ductile, but does not meet the hardness criterion.
46
CHAPTER 12 ELECTRICAL PROPERTIES
12.5
(a) In order to compute the resistance of thiscopper wire it isnecessary to employ Equa tions( 12.2) an d (12.4). Solving for th e r esistance in ter ms of t he conductivity, R =
l
=
A
l A
From Table 12.1, the conductivity of copper is 6.0 × 107 ( -m) −1 , and R =
l
2m
=
A
[6.0 ×
107 ( -m) −1 ]( )
3 × 10−3 m 2
2
= 4.7 × 10−3
(b) If V = 0.05 V then , from Equation (12.1) V
I=
=
R
0.05 V 4.7 × 10−3
= 10.6 A
(c) The current density is just J=
I A
I
=
d
2
10.6 A
=
2
−3
3 × 10
m
2
2
= 1.5 × 106 A /m 2
(d) The electric field is just E = 12.13
V l
=
0.05 V 2m
= 2.5 × 10−2 V/m
(a) The number of free electrons per cubicmeter for copper at room temperature may be computed using Equation (12.8) as n =
=
|e |e
6.0 × 107 ( -m) −1 (1 .602 × 10−19 C)(0.0030 m 2 /V-s)
= 1.25 × 1029 m −3
(b) In order to calculate the number of free electrons per copper atom, we must first determine the number of copper atoms per cubic meter, N Cu . From Equation (5.2) N Cu =
NA
47
A Cu
Note: in the above expression, density is represented by resistivity which is designated by . Thus
N Cu =
in order to avoid confusion with
(6 .023 × 1023 atoms/mol)(8.94 g/cm 3 )(106 cm 3 /m 3 ) 63.55 g/mol
= 8.47 × 1028 m −3
The number of free electrons per copper atom is just
n N 12.18
=
1.25 × 1029 m −3 8.47 × 1028 m −3
= 1.48
This problem asks for us to compute the ro om-temperature conductivity of a two-phase Cu-Sn alloy. It is first necessary for us to determine the volume fractions of the an d ε phases, after which the resistivity (and subsequently, the conductivity) may be calculated using Equation (12.12). Weight fractions of the two phases are first calculated using the phase diagram information provided in the problem. We might represent the phase diagram near room temperature as shown below.
Applying the lever rule to this situation
W = Wε =
Cε − Co Cε − C Co − C Cε − C
= =
37 − 8 37 − 0 8−0 37 − 0
= 0.784 = 0.216
We must now convert these mass fractions into volume fractions using the phase densities given in the problem. ( Note: in th e following expressions, density is represented by in order to avoid confusion with resistivity which is designated by .) Utilization of Equations (10.6a) an d (10.6b)
48
leads to W V =
W
+
Wε ε
0.784 8.94 g/cm 3 0.784 0.216
=
8.94 g/cm 3
+
8.25 g/cm 3
= 0.770
Wε Vε =
ε
W
+
Wε ε
0.216 8.25 g/cm 3 0.784 0.216
=
8.94 g/cm 3
+
8.25 g/cm 3
= 0.230
Now, using E quation (12.12)
=
V
+ ε Vε
= (1 .88 × 10−8 -m)(0.770) + (5 .32 × 10−7 -m)(0.230) = 1.368 × 10−7 -m
Finally, for the conductivity
=
1
12.30
=
1 1.368 × 10−7
-m
= 7.31 × 106 ( -m) −1
(a) In this problem, for a Si specimen, we are given p an d , while h an d e are included in Table 12.2. In ord er to solve for n we must use E quation ( 12.13), which, after rearrangement, leads to n =
=
− p |e |h |e | e
103 ( -m) −1 − (1 .0 × 1023 m −3 )( 1.602 × 10−19 C)(0.05 m 2 /V-s) (1 .602 × 10−19 C)(0.14 m 2 /V-s)
= 8.9 × 1021 m −3
(b) This material is p-type extrinsic since p (1 .0 × 1023 m −3 ) is greater th an n (8 .9 × 1021 m −3 ). 12.38
For this problem, we are given conductivity values at two different temperatures for an intrinsic semiconductor, and are then asked to determine its band gap energy. It is possible, using
49
E quation (12.18), to set up two independent equations with C an d E g as unknowns. At 20 ◦ C ln = C − ln[1.0 ( -m) −1 ] = C −
Eg 2kT Eg (2)(8.62 ×
10−5
eV/atom-K)(293 K)
or C = 19.80E g At 373 K ln[500 ( -m) −1 ] = C −
Eg (2)(8.62 × 10−5 eV/atom-K)(373 K)
6.21 = C − 15.55E g From these two expressions E g = 1.46 eV 12.45* In this problem we are asked to determ ine the magnetic field required to produce a Ha ll voltage of −1.0 × 10−7 V, given th at = 1.5 × 107 ( -m) −1 , e = 0.0020 m 2 /V-s, Ix = 45 A, and d = 35 mm. Combining E quations (12.21) an d (12.23b), and after solving for B z , we get Bz = =
|V H | d
I x e ( |−1.0 × 10−7 V |)[1 .5 × 107 ( -m) −1 ](35 × 10−3 m) (45 A)(0.0020 m 2 /V-s)
= 0.58 tesla
12.52* We want to compute the plate spacing of a parallel-plate capacitor as the dielectric constant is increased from 2.5 to 4.0,while maintainingthe capacitance constant. Combining Equations(12.29) an d (12.30) yields εr εo
C=
A
l
