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The SCIENCE and ENGINEERING of ·MATERIALS Second SI edition
Donald R. Askeland
Solutions manual Solutions supplied by Paul Porgess and Ian Brown Department of Polymers, Metals and Dental Technology, Manchester Polytechnic
Chapman & Hall LONDON • NEW YORK. TOKYO. MELBOURNE. MADRAS
UK
Chapman & Hall, 2-6 Boundary Row, London SEI 8HN
USA
Chapman & Hall, 29 West 35th Street, New York NYlOOOI
JAPAN
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AUSTRALIA
Chapman & Hall Australia, Thomas Nelson Australia, 102 Dodds Street, South Melbourne, Victoria 3205
INDIA
Chapman & Hall India, R. Seshadri, 32 Second Main Road, CIT East, Madras 600 035 First edition 1991
© 1991 Chapman & Hall The Ipswich Book Company ISBN-13:978-0-412-39600-7 DOl: 10.1 007/978-94-009- I 842-9
e-ISBN-13:978-94-009- 1842-9
All rights reserved. No part of this publication may be reproduced or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, or stored in any retrieval system of any nature, without the written permission of the copyright holder and the publisher, application for which shall be made to the publisher. The publisher makes no representation, express or implied, with regard to the accuracy of the information contained in this book and cannot accept any legal responsibility or liability for any errors or omissions that may be made. British Library Cataloguing in Publication Data available
TABLE OF CONTENTS
Solutions to Practice Problems
Chapter 1:
Introduction to Materials
1
Chapter 2:
Atomic Structure
5
Chapter 3:
Atomic Arrangement
11
Chapter 4:
Imperfections in the Atomic Arrangement
37
Chapter 5:
Atom Movement in Materials
50
Chapter 6:
Mechanical Testing and Properties
63
Chapter 7:
Deformation, Strain Hardening, and Annealing
81
Chapter 8:
Solidification and Grain Size Strengthening
92
Chapter 9:
Solidification and Solid Solution Strengthening
106
Chapter 10:
Solidification and Dispersion Strengthening
116
Chapter 11:
Dispersion Strengthening by Phase Transformation and Heat Treatment
126
Chapter 12:
Ferrous Alloys
135
Chapter 13:
Non-Ferrous Alloys
151
Chapter 14:
Ceramic Materials
160
Chapter 15:
Polymers
174
Chapter 16:
Composite Materials
191
Chapter 17:
Electrical Conductivity
204
Chapter 18:
Dielectric and Magnetic Properties
218
Chapter 19:
Optical and Thermal Properties
229
Chapter 20:
Corrosion and Wear
247
Chapter 21
Failure - Origin, Detection, and Prevention
259
Chapter 1
INTRODUCTION TO MATERIALS
1.
The Stealth aircraft is designed so that i t will not be detected by radar. What physical property should the materials used in the Stealth plane possess to meet this design requirement? The materials should be capable of absorbing radiation having the wavelength of radar. In addition, the materials must also have a good strength-to-weight ratio, appropriate corrosion resistance, and other properties that typical aircraft should possess.
2.
Certain materials, such as tungsten carbide, are compounds consisting of both metallic and non-metallic elements. To which category of "materials does tungsten carbide belong? Ceramic materials.
3.
From which category of materials would you select a material best suited for building a vessel to contain liquid steel? Ceramic materials, which often have high melting temperatures and chemical resistance to liquid metals. However, the ceramic must be chosen with care, because many ceramics will react with the metal.
4.
Consider a cement wall reinforced with steel bars. Into which category of materials would you place this reinforced concrete? This is a composite material.
5.
Boron nitride (BN) and silicon carbide (SiC) are important materials in abrasive grinding wheels. In what category of materials do BN and SiC belong? BN and SiC are in the form of small particles and "are often embedded in a polymer to produce the grinding wheel. In what category of materials does the entire wheel belong? The BN and SiC are ceramic materials; because the ceramic particles are embedded in a polymer, the entire wheel is a composite material.
6.
Silicon carbide (SiC) fibers are sometimes mixed with liquid aluminum. After the mixture freezes, a fiber-reinforced composite results. Would you reinforce aluminum with high strength polyethylene fibers in the same manner? Explain your answer. No. The polyethylene fibers are a polymer material and consequently have low melting and degradation temperatures. Introducing liquid aluminum at a temperature above 600 0 C will destroy the polymer fibers.
1
7.
The nose of the space shuttle is composed of graphite (carbon). this information, what type of material would graphite be?
Based on
The nose of the shuttle will experience extremely high temperatures upon re-entering the atmosphere from orbit. Protection therefore requires a very high melting point material, such as a ceramic. Graphite is sometimes considered to be a ceramic, although it is not a combination of metallic and non-metallic elements. 8.
Suppose we would like to make a porous metal filter to keep the oil in our automobile engine clean. Which one of the metal processing techniques listed in Table 1-3 might be used to produce these filters? Powder metallurgy might be an excellent choice. We can compact spherical metal powder particles to a small degree and sinter just long enough so the powder particles are joined together. This will leave interconnected voids between the particles that will allow liquid oil to penetrate but will trap small solid impurities.
9.
Sintering is listed in Table 1-3 as a ceramic processing technique. In which one of the metal processing techniques would you expect sintering also to be used? Sintering is an integral portion of the powder metallurgy process, joining the powder particles together, reducing void space, increasing density, and providing good mechanical properties after the powder metallurgy part has been initially formed by compaction.
10.
Which of the three ceramic processing methods mentioned in Table 1-3 do you think is used to produce glass bottles? The bottles are normally produced by a "compaction" process, in which the glob of hot, viscous glass is introduced into a die and then formed, often using gas pressure.
11.
By which one of the four methods of producing composite materials listed in Table 1-3 would you expect plywood to be made? Plywood is produced by "joining"; the individual plies are joined by adhesive bonding, or glueing.
12.
Injection molding to produce plastic parts most closely resembles which one of the metals processing methods? Injection molding is very similar to die casting, in which pressure is used to force a molten material into a metal die to give the desired shape.
13.
The Voyager is an experimental aircraft that flew around the world non-stop on a single tank of fuel. What type of material do you think made up most of the aircraft? Explain why this type of material was selected. The Voyager was produced primarily from composite materials, including carbon fiber-epoxy and fiberglass materials. These materials provided both the exceptionally light weight and the high strength and stiffness required. 2
14.
United Stages coinage, such as the quarter, appear silvery on the face, but close inspection reveals a reddish color at the edges. Based on your observations, to which one of the five categories of materials should a quarter belong? Explain. The coinage is a composite material composed of high nickel sheets on the two flat surfaces surrounding a core sheet of high copper. the high nickel provides good corrosion resistance and the appropriate silvery appearance, while the high copper in the core minimizes cost. The edge appears reddish because when the coins are stamped from the original sheet, the copper core is partly revealed.
15.
Relays in electrical circuits open and close frequently, causing the electrical contacts to wear. MgO is a very hard, wear resistant material. Why would this material not be suitable for use as the contacts in a relay? The MgO is a ceramic material and consequently acts insulator rather than as a conductor. Although it also will never allow current to flow through the the relay closes very rapidly, the brittle MgO however this latter point is not very important electrical properties.
16.
as an electrical may not wear, it relay. Also, if could fracture; compared to the
What mechanical properties would you consider most important when selecting a material to serve as a spring for an automobile suspension? Explain. The spring must have a high strength in order to support the automobile; i t must have a high modulus of elasticity so little elastic deformation occurs; it must have sufficient ductility so that it can be formed in the first place; it should have reasonably good resistance to corrosion, particularly to salt that might be picked up from the highway during the winter.
17.
The devices used for memory in personal computers typically contain an integrated circuit, electrical leads, a strong, non-conducting base, and an insulating coating. From what material should each of these four basic components be made? Explain your selections. The heart of the integrated circuit should be a semiconducting material such as silicon or GaAs so that electrical signals can be properly processed and information can be stored. The electrical leads must have a high electrical conductivity and might be made of aluminum or gold. Because the base must be both strong and non-conducting, it should be made from a ceramic material. Finally, the insulating coating could be made from either ceramic or polymer material. Polymers are most often used.
3
18.
Automobile bumpers might be made from a polymer material. recommend a thermoplastic of a thermosetting polymer application? Explain.
Would you for this
A thermosetting polymer, due to its network structure, is expected to be very brittle; even slight impacts of the bumper against another car, the end of the garage, or flying rock or gravel could cause it to break. The thermoplastic polymer has better ductility and impact resistance and consequently would be the better choice. 19.
Sometimes a nearly finished part is coined. During coining, a force is applied to deform the part into its final shape. For which of the following could this be done without danger of breaking the part - brass, Al 2 0 3 , thermoplastic polymers, thermosetting polymers, silicon? In order to be coined, the material must possess at least some ductili ty. Of the materials mentioned, Al 2 0 3 is a ceramic and is very brittle, thermosetting polymers are brittle, and silicon is brittle; none of these could easily be coined without a danger of introducing cracks or even. fracture. Both brass and thermoplastic polymers have good ductility and can be deformed.
20.
A scrap metal processor would like to be able to identify different materials quickly, without resorting to chemical analysis or lengthy testing. Describe some possible testing techniques based on the physical properties of materials. He could separate copper-base alloys from other metals by color copper, brass, and bronze are yellow or red. A magnet could be used to identify most iron and steel alloys - with only a few exceptions these are magnetic while most other common alloys that a scrap yard might encounter are not. Austenitic stainless steels could be separated from other stainless steels by the same method. The weight or density might also be used; aluminum and magnesium are lightweight compared to iron, copper, or nickel. Chapter 21 will also describe a variety of non-destructive tests, some of which might be easily adapted by a scrap metal processor.
4
Chapter 2 ATOMIC STRUCTURE
1.
Silicon, which has an atomic number of 14, is composed of three isotopes: 92.21% of the Si atoms contain 14 neutrons, 4.7% contain 15 neutrons and 3.09% contain 16 neutrons. Estimate the atomic mass of silicon. MSi
(0.9221)(14 + 14) + (0.047)(15 + 14) + (0.0309)(16 + 14) 28.1099 g/mol
2.
Titanium, which has an atomic number of 22, is composed of five isotopes: 7.93% of the Ti atoms contain 24 neutrons, 7.28% contain 25 neutrons, 73.94% contain 26 neutrons, 5.51% contain 27 neutrons and 5.34% contain 28 neutrons. Estimate the atomic mass of titanium. MTi
=
(0.0793)(24 + 22) + (0.0728)(25 + 22) + (0.7394)(26 + 22) + (0.0551)(27 + 22) + (0.0534)(28 + 22)
3.
47.9305 g/mol
Bromine, which has an atomic number of 35 and an atomic mass of 79.909 g/mol, contains two isotopes - Br79 and Br8l. Determine the percentage of each isotope of bromine. Let "x" represent the fraction of the Br 79 isotopes and represent the fraction of the Br 81 isotopes. Then 79.909 x
"x-l"
(x)(79) + (1 - x)(81) = 79x + 81 - 81x - 79.909)/(81 - 79) = 0.5455
= (81
Therefore Br contains 54.55% Br79 and 45.45% Br8l. 4.
Silver, which has an atomic number of 47 and an atomic mass of 107.87 g/mol, contains two isotopes - Ag l07 and Ag l09 . Determine the percentage of each isotope of silver. Let "x" represent the fraction of AgI07 isotopes and "x-1" represent the fraction of Agl09 isotopes. Then
=
107.87 (x)(107) + (1 - x)(109) 107x + 109 - 109x x = (109 - 107.87)/(109 - 107) = 0.565
Therefore Ag contains 56.5% Ag l07 and 43.5% Ag l09 .
5
5.
Tin, with an atomic number of 50, has all of its inner energy levels filled except the 4f level, which is empty. From its electronic structure, determine the expected valence of tin. First let's sum the electrons in the first four energy shells: lS2 =
2 electrons
2s 22p 6 =
8 electrons
3s23p63dl0 4s24p64dl04fO
18 electrons 18 electrons 46 electrons
There must be SO - 46 = 4 electrons in the outer energy shell: 5s 2Sp2= 4 electrons = valence of tin 6.
Mercury, with an atomic number of 80, has all of its inner energy levels filled except the Sf and 5g levels, which are empty. From its electronic structure, determine the expected valence of mercury. First let's sum the electrons in the first five energy shells: 2 electrons 8 electrons 3s23p63dl0 = 18 electrons 4s24p64JO 4[4
32 electrons
5s25p6Sdl05f05g0 = 18 electrons 78 electrons There must be 80 - 78
2 electrons in the outer energy shell: 2 electrons = valence of mercury
7.
Calculate the number of atoms in 100 grams of silver. Assuming that all of the valence electrons can carry an electrical current, calculate the number of these charge carriers per 100 grams. (a)
The number of atoms in 100 grams of Ag can be calculated from the molecular weight 107.87 g/mol and Avogadro's number:
number of atoms = (100 g)(6.02 x 10 23 atoms/mol) = 5.58 x 10 23 atoms 107.87 g/mol (b)
From Table 2-2, we expect silver to have a valence of 1. Therefore is all of the valence electrons can carry a current, the number of valence electrons equals the number of atoms in the 100 gram sample, or number of charge carriers
6
5.58
X
10 23 electrons
8.
