Fundamentals of Materials Science and Engineering 5th Ed - Solutions

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C H A PT E R 2

ATOMIC STRUCTURE AND INTERATOMIC BOND ING

2.3 (a) In order to determine the number of grams in one amu of material, appropriate manipulation ofthe amu/atom , g/mol, and atom/mol r elationships is all that is necessary, as

#g/amu = 1 mol

6.023 1023 atoms 1 g/mol

1 amu/atom

= 1.66 1024 g/amu

2.14 (c) This port ion of the problem asks that we deter mine for a K+-Cl ion pair the interato mic spacing(ro ) and the bonding energy (Eo ). From Equation (2.11) for EN

A = 1.436

B = 5.86 106

n = 9

Thus, using the solutions from Prob lem 2.13

ro =

A

nB

1/(1n)

=

1.436

(9)(5.86 106)

1/(19)= 0.279 nm

an d

E o = 1.436

1.436

(9)(5.86 106)

1/(19) + 5.86 106

1.436

(9)(5.86 106)

9/(19)

= 4.57 eV

2.19 The per cent ionic character is a function o f the electrone gativities of the ions XA an d XB accordingto Equation (2.10). The electronegativities of the elements are found in Figure 2.7.

For TiO 2, XTi = 1.5 and XO = 3.5, and therefore,

% IC =

1 e(0.25)(3.51.5)2 100 = 63.2%

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CHAPTER 3

STRUCTURE S OF METALS A ND CERAMICS

3.3 For this problem, we are asked to calculate the volume of a unit cell of aluminum. Aluminum has anFCC crystal structure (Table 3.1). The FCC un it cell volume may be compu ted from Equation (3.4)as

VC = 16R 32 = (16)(0.143 109 m) 32 = 6.62 1029 m3

3.7 This problem calls for a demonstration that the A PF for HCP is 0.74. Again, the A PF is just thetotal 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, the un it cell volume is just the produ ct of the ba se area t imes the cell height, c. This base areais just three times the area of the parallelepiped ACDE shown below.

AB

C D

E

a = 2R

a = 2R

a = 2R

60

30

The area of ACDE is just the length of CD times the height BC. But CD is just a or 2R,an d

BC = 2R cos(30) = 2R

3

2

Thus, the base ar ea is just

A R E A = (3)(CD)(BC) = (3)(2R)

2R32

= 6R 2

3

and since c = 1.633a = 2R(1.633)

VC = (AREA)(c) = 6R 2c

3 = (6R 2

3)(2)(1.633)R = 12

3(1.633)R 3

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Thus,

A PF = VSVC

= 8R3

12

3(1.633)R 3= 0.74

3.12. (a) This portion of the problem asks that we compute the volume of the unit cell for Zr. This volumemay be computed using Equation (3.5) as

VC =nA Zr

NA

Now, for HCP, n = 6 atoms/unit cell, and for Zr, A Zr = 91.2 g/mo l. Thu s,

VC =(6 at oms/unit cell)(91.2 g/mol)

(6.51 g/cm3)( 6.023 1023 atoms/mol)= 1.396 1022 cm3/un it cell = 1.396 1028 m3/un it cell

(b) We are now to compute the values of a an d c, given that c/a = 1.593. From the solution toProb lem 3.7, since a

=2R, then, for H CP

VC =3

3a2c

2

but, since c = 1.593a

VC =3

3(1.593)a3

2= 1.396 1022 cm3/un it cell

Now, solving for a

a =

(2)(1.396 1022 cm3)(3)(

3)(1.593)

1/3

= 3.23 108 cm = 0.323 nm

And finally

c = 1.593a = (1.593)(0.323 nm) = 0.515 nm

3.17 In this problem we are given that iodine has an orthorhombic unit cell for which the a, b, and clattice 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, respec-tively we are to determine the number of atoms in each unit cell. From the definition of theA PF

A PF = VSVC

=n

4

3R 3

ab c

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we may solve for the number of atoms per unit cell, n, a s

n = (APF)abc4

3R 3

= (0.547)(4.79)(7.25)(9.78)(1024 cm3)

4

3(1.77

108 cm) 3

= 8.0 atom s/unit cell

(b) In order to compute the density, we just employ Equation (3.5) as

= nA IabcNA

= (8 a toms/unit cell)(126.91 g/mol)[(4.79)(7.25)(9.78) 1024 cm3/un it cell]( 6.023 1023 atoms/mol)

= 4.96 g/cm3

3.22 This question asks that we generate a three-dimensional unit cell for AuCu3 using the MoleculeD efinition File on the CD-ROM. One set of directions that may be used to construct this unit celland that are entered on the No tepad are as follows:

[DisplayProps]Rotatez=30Rotatey=15

[AtomProps]Gold=LtRed,0.14Copper=LtYellow,0.13

[BondProps]SingleSolid=LtGray

[Atoms]Au 1=1,0,0,GoldAu 2=0,0,0,GoldAu 3=0,1,0,GoldAu 4=1,1,0,GoldAu 5=1,0,1,GoldAu 6=0,0,1,GoldAu 7=0,1,1,GoldAu 8=1,1,1,GoldCu 1

=0.5,0,0.5,Copper

Cu 2=0,0.5,0.5,CopperCu 3=0.5,1,0.5,CopperCu 4=1,0.5,0.5,CopperCu 5=0.5,0.5,1,CopperCu 6=0.5,0.5,0,Copper

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[Bonds]B1=Au1,Au5,SingleSolidB2=Au5,Au6,SingleSolidB3=Au6,Au2,SingleSolidB4=Au2,Au1,SingleSolidB5=Au4,Au8,SingleSolidB6=Au8,Au7,SingleSolidB7=Au7,Au3,SingleSolidB8

=Au3,Au4,SingleSolid

B9=Au1,Au4,SingleSolidB10=Au8,Au5,SingleSolidB11=Au2,Au3,SingleSolidB12=Au6,Au7,SingleSolid

When saving the se instructions, the file name that is chosen should end with a period followed bymdf and the entire file name needs to be enclosed within quotation marks. For example, if onewants 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 coordi-nation number of six is 0.414. Below is shown one of the faces of the rock salt crystal structure inwhich a nions and cations just t ouch along t he e dges, and also the face diagonals.

G H

F

rC

rA

From triangle FGH,

G F = 2rA and FH = G H = rA + rC

Since FGH is a right t riangle

(G H )2 + (FH)2 = (FG)2

or

(r A + rC )2

+ (r A + rC )2

= (2r A )2

which leads to

rA + rC =2rA

2

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Or , solving for rC/rA

rC

rA=

22

1

= 0.414

3.29 This prob lem calls for us to predict crystal structure s for several ceramic materials on the basis ofionic charge a nd ionic radii.