Now, let us use the subscripts 1 an d 2 to denote the initial and final states, respectively. Since C1 = C2 , then εr1 εo
A
=
l1
εr2 εo
A
l2
A nd, solving for l2 l2 =
εr2
l1
εr1
=
(4 .0)(1 mm) 2.5
= 1.6 m m
12.58* (a) We want to solve for the voltage when Q = 3.5 × 10−11 C, A = 160 mm 2 , l = 3.5 mm, and εr = 5.0. Combining E quations (12.27), (12.29), and (12.30) yields Q V
=
εr εo
50
A l
A nd, solving for V V= =
Ql εr εo
A (3 .5 × 10−11 C)(3.5 × 10−3 m)
(5 .0)(8.85 × 10−12 F/m)(160 mm 2 )(1 m 2 /106 mm 2 )
= 17.3 V
(b) For this same capacitor, if a vacuum is used Ql
V=
εo
A (3 .5 × 10−11 C)(3.5 × 10−3 m)
=
(8 .85 × 10−12 F/m)(160 × 10−6 m 2 )
= 86.5 V
(e) The polarization is determined using E quations (12.35) an d (12.6) as P =
(
εo εr
− 1)
V l −12
=
(8 .85 × 10
F/m)(5.0 − 1)(17.3 V )
3.5 × 10−3 m
= 1.75 × 10−7 C/m 2
12.D2 This prob lem asks that we dete rmine the e lectrical conductivity of an 80 wt% Cu-20 wt% Z n alloy at −150◦ C using information contained in Figures 12.8 an d 12.35. In order to solve this problem it is necessary to e mploy Equation (12.9) which is of the form total
=
t
+ i
since it is assumed that the alloy is undeformed. Let us first determine the value of i at room temperature (25◦ C), a value which will be independent of temperature. From Figure ( 12.8), at 25◦ C and for pure Cu, t (25) = 1.75 × 10−8 -m. Now, since it is assumed that the curve in Figure 12.35 was generated also at room temperature, we may take as total (25) at 80 wt% Cu-20 wt% Zn which ha s a value of 5 .3 × 10−8 -m. Thus i
=
total (25)
− t (25)
= 5.3 × 10−8 -m − 1.75 × 10−8 -m = 3.55 × 10−8 -m
Finally, we may determine the resistivity at −150◦ C, total ( −150), by tak ing the resistivity of p ure Cu at −150◦ C from Figure 12.8, which gives us t ( −150) = 0.55 × 10−8 -m. Therefore total ( −150)
=
i
+ t ( −150)
= 3.55 × 10−8 -m + 0.55 × 10−8 -m = 4.10 × 10−8 -m
And , using Equation (12.4) the conductivity is calculated as
=
1
=
1 4.10 × 10−8 -m
51
7 1 = 2.44 × 10 ( -m) −
CHAPTER 13 TYPES AND APPLICATIONS OF MATERIALS
13.5
We are asked to compute the volume percent graphite in a 3.5 wt% C cast iron. It first becomes necessary to compute mass fractions using the lever rule. From the iron-carbon phase diagram (Figure 13.2), the tie-line in the and graphite phase field extends from essentially 0 wt% C to 100 wt% C. Thus, for a 3.5 wt% C cast iron W
=
WG r
=
C Gr
−
CG r
−
Co
−
CG r
Co C
C
−
C
100 − 3.5
=
100 − 0 3.5 − 0
=
=
=
100 − 0
0.965
0.035
Conversion from weight fraction to volume fraction of graphite is possible using Equation (10.6a) as WG r VG r
=
Gr
W
+
W Gr
G r
0.035 2.3 g/cm 3 0.965 0.035
=
7.9 g/cm 3 =
+
2.3 g/cm 3
0.111 or 11.1vol%
13.21* In this problem we are asked to find the maximum temperatures to which magnesia-alumina refractories may be heated before a liquid phase will appear. (a) For a spinel-bonded alumina material of composition 95 wt% Al 2 O 3 -5 wt% MgO we must use Figure 10.22. According to this phase diagram, the maximum temperature without a liquid phase corresponds to the temperature of the eutectic isotherm on the A l2 O 3 -rich side of the phase diagram, which is approximately 2000 ◦ C (3630◦ F) . 13.23* This problem calls for us to compute the mass fractions of liquid for four refractory materials at 1600◦ C. In order to solve this problem it is necessary that we use the SiO 2 -A l2 O 3 phase diagram (Figure 10.24), in conjunction with tie-lines and the lever rule at 1600◦ C. (a) For
Co =
6 wt% Al 2 O 3 the mass fraction o f liquid WL
=
=
Co
−
C SiO 2
CL
−
C SiO 2
6−0 7−0
52
=
WL
0.86
is just
CHAPTER 14 SYNTHESIS, FAB RICATION, AN D PROCESSING OF MATERIALS
14.19 (a) This part of the problem calls for us to construct a r adial hardness profile for a 50 mm (2 in.) diameter cylindrical specimen of an 8640 steel tha t has bee n que nched in mo derat ely agitated o il. In the manner of E xample Prob lem 14.1, the equivalent distances and hardnesses tabulated below were determined from Figures 14.8 an d 14.11. R adial Position
E quivalent D istance, mm (in.)
HRC H ardness
Surface 3/4 R Midradius Center
7 (5/16) 11 (7/16) 14 (9/16) 16 (10/16)
54 50 45 44
The re sulting profile is plotted below.
14.26 (a) Below is shown the logarithm viscosity versus reciprocal of tempe ratur e plot for the borosilicate glass, using the data in Figure 14.16.