Suppose there are 8 x 1013 electrons in 100 grams of germanium that are free to move and carry an electrical current. (a) What fraction of the total valence electrons are free to move? (b) What fraction of the covalent bonds must be broken? (On average, there is one covalent bond per germanium atom and two electrons in each covalent bond. ) (a)
First let's calculate the total number of valence electrons, using the molecular weight of 72.59 g/mole and the valence of germanium, which is 4.
number of atoms
(100 g)(6.02 x 10 atoms/mol) = ~--~~~~~~~~--------~ 72.59 g/mol 23
8.293
number of valence electrons = (4 electrons/atom) (8.293 = 3.7317 x 1024electrons 8 x 10 13
fraction that move (b)
10 24
X
= 2.41
X
10 23
10 23 atoms)
X
10- 11
Because on the average there is one covalent bond per Ge atom, the number of covalent bonds in 100 g is 8.293 X 10 23 • Since each broken covalent bond frees two electrons, the number of broken bonds is half the number of free electrons, or 4 x 10 13 bonds. The fraction of broken bonds is therefore fraction
9.
3.317
X
13
4 X 10 = -----------
8.293
=
4.82 x 10- 11
10 23
X
Compare the number of a toms in one gram of uranium wi th the number of atoms in one gram of boron. Then, using the densities of each (See Appendix A), calculate the number of atoms per cubic centimeter in uranium and boron. (a)
(b)
The molecular weights of uranium and boron are 238.03 g/mol and 10.81 g/mol. In 1 gram of metal, U atoms
(1 g)(6.02 x 10 atoms/mol) = ~~~~~~~~~~~------~ 238.03 g/mol
2.53
X
10 21
B atoms
(1 g)(6.02 x 1023 atoms/mol) 10.81 g/mol
5.57
X
10 22
23
The densities of uranium and boron are 19.05 Mg/m 3 and 2.3 Mg/m 3. The volumes of one gram of U and Bare volume of U volume of B U atoms/cm
B atoms/cm
3 3
3 1 g / 19.05 Mg/m 3 = 5.25 x 10- 8 m 1 g / 2.3 Mg/m 3 = 4.35 x 10- 8 m3 (2.53 x 10 21 atoms) / 0.0525 cm 3 ) (5.57 x 1022 atoms) / 0.435 cm3 )
7
4.82 x 10 22 1. 28 x 1023
10.
Suppose you collect 5 x 1026 atoms of nickel. Calculate the mass in grams and volume in cubic centimeters represented by this number of atoms. See Appendix A for the density. The molecular weight and density of nickel are 58.71 g/mol and 8.902 Mg/m . (5 x 1026 atoms) (58. 71 g/moU mass 48.76 x 103 g 6.02 X 1023
11.
Calculate the volume in cubic centimeters occupied by one mol of gold. See Appendix A for the necessary data. The mass of 1 mol of gold is equal to its atom\c weight, or 196.97 g. The density of gold is 19.302 Mg/m. The volume of one mol is therefore
12.
Suppose you have 15 mols of iron. Calculate the number of grams and the volume in cubic centimeters occupied by the iron. See Appendix A for the necessary data. The mass of 15 mols of iron, which has an atomic weight of 55.847 g/mol is mass = (15 mols) (55. 847 g/mol) = 837.7 g The volume of 15 mols of iron which has a density of 7.87 Mg/m3, is volume
13.
= (837.7
g) / (7.87 Mg/m3 ) (10- 6 cm3/m3 )
= 106.4
cm3
A decorative steel item having a surface area of 93750 mm2 is plated with a layer of chromium 0.125 mm thick. Calculate the number of atoms required to produce the plating. The number of atoms can be determined by calculating the volume of chromium (Cr), then calculating the mass of chromium from its density (7.19 Mg/m 3 ) , and finally calculating the number of atoms from the atomic mass (51.996 g mol-i). Volume
area x thickness
(93i~g cm2) (0'i~5)cm
11. 72 cm3 Mass
volume x density
= 84.26
Number of atoms (
g mass = mass/mol
x number of atoms/mol
84.26 g _1)(6.02 x 1023 atoms mol-i) 51. 996 g mol 8
= 9.76
x 1023
14.
Examine the elements in the IVB and VIIIB columns of the periodic table. As you go to a higher atomic number in each column (as from Ni to Pd to Pt), how does the melting temperature change? Would you expect this, based on the atomic structure? IVB Ti-1668 Zr-1852 Hf-2227
VB
VIB
VIIB
VIIIB
V-1900 Nb-2468 Ra-2996
Cr-1875 Mo-2610 W-3410
Mn-1244 Tc-2200 Re-3180
Fe-1538 Ru-2310 Os-2700
Co-1495 Rh-1963 Ir-2447
Ni-1453 Pd-1552 Pt-1769
For the elements listed, the melting temperature increases for each column as the atomic number increases. 15.
Examine the elements in IA column of the periodic table. As you go to a higher atomic number, how does the melting temperature change? Would you expect this, based on the atomic structure? Is this behavior different from what was observed in the elements in Problem 14? Can you explain this difference? H-gas Li-180.7 Na-97.8 K-63.2 Rb-38.9 Cs-28.6 The melting temperature decreases as the atomic number increases; is opposite the behavior noted in Problem 14.
16.
Determine the formulas of the compounds formed when each of the following metals react with oxygen. (a) calcium, (b) aluminum, (c) germanium, (d) potassium. ... 2s 2 p 4
~
CaO
(b)
AI: ...
3s2 p 1
4 0: ... 2s 2 p
~
Al 0
(c)
2 Ge: ... 4s 2 p
4 0: ... 2s 2 p
~
GeO
(d)
K: ... 4s1
0: ... 2s 2 p 4
~
K2 0
(a)
17.
this
Would
Ca: ... 4s 2
expect
you
elasticity? The
0:
Al 2 0 3
or aluminum
2 3
to
2
have
the
higher
modulus
of
Explain.
ions
in
the
ceramic
Al 2 0 3
are
joined
primarily
by
the
particularly strong ionic bonds; aluminum atoms are joined by metallic bonding, which is normally less strong. We would expect a deeper energy trough in the alumina than in the aluminum, leading to a higher modulus of elasticity for Al 20 3 . The actual values are A1 20 3 : 380 GPa aluminum:
69 GPa
9
18.
Would you expect silicon or nickel to have the higher coefficient of thermal expansion? Explain. Atoms in silicon are joined by covalent bonds, while atoms in nickel are joined by metallic bonds. We expect the covalent bonds to be stronger, leading to a deeper energy trough, a high modulus of elasticity, and a lower coefficient of thermal expansion in the silicon than in the nickel. the actual values are: silicon: 3 x 10- 6 / o C nickel: 13 x 10-6 / o C
19.
The compound GaAs is an important semiconductor material in which the atoms are joined by mixed ionic-covalent bonding. What fraction of the bonding is ionic? The electronegativity of Ga is about 1.8 and the electronegativity of As is about 2.2. From Equation 2-1, the fraction of bonding that is covalent is fcovalent
exp[(-0.25)(2.2 - 1.8)2] = exp [-0.04] = 0.961
The fraction of bonding that is ionic must be 0.039 or 3.9%. 20.
The compound InP is an important semiconductor material in which the atoms are joined by mixed ionic-covalent boding. I f the fraction of covalent bonding is found to be 0.914, estimate the electronegativity of indium. Does your calculated value compare well with what you might expect, based on Figure 2-3? The electronegativity of P from Figure 2-3 is about 2.1. Equation 2-1, letting E be the electronegativity of Indium, fcovalent = 0.914 = In(0.914) = (2.1 - E)2= 2.1 - E E
21.
From
2
exp[(-0.25)(2.1 - E) ] -0.0899 = (-0.25)(2.1 - E)2 0.3596 0.6 1. 5
Would you expect bonding in the intermetallic compound Ca 2Mg to be predominantly ionic or metallic? Explain. The electronegativity of Ca is about 1. 1 and that of Mg is about 1. 3. The electronegativities are relatively the same, so we would expect that bonding might be predominantly metallic.
22.
The electronegativities of both nickel and copper are 1. 8. Would you expect bonding in the intermetallic compound Ni Mg to be more or less metallic than in CuAI? Explain. 2 2
The electronegativities are Ni: 1.8, Mg: 1.3, Cu: 1.8, and AI: 1.5. There is a greater difference in electronegativies between Ni and Mg than there is between Cu and AI. Therefore we would expect bonding in CuAl 2to be more metallic than in Ni 2Mg.
10
Chapter 3 ATOMIC ARRANGEMENT
1.
How many lattice points are unique to the base-centered orthorhombic unit cell? (1/8)(8 corners) + (1/2) (2 faces)
2.
=2
points/cell
Why is there no base-centered tetragonal structure? Draw a lattice this structure, then determine what the actual unit cell is.
for
As the sketch indicates, the base centered tetragonal structure could be redrawn as a simple tetragonal structure.
3.
Why is there no base-centered cubic structure? Draw a lattice structure, then determine what the actual unit cell is.
for
this
As the sketch indicates, the base-centered cubic structure could be redrawn as a simple tetragonal structure.
4.
A material has a cubic unit cell with one atom per lattice point. If a 0
= 4.0786 A and
= 1. 442 A. determine the = 2r = 2.884 ~ 4.0786 0
r
crystal structure.
For simple cubic, a For BCC, For FCC,
5.
a a
0
= 3.33
~
4.0786
= 4r/Y2 = 4.079
9!
4.0786
4r/V3
0
Material is FCC!
A material has a cubic unit cell with one atom per lattice point. a = 5.025 Aand r = 2.176 A, determine the crystal structure. o
For simple cubic, a For BCC,
a
For FCC,
a
0
0
= 2r
4.352
4r/V3
5.025
4r/Y2
6. 156
11
~
5.025 5.025
~
5.025
Material is BCC!
If
6.
Using the atomic radius data in Appendix B, calculate the packing factor for crystalline polyethylene. The lattice parameters for orthorhombic polyethylene are a = 7.41 A, b = 4.94 Aand c = 2.55 A. The atomic radii are o 0 0 rC = 0.77 Aand r H= 0.46 A. From example 3-22, there are 8 H atoms and 4 C atoms per cell
PF
=
(4 C atoms) [4n(0.77)3/ 31 + 8 H atoms) [4n(0.46)3/ 31 (7.41}(4.94)(2.55)
0.117
The unusually low packing factor is due to the restrictions of covalent bonding and partly to the use of the particular atomic radii listed in the Appendix. 7.
The density of lead is 11.36 Mg/m3 , its atomic mass is 207.19 g/mol and the crystal structure is FCC. Calculate (a) the lattice parameter and (b) the atomic radius for lead. (a)
Because lead is FCC, there are 4 atoms/unit cell and we can calculate the lattice parameter from the density equation (4 atoms/cell) (207. 19 x 10- 6
p = 11.36
(a )3 (6.02
o
)
x 1023 )
121. 18 x 10- 30 :. a (b)
X
10-10 m = 4.9485
A
For FCC unit cells r
8.
4.9485
o
= iZ
= (iZ)
a o/4
(4.9485) /4= 1.749
A
The density of tantalum is 16.6 Mg/m3 , its atomic mass is 180.95 g/mol and the crystal structure is BCC. Calculate (a) the lattice parameter and (b) the atomic radius for tantalum. (a)
From the density equation, with 2 atoms per unit cell, p
(2 atoms/cell) (180.95 x 10-6 Mg/mol)
= 16.6
(a )3 (6.02 x 1023 )
o
(ao )3 = 36.215 x 10-30 :. a
(b)
o
=3.3085
x
10- 10
m
= 3.3085 A
For BCC unit cells r
= 13
a /4 o
= (13)
12
(3.3085)/4
1. 433
A
9.
How many unit cells are present in a cubic centimeter of face-centered cubic nickel? The atomic radius of nickel is 1.243 A. The lattice parameter and unit cell volume for FCC nickel are
= 4r/V2 = (4)(1.243)/12 = 3.5163 A = (ao )3 = (3.5163)3 = 43.4768 A3 = 43.4768 3 -24 unit cells/cm = 1 / 43.4758 x 10 = 2.3 x a
o
V
10.
Calculate (a) the volume and (b) the mass of one million unit cells of body-centered cubic iron. The atomic radius of iron is 1.241 A. The lattice parameter and unit cell volume for BCC iron are
= 4r/13 = (4)(1.241)/13 = 2.866 A V = (a )3 = (2.866)3 = 23.541 A3 = 23.541 o
a
o
10- 24 cm3
(a)
The volume of one million unit cells is
(b)
The mass of one million unit cells can be obtained from the density which is 7.87 Mg/m 3. mass
11.
X
= pV =
(7.87 g/cm 3 )(23.541 x 10- 18 cm3 )
A material with a cubic structure has a density of 0.855 Mg/m 3, an atomic
mass of 39.09 g/mol, and a lattice parameter of 5.344 A. If one atom is located at each lattice point, determine the type of unit cell. We would like to find "x", the number of atoms per unit cell, using the density equation. p
(x)(39.09 x 10- ) = 0.855 = --~~~--------~--------6
(5.344 x
10- 10 )3(6.02
x 10 23 )
.. x = 2 a toms/unit ce 11 .. BCC structure 12.
A material with a cubic structure has a density of 10.49 M$/m3, an atomic
mass of 107.868 g/mol, and a lattice parameter of 4.0862 A. If one atom is located at each lattice point, determine the type of unit cell. We would like to find "x", the number of atoms per unit cell, using the density equation.
P
= 10.49
=
-6
(x) (107.868 x 10 ) (4.0862 x 10- 1°)3(6.02 x 10 23 )
.. x = 4 atoms/unit cell .. FCC structure.
13
13.
Antimony has a hexagonal unit cell with a o = 4.307 A and Co = 11.273 X. If its density is 6.697 Mg/m3 and its atomic mass is 121. 75 g/mol, calculate the number of atoms per cell.
v = a o2c 0 cos30 = (4.307)2(11.273)cos30 =181.1 X3 = 181.1 x 10-30m3 We would like to find "x", the number of atoms per unit cell, using the density equation.