(a) For CsI, from Table 3.4

rCs+

rI= 0.170 nm

0.220 nm= 0.773

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

rK+

rI= 0.138 nm

0.220 nm= 0.627

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 dis-tance and bond angle are 0.154 nm and 109.5, respectively. The first thing we need do is todetermine the unit cell edge length from the given CC distance. The drawing below showsthe cubic unit cell with those carbon atoms that bond to one another in one-quarter of the unitcell.

x

y

a

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 = 0.154 nm, the carbon-carbon bond distance.Furthermore, x = a/4, and therefore,

x = a4

= ysin

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O r

a = 4y sin = (4)(0.154 nm)( sin 35.25) = 0.356 nm= 3.56 108 cm

The u nit cell volume, VC, is just a3, that is

VC = a3

= (3.56 108

cm)3

= 4.51 1023

cm3

We must now utilize a modified Equation (3.6) since there is only one atom type. There are eightequivalent atoms per unit cell (i.e., one equivalent corner, three equivalent faces, and four interioratoms), and therefore

= nA C

VC NA

= (8 at oms/unit cell)(12.01 g/g-atom)(4.51 1023 cm3/un it cell) (6.023 1023 atoms/g-atom)

= 3.54 g/cm3

The measured density is 3.51 g/cm3.

3.39 (a) We are asked to compute the den sity of CsCl. Modifying the result ofProb lem 3.4, we get

a = 2rCs+ + 2rCl3

= 2(0.170 nm) + 2(0.181 nm)3

= 0.405 nm = 4.05 108 cm

From Equation (3.6)

=n(A Cs +

A Cl)

VC NA=

n(A Cs +A Cl)

a3NA

For th e CsCl crystal structure, n = 1 formula unit/unit cell, and thus

= (1 formula unit/unit cell)(132.91 g/mol + 35.45 g/mol)(4.05 108 cm) 3/un it cell( 6.023 1023 formula u nits/mol)

= 4.20 g/cm3

(b) This value of the density is greater than the measured density. The reason for this discrepancyis 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 ( VC) 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 cellvolume. From Figure 3.6 there isth e equivalent of one Cs and one Cl ion per unit cell; the ionic radiiof these two ions are 0.170 nm and 0.181 nm, respectively (Table 3.4). Thus, the sphere volume, VS,

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is just

VS =4

3()[(0.170 nm)3 + (0.181 nm)3] = 0.0454 n m3

For CsCl the u nit cell edge length, a, in terms of th e ato mic radii is just

a = 2rCs+ + 2rCl3

= 2(0.170 nm) + 2(0.181 nm)3

= 0.405 nm

Since VC = a3

VC = (0.405 nm)3 = 0.0664 n m3

And, finally the a tomic packing factor is just

A PF = VSVC

= 0.0454 nm3

0.0664 nm3= 0.684

3.50 (a) We are asked for the indices of the two directions sketched in the figure. For direction 1, theprojection 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.

x y z

Projections 0a b/2 cProjections in terms ofa, b, and c 0 1/2 1R eduction to integers 0 1 2E nclosure [012]

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.

y

z

x

[110]

_

[121]

__

[012]_

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.

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We first of all position the origin of the coordinate system at the tail of the direction vector; thenin term s of this new coordinate system

x y z

Projections2a

3b 2c

3

Projections in terms ofa, b, and c

2

3 12

3

R eduction to integers 2 3 2E nclosure [232]

Direction D is a [136] direction, the determination of which is summarized as follows. Wefirst of all position the origin of the coordinate system at the tail of the direction vector; then interms of this new coordinate system

x y z

Projectionsa

6

b

2

c

Projections in terms ofa, b, and c1

6

1

21

R eduction to integers 1 3 6E nclosure [136]

3.56 This prob lem asks that we deter mine the Miller indices for planes that have been drawn within a unitcell. 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 z

Intercepts a b c2

Intercepts in term s ofa, b, and c 1 1 12

Reciprocals of intercepts 1 1 2E nclosure (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 summarizedbelow.

x y z

Intercepts a2

b2

c

Intercepts in term s ofa, b, and c1

2

1

21

R eciprocals of intercepts 2 2 1

E nclosure (221)

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3.59 The (1101) plane in a hexagonal unit cell is shown below.

z

a1

a2

a3

(1101)

_

3.60 This problem asks that we specify the Miller indices for planes that have been dr awn within hexag-onal 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.

a1 a2 a3 z

Intercepts a a a cIntercepts in terms ofas and c 1 1 R ecipr ocals of inter cepts 1 1 0 0E nclosure (1100)

3.61 This problem asks for us to sketch several planes within a cubic unit cell. The (011) and (102) planesare indicated below.

(102)

_

z

x

y

(011)

__

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3.63 This problem asks that we represent specific crystallographic planes for various ceramic crystalstructures.

(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 crys-tallographic 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 (andtherefore the distances between adjacent atoms) are the same (namely 0.40 nm 0.30 nm ), whichare 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 ofanions.

(a) The stacking sequence of close-packed planes of anions for the zinc blende crystal structurewill 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 tetrahedralsites 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 theface-centered cubic crystal structure. For FCC t he linear density of the [100] direction is computedas follows:

The linear density, LD , is defined by the ratio

LD = LcLl

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 tothe atomic radius R according t o a = 2R2 [Equation (3.1)]. Also for this situation, Lc = 2R an dtherefore

LD = 2R2R

2

= 0.71

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3.73* In this problem we are to compute the planar densities of several crystallographic planes for thebody-centered cubic crystal structure. Planar density, PD , is defined as

PD = A cA 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 sameplane. For (110), that portion of a plane that passes through a BCC unit cell forms a rectangle as

shown below.

3

4RR

3

24R

In terms of the atomic radius R, the length of the rectangle base is 4R

23

, whereas the height is

a = 4R3

. Therefore, the area of this rectangle, which is just A p is

A p

= 4R

2

3 4R

3 =16R 2

2

3

Now for the number equivalent atoms within this plane. One-fourth of each corner atom and theentirety of the center atom belong to the unit cell. Therefore, there is an equivalent of 2 atomswithin th e unit cell. Hen ce

A c = 2(R 2)

an d

PD = 2R2

16R 2

2

3

= 0.83

3.80* Using the data for aluminum in Table 3.1, we are asked to compute the interplanar spacings forthe (110) and (221) sets of planes. From the table, aluminum has an FCC crystal structure and anatomic 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)(

2) = 0.4047 nm

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Now, the d110 interplanar spacing may be determined using Equation (3.11) as

d110 =a

(1) 2 + (1) 2 + (0) 2= 0.4047 nm

2= 0.2862 nm

3.84* From the diffraction pattern for -iron shown in Figure 3.37, we are asked to compute the inter-planar spacing for each set of planes that has been indexed; we are also to determine the latticeparameter of Fe for each peak. In order to compute the interplanar spacing and the lattice pa-

rameter we must employ E quations ( 3.11) an d (3.10), respectively. For the first peak which occursat 45.0

d110 =n

2sin = (1)(0.1542 nm)