53
(b) Solving for t he activation energy, Q vis , from the equation given in the problem, we get Q vis
=
RT ln + RT ln A
The activation energy, Q vis , may be computed from this plot according to
Q vis
=
R
ln
1
T
where R is the gas constant, and ln /(1 /T) is the slope of the line that has been constructed. The value o f this slope is 4.36 × 104 . Therefore, Q vis
=
(8 .31 J/mol-K)(4.36 × 104 )
=
362,000 J/mol
14.43 (a) This problem asks that we determine how much adipic acid must be added to 50.0 kg of ethylene glycol to produce a linear chain structure of polyester according to Equation 14.5. Since the chemical formulas are provided in this equation we may calculate the molecular weights of each o f these mater ials as follows: A(adipic)
A(glycol)
=
6(A C ) + 10(A H )
=
6(12.01 g/mol) + 10(1.008 g/mol)
=
2(A C ) + 6(A H )
=
2(12.01 g/mol) + 6(1 .008 g/mol) + 2(16.00 g/mol)
+
+
4(A O ) +
4(16.00 g/mol)
=
146.14 g/mol
2(A O ) =
62.07 g/mol
50000 g
The 50.0kg massof ethylene glycolequals 50,000g or 62.07 g/mol = 805.5mol.Accordingto Equation (14.5), each mo le of adipic acid used re quires one mole of ethylene glycol, which is equivalent to (805.5 mol)(146.14 g/mol) = 1.177 × 105 g = 117.7 kg. (b) Now we are asked for the mass of the resulting polyester. Inasmuch as one mole of water is given off for every mer unit prod uced, this corresponds to 805.5 moles or (805.5 mol)(18.02 g/mol) = 14,500 g or 14.5 kg since the molecular weight of wate r is 18.02 g/mol. The m ass of polyester is just the sum of the masses of the two reactant materials (as computed in part a) minus the mass of water released, or mass(polyester)
=
50.0 k g + 117.7 k g − 14.5 k g
=
153.2 k g
14.D1 A one-inch diameter steel specimen is to be quen ched in modera tely agitated o il. We are t o decide which of five different steels will have surface and center hardnesses of at least 55 and 50 HRC, respectively. In mod erate ly agitated oil, the eq uivalent distances from t he quen ched end for a on e-inch diameter bar for surface and center p ositions are 3 mm (1 /8 in.) and 8 mm (11/32 in.), respectively [Figure 14.11(b)]. The hardnesses at these two positions for the alloys cited (as determined using Figure 14.8) are given below.
A lloy
Surface H ardness (H RC)
1040 5140 4340 4140 8640
50 55 57 56 56
Center H ardness (H RC) 30 47 57 54 52.5
Thus, alloys 4340, 4140, and 8640 will satisfy the criteria for both surface and center hardnesses.
54
14.D5 We are asked to determine the maximum diameter possible for a cylindrical piece of 4140 steel that is to be quenched in moderately agitated oil such that the microstructure will consist of at least 50% martensite throughout the entire piece. From Figure 14.8, the equivalent distance from the quenched end of a 4140 steel to give 50% martensite (or a 42.5 HRC hardness) is 26 mm (1–1/16 in.). Thus, the quenching rate at the center of the specimen should correspond to this equivalent distance. Using Figure 14.11(b), the center specimen curve tak es on a value of 26 mm (1–1/16 in.) equivalent distance at a diameter of about 75 mm (3 in.).
55
CHAPTER 15 COMPOSITES
15.4
This problem asksfor the maximum and minimum thermal conductivityvaluesfor a TiC-Co cermet. Using a modified form of E quation ( 15.1) the maximum thermal conductivity kc ( u) is calculated as k c (u )
=
k m V m
=
(69W/m-K)(0.15) + (27 W/m-K)(0.85)
+
k p V p
k Co V Co
=
+
k TiC V TiC =
33.3W/m-K
The minimum thermal conductivity kc ( l) will be k c (l)
=
= =
k Co k TiC V Co k TiC
+
V TiC k Co
(69 W/m-K)(27 W/m-K) (0 .15)(27 W/m-K)
+
(0 .85)(69 W/m-K)
29.7 W/m-K
15.12 This problem asks for us to det ermine if it is possible to produ ce a continuou s and oriented ara mid fiber-epoxy matrix composite having longitudinal and transverse mo duli of elasticity of 57.1 GPa and 4.12 GPa, respectively, given that the modulus of elasticity for the epoxy is 2.4 GPa. Also, from Table 15.4 the value of E for aramid fibers is 131 GPa. The approach to solving this problem is to calculate two values of V f using the data and E quations (15.10b) an d (15.16); if they are the same then this composite is po ssible. For the longitudinal modulus E cl , E cl
=
E m [1 − V fl ] + E f V fl
57.1 GPa
=
(2 .4 GPa)[1 − V fl ] + (131 G Pa)V fl
Solving this expression for V fl yields V fl = 0.425. Now, repeating this procedure for the transverse modulus E ct
=
4.12 GPa
=
Solving this expression for composite is possible.