P = 6.697 =
-6
(x) (121.75 x 10 ) (181.1 x 10-3°)(6.02 x 1023 )
x = 6 atoms/unit cell 14.
One of the forms of plutonium has a face-centered orthorhombic structure, with a = 3.159 X, b = 5.768 X, and c = 10.162 X. The density of Pu is 17.14 Mg/m3 and the atomic mass is 239.052 g/mol. Determine (a) the number of atoms per cell and (b) the number of atoms at each lattice point. (a)
From the density equation we can calculate the number of atoms per cell "x". p
= 17.14 =
:. X
(b)
15.
(x) (239.052 x 10- 6
)
----:.::..::.:.---=..:::..::..:..:....:..::.::.....::..::....=-:~------::-:,---
(3.159) (5. 768) (10. 162) (10- 30 ) (6. 023 x 10 23 )
= 8 a toms/unit ce 11
There are 4 lattice points/unit cell in the face-centered orthorhombic crystal structure. Therefore there must be 2 atoms per lattice point.
Prasiodymium has a special hexagonal structure with 4 atoms per uni t cell; the lattice parameters are a = 3.6721 Aand c = 11.8326 X, while o the atomic radius is 1.8360 X. Calculate the packing factor of Pro V unit cell
Vatom PF
(3.6721)3(11.8326)cos30 = 138.18 4nr 3/3
X3
= (4) (n) (1.8360)3/3 = 25.9243 X3
(4 atoms/cell) (25. 9243 138.18 X3 /ce11
14
X3/atom)
0.75
16.
Gadolinium has a HCP structure just below 12600 C with a = 3.6745 Aand c = 5.825 A. Just above 12600 C, Gd transforms to a BCC structure with a = 4.06 A. Calculate the percent volume change when Gd cools from the BCC to the HCP structure. Does the metal expand or contract during cooling? Below 12600 C: Above 12600 C:
VHCP VBCC
= (3.6745) 2 (5.8525)cos30 = 68.4335 = (4.06) 3 = 66.9234 03 A
93
A
Both unit cells contain 2 atoms, so we can directly compare the two volumes. %change 17.
66.9234 - 68.4335 x 100 66.9234
-2.26% expansion
Lanthanum has a FCC structure just below 865°C with a = 5.337 A, but has a BCC structure with a = 4.26 A just above 865°C. Calculate the percent volume change when La heats from the FCC to the BCC structure. Does the metal expand or contract during heating?
VFCC VBCC
Below 865°C: Above 865°C:
= (5.337) 3 = 152.02 A93 = (4.26) 3 = 77.309 A93
But the FCC structure contains 4 atoms/cell while the BCC structure contains only 2 atoms/cell. To compare the volume of equal numbers of atoms, we should use two BCC cells. 152.02 - 2(77.309) x 100 152.02
%change
18.
-1.71% expansion
Lanthanum has a special HCP structure just below 325°C and the FCC structure just above 3250 C. At 3250 C, the lattice parameters for the HCP structure are a = 3.779 A and c = 12.270 X; the lattice parameter for the FCC structure at this temperature is 5.303 X. Lanthanum has a density of 6.146 Mg/m3and an atomic mass of 138.9055 g/mol. (a) Calculate the number of atoms in the special HCP unit cell. (b) Calculate the percent volume change when the FCC form of La transforms to the HCP structure on cooling. Does the metal expand or contract during cooling? (a)
The volume of the special HCP structure is V
(3.779)2(12.270)cos30
p
6.146 Mg/m3 =
~
(b)
x
=4
(x)
= 151.75 A3= 151.75
(138. 9055 x
1(f6
Mg/moll
(151.75 x 10- 3°)(6.02 x 1023 )
atoms/cell
X3
Below 325 0 C:
151. 75
Above 325 C:
(5.303)3
Both structures have compared directly.
4 atoms/cell
0
%change
x 10- 30 m3
=
= 149.13 X3
149.13 - 151.75 x 100 149.13 15
so
the
volumes
-1.76% expansion
can
be
19.
At 1450o C, thorium changes from one type of cubic unit cell to a different cubic cell, with a 0.5% decrease in volume during heating. Below 1450oC, the lattice parameter is 5.187 X while the lattice parameter of the higher temperature form is 4.11 X. What is the ratio between the number of atoms in the unit cell of the high-temperature from the the number of atoms in the unit cell of the low-temperature form of Th?
= (5.187)3 = 139.556 X3 3 Vhigh= (4.11)3 = 69.4265 X
v low
We really do not need to consider the 0.5% volume difference. By inspection of the volumes of the unit cells, it is apparent that the high temperature form of thorium must contain half the number of atoms as the low temperature form. atoms in high T form atoms in low T form 20.
a-Mn has a cubic structure with a
1
=2 8.931A and a density of 7.47 Mg/m 3 •
o
~-Mn has a different cubic structure with a o = 6.326 AOand a density of 7. 26 ~/m3. -r-Mn has a tetragonal structure with a = 3.784 A and c =
9.40 X and a density of 7.21 Mg/m3 • The atomic mass of manganese is 54.9380 g/mol. (a) Calculate the number of atoms in each of the three polymorphic forms of manganese. (b) Assuming that the radius of the Mn atom is 1.12 A in all three forms, calculate the packing factor for each of the three unit cells. (a)
a-Mn: :. x
(b)
Mg/m3
=
(x)(54.938 x 10-6 ) (8.931 x 10-1 °)3(6.02 x 1023 )
p
7.26 Mg/m3
(x) (54. 938 x 10- 6 ) (6.326 x 10- 1°)3(6.02 x 1023 )
= 20
-r-Mn:
:. x
= 7.47
= 58
~-Mn:
:. x
p
P
7.21 Mg/m 3
(x)(54.938 x 10-6 ) (9.4)(3.784)2
X
= 10
The packing factors are approximately
vatom = ~nr3 3 4(n) (1. 12 x 10- 1°)3 /3
5.8849 x 10- 30m3 PF(a-Mn) =
(58) (5. 8849 x 10-30 ) -'.....-....:....:---~-=---'(8.931 x 10-1°)3
16
0.479
(10- 3°) (6. 02 x 102~)
PF(/3-Mn) =
= 0.465
(6.326 x 10- 1 °)3 (10)(5.8849 x 10-30
PF(7-Mn)
21.
20(5.8849 x 10- 3 °)
)
-'---'-'---'--'~"--'-:=
(9.40)(3.784)2
X
10- 30
= O. 437
Determine the Miller indices for the directions in the cubic unit cell shown in Figure 3-35
[221] [101] c: (1, 0, 1) - (0, 1, 0) [1111 1, -1, 1 D: (3/4, 1, 0) - (1, 0, 213) = -1/4, 1, -2/3 = [3 12
A: B:
22.
A: C: D:
0, 112, 1)
0, 3/4,
1, -1/2, 0
(0, 1, 1)
0, 1, 0)
(0, 1/3, 1) (1/4, 1, 0)
-1, -2/3, 1
(0, 0, 0)
0, 1/4,
0)
114, 1, 0 1) = 0,
1/2, -1
[210] [323] [140] [012]
Determine the Miller indices for the planes in the cubic unit cell shown in Figure 3-37. A: B:
-1
l/y
= 1/2, = 2,
x = "', l/x 0,
y l/y 1/y
= 0, = 0,
x l/x
= -1,
l/x = 0,
c:
24.
8]
Determine the Miller indices for the directions in the cubic unit cell shown in Figure 3-36. B:
23.
(1, 0, 0) - (0, 1, 1/2) = 1, -1, -1/2 (0, 112, 1) - (1, 1/2, 0) = -1, 0, 1
X
= "',
1/x = 0, l/x = 0,
Y
= "',
z liz
z
liz liz
1 1
= 2/3 = 3/2
=3
y = -2/3, z = -1 l/y = -3/2,l/z -1 l/y = -3, liz = -2
(121)
(003)
(032)
Determine the Miller indices for the planes in the cubic unit cell shown in figure 3-38.
A:
B:
C:
x
= -1, = -1, = -3,
x
= 2,
l/x l/x l/x
y
= 3/4, = 4/3, = 4.
z liz liz
-1/2 -2 -6 (346)
y
= 1, = 1, = 2,
z liz liz
1/4 4 (128) 8
Y
= -1, = -1,
z liz
114 (014) 4
l/y l/y
l/x
=
1/2,l/y 1, l/y
x l/x
=
01),
= 0,
l/y
17
25.
Determine the Miller indices for the directions in the hexagonal unit cell in Figure 3-39. using the three-digit system.
A: B:
c: D:
26.
A:
c:
D:
-4. -8. 3 O. -1. -2
[210] [483] [012] [121]
(0. (1. (0. (1.
a
-1. 1) - (1. 1. II
-1. -2.
O. 1) -
(0.
1. O. 1 O. 2. 112
O. 0) -
(112.
o.
0)
1. 112) - (0. -1. 0)
112. -1.
1. 0)
[120] O. 4. 1
a
1. -2.
a
[lOll [04ll
[120]
Determine the Miller-Bravais indices for the planes in the hexagonal unit cell in Figure 3-41. A:
a l/a
B:
C:
a
1
1
1
a l/a
2
l/a
1
a
2
lIa a
1
lIa 28.
2. 1. a -1. -2. 3/4. O. -112. -1 1. 2. -1
Determine the Miller indices for the directions in the hexagonal unit cell in Figure 3-40. using the three-digit system. B:
27.
(1. o. 0) - (-1. -1. 0) (0. -1. 3/4) - (1. 1. 0) (1. 1/2. 0) - (1. 1. II (0. 1. 0) - (-1. -1. II
2
a
a
l/a
3
3
a
c = 1
00
=a
3
00
-2
2
=
3
-1/2
2
l/a
1
a
l/a
2
a
1
=-1 =-1
-1
c
-1
llc
c
3
l/a
=1
llc
= -1
=
(1011)
00
a
lIc
3
(1101 )
(1210)
Determine the Mi ller-Bravai s indices for the planes in the hexagonal unit cell in Figure 3-42.
A:
a
1
l/a
B:
a l/a
c:
a lIa l/a
1
1
= =a
00
=
1 1 1 1
=
a l/a
2
1
a
a
l/a
00
a l/a
2
2
3
3
1/a
2
l/a
3
a
2
2
a l/a
2
00
a a
=
3
=
00
a a
18
a l/a l/a
3 3
3
= -1 -1
c llc
= 1/3 =3 =
1/2
-1
c
-1
lIc
00
c
= 2/3
lie
3/2
llc
3
a a
(0113)
2
(0112)
(0003)
29.
Sketch the following directions and planes within a cubic unit cell. z
a. [112]
b. [310]
c. [111]
d. [101]
e. [041]
f.
crOll
h. (111)
1. (013)
J. (12ll k. (20ll
l. (120)
g.
(111)
[203]
y
x z
z
/--=--+---7- y x
30.
(201 )
y
x
Sketch the following directions and planes within a cubic unit cell.
[lID]
a. [123]
b.
d. [131]
e. [1211
f.
g. (220)
h. (301)
i. (112)
j. (011)
k.
(421)
l. (141)
1 2
3'3,1
z
c. [010] [134]
x z
z
(301)
y
x
31.
(220)
z
y
x
Draw the (111) plane and identify the six directions that plane in a cubic lattice. z
~--+--+-y
x
19
[lOll
[101]
[011]
[OIl]
[lID]
[110]
that
lie
in
32.
Draw the (110) plane and identify the four directions of the that lie in that plane in a cubic lattice.
form
z
[111]
611]
[111]
[111]
y
x
33.
How many planes of the form {131} are found in a cubic system? Would you give the same answer if we used a tetragonal or orthorhombic system? Explain. For the cubic system. there are 12 unique planes of the form or 24 planes if the negatives are included. (13) (131)
(311)
(113)
(131) (311)
(13) (131) (311)
{131}.
(113)
(31) (311)
In the tetragonal system. planes of the form {131} would only include 2/3 of those listed above. giving 8 unique planes or 16 planes if the negatives are included. (131) (311)
(131 ) (311 )
(131) (311)
(131)
(311)
In the orthorhombic system. only 4 unique planes. or 8 negatives are included. belong to planes of the form {131}.
if
the
How many planes of the form {123} are found in a cubic system? What the indices of the planes of the form {123} in a tetragonal system?
are
(31) 34.
(131 )
(31)
(131)
In the cubic system. there are 24 planes of the form planes if the negatives are included. (23) (132) (213) (231) (312) (321)
(123) (132) (213) (231) (312) (321)
(123) (32) (213) (231 ) (312) (321)
(123) (213)
(23) (213) 20
or
48
(123) (132) (213) (231) (312) (321)
In the tetragonal system. there are 8 unique planes {123}. or 16 planes if the negatives are included. (123) (213)
{123}.
(23) (213)
of
the
form
35.
How many directions of the form are found in a cubic system? Would you give the same answer if we used a tetragonal or orthorhombic system? Explain. In a cubic system there are 48 directions of including the following plus their negatives. [123] [132] [213] [231] [312] [321]
[123] [132] [213] [231] [312] [321]
[123] [132] [213] [231] [312] [321]
the form
,
[123] [132] [213] [231] [312] [321]
In tetragonal systems, there are 16 directions of the form , including the following plus their negatives. [123] [213]
[123] [213]
[123] [213]
In orthorhombic systems, there are 8 directions of the form , including the following plus their negatives. [123] 36.