(2)

sin

45.0

2

= 0.2015 nm

A nd

a = dhkl

(h )2 + (k )2 + (l) 2 = d110

(1) 2 + (1) 2 + (0) 2

= (0.2015 nm )

2 = 0.2850 nm

Similar computations are made for the other peaks which results are tabulated below:

Peak Index 2 dhkl (nm) a (nm)

200 65.1 0.1433 0.2866211 82.8 0.1166 0.2856

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CHAPTER 4

POLYMER STRUCTUR ES

4.4 We are asked to comput e the numbe r-average degree of polymerization for polypropylene, given thatthe n umber-average molecular weight is 1,000,000 g/mol. The mer molecular weight of polypropyleneis just

m = 3(A C ) + 6(A H )= (3)(12.01 g/mol) + (6)(1.008 g/mol) = 42.08 g/mol

If we let nn represent the number-average degree of polymerization, then from Equation (4.4a)

nn =Mn

m= 10

6 g/mol

42.08 g/mol= 23,700

4.6 (a) From the tabulated data, we are asked to compute Mn, the number-average molecular weight.This is carried out below.

Molecular wt

R ange Mean Mi xi xiMi

8,00016,000 12,000 0.05 60016,00024,000 20,000 0.16 320024,00032,000 28,000 0.24 672032,00040,000 36,000 0.28 10,08040,00048,000 44,000 0.20 880048,00056,000 52,000 0.07 3640

Mn =

xiMi = 33,040 g/mol

(c) Now we are asked to compute nn (the number-average degree of p olymerization), using theEquation (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 =Mn

m= 33040 g/mol

42.08 g/mol= 785

4.11 This problem first of all asks for us to calculate, using Equat ion (4.11), the average total chainlength, L, for a linear polytetrafluoroethylene polymer having a number-average molecular weightof 500,000 g/mol. It is necessary to calculate the numbe r-average de gree of po lymerization, nn, usingEquation (4.4a). For PTFE, from Table 4.3, each mer unit has two carbo ns and four fluorines. Thus,

m=

2(AC

)+

4(AF

)

= (2)(12.01 g/mol) + (4)(19.00 g/mol) = 100.02 g/mol

an d

nn =Mn

m= 500000 g/mol

100.02 g/mol= 5000

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which is the number of mer units along an average chain. Since there are two carbon atoms per merunit, there are two CC chain bonds per mer, which means that the total number of chain bonds inthe molecule, N, is just (2)(5000)= 10,000 bonds. Furt hermo re, assume that for single carbo n-carbonbonds, d = 0.154 nm and = 109 (Section 4.4); therefore, from Equation (4.11)

L = Nd sin

2

= (10,000)(0.

154 nm)

sin109

2 = 1254 nm

It is now possible to calculate the average chain end-to-end distance, r, using Equa-tion (4.12) as

r = d

N = (0.154 nm)

10000 = 15.4 n m

4.19 For a poly(styrene-butadiene) alternating copolymer with a number-average molecular weight of1,350,000 g/mol, we are asked to determine the average number of styrene and butadiene mer unitsper molecule.

Since it is an alternating copolymer, the number of both types of mer units will be thesame. 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, whilethe butadiene mer consists of four carbon atoms and six hydrogen atoms. Therefore, the styrene-butadiene 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 =Mn

m= 1350000 g/mol

158.23 g/mol= 8530

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 tocompute the density of totally crystalline polyethylene. In orde r t o solve this problem it is necessaryto 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, VC is the unit cell volume, which is just the product of the three unit cell edge lengths inFigure 4.10. Thus,

= nAVC NA

= (2 m ers/uc)(28.05 g/mol)(7.41 108 cm)(4.94 108 cm)(2.55 108 cm)/uc(6.023 1023 mers/mol)

= 0.998 g/cm3

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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 employEquation (5.1). As stated in the problem, QV = 0.55 eV/atom. Thus,

NVN

= expQ VkT

= exp 0.

55 eV/atom(8.62 105 eV/atom-K)(600 K)

= 2.41 105

5.4 This problem calls for a det ermination of the number of atoms per cubic meter of aluminum. Inorder to solve this problem, one must employ Equation (5.2),

N = NAA lA A l

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,

N = (6.023 1023 atoms/mol)(2.71 g/cm3)

26.98 g/mol

= 6.05 1022 atoms/cm3 = 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 isat 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 unitcell edge length (a) and the radii of the two host atoms that are located on either side of the site ( R);

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that is

2r = a 2R

However, for FCC a is related to R according t o Equation (3.1) as a = 2R

2; therefore, solving forr gives

r=

a

2R

2 =2R

2

2R

2 =0.41R

5.10 (a)For Li+ substituting for Ca2+ inCaO, oxygen vacancieswould be created.For each Li+ substitutingfor Ca2+, one positive charge is removed; in order to maintain charge neutrality, a single negativecharge may be removed. Negative charges are eliminated by creating oxygen vacancies, and for everytwo 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 gtin and 65 g of lead. The concentration of an element in an alloy, in atom percent, may be computedusing Equation (5.5). With this problem, it first becomes necessary to compute the number of molesof both Sn and Pb, for which Equation (5.4) is employed. Thus, the numbe r o f moles of Sn is just

nmSn =m

Sn

A Sn =98 g

118.

69 g/mol = 0.

826 mol

Likewise, for Pb

nmPb =65 g

207.2 g/mol= 0.314 mol

Now, use ofEquation (5.5) yields

CSn =nmSn

nmSn + nmPb 100

=0.826 mol

0.

826 mol + 0.

314 mol 100 = 72.

5 at%

Also,

CPb =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 thatthe resultant alloy will contain 1.55 1022 Nb atoms per cubic centimeter. To solve this problem,employment ofEquation (5.18) is necessary, using the following values:

N1 = NNb = 1.55 1022 atoms/cm 3

1 = Nb = 8.

57 g/cm3

2 = V = 6.10 g/cm3

A 1 = A Nb = 92.91 g/molA 2 = A V = 50.94 g/mol

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Thus

CNb =100

1 + NAVNNb A Nb

VNb

= 100

1

+

(6.023 2023 atoms/mole)(6.10 g/cm3)(1.55

1022 atoms/cm 3)(92.91 g/mol)

6.10 g/cm3

8.57 g/cm3= 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 vectorand are asked to give Burgers vector representations for specific crystal structures, and then tocompute Burgers vector magnitudes.