V ft
E ct
E m E f [1 − V ft ]E f + V ft E m (2 .4 GPa)(131 GPa) [1 − V ft ](131 GPa)
leads to
V ft =
+
V ft (2 .4 GPa)
0.425. Thus, since
Vfl
an d
V ft
are equal, the proposed
15.17 The problem stipulates that the cross-sectional area of a composite, A c , is 320 mm 2 (0.50 in.2 ), and the longitudinal load, Fc , is 44,500 N (10,000 lb f ) for the composite described in Problem 15.11. (a) First, we are asked to calculate the F f Fm O r,
Ff =
=
E f V f E m Vm
Ff /Fm
=
ratio. According to E quation (15.11)
(131 GP a)(0.30) (2 .4 GPa)(0.70)
=
23.4
23.4Fm
(b) Now, the actual loads carried by bot h phases are called for. Since F f + F m 23.4F m
+
Fm
=
Fc
=
44,500 N
56
=
44,500 N
which leads to Fm
=
1824 N (410 lb f )
F f = 44,500 N
−
1824 N
42,676 N (9590 lb f )
=
(c) To compute the stress on each of the phases, it is first necessary to know the cross-sectional areas of both fiber and matrix. These are determined as A f = V f A c Am
=
(0 .30)(320 mm 2 )
=
VmA c
=
=
(0 .70)(320 mm 2 )
96 mm 2 (0 .15 in .2 )
=
224 mm 2 (0 .35 in .2 )
Now, for the stresses, f =
m =
F f A f Fm Am
42676 N
=
(96 mm 2 )
=
445 MPa (63,930 psi)
=
1824 N (224 mm 2 )
=
8.14 MPa (1170 psi)
(d) The strain on the composite is the same as the strain on each of the mat rix and fiber phases, as
εm
εf
=
=
m
Em f
E f
=
=
8.14 MPa 2.4
× 103
MP a
445 MPa 131
× 103
MP a
=
3.39 × 10−3
=
3.39 × 10−3
15.21 In this problem, for an aligned glass fiber-epoxy matrix composite,we are asked to comput e the longitudinal t ensile stren gth given the following: the average fiber diameter (0.010 mm), the average fiber length (2.5 mm), the volume fraction of fibers ( 0.40), th e fiber fracture strength (3500 MPa), th e fiber-matrix bond strength (75 MPa), and the matrix stress at composite failure (8.0 MPa). It is first necessary to compu te th e value of the critical fiber length u sing Equation (15.3). If the fibe r length is much greater than lc , then we may dete rmine cl∗ using Equation (15.17), o therwise, use of either Eq uation (15.18) or (15.19) is ne cessary. Th us, ∗
lc
=
f d
=
2c
(3500 MPa) (0.010 mm) 2(75 MPa)
Inasmuch as l > lc (2 .5 mm > 0.233 mm), but tion (15.18) is necessary. Therefore, ∗
∗
cd = f V f
1−
lc 2l
+ m (1 −
=
(3500 MPa)( 0.40) 1 −
=
1340 MPa (194,400 psi)
l
=
0.233 mm (0.0093 in .)
is not much greater than lc , then use of Equa-
V f ) 0.233 mm
(2)(2.5 mm)
+
(8 .0 MPa)(1 − 0.40)
15.D1 In or der t o solve this problem, we want to m ake longitudinal elastic modulus and ten sile strength computations assuming 50 vol% fibers for all three fiber materials, in order to see which meet the stipulated criteria [i.e., a minimum elastic modulus of 50 GPa (7.3 ×106 psi), and a minimum tensile strength of 1300 MPa (189,000 psi)]. Thus, it becomes necessary to use E quations (15.10b) ∗ an d (15.17) with V m = 0.5 and V f = 0.5, E m = 3.1 GPa, and m = 75 MPa.
57
For glass, E f = 72.5 GPa and E cl
=
E m (1 − V f ) + E f V f
=
(3 .1 GPa)(1 − 0.5)
+
∗
f =
3450 MPa. The refore,
(72 .5 GPa)(0.5)
=
37.8 GPa (5.48 × 106 psi)
Since t his is less than the specified m inimum, glass is not an acceptab le candidate. For carbon (PAN standard-modulus), E f = 230 GPa and f ∗ = 4000 MPa (the average of the range of values in Table B.4), thus E cl
=
(3 .1 GPa)(0.5)
+
(230 GP a)(0.5)
116.6 GPa (16.9 × 106 psi)
=
which is greater than the specified minimum. In addition, from E quation (15.17) ∗
cl = m (1 −
=
V f )
∗
+ f V f
(30 MPa)(0.5) + (4000 MPa) (0.5)
=
2015 MPa (292,200 psi)
which is also greater than the minimum. Thus, carbon (PAN standard-modulus) is a candidate. For aramid, E f = 131 G Pa and f ∗ = 3850 MPa (the average of the range of values in Table B.4), thus E cl
=
(3 .1 GPa)(0.5)
+
(131 GP a)(0.5)
=
67.1 GPa (9.73 × 106 psi)
which value is greater than the m inimum. Also, from Eq uation (15.17) ∗
cl = m (1 −
=
V f )
∗
+ f V f
(50 MPa)(0.5) + (3850 MPa) (0.5)
=
1950 MPa (283,600 psi)
which is also greater than the minimum strength value. Therefore, of the three fiber materials, both the carbon (PAN standard-modulus) and the ar amid meet b oth minimum criteria. 15.D3 This problem asks us to determine whether or not it is possible to produce a continuous and oriented glass fiber-reinforced polyester having a tensile strength of at least 1400 MPa in the longitudinal direction, and a maximum specific gravity of 1.65. We will first calculate the minimum volume fraction of fibers to give the stipulated tensile strength, and then the maximum volume fraction of fibers po ssible to yield the maximum per missible specific gravity; if there is an overlap of these two fiber volume fractions then such a composite is possible. With regard to tensile strength, from E quation (15.17) ∗
cl = m (1 −
1400 MP a
=
V f ) + f ∗ V f
(15 MPa)(1 − V f )
+
(3500 MPa) (V f )
Solving for V f yields V f = 0.397. Therefore, V f > 0.397 to give the minimum desired tensile strength. Now, upon considerat ion of the specific gravity, , we em ploy the following relationship: c = m (1 −
1.65
=
V f ) + f V f
1.35(1 − V f ) + 2.50(V f )
A nd, solving for V f from this expression gives V f = 0.261. Therefore, it is necessary for V f < 0.261 in order to have a composite specific gravity less than 1.65. Hence, such a composite is not possible since there is no overlap of the fiber volume fractions as comput ed using the t wo stipulated criteria.