[123]
How many directions of the form are found in a cubic system? What are the indices of the directions of the form in a tetragonal system? There are 24 directions of the form including the following and their negatives. [221] [122] [212]
[221] [122] [212]
[221] [122] [212]
in
cubic
systems,
[221] [122] [212]
In a tetragonal system, there are 8 directions of the form , including the following and their negatives. [221] 37.
[221]
[221]
[221]
What are the indices of the planes of the form {412} in an orthorhombic system? Planes of the form {412} in orthorhombic following 4 planes plus the 4 negatives. (412)
38.
systems
include
the
(412)
What are the indices of the directions of the form , in an orthorhombic system? Directions of the form include the following 4 directions plus the 4 negatives. [121]
[121]
[121]
21
39.
Determine whether the [101] direction in a tetragonal unit cell with a c/a ratio of 1.5 is perpendicular to the (101). plane. If it is not perpendicular, calculate the angle between the direction and the plane. The sketch shows the [101] direction and (101) plane in a tetragonal unit cell with a= 2 and c = 3 (giving a c/a ratio of 1.5). The second sketch shows the trace that the plane and direction make on the (010) face of the cell. z
tan(et/2) et/2
1. 5/1. 0
1.5
56.30 112.60
CIt
Obviously the direction and plane are not perpendicular to one another, as they would be in a cubic system.
40.
Determine whether the [110] direction in an orthorhombic unit cell with a = 3 A, b = 4 A, and c = 5 Ais perpendicular to the (110) plane. If it is not perpendicular, calculate the angle between the direction and the plane. z The sketches show the plane and direction in the unit cell and a view of the (001) plane which contains the trace of the plane and direction. tan(et/2) et/2 IX
41.
2/1. 5
53.130 106.260
k--t--::::;r#-- Y
1. 3333 x
Draw the plane in a cubic system that passes through the coordinates I, I, 0; 0, I, 1; and 0, 0, 1. What are the Hiller indices of this plane? z
x l/x
1
1
Y
l/y
co
o
(101)
r----I+-+--
Y
x
22
Z
l/z
1 1
42.
Draw the plane in a cubic system that passes through the coordinates 1, 1, 0; 0, 0, 1; and 0, 1, O. What are the Miller indices of this plane? z
x l/x
=m y = 1 = ally = 1
z liz
1
1
(all) )-->',.--H'JIJ-- Y
x
43. Draw the plane in a cubic system that passes through the coordinates 1, 0, 1; 1/2, 0, 1; and 1, 1/2, O. What are the Miller indices of the plane? z
x l/x
= m y = 1/2 z = ally = 2 liz
1 1
(021)
Jr-+-+---;f- y
x
44.
Draw the plane in a cubic system that passes through the coordinates 1, 0, 0; 0, 0, 1; and 1/2, 1, 1/2. what are the Miller indices of this plane? z
x l/x
1 1
y lly
=m a
Z
liz
1 1
(010) J----+hf-y
45.
x
In the four-digit system for finding the indices for a direction in HCP unit cells, is [110] equal to a [1120]? Show this by constructing the path from the tail to the head of the direction. c The sketch below shows the [110] Direction, which is actually the negative a axis. Let's let the origin be one of the p5ints on our [1120]direction,. If we start at the origin and move 1 lattice parameter in the a direction, 1 lattice parameter in the a direction, and -2 lattice parameters in th~ a direction, we have a second point on the d~rection. This point and the origin form a direction that also is the negative a direction and is identical in direction t5 the [110] direction. 23
+
I
\-z \
\
,
~ to]:[f
t2.0]
46.
In the four-digit system for finding the indices for a direction in HCP unit cells, is the [100] equal to a [2110]7 Show this by constructing the path from the tail to the head of the direction. The [100] direction is the at direction.
t
Let's let the origin be one point on the [2110] direction. We move 2 lattice parameters in the a l direction, -1 lattice parameters
in
the
a2
direction,
and
-1
lattice parameter in the a 3 direction and produce a point 3 lattice parameters along the at direction. The [2110] direction also
,
lies on the at axis and is identical to the -I
[110] direction.
r ---- #'
,'.z
-I'
~OOl" c%.iiol
47.
In the four-digit system for finding the indices for a direction in HCP unit cells, is the [011] equal to a [1213]7 Show this by constructing the path from the tail to the head of the direction. The [011] direction is shown in the sketch and lies in the plane formed by the a 2 and c axes. Let's let one point on the [1213] direction lie at the origin we move -1 lattice parameter in the a 1 direction, 2 lattice parameters in the a 2 direction,
-1
lattice parameter in the a 3 direction, and 3 lattice parameters in the c direction. This puts us at coordinates indicated by +3 in the a 2 direction and +3 in the c direction and gives a direction that is identical to the [011].
,., 1
, I
I
}t--+:HlIIl-- - -
_._l.. - --- - .: -I
0..
24
\~
48.
Is the [1210] direction in an HCP unit cell perpendicular to the (1210) plane? Draw each and verify your answer. First we need to construct the direction and plane in the unit cell. We let the origin be one point on [1210]. We move 1 lattice parameter in the a 1 direction, -2 lattice
c
parameters in the aa direction and 1 lattice parameter in the a 3 direction, giving a second point that lies 3 lattice parameters in the negative aa direction. The [1210] direction is the same as the negative aa axis. The intercepts for the plane will be a 1 = 1; a
= -1/2, a 3
= 1;
c = m.
ct.
The plane shown in
a the sketch satisfies these four intercepts. It shol!.1d be aeparent by simple inspection that [1210] ~ (1210).
49.
Calculate the linear density of a line in the [111] direction in (a) simple cubic, (b) body-centered cubic, and (cl face-centered cubic unit cells, assuming a lattice parameter of 4.0 A in each case. (al
In simple cubic, atoms are located at corners of the cube. If we start at the origin and move in the [111] direction, or body diagonal, we do not encounter another lattice point until coordinates 1, 1, 1. repeat distance
= iJao = (iJl(4
Al
= 6.9282
x 10-8 cm
linear density = l/repeat distance = 1.443 x 107 points/cm (bl
When we start at the origin of BCC and move in the [111] direction we encounter the body centered atom at 1/2. 1/2. 1/2. repeat distance
= iJao/2 = (iJl(4
Al/2
= 3.4641
x 10-8 cm
linear density = l/repeat distance = 2.887 x 107 points/cm (cl
In FCC, we again do not encounter a second lattice point until coordinates 1, 1, 1, just as in simple cubic. repeat distance linear density
25
= 6.9282
= 1.443
x 10-8 cm
x 107 points/cm
50.
Calculate the linear density of a line in the [110] direction in (a) simple cubic, (b) body-centered cubic, and (c) face-centered cubic unit cells, assuming a lattice parameter of 4.0 X in each case. (a)
As we move in a simple cubic structure from the origin along the [110], or face diagonal, we do not encounter another lattice point until coordinates 1, 1, O. repeat distance = iZa
o
5.6569
= (iZ)(4 X)
linear density = l/repeat distance (b)
linear density
107 points/cm
X
5.6569
= (iZ)(4 X)
= l/repeat
linear denSity
X
10- 8 cm
1.768 x 107 points/cm
distance
As we move along the [110] direction in another lattice point ·at 1/2, 1/2, o. repeat distance
51.
10-8 cm
In BCC, the situation is identical to that in simple cubic along the [110] direction. repeat distance = iZa
(c)
1.768
X
FCC,
we encounter
= iZa/2 = (iZ)(4 X)/2 = 2.8284 x 10-8 cm = l/repeat distance = 3.536 x 107 points/cm
Calculate the packing fraction in the [111] direction in (a) simple cubic (b) body-centered cubic, and (c) face-centered cubic unit cells. In which, if any, of these structures is the [111] direction a close-packed direction? (a)
Simple cubic: Length iJa
(b)
BCC:
FCC:
= iJ(2r) = 3. 464r
= 2r
or PF
/ 3. 464r
0.577
There are four atomic radii along the [111]
Length (c)
o
There are two atomic radii along the [111]
= iJ(4r/iJ) = 4r
or PF
= 4r
/ 4r
= 1.00
There are two atomic radii along the [111]
Length
= iJ(4riZ)
4.899r
or PF
= 2r
/ 4. 899r
= 0.408
The [111] direction is a close packed direction in the BCC structure 52.
Calculate the packing fraction in the [110] direction in (a) simple cubic, (b) body-centered cubic, and (c) face-centered cubic unit cells. In which, if any, of these structures is the [110] direction a close-packed direction? (a)
Simple cubic: Length iZa o
(b)
BCC:
There are two atomic radii along the [110]
= iZ(2r) = 2. 828r
or PF
= 2r
/ 2. 828r
There are two atomic radii along the [110] 26
0.707
Length (c)
FCC:
= iZ(4r/iJ) = 3.266r
or PF
= 2r
I 3.266r
= 0.612
There are four atomic radii along the [110]
Length
= iZ(4riZ) = 4.r
or PF
= 4r
I 4r
= 1.00
The [111] direction is a close packed direction in the FCC structure 53.
Calculate the packing fraction of a (111) plane in (a) a simple cubic, (b) a body-centered cubic, and (c) a face-centered cubic unit cell. In which, i f any, of these structures is the (111) plane a close-packed plane? Sketches of the atoms centered on the (111) plane in each unit cell are shown.
11
(a)
Simple cubic: Only 1/6 of each corner atom is included in the plane within the cell. (iZa 12)(iJa liZ)
= 0.866a2 = 0.866(2r)2 = 3. 464r2
0 0 0
Aatoms = (3 corners) (1/6 atomlcorner)nr 2 = nr2/2 PFSC = [nr 212] 1 3.464r2 = 0.453 (b)
Body-Centred cubic: (Note that the (111) plane does not pass through the center atom! ) A
111
= 0.866(4r/iJ) = 4.619r 2
Aatoms PF (c)
= (3
= (nr 2 /2)
corners) (1/6)nr I 4. 619r 2
= nr 212
2
= 0.34
Face-centered cubic: (Note that the (111) plane also bisects an atom along each of the three edges) _'" 2 Alll = 0.866(4r/v2)
= 6.928
r2
2 Aatoms = [(3 corners)(1/6) + (3 edges)(1/2)]nr
PF
= 2nr2
I 6. 928r 2
= 2nr2
= 0.907
The (111) plane is a close packed plane in FCC unit cells.
27
54.
Calculate the packing fraction of a (110) plane in (a) a simple cubic, (b) a body-centered cubic, and (c) a face-centered cubic unit cell. In which, if any, of these structures is the (110) plane a close-packed plane?
(a)
Simple cubic A
110
""fio.'
= iZa Z = iZ(2r)z = 5. 657rZ 0
Z
Aatoms = (4 corners)(1/4 atom/corner)nr = nr
Z
PF = nrZ / 5.657rZ = 0.555 (b)
Body-centered cubic: A
110
= iZa Z = iZ(4r~)Z = 7. 542rz 0
A = [(4 corners)(1/4) + 1 centre]nr z = nrz atoms PF =2nr z / 7. 542rz = 0.833 (c)
Face-centered cubic: Z
A110 = iZa o = iZ(4r/iZ)z = 11.314rz Aatoms = [(4 corners)(1/4) + (2 edges) (1/2)]nrZ
2nr Z
PF = 2nr Z / 11.314rZ = 0.555 55.
Calculate the planar density on a (111) plane in (a) simple cubic, (b) body-centered cubic, and (c) face-centered cubic unit cells assuming a lattice parameter in each case of 4.0 A. Referring to the sketches in problem 53. A
111
(a)
= 0.866a Z = 0.866(4 x 10-8 )z = 13.856 x 10-16 0
Simple cubic: The number of atoms located on the is (3 corners)(1/6 atom/corner) = 0.5 atom. PO = 0.5 atom / 13.586 x 10-16
(b)
=
3.61
BCC: The number of atoms located on the corners)(1/6 atom/corner) = 0.5 atom PO = 0.5 atom / 13.856 x 10-16 = 3.61
(c)
X
X
Z
Cia
(111)
plane
1014 atoms/cmZ (111)
plane is (3
1014 atoms/cmZ
FCC: The number of atoms located on the (111) plane is (3' corners) (1/6 atom/corner) + (3 edges)(1/2 atom/edge) = 2 atoms PO = 2 atoms / 13.856 x 10-16 = 14.43
28
x
1414 atoms/cmZ
56.
Calculate the planar density of a (110) plane in (a) simple cubic, (b) body-centered cubic, and (c) face-centered cubic unit cells assuming a lattice parameter in each case of 4.0 A. Referring to the sketches in problem 54:
= .f2a'2.0 = .f2(4
A
110
(a)
Simple cubic:
PD (b)
BCC:
=1
x 10-8 )'2.
= 22.627
x 10-16 cm'2.
There is one atom located on the (110 plane)
atom / 22.627 x 10-16
= 4.42
There are two atoms located on the (110) plane.
PD = 2 atoms / 22.627 x 10-16 = 8.84 (c)
FCC:
101 ' atoms/cm'2.
X 101 ' atoms/cm'2.
Calculate the llnear densities in the [110] and [101] directions in a face-centered tetragonal unit cell with a = 4.0 Aand 6 = 6.0 A. (110):
The [110] passes through a face-centered lattice point
repeat distance
LD [101]:
=1
/ repeat
LD
= (1/2) (.f2)a = (1/2)(.f2)(4) = 2.828 distance = 3.54 x 107 points/cm
=1
=
(1/2)(..4'2. + 6'2.)
/ repeat distance
= 2.774
= 3.6055
10-8 cm
and
A and
(101) planes c = 6.0 A.