(a) The Burgers vector will point in that direction having the highest linear density. From Problem3.70 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=

0.1431 nm (Table 3.1) an d a=

2R

2=

0.4047 nm

[Equation (3.1)]; therefore

b = a2

h2 + k2 + l2

= 0.4047 n m2

(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 foran A STM grain size of 4. From Equation (5.16), n = 4, and

N = 2(n1) = 2(41) = 23 = 8

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C H A PT E R 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 edensities of carbon and iron (from inside the front cover of the book) are 2.25 and 7.87 g/cm 3. For0.012 wt% C

CC =

CCCC

C

+ CFeFe

103

=

0.0120.012

2.25 g/cm3+ 99.988

7.87 g/cm3

103

= 0.944 kg C/m3

Similarly, for 0.0075 wt% C

CC =

0.00750.0075

2.25 g/cm3+ 99.9925

7.87 g/cm3

103

= 0.590 kg C/m3

Now, using a form ofEquation (6.3)

D = J

xA xBCA CB

= (1.40 108 kg/m2-s) 103 m

0.944 kg/m3 0.590 kg/m3

= 3.95 1011 m2/s

6.13 This problem asks us to compute the nitrogen concentration (Cx) at the 1 mm position after a 10 hdiffusion time, when diffusion is non steady-state. Fro m Equation (6.5)

Cx Co

Cs Co =

Cx 0

0.1 0 =1

er f

x

2D t= 1 er f

103 m

(2)

(2.5 1011 m2/s)( 10 h )( 3600 s/h )

= 1 erf(0.527)

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Using data in Table 6.1 and linear interpolation

z erf(z)

0.500 0.52050.527 y0.550 0.5633

0.527

0.500

0.550 0.500 =y

0.5205

0.5633 0.5205from which

y = erf(0.527) = 0.5436

Thus,

Cx 00.1 0 = 1.0 0.5436

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 of0.45 wt% at a point 5 mm from the surface. From Equat ion (6.6b),

x2

D t= constant

But since the temperature is constant, so also is D constant, and

x2

t= constant

or

x21

t1 =x2

2

t2

Thus,

(2.5 mm)2

10 h= (5.0 mm)

2

t2

from which

t2 = 40 h

6.21 (a) Using Equation (6.9a), we set up two simultaneous equations with Qd an d Do as unknowns.Solving for Qd in terms of temperatures T1 an d T2 (1273 K and 1473 K) and D 1 an d D 2 (9.4 1016and 2.4 1014 m2/s), we get

Q d = Rln D 1 ln D 21/T1 1/T2

= (8.31 J/mol-K)[ln(9.4 1016) ln(2.4 1014)]

1/(1273 K) 1/(1473 K)= 252,400 J/mol

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Now, solving for D o from Equation (6.8)

D o = D 1 exp

Q d

RT 1

= (9.4 1016 m2/s) e xp

252400 J/mol

(8.31 J/mol-K)(1273 K)

= 2.2 105 m2/s(b) Using these values ofD o an d Qd, D at 1373 K is just

D = (2.2 105 m2/s) e xp 252400 J/mo l

(8.31 J/mol-K)(1373 K)

= 5.4 1015 m2/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-hheat treatment at 1000 K, the concentration of A in B is 3.2 wt% at the 15.5-mm position. Afteranother heat treatment at 800 K for 30 h, we are to determine at what position the composition willbe 3.2 wt% A . In order to make this determination, we must employ Equation (6.6b) with t constant.That is

x2

D= constant

O r

x2800D 800

= x21000

D 1000

It is necessary to compute both D800 an d D1000 using Equation (6.8), as fo llows:

D 800 = (1.8 105 m2/s) e xp 152000 J/mo l

(8.31 J/mol-K)(800 K)

=2.12

1015 m2/s

D 1000 = (1.8 105 m2/s) e xp 152000 J/mo l

(8.31 J/mol-K)(1000 K)

= 2.05 1013 m2/s

Now, solving for x800 yields

x800 = x1000

D 800

D 1000

= (15.5 mm)

2.12 1015 m2/s2.05

1013 m2/s

= 1.6 m m

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CHAPTER 7

MECHA NICAL PROPERTIES

7.4 We are asked to compute the maximum length of a cylindrical titanium alloy specimen that isdeformed elastically in tension. For a cylindrical specimen

A o =

do2

2

where do is the original diameter. Combining Equations (7.1), (7.2), an d (7.5) and solving for loleads to

lo =Ed2ol

4F

=(107 109 N/m 2)()( 3.8 103 m) 2(0.42 103 m)

(4)(2000 N)

= 0.25 m = 250 mm (10 in.)

7.9 This problem asks that we calculate the elongation l of a specimen of steel the stress-strainbehavior of which is shown in Figure 7.33. First it becomes necessary to compute the stress when aload of 23,500 N is applied as

=F

A o=

F

do

2

2 = 23500 N

10 103 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 = 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 cylindricalspecimen of aluminum. U sing Equations (7.1), (7.2), an d (7.5)

F

d2o

4

= E llo

O r

l =4Flo

d2o E

=(4)(48,800 N)(200 103 m)

()(19 103 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)

= x

z

= d/do

l/lo

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From Table 7.1, for Al, = 0.33. Now, solving for d yields

d = ldo

lo=

(0.33)(0.50 mm)(19 mm)

200 mm

= 1.6 102 mm (6.2 104 in.)

The d iameter will decrease.

7.16 This problem asksthat we compute Poissons ratio for the metal alloy. From Equations(7.5) an d (7.1)

z =

E=

F/A o

E=

F

do

2

2E

=4F

d2o E

Since the tran sverse strain x is just

x =d

do

and Poissons ratio is defined by Equation (7.8) then

= x

z

= d/do

4F

d2o E

= dodE4F

= (8 103 m) (5 106 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. Thefirst calculation ne cessary is that of the applied stress using Equation (7.1), as

=

F

A o =

F

do

2

2 = 5000 N

6 103 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 2.0 103.From the definition of strain, Equation (7.2),

l = lo = (2.0 103)(50 mm) = 0.10 mm (4 103 in.)

(b) In order to determine the reduction in diameter d, it is necessary to use Equation (7.8) an dthe definition of lateral strain ( i.e., x = d/do ) as follows:

d = do

x = do

z = (6 mm)(0.30)(2.0 10

3

)= 3.6 103 mm (1.4 104 in.)

7.27 This problem asks us to determine the d eformation character istics of a steel specimen, the stress-strain 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 thestress, then locate it on the stress-strain curve, and, finally, note whether this point is on the elastic

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or plastic region. Thus,

=F

A o=

44500 N

10 103 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 beboth elastic and plastic.

(b) This portion of the problem asks us to compute the increase in specimen length. From thestress-strain curve, the strain at 565 MPa is approximately 0.008. Thus, from Equation (7.2)

l = 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-lengthdata, 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.

0.0120.0100.0080.0060.0040.0020.000

0

100

200

300

Strain

Stre

ss(MPa)

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(b) The elastic modulus is the slope in t he linear elastic region as

E =

=

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-straincurve 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 one-hundred. The total fracture strain at fracture is 0.165; subtracting out the elastic strain (which isabout 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

U r =

2y

2E

which, using data computed in the problem yields a value of

U r =

(285 MPa)2

(2)(62.5 103 MPa) = 6.5 105

J/m3

(93.8 in .-lbf/in.3

)

7.32 This prob lem asks us to calculate the modu li of resilience for the m aterials having the stress-strainbehaviors shown in Figures 7.12 and 7.33. According to Equation (7.14), the modu lus of resilienceU r is a function o f the yield strength an d th e mo dulus of elasticity as

U r =

2y

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

U r = (250 MPa)2

(2)(93.9 103 MPa)= 3.32 105 J/m 3 (47.6 in .-lbf/in.