58
CHAPTER 16 CORROSION AND D EGRA DATION OF MATERIALS
16.5 (a) We are asked to compute t he voltage of a nonstand ard Cd-Fe electrochemical cell. Since iron is lower in the emf series (Table 16.1), we will begin by assuming that iron is oxidized and cadmium is reduced, as Fe
+ Cd 2+ −→ Fe 2+ + Cd
an d 2
V
= [−0.403V − ( −0.440V)] − = −0.031V (b) Since the
V
+
[Fe ] log = (V ◦Cd − V ◦Fe ) − 0.0592 2 [Cd 2+ ] 0.0592 2
log
0.40 2
× 10−3
is negative, the sponta neous cell direction is just the r everse of that above, or Fe 2+
+ Cd −→ Fe + Cd 2+
16.13 This problem calls for us to compute the t ime of submersion of a steel piece. In order to solve this problem, we must first rearrange E quation (16.23), a s t
KW = A(CPR)
Thus, t
(534)(2.6 × 106 mg) = (7.9 g/cm 3 )(10 in .2 )(200 mpy)
= 8.8 × 104 h = 10 yr 16.20 (a) This portion of the problem asks that we compute the rate of oxidation for Pb given that bo th the oxidation and reduction reactions are controlled by activation polarization, and also given the polarization data for both lead oxidation and hydrogen reduction. The first thing necessary is to establish relationships of the form of Equation (16.25) for the potentials of both oxidation and reduction reactions. Next we will set these expressions equal to one another, and then solve for the value of i which is really the corrosion current density, ic . Finally, the corrosion rate may be calculated using E quation (16.24). The t wo pote ntial expressions are as follows: For hydrogen reduction VH
= V (H + /H ) + H log 2
i
io H
And for Pb oxidation V Pb
= V (Pb/Pb +) + Pb log 2
59
i
io Pb
Setting V H logic
=
=V
Pb
and solving for log i (log ic ) leads to
1 Pb
− H
[V (H + /H 2 )
1
= 0.12 − (−0.10) = −7.809
[0
− V (Pb/Pb +) − H logio + Pb logio 2
H
Pb
]
− (−0.126) − ( −0.10) {log(1.0 × 10−8) } + (0 .12) {log(2 × 10−9 )}]
Or ic
= 10−7 809 = 1.55 × 10−8 A/cm 2 .
And from E quation (16.24) r
= nFic =
1.55
× 10−8 C/s-cm2 = 8.03 × 10−14 mol/cm2-s
(2)(96500 C/mol)
16.34 For this problem we are given, for three metals, their densities, chemical formulas, and oxide densities, and are asked to compute the Pilling-Bedworth ratios, and then specify whether or not the oxide scales that form will be pro tective. The general form of the equation used to calculate this ratio is Eq uation (16.33) [or E quation (16.32)]. For t in, oxidation occurs by the reaction Sn
+ O 2 −→ SnO 2
and therefore P-B ratio
= AA SnO Sn 2
Sn SnO 2
(150.69 g/mol)(7.30 g/cm 3 )
= (118.69 g/mol)(6.95 g/cm 3) = 1.33 Hence, the film would most likely be protective since the ratio lies between one and two. 16.36 For this problem we are given weight gain-time data for the o xidation of Cu at an elevated te mperature. (a) We are first asked to determine whether the oxidation kinetics obey a parabolic, linear, or logarithmic rate expression, expressions which are described by Equations (16.34), (16.35), and (16.36), respectively. One way to make this determination is by trial and error. Let us assume that the parab olic relationship is valid; that is, from E quation (16.34) W2
= K1t + K2
which means that we may establish three simultaneous equations using the three sets of given W an d t values, then using two combinations of two pairs of eq uations, solve for K1 an d K2 ; if K1 an d K2 have the same values for both solutions, then the kinetics are parabolic. If the values are not identical then the other kinetic relationships need to be explored. Thus, the three equations are (0 .316) 2
= 0.100 = 15K 1 + K 2 (0 .524) 2 = 0.275 = 50K 1 + K 2 (0 .725) 2 = 0.526 = 100K 1 + K 2 60
From the first two equations K1 5 10−3 an d K2 0.025; these same two values are obtained using the last two equat ions. Hen ce, the o xidation r ate law is parabolic.
= ×
=
(b) Since a parabolic relationship is valid, this portion of the problem calls for us to determine after a total time of 450 min. Again, using E quation (16.34) and the values of K1 an d K2 W2
Or
W
=
√
2.28
= K1t + K2 = (0.005)( 450 min) + 0.025 = 2.28
= 1.51 mg/cm 2 .