"in
a
There are (4 corners) (1/4) +(2 faces) (1/2) or two atoms located on the (110) plane A110
PD (101):
X
x 107 points/cm
Calculate the planar densities in the (110) face-centered tetragonal unit cell with a = 4.0 (110):
x 10-8
The [101] passes through a face-centered lattice point
repeat distance
58.
X
There are two atoms located on the (110) plane.
PD = 2 atoms / 22.627 x 10-16 = 8.84 57.
101 ' atoms/cm'2.
X
= .f2ac
= (.f2) (4) (6) 33.94 x 10-16 cm'2.
2 atoms / 33.94 x 10-16 5.89 X 1014 atoms/cm'2.
"
There are (4 corners)(1I4) + (2 faces)(1/2) or two atoms located on the (101) plane A101
PD
= a./a'2.
+ c'2.
= 28.84
= (4) (/4'2.
X 10-16
+ 6'2.
cm'2.
2 atoms/ 28.84 x 10-16
= 6.93
X
1014 atoms/cm'2. 29
(101)
59.
Calculate the linear densities in the [100], [010], [110] and [0011 directions of a base-centered orthorhombic unit cell with a = 3.0 b = 5.0 A, and c = B.O-A.
A,
[100]: LD [010]: LD [110]:
= l/repeat
repeat distance
= l/repeat
[001]: LD 60.
=b =5
distance
X
= 2.0
f
10-8 cm
x 107 points/cm
There are two points along the [110]
•
(1/2)/a2 + b 2 = (1/2) ,h2 + 52
repeat distance LD
= a = 3 x 10-8 cm distance = 3.33 x 107 points/cm
repeat distance
2.9155
= l/repeat
10-8 cm
X
distance
repeat distance = c
= l/repeat
distance
= 3.43
x 107 points/cm
B X 10-8 cm
= 1.25
x 107 points/cm
Calculate the planar densities in the (100), (010), (110), and (00l) planes of a base-centered orthorhombic unit cell with a = 3.0 A , b = 5.0 Aand c = B.O A. (100):
There is one atom located on the plane contribute) \00
= bc = (5) (B)
1 atom /ti0 x 10 2 2.5 x 10 atoms/cm
There is one atom located on the plane (only corners contribute) A010
= ac
= (3) (B)
= 24 x 10-16 cm2
1 atom / 24 x 10-16 4.17 X 1014 atoms/cm2
PO (110)
x 10-16 cm2
-16
PO (010):
= 40
(only corners
~>.:
!l .- . . . 8
,:,
'.,-'
.
.
,', .
3
There are two atoms located on the plane (corners and half of each base centered atom contribute) A
110
PO
~ ~ = cVa- + b- = (B)"'3- + 5-
46.65 x 10-16 cm2
=2 =
atoms / 46.65 x 10-16 4.29 X 1014 atoms/cm2
30
(001) :
There are two atoms located on the plane (corners and one base centered atom contribute)
= ab =
AOO1
= 15
x 10-16 cm2
2 atoms I 15 x 10-16 13.33 X 1014 atoms/cm2
PD 61.
(3) (5)
Calculate the interplanar spacing between the following planes in gold (see Appendix A for the lattice parameter). (a)
(12i)
(b)
(201
(c)
(li2)
The lattice parameter for gold is 4.0786
62.
d
(b)
d
(c)
d
(d)
d
A 0.890
A
= 4.0786 I /z2 + 02 + 12 = 1. 824
A
4.0786 I
201
~2
(~21)
112 321
+ 22 + 42
4.0786 I ~2 + 12 + 22
1. 665
A
4.0786 I /32 + 22 + 12
1. 090
A
The interplanar spacing between (231) planes is found to be 0.89 A. Calculate the lattice parameter if the material has a cubic crystal structure. d
63.
(a)
(d)
231
a I /z2 + 32 + 12
= 0.89
or a
= 0.89V14 = 3.33 A
Show that the radius ratio for an atom or ion that just fits into a tetrahedral interstitial site without disturbing the surrounding atoms or ions is 0.225. Let "R" be the radius of the normal atoms and "r" be the radius of the interstitial site. The atoms in a tetrahedron actually touch one another along the face diagonal. Along the body diagonal or the tetrahedral cube 2r + 2R 2r rlR
64.
= v'3a = v'3VZR = v'6R
v'6R - 2R
= 0.4495R
= 0.225
Show that the radius ratio for an atom or ion that just fits into a triangular interstitial site without disturbing the surrounding atoms or ions is o. 155 cos30 = R I r (r + R)
= 0.866r + 0.866R rlR = 0.134 I 0.866
0.866
R
0.155 31
65.
List the coordinates for all six of the octahedral sites that lie in a BCC unit cell. How many of these sites belong uniquely to one BCC cell? The six octahedral sites are the faces of the cube. 1/2,
1,
0 112, 1/2
1/2,
0, 1/2,
112, 1,
112
Each is shared with another uniquely to one cell. 66.
112, O. 112 112, 112, 1
1/2
cell,
so
only
three
sites belong
Using the ionic radii given in Appendix B, determine the coordination number expected for the following compounds. (a)
FeO
(b)
CaO
SiC
(c)
(d)
PbS
(e)
BO
2 3
From Appendix B, we find the ionic radii.
67.
(a)
r Fe
0.74, rO
1. 32, rFe/rO
CN
=6
(b)
rCa
0.99, rO
.1. 32,
CN
8
(c)
r Si
0.42, rC
0.16, rC/rSi
0.38
CN
4
(d)
rpb
0.84, rS
1. 84, rpb/r S
0.46
CN
6
(e)
rB
0.23, rO
1. 32, rB/r O
0.17
CN
3
= 0.56 rCa/rO = 0.75
Using the ionic radii given in Appendix B, determine the coordination number expected for the following compounds. (a)
Al 0 2
3
(b)
no
2
(c)
MgO
(d)
SiO
(e)
2
CuZn
From Appendix B, we find the ionic radii:
68.
(a)
r AI
0.51,
rO
1. 32,
rAl/rO
0.39
CN
4
(b)
rn
0.68,
rO
1. 32,
rn/rO
0.52
CN
6
(c)
r Mg
0.66,
rO
1. 32,
rMlr o
0.50
CN
6
(d)
rCu
0.96, r Zn
= 0.74,
rcu/rZn
0.77
CN
8
Based on the ionic radius ratio and the necessity for charge balance, which of the cubic structures discussed in the text would you expect CdS to possess? From the ionic radii, rCd
= 0.97,
rs
= 1.84,
rCd/rS
= 0.527
CN
=6
The two ions have equal but opposite charges, and with the coordination number of 6, a sodium chloride structure seems likely.
32
69.
Based on the ionic radius ratio and the necessity for charge balance which of the cubic structures discussed in the text would you expect CoO to possess? From the ionic radii, rCo
= 0.72,
rO
= 1.32,
= 0.545
rCo/rO
CN
=6
The ions have equal but opposite charges and the coordination number is six, making the sodium chloride structure a possibility. 70.
The compound NiO has the sodium chloride crystal structure. Based on the (b) the data in Appendix B, calculate (a) the lattice parameter, density, and (c) the packing factor for NiO. The ionic radii are r Ni (a)
and rO
= 1.32
the ions touch along the [100] direction, so
= 2r Ni
ao (b)
= 0.69
MNi
+ 2rO
= 58.71
= 2(0.69)
g/mol, MO
+ 2(1.32)
= 16
= 4.02 A
g/mol, and there are 4
nickel and 4 oxygen ions per unit cell. p
(cl
PF
PF
71.
=
(4)(58.71) + (4)(16) (4. 02 x 10- 8 )3(6. 02 x 10 23 )
= 7.64
g/cm3= 7.64 Mg/m3
4(4n/3)r Ni 3 + 4(4n/3)rO 3 (4.02)3 4(4n/3) (0. 69)3 + 4(4n/3)
(1.
32)3
(4.02)3
The compound Te0 2 has the fluorite structure. Determine (a) the number of tellurian and oxygen ions in each unit cell, (b) the expected the lattice coordination number based on the radius ratio, (c) parameter, and (d) the packing factor for the compound. (a)
From inspection of the fluorite crystal structure and the formula of the compound, there must be 4 tellurium ions and 8 oxygen ions.
(b)
The ionic radii are r Te
= 2.11,
rO
= 1.32,
rO/r Te
= 0.626
CN
=6
Note, however, that the expected coordination number is not achieved by the fluorite crystal structure due to the requirements of the charge balance.
33
(c)
The ions touch along the body diagonal. There are 4 tellurium ionic radii and 4 oxygen ionic radii along the body diagonal, al though not all of the positions are actually occupied by ions. Therefore v'3a
(d)
4r Te + 4rO
4(2.11) + 4(1.32)
13.72 I v'3
7.92
4(4nI3)r Te
PF
3
= 13.72
A
3
+ 8(4nI3)r o
(7.92)3 4(4nI3) (2. 11)3 + 8 (4nI3) (1. 32)3
PF
72.
o
0.472
(7.92)3
One of the forms of BeO has the zinc blende structure at high (b) the density, temperatures Determine (a) the lattice parameter, and (c) the packing factor for the compound. From Appendix B, r Be and MO = 16 g/mol. (a)
= 0.35
A,
= 1.32
ro
A,
= 9.01
MBe
g/mol,
The ions touch along the body diagonal of the zinc blende unit cell. There are 4 beryllium and 4 oxygen ionic radii along this distance. v'3a a
(b)
p
(c)
PF
o o
4r Be + 4rO
= 4(0.35)
6.68 I v'3
= 3.857 A
+
4(1.32)
4(9.01) + 4(16)
6.68
A
2.896 g/cm3
(3.857) x 10- 6 )3(6.02 x 10 23 ) 3
4(4nI3)r Be 3 + 4(4nI3)r o (3.857)3
4(4nI3)(0.35)3 + 4(4nI3)(1.32)3
PF
73.
0.684
(3.857)3
Which of the cubic structures discussed in the text would you expect CsBr to possess, based on its expected coordination? Calculate (a) the lattice parameter, (b) the packing factor, and (c) the density of the compound. From Appendix rCs/rBr
(a)
B, rcs
= 1.67
1.
I 1.96
A and
r Br
= 0.852
CN
67
1. 96
o
A.
8
The ions touch along the body diagonal. there are 2 Cs ionic radii and 2 Br ionic radii along this distance. v'3a a
o o
2rCs + 2r Br 7.26 I v'3
= 2(1.67)
= 4.1917 A
34
+
2(1.96)
= 7.26
(b)
There is one Cs ion and one Br ion per unit cell. PF '" PF
(4n/3)r Cs3 + (4n/3)r Br3
---=-=------==(4.1917)3
(4n/3)(1.67)3 + (4n/3)(1.96)3
132.91 g/mol and MBr
79.909 g/mol
132.91 + 79.909 '" 4.S g/cm3 (4.1917 x 10- 8 )3(6.02 x 1023 )
p
74.
0.693
(4.1917)3
4.S Mg/m3
Germanium has the diamond cubic structure with a lattice parameter of 5.6575 A. Calculate the size of the germanium atom in the unit cell. Does this best match up with germanium's atomic radius or ionic radius? The atoms in diamond cubic touch along the body diagonal. there are S atomic radii (although not all are actually occupied by atoms) along this distance. r
V3a o '" Sr
= V3a o /S = V3(5.6575)
/ S
= 1.2249 A
The atomic radius listed in Appendix B is 1. 225 A and the ionic radius is 0.53 A. It appears that we should use the atomic rather than the ionic radius. 75.
Calculate the fraction of the [111] direction and the fraction of the (111) plane actually covered by atoms in the diamond cubic cell. There is space along the [111] direction for S atomic radii. although only 2 atoms (4 atomic radii) are actually located along this distance. Therefore fraction [111]
= 4r/Sr = 0.50
The area of the (111) plane is 0.S66a 2 o
a
= Sr / V3 = O. S66(Sr 111
o
A
/ V3)2
In diamond cubic,
= lS.4747r 2
Only the corners and face-centered atoms lie on the (111) plane: A atoms
=
[(3 corners) (1/6) + (3 edges) (1/2)]nr 2
fraction (111) = 2nr2 / lS.4747r 2
35
= 0.34
= 2nr2
76.
Calculate the fraction of the [111] direction and the fraction of the (111) plane actually covered by sodium ions in the NaCI unit cell. Let's assume that the Na + ions are located at normal FCC lattice points. In the NaCI structure, ions touch along the edge of the cell, or ao
2r Na + 2r CI
= 2(0.97)
+ 2(1.81)
= 5.56
X
There are only two Na+ionic radii along the [111] direction of the unit cell. fraction
= 2r
/ V3a
o
= 2(0.97)
/ V3(5.56)
= 0.201
Note: The [111} also passes through a CI ions at the center of the cell so the total fraction covered by ions is
The area of the (111) plane' is A
= 0.866a 0 2 = 0.866(5.56)2 = 26.77
~a
ions
111
[(3 corners)(1/6) + (3 edges) (1/2)]rrr Na 2
= 2rr(0.97) 2 = 5.912
2rrrNa2 fraction 77.
X2
= 5.912
/ 26.77
92
A
= 0.221
How many oxygen and silicon ions are present in cristobalite? Using the ionic radii in Appendix B, calculate (a) the lattice parameter, (b) the packing factor, and (c) the density of cristobalite. Silicon ions are located at face centered positions and at four tetrahedral locations, giving 8 Si ions per cell. There must therefore be 16 0 ions per cell, since the formula is Si02. (a)
The ionic radii are r Si = 0.42
X
and rO = 1.32
X.