3)

7.41 For this problem, we are given two values ofT an d T, from which we are asked to calculate thetrue stress which produ ces a true plastic strain of 0.25. Employing Equation (7.19), we may set uptwo 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) log(60,000)log(0.1) log(0.2)

= 0.263

log K = 4.96 or K = 91,623 psi

Thus, for T = 0.25

T = K(T )2= (91,623 psi)(0.25)0.263 = 63,700 psi (440 MPa)

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7.45 This problem calls for us to utilize the appropriate data from Problem 7.29 in order to determinethe values of n an d K for this material. From Equ ation (7.32) the slope and intercept of a log Tversus log T plot will yield n and log K, respectively. H owever, Equation (7.19) is only valid inthe r egion of plastic deformation t o the point of ne cking; thus, only the 7th, 8th, 9th, and 10th da tapoints may b e ut ilized. The log-log plot with t hese dat a po ints is given b elow.

- 1 .2-1 .4-1 .6-1 .8-2 .0-2 .2

2.46

2.48

2.50

2.52

2.54

2.56

2.58

2.60

log true strain

logtruestress(MPa)

The slope yields a value of 0.136 for n, whereas th e inter cept gives a value of 2.7497 for log K, andthus K = 562 MPa .

7.50 For this problem, the load is given at which a circular specimen of aluminum oxide fractures when

subjected to a three-point bending test; we are then are asked to determine the load at which aspecimen of the same ma terial having a square cross-section fractures. It is first necessary to com-pute the flexural strength of the alumina using Equation (7.20b), and then, u sing this value, we ma ycalculate th e value ofFf in Equation (7.20a). From Equation (7.20b)

fs =FfL

R 3

=(950 N)(50 103 m)

()( 3.5 103 m) 3= 352 106 N/m 2 = 352 MPa (50,000 psi)

Now, solving for Ff from Equation (7.20a), realizing t hat b = d = 12 mm, yields

Ff =

2fsd3

3L

=(2)(352 106 N/m 2)(12 103 m) 3

(3)(40 103 m)= 10,100 N ( 2165 lbf)

7.54* (a) This part of the prob lem asks us to determine the flexural strength of nonporous MgO assum-ing that the value of n in Equation (7.22) is 3.75. Taking natural logarithms of both sides of

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Equation (7.22) yields

ln fs = ln o nP

In Table 7.2 it is noted that for P = 0.05,fs = 105 MPa. For the nonporous material P = 0 and,ln o = ln fs. Solving for ln o from the above equation gives and using these data gives

ln o = ln fs + nP

= ln(105 MPa ) + (3.75)(0.05) = 4.841

or

o = e4.841

= 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 ofEquation (7.22) and solving for P leads to

P =ln o ln fs

n

=ln(127 MPa ) ln(62 MPa)

3.75

= 0.19 or 19 vol%

7.65 This problem calls for estimations of Brinell and R ockwell hardnesses.

(a) For the brass specimen, the stress-strain be havior for which is shown in Figure 7.12, the tensilestrength is 450 MPa (65,000 psi). From Figure 7.31, the hardness for brass corresponding to thistensile strength is about 125 HB or 70 HRB.

7.70 The working stresses for the t wo alloys, the stress-strain behaviors of which are shown in Figures7.12 an d 7.33, are calculated by dividing the yield stren gth by a factor of safety, which we will taketo be 2. For the brass alloy (Figure 7.12), since y = 250 MPa (36,000 psi), the working stress is125 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 psi).

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CHAPTER 8

DE FORMATION A ND STRENGTHENING MECHAN ISMS

8.7 In th e man n er o f Figure 8.6b, we are to sketch the atomic packing for a BCC {110} type plane, andwith arrows indicate two d ifferent 111 type directions. Such is shown below.

8.10* We are asked to compute the Schmid factor for an FCC crystal oriented with its [100] directionparallel 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.

x

y

z

[100]

[111]

[110]

_

The a ngle between the [100] and [110] directions, , is 45. For the (111) plane, the anglebetween its normal ( which is the [111] direction) and the [100] direction, , is tan1( a

2

a) = 54.74;

therefore

cos cos = cos(45) cos(54.74) = 0.4088.20 We are asked to determine the grain diameter for an iron which will give a yield strength of 205 MPa

(30,000 psi). The best way to solve this problem is to first establish two simultaneous expressions ofEquation (8.5), solve for o an d ky, and finally determine the value of d when y = 205 MPa. Thedata pertaining to this problem may be tabulated as follows:

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y d (mm) d1/2 (mm)1/2

135 MPa 5 102 4.47260 MPa 8 103 11.18

The two equations thus become

135 MPa = o + (4.47)ky260 MPa = o + (11.18)ky

which yield the values, o = 51.7 MPa and ky = 18.63 MPa(mm)1/2. At a yield strength of 205 MPa

205 MPa = 51.7 MPa + [18.63 MPa(mm)1/2]d1/2

or d1/2 = 8.23 (mm)1/2, which gives d = 1.48 102 mm .

8.25 This problem stipulates that two previously undeformed cylindrical specimens of an alloy are to bestrain hard ened by reducing their cross-sectional area s. For one specimen, the initial and de formedradii are 16 mm an d 11 mm, r espectively. The second specimen with an initial rad ius of 12 mm is tohave the same deformed hardness as the first specimen. We are asked to compute the radius of thesecond specimen after deformation. In order for these two cylindrical specimens to have the samedeformed hardness, they must be deformed to the same percent cold work. For the first specimen

% CW = A o A dA o

100 = r2o r2dr2o

100

= (16 mm)2 (11 mm)2

(16 mm)2 100 = 52.7%CW

For the second specimen, the deformed radius is computed using the above equation and solvingfor rd as

rd=

ro1 % CW

100

= (12mm)

1 52.7%CW100

= 8.25mm

8.27 This problem calls for us to calculate the p recold-worked radius of a cylindrical specimen of copperthat has a cold-worked ductility of 25%EL. From Figure 8.19(c), copper that has a ductility of25% E L will have experienced a deformation of about 11% CW. For a cylindricalspecimen, Equation(8.6) becomes

% CW =r2o r2d

r2o

100

Since rd = 10 mm (0.40 in.), solving for ro yields

ro =rd

1 % CW100

= 10mm1 11.0

100

= 10.6 mm (0.424 in.)

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8.35 In this problem, we are asked for the length of time required for the average grain size of a brassmaterial to increase a specified amount using Figure 8.25.