61
W
CHAPTER 17 THERMAL PROP ERTIES
17.4
(a) For aluminum, Cv at 50 K may be approximated by Equation (17.2), since this temperature is significantly below the Debye temperature. The value of Cv at 30 K is given, and thus, we may compute the constant A as
A =
Cv T3
=
0.81 J/mol-K (30K) 3
= 3 × 10−5 J/mol-K 4
Therefore, at 50 K C v = AT 3 = (3 × 10−5 J/mol-K 4 )(50 K) 3 = 3.75 J/mol-K an d cv = (3 .75 J/mol-K)(1 mol/26.98 g)(1000 g/kg) = 139 J/kg-K (b) Since 425 K is above the Debye temperature, a good approximation for Cv is C v = 3R = (3)(8.31 J/mol-K) = 24.9 J/mol-K
And, converting this to specific heat cv = (24 .9 J/mol-K)(1 mol/26.98 g)(1000 g/kg) = 925 J/kg-K 17.14 This problem asks for us to determine the temperature to which a cylindrical rod of tungsten 10.000 mm in diameter must be heated in order for it to just fit into a 9.988 mm diameter circular hole in a plate of 316 stainless steel, assuming that the initial temperature is 25 ◦ C. This requires the use of Equation (17.3a), which is applied to the diameters of the rod and hole. That is d f − d o do
=
l
(T f − T o )
Solving this expression for df yields d f = d o [1 + l (T f − T o )] Now all we need do is to establish expressions for df (316 stainless) and df (W), set them equal to one another, and solve for Tf . According to Table 17.1, l (316 stainless) = 16.0 × 10−6 ( ◦ C) −1 an d −6 ◦ l (W ) = 4.5 × 10 ( C) −1 . Thus d f (316 stainless) = d f (W ) (9 .988 mm)[1 + {16.0 × 10−6 ( ◦ C) −1 }(T f − 25◦ C) ] = (10 .000 mm)[1 + {4.5 × 10−6 ( ◦ C) −1 }(T f − 25◦ C) ]
Now solving for Tf gives Tf = 129.5◦ C.
62
17.24 This problem asks that we treat a porous mater ial as a composite wherein one of the phases is a pore phase, and that we estimate upper and lower limits for the room-temperature thermal conductivity of a magnesium oxide material having a 0.30 volume fraction of pores. The upper limit of k (kupper) may be determined using E quation ( 15.1) with therm al conductivity substituted for the elastic modulus, E. From Table 17.1, the value of k for M gO is 37.7 W/m-K, while for still air in the pore pha se, k 0.02 W/m-K. Thu s =
k u pper = V p k air + V MgO k MgO = (0 .30)(0.02 W/m-K) + (0 .70)(37.7 W/m-K) = 26.4 W/m-K
For the lower limit we employ a modification of E quation (15.2) as k lower = =
k air k MgO V p k MgO + V MgO k air (0 .02 W/m-K)(37.7 W/m-K) (0 .30)(37.7 W/m-K) + (0 .70)(0.02 W/m-K)
= 0.067 W/m-K
17.29 We want to heat the copper wire in order to reduce the stress level from 70 MPa to 35 MPa; in doing so, we reduce the stress in the wire by 70 MPa − 35 MPa = 35 MPa, which will be a compr essive stress ( i.e., = −35 MPa). Taking a value for E of 110 GPa (Table 7.1) and solving for Tf from Eq uation (17.8) T f = T o −
E 1
= 20◦ C −
−35 MPa
(110 × 103 MPa)[17 × 10−6 ( ◦ C) −1 ]
= 20◦ C + 19◦ C = 39◦ C (101◦ F)
17.D 1 This problem stipu lates tha t 1025 stee l railroad t racks are laid at a tem per atu re of 10°C. We a re asked to determine the hottest possible temperatur e that can be tolerated without the introduction of ther mal stresses if the rails are 11.9 m long, and for a joint space of 4.6 mm. For these r ailroad tra cks, each end is allowed to expan d one -half of the joint space distan ce, or th e track ma y expan d a tot al of this distance (4.6 mm). Equation (17.3a) is used to solve for Tf , where l for the 1025 steel is foun d in Table 17.1. Thus, T f = =
l l
lo
+ To
4.6 × 10−3 m [12.0 × 10−6 ( ◦ C) −1 ](11.9 m)
+ 10◦ C
= 32.2◦ C + 10◦ C = 42.2◦ C (108◦ F)
63
CHAPTER 18 MAGNETIC PROPERTIES
18.1 This problem concerns a coil of wire 0.20 m long that has 200 turns and carries 10 A . (a) We may calculate the m agnetic field strength genera ted b y this coil using Equation (18.1) as H
=
=
NI l (200 turns)(10 A) 0.2 m
=
10,000 A-turns/m
(b) In a vacuum, the flux density is determined from Equation (18.3). Thus, Bo
= o H =
(1 .257 × 10−6 H/m)( 10,000 A -turns/m)
=
1.257 × 10−2 tesla
(c) When a bar of titanium is positioned within the coil, we must use an expression that is a combination of E quations ( 18.5) an d (18.6) in order t o compute the flux density given the magnetic susceptibility. Inasmuch as m = 1.81 × 10−4 (Table 18.2), then B
= o H + o M = o H + o m H = o H (1 + m ) =
(1 .257 × 10−6 H/m)( 10,000 A-turn s/m)(1 + 1.81 × 10−4 )
∼ =
1.257 × 10−2 tesla
which is essentially the same result as part (b). This is to say that the influence of the titanium bar within the coil makes an imperceptible difference in the magnitude of the B field. (d) The magnetization is computed from Equation (18.6): M
= m H =
(1 .81 × 10−4 )(10,000 A-turns/m)
=
1.81A/m
18.4 For this problem, we want to convert the volume susceptibility of silver (i.e., 2.38 × 10−5 ) into other systems of units. For t he mass susceptibility m
(kg)
=
=
m (kg/m 3 ) −2.38 ×
10.49
10−5
× 103
kg/m 3
= −2.27 ×
10−9
For th e a tomic susceptibility m (a ) = m =
(kg) × [atomic weight (in kg)]
( −2.27 × 10−9 )( 0.10787 kg/mol)
= −2.45 ×
10−10
18.6 This problem stipulates that t he magnet ic flux density within a bar of some mater ial is 0.435 tesla at an H field of 3.44 × 105 A/m.