The
ions touch along a body diagonal, with 8 silicon radii and 8 oxygen radii along that length. V3a (b)
PF PF
(c)
o
= 8(rSi
+ rO)
= 8(0.42
+ 1.32)
13.92
a
o
= 8.037 A
(8 Si ions) [4rr(0.42)3/3 ] + (160 ions) [4rr( 1. 32)3/3 ] (8.037)3
= 156.628
/ 519.1369
= 0.302
(8 Si ions)(28.08) + (16 0 ions) (16) p = ~------~~--~--~--------~~ (8.037 X 10-8 )3 (6.02 X 1023 )
36
1. 538 g/cm3 1. 538 Mg/m3
Chapter 4 IMPERFECTIONS IN TIlE ATOMIC ARRANGEMENT 1.
Calculate, using the data in Appendix A, the c/a ratios for magnesium, cadmium, rhenium and beryllium. In which of these metals is slip expected to occur easily on the basal planes? In which is slip expected to occur on the prismatic planes? Mg: Cd: Rh:
Be:
2.
c/a c/a c/a c/a
5.209 / 3.2087 5.6181 / 2.9793 4.458 /2.760 3.5842/2.2858
1.623 1.886 1.615 1.568
basal plane slip basal plane slip prismatic plane slip prismatic plane slip
Determine the Miller indices of the slip directions on the (111) plane in an FCC unit cell. [101] [110] [011]
3.
[101] [110]
[011]
Determine the Miller indices of the slip directions on the (101) plane in a BCC unit cell. z
4.
Determine the Miller indices of the slip planes in FCC unit cells that include the (110) slip direction. (111) (111)
x
37
5.
Determine the Miller indices of the {110} slip planes in Bee unit cells that include the [111] slip direction. (110) (101) (011)
6.
Does the [111] slip direction lie in the (112) slip plane in the BCC unit cell? Show by suitable drawings .
~,
. 7.
(110) (101) (011)
Yes Note that the direction drawn in the sketch is: 0, 0, 1/2, -1/2, 112, 0 = -1/2, -1/2, 112 or -I, -I, I, = [111]
Does the [111]slip direction lie in the (123) slip plane in the Bee unit cell? Show by suitable drawings. 1:
No Simple inspection reveals does not lie in the plane.
8.
the direction
Using the three-digit form for indices in the HCP unit cell, determine the slip directions the (0111) slip plane.
[100]
9.
that
[100]
Using the thr~e-digit form for indices in the HCP unit cell, determine the slip directions in the (1010) slip plane.
[010]
38
[010]
10.
Verify that \he planar density of the {112} planes in BCC iron is 9.94 x 10 atomslcm, as used in Example 4-3. The (112) plane is shown in the cubic cell and is also isolated to better illustrate the dimensions and atom locations. The three corners of tpe plane contribute a total of 1/2 atom to the plane; one edge contributes 1/2 atoms to the plane; and the center contributes 1 atom to the plane. atoms
= 1/2
+ 1/2 + 1
=2
atoms.
= (112) (2v'2a Hv'3a ) = v'6a
A
112
= 2.866 A o A = v'6(2.866 x 112
0
0
0
2
a
PD
11.
112
=2
= 2. 012
10- 1°)2
atoms I 2. 012
X
X
10-15 cm2
= 2. 012
10-19 m
= 9.94
10-15 cm2
X
1014 atoms/cm2
X
Determine the length of the Burger's vector in BCC tungsten.
A
3.1652 The
Burger's
vector
is
the
repeat
distance
rd 111
along
the
direction, which is half of the body diagonal. b
111 2.7411
b 12.
=
rd
(1/2)v'3a 10- 10
X
=
0
(1/2) (v'3) (3.1652 x 10-10)
m
Compare the planar density and interplanar spacings for the Ull} and {1l0} planes in FCC aluminum. Based on your calculations, on which planes would slip occur? a
o
= 4.04958 A
For (110): atoms A
110
PD
110
=2
I
v'2a
0
2
For (111) atoms
A
111
= (4 corners)(1/4) + = (v'2a 0 )( (a0 ) = v'2a 0 2 =2 =
I
(2 edges) (1/2)
v'2(4. 04958 x 10- 1°)2
= 8.625
X
(3 corners) (1/6) + (3 edges) (1/2) (1/2)(v'3a 1v'2)(v'2a) 0
39
0
0.866a
0
2
2
1018 pOints/m2
=2
PD
111
= 2 / 0.866a0 2 = 2 / (0.866) (4. 04958) = 14.083 X 10 18 points/m2
Slip will density. 13.
occur on
the
(111)
x 10- 1 °)2
planes due
to
the higher planar
When BCC iron is in the softest condition, the dislocation density is about 106cm/cm3; large amounts of deformation of the iron increase the dislocation density to 10 12 cm/cm3. How many grams of iron are necessary to produce 1600 kilometers of dislocation in (a) soft iron and (b) deformed iron? Firstly, find the volume of iron containing dislocation, and hence the mass of iron.
the
length
of
km = 1600 x 1000 m x 1000 x 100 cm = 1.6 x 108 cm density of iron, p = 7.87 Mg m3 = 7.87 g cm- 3 length of dislocation 1
14.
= 1600 = 1600
(a)
mass
[1.6 X 10 8 cm/10 6 cm cm- 3 ] [7.87 g cm- 3 ] 1259 g
(b)
mass
[1.6 x 10 8 cm/10 12 cm cm- 3 ] [7.87 g cm- 3 ] = 0.00126 g
The dislocation density in an aluminum sample is found to be 5 x Calculate the total length of dislocations in 100 g of 10 7 cm/cm3. aluminum (see Appendix A for the necessary data). How many kilometers of dislocation are present in the sample? p 1
2.669 g/cm3 (100 g)(5 x 107 cm cm3 )
1. 87
2.669 g/cm 3
X
109 cm
18730 km 15.
The circumference of the earth is roughly 38,600 kilometers. If dislocations totalling this length were placed into one cubic centimeter, what would be the dislocation density? Dislocation density
= length
of dislocations volume
(38.600 km)(1000 m/km) (100 cm/m) 1 cm3
3.86
40
X
109 cm/cm3
16.
Suppose that a single crystal of an FCC metal is oriented so that the [0011 direction is parallel to an applied stress of 20 HPa. Calculate the resolved shear stress acting on the (111) slip plane in the [110], [011] and [101] slip directions. Which slip system(s) will become active first? ~
= 54. 76°cos
~
= 0.577
=~
T
cos ~ cos A
[110]: A = 90°, cos A = 0 T = (20)(0)(0.577) = 0 HPa [011]: A T
[101]: A T
45°, cos A = 0.707 = (20)(0.707)(0.577) = 8.16
HPa
45°, cos A = 0.707 = (20)(0.707)(0.577) = 8.16
HPa
Slip begins first on [011] and [101]; no slip occurs on [110]. 17.
Suppose that a single crystal of an FCC metal is oriented so that the [001] direction is parallel to the applied stress. If the critical resolved shear stress required for slip is 1.5 HPa, calculate the magnitude of the _appliec!. stress reguired to cause slip to begin on the (111) plane in [110], [011] and [101] slip directions. ~
= 54.76°
[lio]: A ~
[011]: A ~
cos ~
= 90°
= 0.577 cos A
~
= T/cos
=0
= 1.5/(0)(0.577) = ~
~ cos A
HPa
= 45°, cos A = 0.707 = 1.5/(0.707)(0.577)
3.67 HPa
[101]: A = 45°, cos A = 0.707 ~ = 1.5/(0.707)(0.577)
= 3.67
HPa
Slip never occurs in the [110] direction. 18.
Suppose that a single crystal of a BCC metal is oriented so that the [001] is parallel to an applied stress of 80 HPa. Calculate the resolved shear stress acting on the (110), (011) and (101) planes in the [111] slip directions. Which slip system(s) will become active first?
A = 54.76°, cos A = 0.577 (110)
~ T
= 90°,
cos ~
=~
T
=0
= (80)(0.577)(0) = 0
(010): ~ = 45°, cos ~ = 0.707 T = (80)(0.577)(0.707) (101): ~ T
= 45°,
cos ~
= 0.707
cos ~ cos A
HPa 32.6 HPa
= (80)(0.577)(0.707) = 32.6
HPa
Slip begins first on the (011) and (101) slip planes.
41
19.
Suppose that a single crystal of a BCC metal is oriented so that the [001] direction is parallel to the applied stress. If the critical resolved shear stress required for slip is 58.6 MPa, calculate _the magnitude of the applied stress to cause slip to begin in the [1111 direction on the (110), (Olll and (10ll slip planes. A = '54.76°
u
= cos ~ cos A
(11)):
~ = 90°, cos ~ = 0, u = 58.6/(0.577)(0) = m MPa
(011):
~ = 45°, cos ~ = 0.707, u =-58.6/(0.577)(0.707)
(101): 20.
cos A = 0.577
143.6 MPa
~ = 45°, cos ~ = 0.707,
u = 58.6/(0.577)(0.707) = 143.6 MPa
Suppose that a single crystal of an FCC metal is oriented so that the [0011 direction is parallel to an applied stress. When the applied stress is 48 MPa, a dislocation on the (111) plane just begins to move in the [011]direction. Calculate the critical resolved shear stress in this material. Based on this result, do you suspect that there are numerous lattice imperfections in the metal? cos ~ cos A
0.577 0.707
T = TCRSS U cos ~ cos A = 48(0.577)(0.707) = 19.6 MPa Because TCRSS »
0.34 - 0.69 MPa
there must be many lattice imperfections difficult for the dislocations to move. 21.
which
make
it
more
Suppose that a single crystal of a BCC metal is oriented so that the [001] direction is parallel to an applied stress. When the applied stress is 93 MPa, a dislocation on the (101) plane begins to move in the [111] direction. Calculate the critical resolved shear stress in this material. Based on this result, do you suspect that there are numerous lattice imperfections in the metal? cos A cos ~ T = TCRSS
0.577 0.707
cos ~ cos A 93(0.707)(0.577) = 37.9 MPa
U
This is a typical value for the critical resolved shear stress in BCC metals; there are probably few lattice imperfections that are capable of interfering with slip.
42
22.
Suppose a single crystal of a hexagonal~close packed metal is oriented so that the [001] direction is parallel to an applied stress. Can slip occur in the basal plane? Can slip occur in the prismatic planes such as (1010)? Explain your answer to both questions. In the basal plane. A = 90°; in the prismatic planes. Consequently slip will not occur in either plane.
23.
+= 90°.
FCC aluminum has a density of 2.695 Mg/m3 and a lattice parameter of 4.04958 A. Calculate (a) the fraction of the lattice points that contain vacancies and (b) the. total number of vacancies in a cubic centimeter of aluminum.
Using the density equation. 2.695 x 106 /m3 = (x atoms/cell) (26.981 g/mol) g (4.04958 x 10- 10 m)3(6.02 x 10 23 atoms/mol) x
= 3.9933
Al atoms/cell
vacancies/cell (a) (b)
24.
=4
- 3.9933 = 0.0067 vacancies/cell
0.0067 vacancies/cell fraction = 4.0000 lattice points/cell = 0.001675 vacancies/cm3 = (0.0067 vacancies/cell) =.1 x 1020 vacancies/cm3 (4.04958 x 10- 8 cm)3
HCP magnesium has a density of 1.735 Mg/m3 and lattice parameters of a = 3.2087 A and c = 5.209 A . Calculate (a) the average number of o
0
atoms per lattice point in the unit cell and (b) vacancies in a cubic centimeter of magnesium.
the total number of
Using the density equation: 1.735 x 106 =
(x atoms/cell) (24.312 g/mol) (46.4455 x 10- 29 m3 )(6.02 x 10 23 atoms/mol)
x = 1.99535 Mg atoms/cell (a)
average Mg atoms/lattice point
(b)
vacancies/cell
=2
- 1.99535
= 1.99535
= 0.00465 =1 x
vacancies/cm3 = ____0_._0_0_4_6_5_____ 4.64455 x 10-24m3
43
/ 2
1026 m3
0.99767
25.
If one atom is missing The lattice parameter of BCC caesium is 6.13 A. from one out of 800 unit cells, calculate (a) the number of vacancies per cubic centimeter and (b) the density of caesium.
= (1
(a)
vacancies/cm3
(b)
In 800 cells of a BCC metal, there are 1600 lattice points. Since one atom is missing in 800 cells, the average number of Cs atoms per cell is
vacancy / 8000 cells) (6.13 x 1O-B cm )3
(1599 / 1600)(2 lattice points/cell) p
26.
X
= 1.99875
= ~9875
atoms/cell) (132.91 g/mol) (6.0849 x 10- B cm)3(6.02 x 1023 atoms/mol)
101B
Cs atoms/cell 1.9157 g/cm3
The lattice parameter of FCC strontium is 6.0849 A. If one atom is missing for each 1,500 strontium atoms, calculate (a) the density of strontium and (b) the number of vacancies per gram. (a)
The number of Sr atoms/cell is (1499/1500)(4 lattice points/cell) p
(b)
=
= 3.99733
Sr atoms/cell
(3.99733 atoms/cell) (87.62 g/mol) -B 3 23 (6.0849 x 10 cm) (6.02 x 10 atoms/mol)
= 2.582
M /m3 g
vacancies/m3 = (1 vacancy/1500 atoms) (4 atoms/cell) (6.0849 x 10- 10 m)3 0.11836 x 1026 1026 vacancies/m3 ) 2.582 x 106 g/m 3
(0.11836
vacancies/gram
27.