(a) At 500C, the time n ecessary for the average grain diameter to increase from 0.01 to 0.1 mm isappro ximately 3500 min.

(b) At 600C 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 fortwo 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 itis possible to solve for the two constants TS an d A . These equ ations are as follows:

107 MPa = TS A

40000 g/mol

170 MPa = TS A

60000 g/mol

Thus, the values of the two constants are TS = 296 MPa a nd A = 7.56 106 MPa-g/mol. Substi-tuting these values into an equation for which Mn = 30,000 g/mol leads to

TS = TS A

30000 g/mol

= 296 MPa 7.56 106 MPa-g/mol

30000 g/mol

= 44 MPa

8.54 This prob lem asks that we compute t he fraction of possible crosslink sites in 10 kg of polybutadienewhen 4.8 kg of S is added, assuming that, on the average, 4.5 sulfur atoms participate in eachcrosslink bond. G iven the butadiene mer unit in Table 4.5, we may calculate its molecular weightas follows:

A(butadiene) = 4(A C ) + 6(A H )= (4)(12.01 g/mol) + 6(1.008 g/mol) = 54.09 g/mol

which means that in 10 kg of butadiene there are10000 g

54.09 g/mol= 184.9 mol.

For the vulcanization of polybutadiene, there ar e two possible crosslink sites per meron efor each of the two carbon atoms that are doubly bonded. Furthermore, each of these crosslinksforms a bridge between two mers. Therefore, we can say that there is the equivalent of one crosslinkper mer. Therefore, let us now calculate the number of moles of sulfur (nsulfur) that react with thebutad iene, by taking the mole ra tio of sulfur t o but adiene, and th en dividing this ratio by 4.5 atomsper crosslink; this yields the fraction of p ossible sites th at are crosslinked. Thus

nsulfur=

4800 g

32.06 g/mol =149.7 mol

A nd

fraction sites crosslinked =149.7 mol

184.9 mol

4.5= 0.180

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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 givea minimum B rinell hardness of 225 and a d uctility of at least 12% EL . According to Figure 7.31, aBrinell 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 least13% CW is necessary. Finally, if we cold work t he steel to 13% CW, then the ductility is reduced toonly 14% EL from Figure 8.19(c).Therefore,it is possible to meet both of these criteria by plasticallydeforming the steel.

8.D6 This problem stipulates that a cylindrical rod of copper or iginally 16.0 mm in diameter is to be coldworked by drawing; a cold-worked yield strength in excess of 250 MPa and a ductility of at least12%EL are required, whereas the final diameter must be 11.3 mm. We ar e to explain how this isto be accomplished. Let us first calculate the percent cold work and attendant yield strength andductility if the dr awing is carried out without interrup tion. From Equation (8.6)

% CW =

do

2

2

dd

2

2

do

2

2 100

=

16 mm

2 2

11.3 mm

2 2

16 mm

2

2 100 = 50%CW

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), whichis insufficient.

Instead of perform ing the dr awing in a single operat ion, let us initially draw some fractionof the total deformation, then anneal to recrystallize, and, finally, cold work the mater ial a secondtime 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 of250 MPa. Similarly, a m aximum of 23% CW is possible for 12% E L [Figure 8.19(c)]. The averageof 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 =

d o2

2

11.3

2

2

d o2

2 100

A nd, solving for do yields do = 12.8 mm (0.50 in.).

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CHAPTER 9

FAILURE

9.7 We are asked for the critical crack tip radius for an Al2O 3 material. From Equation (9.1b)

m = 2o at 1/2

Fracture will occur when m reaches the fracture stren gth of the mater ial, which is given as E/10;thus

E

10= 2o

a

t

1/2

O r, solving for t

t =400a2

oE 2

From Table 7.1, E = 393 GPa, and thus,

t =(400)(2 103 mm)(275 MPa)2

(393 103 MPa)2

= 3.9 107 mm = 0.39 nm

9.8 We may determine the critical stress required for the propagation of a surface crack in soda-limeglass using Equat ion (9.3);takingthe value of69GPa (Table 7.1) as the modulusof elasticity, we get

c =

2Es

a

=

(2)(69 109 N/m 2)( 0.30 N/m)()( 5 105 m) = 16.2 10

6 N/m 2 = 16.2 MPa

9.12* 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 appliedstress is 100 MPa (14,500 psi). In order to determine the stress concentration it is necessary toconsult Figure 9.8c. From the geometry of the specimen, w/h = (25 mm)/(20 mm) = 1.25; fur-thermore, 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

Kt value at r/h = 0.15 is 1.7. An d since Kt = mo

, then

m = Kto = (1.7)(100 MPa ) = 170 MPa (24,650 psi)

9.15* Thisproblem callsfor us to determine the value ofB, the minimum compon ent thicknessfor whichthe condition of plane strain is valid using Equation (9.12) for th e m etal alloys listed in Table 9.1.

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For t he 2024-T3 aluminum alloy

B = 2.5

Klc

y

2= (2.5)

44 MPa

m

345 MPa

2= 0.041 m = 41 mm (1.60 in.)

For the 4340 alloy steel temper ed at 260C

B = (2.5)50 MPa

m

1640 MPa2 = 0.0023 m = 2.3 mm (0.09 in.)

9.19 For this problem, we are given values ofKlc,, a n d Y for a large plate and are asked to determinethe m inimum length of a surface crack that will lead t o fracture. A ll we nee d do is to solve for acusing Equation (9.14); therefore

ac =1

Klc

Y

2= 1

55 MPa

m

(1)(200 MPa)

2= 0.024 m = 24 mm (0.95 in.)

9.26 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.

0- 5 0-100-150-200

0

20

40

60

80

100

120

140

Temperature, C

I

mpactEnergy,

J

(b) This port ion of the prob lem asks us to deter mine the ductile-to-brittle transition temp eratu reas that temperatur e corresponding to the average of the maximum and minimum impact energies.From these data, this average is

Average =124 J

+6 J

2 = 65 J

As indicated on the plot by the one set of dashed lines,t he ductile-to-brittle tran sition temperat ureaccording to this criterion is about 105C.(c) Also as noted on the plot by the other set of dashed lines, the ductile-to-brittle transitiontemperature for an impact energy of 80 J is about 95C.

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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 ofmax an d min using Equation (7.1). Thus

max =Fmax

A o= Fmax

do

2

2

= 7500 N()

8.0 103 m

2

2 = 150 106 N/m 2 = 150 MPa (22,500 psi)

min =Fmin

do

2

2

= 7500 N

()

8.0 103 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 min

2= 150 MPa (150 MPa )

2= 150 MPa (22,500 psi)

From Figure 9.46 for t he r ed br ass, the n umber of cycles to failure at this stress amplitude is about1 105 cycles.

9.33 T his p rob 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.

1098765

100

200

300

Log cycles to failure

Stres

samplitude,

MPa

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(b) As indicated by one set of dashed lines on the plot, the fatigue strength at 5 105 cycles[log(5 105) = 5.7] is about 250 MPa.(c) As noted by the other set of dashed lines, the fatigue life for 200 MPa is about 2 106 cycles(i.e., the log o f the lifetime is abou t 6.3).