64
(a) We are first of all asked to compute the magnetic permea bility of this mater ial. This is possible using Equation (18.2) as =
B H
0.435 tesla
=
3.44
× 105
A /m
=
1.2645 × 10−6 H /m
(b) The magnet ic susceptibility is calculated as
m = =
−
1
=
o
1.2645 × 10−6 H /m 1.257 × 10−6 H /m
−
1
6 × 10−3
18.27 (a) The B -H data provided in the problem are plotted below. 1.6
1.2
) a l s e t (
0.8
B
0.4
0.0 0
200
400
600
800
1000
1200
H (A/m)
(b) This portion of the problem asks for us to determine values of the initial permeability and initial relative p ermea bility. The first four data points are plotted below. 0.3
0.2 ) a l s e t ( B
0.1
µ
i
0.0 0
10
20
30
H (A/m)
65
40
50
60
The slope of the initial portion of the curve is i =
B
(as shown), is
(0 .15 − 0) tesla
=
H
i
(50 − 0) A/m
=
3.0 × 10−3 H /m
A lso, the initial relative perme ability [Equation (18.4)] is just ri =
i
=
o
3.0 × 10−3 H /m
=
1.257 × 10−6 H /m
2400
(c) The maximum permeability is the tangent to the B -H curve having the greatest slope; it is drawn on the plot below, and designated as (max) . 1.6
µ(max)
1.2
) a l s e t (
0.8
B
0.4
0.0 0
200
400
600
800
1000
1200
H (A/m)
The value of (max) is (max) =
B H
=
(1 .3 − 0.3) tesla (160 − 45) A-m
=
8.70 × 10−3 H /m
18.32 (a) Given Eq uation (18.12) and the data in Table 18.7, we are asked to calculate the critical magnetic fields for tin at 1.5 and 2.5 K. From the table, for Sn, TC = 3.72 K and B C (0) = 0.0305 te sla. Thu s, from Equation (18.3) H C (0)
=
B C (0) o
=
0.0305 tesla 1.257 × 10−6 H /m
=
2.43 × 104 A /m
Now, solving for H C (2 .5) using E quation (18.12) yields
H C (T )
H C (2 .5)
=
=
H C (0) 1 −
4
T2 T 2C
(2 .43 × 10 A/m) 1 −
66
(2 .5 K ) 2 (3 .72
K) 2
=
1.33 × 104 A /m
(b) Now we are to determine the temperature to which lead must be cooled in a magnetic field of 20,000 A/m in orde r for it t o be superconductive. The value of H C (0) must first be determined using B C (0) given in the table (i.e., 0.0803 tesla); thus H C (0) Since TC
=
=
B C (0)
=
o
0.0803 t esla 1.257 × 10−6 H /m
=
6.39 × 104 A /m
7.19 K we may solve for T using Equation (18.12) as
T
=
=
TC
1−
H C (T ) H C (0)
(7 .19 K) 1 −
20000 A /m 63900 A /m
67
=
5.96 K
CHAPTER 19 OPTICAL PROPERTIES
19.9 We want to compute the velocity of light in calcium fluoride given that εr 2.056 and m 1.43 10−5 . The velocity is determined using Equation (19.8); but first, we must calculate the values of ε an d for calcium fluoride. According to E quation (12.30)
−
=
×
=
ε
εr εo
=
= (2 .056)(8.85 × 10−12 F/m) = 1.82 × 10−11 F/m
Now, ut ilizing E quations (18.4) an d (18.7)
= o ( m + 1) = (1 .257 × 10−6 H/m)(1 − 1.43 × 10−5 ) = 1.257 × 10−6 H /m
And, finally v
= √ 1 ε
1
=
(1 .82
× 10−11 F/m)(1.257 × 10−6 H/m)
= 2.09 × 108 m/s 19.11 This problem asks for us, using data in Table 19.1, to estimate the d ielectric constants for silica glass, soda-lime glass, PTFE , polyethylene, and p olystyrene, and t hen to compa re t hese values with t hose cited in Table 12.4 and briefly explain any discrepan cies. From E quation (19.10) εr
Thus, for fused silica, since n
= 1.458 εr
= n2
= (1 .458) 2 = 2.13
When we compare this value with that given in Table 12.4 at a frequency of 1 MHz (i.e., εr 3.8) there is a significant discrepancy. The reason for this is that, for this material, an ionic component to the dielectric constant is present at 1 MHz, which is absent at frequencies within the visible electromagnetic spectrum, frequencies which are on the order 10 9 MHz. This effect may be noted in Figure 12.32.