5.43
X
4.552
X
101B
A BCC alloy of tungsten containing substitutional atoms of vanadium has a density of 16.912 Mg/m3 with a lattice parameter of 3.1378 the fraction of vanadium atoms in the alloy Let Fy be the fraction of Y atoms;
A.
Calculate
then (1 - fy) is the fraction of
tungsten atoms. The molecular weight of Y is 50.941 g/mol and that of W is 183.85 g/mol. From the density equation 6 3 16.912 x 10 g/m fy
(2
atoms/cell)[(~
)(50.941) + (1 - f y )(183.85)
(3.1378 x 10-10 m)3(6.02 x 1023 atoms/mol) 0.2
44
28.
An FCC alloy of co~per containing substitutional atoms of tin has a density of 7.17 Mg/m and a lattice parameter of 3.903 A. Calculate the fraction of tin atoms in the alloy.
Let fSn be the fraction of Sn atoms;
then
(1 -
fSn) is the fraction
of copper atoms. The molecular weight of tin is 118.69 g/mol and that of copper is 63.54 g/mol. From the density equation (4 atoms/cell) [(fSn )(118.69) + (1 - f Sn )(63.65») 7.727 x 10 6 g/m 3 ; ------------~~~~------~~----------~- (3.903 X 10- 10 m)3(6.02 x 10 23 atoms/mole)
29.
Suppose that when one-third of the atoms in RCP magnesium are replaced by cadmium atoms, the lattice parameters of the alloy are a ; 3.133 A and o 5.344 A. Calculate the expected density of the alloy. c o
MWMg ; 24.312 g/mol Vcell; (3.133 x 10
3
m) (5.344 x 10
-10
m)cos30; 4.54274 x 10
(2 atoms/cell) [(0.667)(24.312) + (0.333)(112.4») (4.54274 x 10- 29 m3 )(6.02 x 10 23 atoms/mol)
p ;
30.
MWCd ; 112.4 g/mol
-10
-29
3
m
= 3.924
Mg/m3
Body centered cubic iron has a lattice' parameter of 2.868 A after a carbon atom enters one interstitial site in every twentieth unit cell. Estimate (a) the density of the iron-carbon alloy and (b) the packing factor for the structure, assuming r Fe = 1.241 Aand rC = 0.77 A. (a)
When we use the density equation, we must add onto the usual 3 Fe atoms per cell a factor of 1/20 of a carbon atom per cell. MW Fe p
(b)
= 55.847
g/mol
MWC
12 g/mol
(2 Fe/cell) (55. 847) + (1/20 C/cell)(12) (2.868 x 10- 10 m)3(6.02 x 1023 atoms/mol)
7.907 Mg/m 3
In each unit cell, there are 2 Fe atoms and 0.05 C atoms. PF
2(4/3)(n)(1.241 A)3 + (0.05)(4/3)(n)(0.77 A)3
(2.868 A)3
45
0.6828
31.
Carbon atoms enter interstitial positions in FCC nickel. producing a lattice parameter of 3.5198 A and a density of 8.955 Mg/m3 Calculate (a) the atomic fraction of carbon atoms in the nickel and (b) the number of unit cells you would have to examine to find one carbon atom. (a)
The unit cell will contain 4 nickel atoms and "x" carbon atoms. From the density equation. using MWNi = 58.71 g/mol and MWC
= 12
g/mol.
8.955 x 106 235.08
= (4
= 234.84
Ni atoms/cell)(58.71) + (x C atoms/cell) (12) (3.5198 x 10- 10 m)3(6.02 x 10 23 atoms/mol) + 12x
or
fraction carbon atoms/cell (b)
32.
x
= 0.2
carbon atom per cell
0.02 atoms/cell 4 Ni atoms/ cell
0.005
Since there is 0.02 carbon atoms/cell. then 1/0.02 cells would be required to find one carbon atom.
50 unit
Would you expect a Frenkel defect to change the lattice parameter or density of MgO? Explain. Yes; the misplaced ion will strain the lattice. causing the lattice parameter to increase and therefore reducing the density.
33.
Suppose one Schottky defect occurred in every tenth unit cell of NaCI. producing a lattice parameter of 5. 57A.I Calculate the density of the sodium chloride. Although normally there are 4 sodium ions and 4 chlorine ions per cell. the Schottky defects in every tenth unit cell produce. on the average. (39/40) (4) sodium ions and (39/40) (4) chloride ions per cell. MWNa p
22.99 g/mol
(5.57
X
MWCI
= 35.453
g/mol
(39/40)(4)(22.99 + 35.453) 10- 10 m)3(6.02 x 10 23 ions/mol)
46
2.191 Mg/m3
34.
Suppose th3e lattice parameter of CsCI is 4.0185 A and the density is 4.285 Mg/m. Calculate the number of Schottky defects per unit cell. Let "x" be the average number of Cs and CI ions in the unit cell. MWCs 4.285
= 132.91 X
106
g/mol
MWCI
= 35.453
g/mol
= ____-'x:.c. :. :(1:.. :3c.2::c:..• . :. 9.:. 1::-+---=3.::.5.:.....4.:..5:...:3:....:)---=-=--_ __ (4.0185 x 10
x
10
m)3(6.02 x 1023 ions/mol)
0.99424 ions/cell
There should be 1 of each ion in the caesium chloride structure. The number of Schottky defects is therefore defects/cell 35.
=1
- 0.99424
= 0.00576
Suppose the (111) plane is parallel to the surface of an FCC metal. is the coordination number for each atom at the surface?
What
At the surface, 6 atoms would touch a central atom at the surface, 3 more atoms from beneath the surface would contact the central atom, but the other 3 atoms are not present. Therefore the coordination number would be 9. 36.
How many grains would be The ASTM grain size number for a metal is 6. observed per square inch in a photograph taken at a magnification of 100? How many actual grains are present per square inch? N
= 2n - 1 = 26 - 1 = 25 = 32
total grains 37.
grains at 100x
= 32(100)2 = 320,000
grains
Twelve grains per square inch are counted in a photograph taken at a magnification of 500. Calculate the ASTM grain size number. Is this a coarse, medium, or fine grain size? At 100x, the number of grains N would be N
In(300)
= 12(500/100)2 = 300 = 2n- 1 (n-l)(ln(2)
or
This is a fine grain size.
47
5.7038
= (n-1)(0.693)
or n
9.23
38.
Eighteen grains per square inch are counted in a photograph taken at a magnifIcation of 75. Calculate the ASTM grain size number. Is this a coarse, medium or fine grain size? At 100x, the number of grains N would be
N = 18(75/100)2 In(10.125)
= 10.125 = 2n- 1
= (n-l)ln(2)
or
2.315
= (n-l)(0.693)
or
n
= 4.34
This is medium grain size. 39.
Figure 4-23 shows the microstructure of a material at a magnification of 100. Estimate the ASTM grain size number. From the photograph, there are approximately 7 grains/in2.
= 2n - 1
7
In(7) 40.
=
(n-l)ln(2)
or 1.95
(n-l)(0.693)
or n
Figure 4-24 shows the microstructure of a material at a magnification of 500. Estimate the ASTM grain size number. There are approximately 25 grains/in2 at 500x. N
In(625) 41.
3.8
Therefore at 100x
= 25 (500/100) 2 = 625 = 2n - 1
= (n-l)ln(2)
or 6.44
= (n-l)(0.693)
or n
10.3
Calculate the angle 9 of a small angle grain boundary in BCC iron when the dislocations are 7500 A apart. The lattice parameter for BCC iron is 2.866 A. The repeat distance of Burger's vector is half of the body diagonal, or b
= (1/2)(13)(2.866) = 2.482 A = 2.482/7500 = 0.0003309 = 0.01896 or 9 = 0.038°
sin(9/2) 9/2 42.
A small angle grain boundary is tilted 0.75° in FCC copper. the average distance between the dislocations.
Calculate
The lattice parameter is 3.6151 A. The repeat distance of Burger's vector is half of the face diagonal, or b
= (1/2)(i2)(3.6151) = 2.556 A
If "x" is the distance between dislocations, then sin(0.7512)
2. 556/x
0.006545
2.556/x 48
or
x
390
A
43.
Suppose that 2 Fe3 + ions are substituted for normal Fe2 + ions in FeO. What other changes in the atom arrangement (creation of vacancies etc) would be required to maintain the proper charge balance? We must create a Fe 2 + vacancy. The addition of two Fe 3 + ions introduces a charge of +6; removal of the two Fe 2 + ions only removes However. if three Fe 2 + ions are removed. then a a charge of +4. total of +6 charge is removed~+the charge is balanced. but a vacancy is created where the third Fe ion was removed.
44.
Suppose an Fe 2+ ion is substituted for an Na+ ion in NaCl. Explain why you would expect that the charge balance would be maintained by forming a vacancy rather than by adding another Cl- ion to the lattice. The large Cl- ion would be added as an interstitial ion and would create a larger amount of lattice strain than would be caused by the formation of a vacancy.
49
Chapter 5 ATOM MOVEMENT IN MATERIALS
1.
When a force is applied at 300°C, each inch of a metal stretches at the rate of 0.025 mmlmin; at 400°C, the rate of stretching is 0.09375 mmlmin; and at 500°C, the rate is 0.25 mmlmin. Calculate (a) the activation energy for the stretching process and (b) the constant co' What are the units for each? (a)
From the equation rate = coexp(-QlRT): 0.025 _ c oexp[-Q/(8. 314) (300 + 273)] 0.25 - c exp[ 01(8.314)(500 +2731)] o
0.1 = exp [(-Q) (0.0002099 - 0.0001556)] 0.1 = exp [-Q(0.0000543)] In(O.l) = -2.302585 =.-0.00005430 :. Q = 42,405 J/mol
0.025 = c oexp[-42,405/(8.314) (573)] 0.025 = c oexp(-8/9012792) 0.025 = c o (0.0001362) c 2.
o
= 183.55 mmlminute
The fraction of the lattice points containing vacancies in copper is 2.24 x 10-15 at 100°C but is 2.42 X 10-6 at 700°C. Calculate (a) the activation energy required to form a vacanc~, (b) the fraction of the lattice points that contain vacancies 5 C below the melting point (l085°C), and (c) the number of vacancies per unit cell 5°C below the melting point. (a)
2.24 x 10-15 2.42 x 10- 6 9.256 x 10-10
_ -
exp!-Q/(8.314)(373)] exp[-Q/(8.314)(973)]
= exp!-Q(0.0003224
- 0.0001236)]
In(9.256 x 10-19 ) = -20.80056 = -0.0001988Q Q = 104,631 J/mol (b)
To find the constant c 2.42
10- 6 = c
X
c ..
c
o
o
o
exp[-104,631)/(8.314) (973)]
o
exp(-12.934143 = c (2.42 x 10-6 0
1. 0
50
)
At 5·C below the metling point, or 1080·C, the fraction 'f' is f
1.0 exp [-104,631/(8.314)(1353)] 1.0 exp(-9.3014908) 9. 12
(c)
10- 5 vacancies/ lattice point.
X
In FCC copper there are 4 lattice points/cell (9.12 x 10-5 ) (4)
:. vacancies/cell
36.48 x 10-5 3.
The diffusion coefficient for Al in Al 0 is 7.48 x 2 3
10-23 m2 /s
at
1000·C
and is 2.48 x 10-14 m2/s at 1500·C. Calculate the activation energy the diffusion constant D . o 7.48 X 10-23 D o[exp -Q/(8.314)(1273)] 2.48 x 10-14 3.016 x 10- 9
Do[exp - Q/(8.314)(1773l]
= exp[-Q(0.0000944
x 0.0000678)]
= exp[-Q(0.0000266)]
= -19.619334 = -Q(0.0000266)
In(3.016 x 10- 9 )
= 737,569
Q 2.48 x 10-14
and
J/mol
= D Jxp[-737,569/(8.314)(1773)]
= Doexp(-50.036152)
2.48 x
4.
10- 14
= Do (1. 8602
x 10- 22 )
The diffusion coefficient for Ni in MgO is 1.23 and 1.45 x 10- 14 m2 /s at 1800·C. Calculate the the diffusion constant D .
X
o
10- 16
D J xp[-Q/(8.314)(1473)]
1.45 x 10- 14
Do exp[-Q/(8.314) (2073)]
1.23
X
8.427 x 10- 3
= exp[-Q(0.0000816
- 0.000058)]
= exp[-Q(0.0000236)] In(8.4827 x 10- 3 )
= -4.7697 = -Q(0.0000236)
Q = 202, 107 J/mol 1.23 x 10-16
=
D exp[-202,107/(8.314)(1473)] o
51
10- 16 m2 /s at 1200·C activation energy and
1.23 X 10-16 = (6.8036 x 10-8)D :. DO
5.
= 1. 808
x 10
-8
o 2 m Is
Estimate the activation energy for self-diffusion in titanium. From Figure 5-S the activation energy is approximately 314,000 J/mol because the melting point is 166SoC.
6.
293,000
Would you expect the activation energy for self-diffusion in which has the covalent bond, to follow the curve in Figure 5-8?
silicon, Explain.
No. We would expect a higher activation energy for silicon the strong covalent bonds. 7.
to
due
to
Suypose a 0.1 mm thick wafer of germanium contains one8 gallium atom per 10 Ge atoms on one surface and 1000 Ga atoms per 10 Ge atoms at the other surface. Calculate lhe concentration gradient (a) in atomic percent/m and (b) in atoms/m .m. The lattice parameter for Fe is 5. 66A. (1 Ga/108 Ge)(100) = 0.000001 at % Ga c2
(10 3 Ga/10 8 Ge)(100)
= 0.001
at % Ga
dc _ (0.000001 - 0.001) dx (0.1)(10-3) -9.999 at % Ga/m (b)
Vcell
= (0.566 x
1
c2
1 2.2665
= 0.18132
X
0.441
10-21
1000 2.2665 x 10-21
X
10-27 m3
0.1
X
1021 Ga atoms/m3
441. 21 x 1021 Ge atoms/m3
(0.441 - 441.21) x 1021
8.