9.34 We are asked to compute the maximum torsional stress amplitude possible at each of several fa-tigue lifetimes for the brass alloy, the fatigue behavior of which is given in Problem 9.33. For eachlifetime, 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) = 7.9 108 cy-cles. The stress amplitude corresponding to this lifetime is about 130 MPa.

(c) Fatigue lifetime = (24 h) (60 min/h) (1200 cycles/min) = 2.2 106 cycles. The stress amplitudecorresponding to this lifetime is about 195 MPa.

9.48 This problem asks that we determine the total elongation of a low carbon-nickel alloy that is ex-posed t o a tensile stress of 40 MPa (5800 psi) at 538C for 5000h; the instantaneous and primarycreep elongations ar e 1.5 mm ( 0.06 in.).

From the 538C line in Figure 9.43, the steady-state creep rate, s, is about 0.15% /1000 h(or 1.5 104% /h) at 40 MPa. The steady-state creep strain, s, therefore, is just the product ofs and time as

s = s (time)= (1.5 104% /h)(5000 h) = 0.75% = 7.5 103

Strain and elongation are related as in Equat ion (7.2); solving for the steady-state elongation,ls, leads to

ls = los = (750 mm)(7.5 103) = 5.6 mm (0.23 in.)

Finally, the total e longation is just th e sum o f this ls and the total of both instantaneous and pri-mary 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 s versus log plot yields the value ofn in Equation (9.33);that is

n = log s log

We are asked to determine the values ofn for the creep data at the three temperatures in Fig-ure 9.43. This is accomplished by taking ratios of the differences between two log s and logvalues. Thus for 427C

n = log s

log =log(101)

log(102)

log (85 MPa) log( 55 MPa) = 5.3

and for 538C

n = log s log

= log(1.0) log(102)

log (59 MPa) log( 23 MPa) = 4.9

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9.55* This problem gives s values at two different temperatures and 70 MPa (10,000 psi), and the stressexponent 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 ofEquation (9.34) yields

ln s = ln K 2 + n ln Q c

RT

With the given data there are two unknowns in this equation namely K2 an d Qc. Using the dataprovided in the problem we can set up two independent equations as follows:

ln[1.0 105 (h )1] = ln K 2 + (7.0) ln(70 MPa ) Q c

(8.31 J/mol-K)(977 K)

ln[2.5 103 (h )1] = ln K 2 + (7.0) ln(70 MPa ) Q c

(8.31 J/mol-K)(1089 K)

Now, solving simultaneo uslyfor K2 an d Qc leadsto K2 = 2.55 105 (h )1 an d Qc = 436,000 J/mol.Thus it is now possible to solve for s at 50 MPa and 1250 K using Equation (9.34) as

s = K2n exp Q cRT

s = [2.55 105 (h )1](50MPa)7.0 exp

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 occurfor a flat plate given an expression from which Y(a/W) may be determined, the internal cracklength, 2a (20 mm), the plate width, W (90 mm), and the value of (375 MPa). First we mustcompute the value ofY(a/W) using Equation (9.10), as follows:

Y( a/W) =

Wa

ta naW

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

Klc = Y( a/W)a

= (1.021)(375 MPa)

()(10 103 m) = 67.9 MPa

m (62.3 ksi

in.)

9.D7* We are asked in this prob lem to estimate the maximum tensile stress that will yield a fatigue lifeof 2.5 107 cycles, given values ofao , ac, m, A , and Y. Since Y is indepe ndent of crack length wemay utilize Equation (9.31) which, upon integration, takes the form

Nf=1

Am/2()m Ym

acao

am/2da

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And for m = 3.5

Nf =1

A1.75()3.5Y3.5

acao

a1.75da

= 1.33A1.75()3.5Y3.5

1

a0.75c 1

a0.75o

Now, solving for from this expression yields

=

1.33

NfA1.75Y3.5

1

a0.75o 1

a0.75c

1/3.5

=

1.33

(2.5 107)( 2 1014)()1.75(1.4)3.5

1

(1.5 104)0.75 1

(4.5 103)0.751/3.5

= 178 MPa

This 178 MPa will be the maximum tensile stress since we can show that the minimum stressis a compressive onewhen min is negative, is taken to be max. If we take max = 178MPa, and since m is stipulated in the problem to have a value of 25 MPa, then from Equa-tion (9.21)

min = 2m max = 2(25 MPa) 178 MPa = 128 MPa

Therefore min 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 be5 years and 20 years when an 18-8 Mo stainless steel component is subjected to a temperature of500C (773 K). It first becomes necessary, using the specified temperature and times, to calculatethe values of the Larson-Miller parameter at each temperature. The values oftr correspondingto 5 and 20 years are 4.38 104 h a n d 1.75 105 h, respectively. Hence, for a lifetime of 5years

T(20 + logtr ) = 773[20 + log(4.38 104)] = 19.05 103

And for tr = 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-yearlifetimes are approximately 260 MPa (37,500 psi) and 225 MPa (32,600 psi), respectively.

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CHAPTER 10

PHASE D IAGRAMS

10.5 This problem asks that we cite the phase or phases present for several alloys at specified temper-atures.

(a) For an alloy composed of 90 wt% Zn-10 wt% Cu and at 400C, from Figure 10.17, an d

phases are present, and

C = 87 wt% Zn-13 wt% Cu

C = 97 wt% Z n-3 wt% Cu

(c) For an alloy composed of 55 wt% Ag-45 wt% Cu and at 900C, from Figure 10.6, only theliquid 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 inProblem 10.5.

(a) For an alloy composed of 90 wt% Zn-10 wt% Cu and at 400C, an d phases are present,

an d

Co = 90wt% Zn

C = 87wt% Zn

C = 97wt% Zn

Therefore, using modified forms ofEquation (10.2b) we get

W =C Co

C C=

97 90

97 87= 0.70

W =Co C

C C=

90 87

97 87= 0.30

(c) For an alloy composed of 55 wt% Ag-45 wt% Cu and at 900C, since only the liquid p hase ispresent, then WL = 1.0.

10.9 This problem asks that we determine the phase volume fractions for the alloys and temperaturesin Problem 10.5a, b, and c. This is accomplished by using the technique illustrated in ExampleProb lem 10.3, and t he r esults of Problem 10.7.