=
19.19 In this problem we are asked to calculate the fraction of nonreflected light transmitted through a 20 mm thickness of transparent material, given that the fraction transmitted through a 10 mm width is 0.90. From E quation (19.18), the fraction of nonr eflected light transmitted is just IT /Io . Using this expression we must first determine the value of as
=− 1 x
=−
I
Now, solving for IT when o
x
IT
ln
Io
1
10 mm
ln(0.90)
= 1.05 × 10−2 mm −1
= 20 mm I T I o
exp[ (1 .05
−
= exp( −x)
× 10−2 mm −1)(20 mm)] = 0.81 68
19.30 This problem asks for the difference in energy between metastable and ground electron states for a rub y laser. The wavelength of the radiation em itted by an electron t ransition from the m etastable to ground state is cited as 0.6943 m. The difference in energy between these states, E , may be determined from Equation (19.6), a s E
= h = hc −15 × 108 m/s) = (4 .13 × 106.943eV-s)(3 × 10−7 m = 1.78 eV
69
CHAPTER 20 MATERIALS SELECTION A ND D ESIGN CONSID ERATIONS
20.D3 (a) This portion of the problem asks that we derive a p erformance index expression for strength analogous to E quation ( 20.9) for a cylindrical cantilever beam th at is stressed in the manne r shown in the accompanying figure. The stress on the unfixed end, , for an imposed force, F, is given by the expression [Equ ation (20.24) in the textbook]
= FLI r
(20.D1)
where L an d r are the rod length and radius, respectively, and I is the moment of inertia; for a cylinder the expression for I is provided in Figure 7.18:
I
= 4r
4
(20.D2)
Substitution for I into Equation (20.D1) leads to
= 4FL r 3
(20.D3)
Now, the mass m of some given quantity of material is the product of its density ( ) and volume. Inasmuch a s the volume of a cylinder is just r2 L, then m
= r 2 L
(20.D4)
From this expression, the radius is just
r
=
m
(20.D5)
L
Inclusion of Equation (20.D5) into Equation (20.D3) yields 1/2
5/2 3/2
= 4F mL3/2
(20.D6)
A nd solving for the mass gives m
= (16 F 2L 5) 1/3 2/3
(20.D7)
To en sure tha t th e bea m will not fail, we re place stress in Equation (20.D7) with t he yield strength ( y ) divided by a factor of safety (N) as m
= (16 F 2L 5N 2) 1/3
70
2/3
y
(20.D8)
Thus, the best mat erials to be used for t his cylindrical cantilever beam when strength is a consideration are those having low 2/3 ratios. Furthermore, the strength performance index, P, is just the y reciprocal of this ratio, or 2/3
P
=
y
(20.D9)
The second portion of the problem asks for an expression for the stiffness performance index. Let us begin by consideration of Equation (20.25) which relates , the elastic deflection at the unfixed end, to the force ( F), beam length (L), the modulus of elasticity ( E), and moment of inertia (I) as 3
= FL 3E I
(20.25)
Again, Equ ation (20.D2) gives an expression for I for a cylinder, which when substituted into Eq uation (20.25) yields
4FL 3
(20.D10)
= 3E r 4
And, substitution of the expression for r [Equation (20.D5)] into Equation (20.D10), leads to
= =
4FL 3 3E
m
4
L
4FL 5 2
(20.D11)
3E m 2
Now solving this expression for the mass m yields
m
=
4FL 5 3
1/2
√
E
(20.D12)
Or, for this cantilever situation, the mass of material experiencing a given deflection produced by a specific force ispro portional to the √ ratio for that material.A nd, finally, the stiffness performa nce E index, P, is just the reciprocal of thisr atio, or
P
√ =
E
(20.D13)
(b) He re we are asked to select those metal alloys in the datab ase that have stiffness performa nce indices greater than 3.0 (in SI units). ( Note: for this performance index of 3.0, density has been taken in terms of g/cm3 rather than in the SI units of kg/m 3 .) Seventeen metal alloys satisfy this criterion; they and their E / values are listed below, and r anked from highest to lowest value.
√
71
A lloy
√
E
Condition
A Z 31B Mg A Z 31B Mg A Z 91D Mg 356.0 A l 356.0 A l 356.0 A l 6061 A l 6061 A l 6061 A l 2024 A l 2024 A l 2024 A l 1100 A l 1100 A l 7075 A l 7075 A l 7075 A l
R olled E xtruded A s cast A s cast, high production A s cast, custom T6 O T6 T651 O T3 T351 O H 14 O T6 T651
3.790 3.790 3.706 3.163 3.163 3.163 3.077 3.077 3.077 3.072 3.072 3.072 3.065 3.065 3.009 3.009 3.009
√
(c) We are now asked to do a cost analysis on the above alloys. Below are tabulated the / E ratio, the relative material cost ( c¯ ), and the product of these two parameters; also those alloys for which cost data are provided are ranked, from least to most expensive.
A lloy A Z 91D Mg 6061 A l 356.0 A l 6061 A l A Z 31B Mg 1100 A l A Z 31B Mg 7075 A l 2024 A l 356.0 A l 356.0 A l 2024 A l 1100 A l 2024 A l 6061 A l 7075 A l 7075 A l
Condition
√
c¯
0.2640 0.3250 0.3162 0.3250 0.2640 0.3263 0.2640 0.3323 0.3255 0.3162 0.3162 0.3255 0.3263 0.3255 0.3250 0.3323 0.3323
5.4 7.6 7.9 8.7 12.6 12.3 15.7 13.4 14.1 15.7 16.6 16.2 – – – – –
E
A s cast T6 A s cast, high production T651 E xtruded O R olled T6 T3 A s cast, custom T6 T351 H 14 O O O T651
c¯
√
E
1.43 2.47 2.50 2.83 3.33 4.01 4.14 4.45 4.59 4.96 5.25 5.27 – – – – –
It is up to the student to select the best metal alloy to be used for this cantilever beam on a stiffness-per-mass basis, including th e element of cost, and other relevant considerat ions.
72