(a) This is a Cu-Zn alloy at 400C, wherein

C = 87 wt% Zn-13 wt% Cu

C=

97 wt% Zn-3 wt% CuW = 0.70

W = 0.30

Cu = 8.77 g/cm3

Zn = 6.83 g/cm3

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Using these data it is first necessary to compute the densities of the an d phases usingEquation (5.10a). Thus

=100

CZn ()

Z n+

CCu( )

Cu

=100

87

6.83 g/cm3+

13

8.77 g/cm3

= 7.03 g/cm3

=100

CZn ()

Zn+

CCu()

Cu

=100

97

6.83 g/cm3+

3

8.77 g/cm3

= 6.88 g/cm3

Now we may determine the V an d V values using Equation 10.6. Thus,

V =

W

W

+

W

=

0.70

7.03 g/cm3

0.70

7.03 g/cm3+

0.30

6.88 g/cm3

= 0.70

V =

W

W

+

W

=

0.30

6.88 g/cm3

0.70

7.03 g/cm3+

0.30

6.88 g/cm3

= 0.30

10.12 (a) We are asked to deter mine how much sugar will dissolve in 1500 g of water a t 90C. From thesolubility limit curve in Figure 10.1, at 90C the maximum concentration of sugar in the syrup isabout 77 wt% . It is now possible to calculate th e ma ss of sugar using Equation (5.3) as

Csugar (wt%) =msugar

msugar + mwater 100

77wt% =msugar

msugar + 1500 g 100

Solving for msugar yields msugar = 5022 g.

(b) Again using this same plot, at 20C the solubility limit (or the concentration of the saturatedsolution) is about 64 wt% sugar.

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(c) The mass of sugar in this saturated solution at 20C (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

msugar = msugar m

sugar = 5022 g 2667 g = 2355 g

10.21 Up on cooling a 50 wt% Pb-50 wt% Mg alloy from 700C and utilizing Figure 10.18:

(a) The first solid phase forms at the temperature at which a vertical line at this compositionintersects the L-(+ L) phase boundaryi.e., about 550C;

(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 550Ci.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 isothermi.e., about 465C;

(d) The composition of the last liquid phase remaining prior to complete solidification corre spondsto the eutectic compositioni.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-70wt% Pb alloy and asked to estimate the temperature of the alloy. Using the appropriate phasediagram, Figure 10.7, by trial and error with a ruler, a tie line within the + L phase region thatis divided in h alf for an alloy of this composition exists at abou t 230C.

(b) We are now asked to determine the compositions of the two phases. This is accomplishedby noting the intersections of this tie line with both the solidus and liquidus lines. From theseintersections, 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 thatconsists of mass fractions W = 0.60 and W = 0.40. Such a n alloy is not possible, based on thefollowingar 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 roomtemperature. At just below the eutectic, C = 8.0 wt% Ag and C = 91.2 wt% A g; thus,

W =C Co

C C=

91.2 50

91.2 8= 0.50

W = 1.0 W = 1.0 0.5 = 0.50

Furthermore, at room temperature, C = 0 wt% A g a nd C = 100 wt% Ag; employment ofE qua-tions (10.1) an d (10.2) yields

W =C Co

C C=

100 50

100 0= 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 ofW = 0.60 and W = 0.40 as called for in the prob lem.

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10.35* This problem asks that we d eterm ine the composition of a Pb-Sn alloy at 180C given th at W =0.57 and We = 0.43. Since there is a primary microconstituent present, then we know that thealloy composition, Co , is between 61.9 and 97.8 wt% Sn (Figure 10.7). Furthermore, this figurealso indicates th at C = 97.8 wt% Sn and Ceutectic = 61.9 wt% Sn. Applying the appropriate leverrule expre ssion for W

W =Co Ceutectic

C Ceutectic=

Co 61.9

97.8 61.9= 0.57

and solving for Co yields Co = 82.4 wt% Sn.

10.47* We are asked to specify the value ofF for Gibbs phase rule at point B on the pressure-temperaturediagram for H 2O. 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 noncompositionalvariables, is 2viz. temperature and pressure. Thus, the phase rule now becomes

P + F = 1 + 2 = 3

O r

F = 3 P

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 = 1), or

F = 3 P = 3 1 = 2

which means that bot h temper ature and pressure are necessary to define the system.

10.54 This problem asks that we compute the carbon concentration of an iron-carbon alloy for whichthe fraction of total ferrite is 0.94. Application of the lever rule [of the form of Equation (10.12)]yields

W = 0.94 =CFe3C C

o

CFe3C C=

6.70 Co6.70 0.022

and solving for Co

Co = 0.42 wt% C

10.59 This prob lem asks that we deter mine the carbon concentra tion in an iron-carbon alloy, given themass fractions of proeutectoid ferrite and pearlite. From E quation (10.20)

Wp = 0.714 =Co 0.022

0.74

which yields Co = 0.55 wt% C.

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10.64 This problem asks if it is possible to have an iron-carbon a lloy for which W = 0.846 and WFe3C =0.049. In order to ma ke t his determination, it is necessary to set up lever ru le expressions for th esetwo mass fractions in terms of the alloy composition, then to solve for the alloy composition ofeach; if both alloy composition values are equal, then such an alloy is possible. The expression forthe mass fraction of total ferrite is

W =CFe3C Co

CFe3C C=

6.70 Co

6.70 0.022= 0.846

Solving for this Co yields Co = 1.

05 wt% C. Now for WFe3C we ut ilize E quation (10.23) as

WFe3C =C

1 0.76

5.94= 0.049

This expression leads to C1= 1.05 wt% C. And, since Co = C

1, this alloy is possible.

10.70 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 =Co 0.022

0.74=

0.20 0.022

0.74= 0.24

W =0.76 Co

0.74=

0.76 0.20

0.74= 0.76

Now, we compute the B rinell hardness of the alloy as

H Balloy = H BW + H Bp Wp

= (80)(0.76) + (280)(0.24) = 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 650C (1200F) .

(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 3C) phase boundary is at a negligible carbon concentration. Modi-fying Equation (10.21) leads to

W =0.62 Co0.62 0

=0.62 0.20

0.62= 0.68

Likewise, using a m odified E quation (10.20)

Wp =Co 0

0.62 0=

0.20

0.62= 0.32

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CHAPTER 11

PHA SE TRA NSFORMATIONS

11.4 This problem gives us the value ofy (0.40) at some time t (200 min), and also the value ofn (2.5)for the recrystallization of an alloy at some temperature, and then asks that we determine therate of recrystallization at this same tempe ratur e. It is first ne cessary to calculate t he value ofk inEquation (11.1) as

k= ln(1 y)

tn

=

ln(1 0.4)

(200 min)2.5= 9.0 107

At this point we want to compute t0.5, the value of t for y = 0.5, also using Eq uation (11.1).Thus

t0.5 =

ln(1 0.5)k

1/n

=

ln(1 0.5)

9.0 107

1/2.5= 226.3 min

And , therefore, from Equation (11.2), the rate is just

rate =1

t0.5=

1

226.3 min= 4.42 103 (min)1

11.7 This problem asks us to consider the percent recrystallized versus logarithm of time curves forcopper shown in Figure 11.2.

(a) The rates at the different temperatures are determined using Equation (11.2), which ra tes aretabulated below:

Temperature (C) R ate (min)1

135 0