X
(l08 Ge) (0.18132 x 10-27 ) 8 Ge atoms/cell 2.2665 X 10-21 m3
V(10 8 atoms)
c
10-9 )3
10-3
-4.408 x 10 27 atoms/m3 . m
One surface ~f an 0.5 mm thick wafer of germanium ~ontains 5 gallium atoms per 10 Ge atoms. How many gallium atoms per 10 Ge atoms must be in*t0duced at the other surface to give a concentration gradient of -2 x 10 atoms/m3.m? The lattice parameter for Ge is 5.66 A. -28 3 ) = 1. 8132 x 10 m 5 28 V(10S Ge atoms) = (0 ) (1. 8132 x 10- ) Vcell = (0.566 x 10
-9 3
8
2.2665 52
X
10-24 m3
512.2665 x 10-24 y/2.2665 x 10-24 [(y - 5)/2.2665 x 10- 24 )10.5 -2 X 1027 atoms/m3.m (y -
5)
103
X
-2.2665
.. y = 2.7335 Ga atoms/lOs Ge atoms 9.
wm
A 0.01 thick foil of iron separates a gas 2fontaining ~ x 1028 H H atoms/m. If the atoms/m from another chamber containing 6 x 10 system is operating at 1000·C and the iron is FCC, calculate (a) the concentration gradient of hydrogen through the foil and (b) the flux of hydrogen atoms through the foil.
(a)
the concentration gradient is:
=6
x Hj32
Using D for H in FCC Fe at 1000·C
= 1.07
fJ.x
(b)
atoms/ffi. m X
10-8 m2s- 1.
The flux is: -DfJ.c/fJ.x - (1.07 x 10-8 )(-9.9999 x 1032 ) 10.69 X 10 24 atoms/(m2 s)
J
10.
22 28 x 10 - 1 x 10 3 0.01 x 10-
= -9.9999
fJ.c
Nitrogen is held in a pressure chamber by a BCC iron foil only 0.2 mm thick. The concentrgtion of 3nitrogen is 3 x 1026 atoms/m3 on one side of the foil but 5 x 10 atoms/m on the other side. Calculate the flux of nitrogen atoms through the foil at 750·C. 'c/'x = 5 x 1016 - 3 X lcf 6 = -1. 5 x 10 30 atoms/m. 3 m u u 0.2 x 10- 3 D for N in BCC Fe at 750·C
5.784. x 10
-11
2-1 ms
- D fJ.c/fJ.x
.. J
- (5.784 x 10- 11 )(-1.5 x 1030 ) 8.676 11.
X
1019 atoms/(m2 s)
the concent[ation gradient required .to produ~e a flux of hydrogen atoms/m . s through a BCC iron fOll at 800 C.
Cat~ulate
10
D for C in BCC Fe at 800·C 10 x 109 atoms/m2 /s fJ.c/fJ.x
= -4.5095
= 2.217
= -(2.217
X
x 10-8 ) fJ.c/fJ.x
x 1017 atoms/m3.m
53
10-8 m2s- 1
10
x
12.
An FCC iron structure is to be manufactured that will allow no more
than the 100 x 104 H atoms/m2 to pass through it in one minute at 950·C. composition of the hydrogen is 1 x 1026H atoms/m3 on one side of the foil and 5 x 1023 H atoms/m3 on the other side. Calculate (a) the concentration gradient and (b) the minimum thickness of the iron required. J
= 106
D for H in FCC Fe at 166.67
= 166.67 atoms/(m2 s) 950·C = 9.09 x 10- 9 m2 /s
atoms/(m 2 minute)
= -(9.09
x 10-9 ) ~c
~x
~~ = -1. 8335 x 1012 atoms/m3 .m
= -9.95
~c
5 X 1023
~
-(9.95 x 10 25 )/(-1.8335 x 1012 )
1 X 1026
-
X
1025 atoms/m3
= 5.43
X
1013 m
:. minimum thickness is impr.actical. 13.
Determine the maximum allowable temperature that will produce a less than 950 x 10 4 N atoms/(m2 s) through a BCC iron foil the concentration gradient is -3 x 1024 atoms/m3 .m. 950
X
104 atoms/(m2 s) = -(0.0047 x 10-14 )exp [ - (8.314) 76571 (T) ] x (-3 950
(0.0047 x
X
10 4
10-4 )(3
X
of
1024 )
exp (- 9209.86/T)
x 10 24 )
In(6.7375 x 10- 12 )
flux when
= -25.723319
-9209.86/T
:. T = 358 K = 85·C
14.
Suppose a diffusion couple is produced between pure nickel and pure copper at 1000·C. Will the interface tend to move toward the pure nickel or the pure copper side? Explain using suitable calculations. For Ni diffusing in Cu DNi
-4 [ 242,265] -4 (2.3 x 10 )exp - (8.314)(1273) = (2.3 x 10 )exp(-22.890343)
(2.3 x 10- 4 )(1.1451 x 10-1°) 2.6338 x 10- 14 m2 /s For Cu diffusing in Ni DCu
= (0.65
x 10
-4
[ 257, 328 ] )exp -(8.314)(1273)
54
(0.65 x 10- 4 )exp(-24.313576) (0.65 x 10-4 )(2.7589 x 10- 11 ) 1. 7932 x 10- 15 m2 /s DNi > DCu
15.
:. more nickel will move towards the copper.
A carburizing process is done to a 0.15% C steel by introducing 1.1% C at the surface at 1000°C, where the iron is FCC. Calculate the carbon content at 0.1 mm, 0.5 mrn and 1 mm beneath the surface after 1 h. D
= 0.23
x 10
1.1 - x 1. 1 - 0.5
-4
[ 137,660 ] exp (8.314)(1273)
= erf
--:;::::========= Y
2v15.1637 x 10-11 (3600)
erf (1159.68 y) x
~
= 0.1
For y
1.1 - (0.95) erf (1159.68 y) mm
= 0.1
x 10- 3 m
erf[(1159.68)(0.1 x 10- 3 )]
= erf
(0.11597)
= 0.1 ~
x
= 1.1
= 0.5
For y
= 1.005%
- [(0.95)(0.1)]
mrn
0.5 x 10- 3 m
erf[(1159.60)(0.1 x 10- 3 )] ~
For y
= 1.1 = 1 mm = 1 x
[(0.95)(0.6)]
16.
= 1. 1
= erf (0.5798) = 0.6 = 0.53% C
x 10- 3 m
erf[(1159.68)(1 x 10- 3 )] :. X
C
= erf
- [( O. 95) (0. 9)]
(1.1597)
= O. 245%
0.9
C
A carburizing process is done to a 0.0% C steel by introducing 1.0% C at the surface. Calculate the carbon content 1 mm beneath the surface after
holding at 912°C for 1 h if (a) the iron is FCC and (b) the iron is Explain the difference that the structure makes. (a)
DFCC
= 0.23
x 10
-4
[-137 660 ] exp (8.314i(1185)
(0.23 x 10- 4 ) exp (- 13.9727) 1.9654 x 10-11 m2 s-l
55
BCC.
1 - x
r=ci
.. x
(b)
= erf
{
= erf
(1.88)
= 0.02%
0.001 } I 2 v1.9654 x 10-11 (3600)
= 0.98
C
DBCC = (0.011 x 10
-4
[ 87,450 ] ) exp -(8.3147)(1187)
(0.011 x 10- 4 ) exp(-8.8763) 1. 536 x 10- to m2 Is 1 - x - = erf
{
1 - 0
= erf
(0.6724)
= 0.38%
:. X
0.001 } 2 ~.5364 x 10-1 °(3600) 0.62
C
Diffusion occurs more rapidly in BCC iron due to the more open crystal structure. 17.
We would like to produce 0.45% C at a distance surface of a steel part by carburizing. If the 0.15% C and 1.2% C is introduced to the carburizing take at 1100°C. Assume the iron is D
= 0.23
137,660 ] exp [ -(8.314)((1373)
1. 332 x 10-10 1.2 - 0.45 1.2 - 0.15 0.7143 :. t 18.
= 0.23
of 2.5 mm beneath the steel originally contains surface, how long will FCC.
exp (-12.05945)
m2 /s
= erf
2.5
[ 2
X
It. 332
10-3
]
x 10- 10 t
= erf(108.324/vt)
= 239475 = 6.65
hours
We would like to produce 0.3% C at a distance of 1.5 mm beneath the surface of a steel part by carburizing. If the steel originally contains 0.10% C and 1.0% C is introduced to the surface, what carburizing temperature is required if the treatment is to be accomplished in 0.33 hours? Assume the iron is FCC.
! = ~:t = 0.7778
-3
erf
[1. 5 x 10 2v'Dt
56
]
1. 5
0.8 .. 9.375
X
.. 8.8
10-7
X
10- 4
X
10-3
2v'Dt
= v'Dt
= Dt
.. D = (8.8 x 10-7 )/(0.33)(3600) = 7.398 x 10- 10 m2 /s
7.398 x 10-10
= 0.23
1~:10)
1600 K
19.
10- 4 exp [
= In
In [7.398 x 0.23 x 10 .. T
X
-~~~i:~~)]
(3.21 x 10-5 )
=-
10.3446
137660 8.314(T)
1328·C
We would like to produce 0.2% C at a distance of 1 mm beneath the surface of a 0.01% C steel part by carburizing. If we carburize at 975°C for 10 h, how much carbon must we expose to the surface of the part? Assume the steel is FCC during treatment. 137,660] (0.23 x 10- 4 ) exp [ - (8.314)(1248)
D
(0.23 x 10- 4 ) exp (-13.26732) 3.979 x 10-11 t
=
(10) (3600)
x - 0.2 x - 0.01
m2 /s
= 36,000
1. 0 x 10
erf
0.4 (x - 0.001)
.. 0.6x
0.196
= 0.327%
]
= 0.4
.. x - 0.2
.. x
-3
[ 2 /c3.979 x 10- 11 )(36,000)
erf (0.42)
20.
s
= 0.4x
- 0.004
C
We would like to produce 0.12% N at a distance of 4 mm beneath the surface of a steel containing 0.002% N by introducing 0.35% N to the surface of the steel. How long would this nitriding require if it were done at 700°C? Assume the iron is BCC. -4
76570 ] - (8.314)(973) ) exp (-9.46543)
D = (0.0047 x 10 ~4exp :. D
=
(0.0047 x 10
[
57
0.35 - 0.12 0.35 - 0.002 = 0.661
• J'Lt .. Yl;
:. t
= 331. 42
0.7
2.2416
X
105 s
62.27 hours 21.
During nitriding of a BCC steel containing 0.005% N, 0.45% N is introduced to the surface at 650°C for 4 h. Calculate the nitrogen content at a distance of 2 mm beneath the surface of the steel. D = (0.0047 x 10-4 ) exp
(-
76570 ) (8.314)(923)
(0.0047 x 10- 4 ) exp (-9.97808) 2.18 x 10-11 m2 t
S-l
4 x 3600 = 14,400 s
0.45 - x 0.45 0.005
erf (
= erf 0.45 - x
2
X
10-3
2 yf(2.18 x 10- 11 )(14,400)
(1.7848)
)
0.97
(0.97) (0. 445)
x = 0.018% N
22.
Calculate the nitriding temperature when a BCC steel containing 0.075% N is exposed to 0.25% N for 18 h, producing 0.12% N at a distance of 8 mm beneath the surface. t
= (18)(3600) = 64,800
0.25 - 0.12 0.25 - 0.0075
s 3 erf (0.8 x 10- ) 2m
0.7429
6 = erf (1. 5~3 x 10- )
.. 0.78
58
1. 5713 x 10- 6 Vi)
:. D
=
:. 4.0584 x 10 -12
=-
12
23.
= 790
K
X
10-12 m2 /s
(0.0047 x 10-4 ) exp [76,570] - (8.314) (T)
In ( 4.0584 x 10- _ 4 ) 0.0047 x 10 ~ T
= 4.0584
11.659699
=-
76,570 8.314(T)
517·C
Decarburization of a steel occurs when carbon diffuses from the steel to the surface and enters the atmosphere. How long will it take for a 0.60% C steel surface to decarburize below 0.10% C for a depth of 2 mm if the FCC steel is held at 1250·C in an atmosphere containing zero carbon? D
=
137,660 (0.23 x 10- 4 ) exp [ - (8.314) (523)
]
(0.23 x 10- 4 ) exp (-10.8717) 4.3672 x 10- 10
o-
0.1 0 - 0.6
erf
m2 /s
2
x 10- 3
]
[2/4.3672 x 10- 1O (t)
.. 0.1667
erf [47.852 ) 2v'Dt
.. 0.15 vt
= 47.852
:. t = 101,768 s = 28.27 hours 24.
Calculate the distance below the surface of 0.80% C steel that, after exposure to air for 100 h at 1200·C, is decarburized to 0.2% C or less. The steel is FCC during this process. D
= (0.23
x 10
-4
) exp
(
137660) - (8.314)(1473)
0.23 x 10- 4 exp (-11.24075) 3.01948 x 10- 10 m2 /s t
360,000 s
.. Dt = 0.0001087 and v'Dt
= 0.010426
oo-
= 0.25 = erf (2(0~010426»)
0.2 0.8
.. 0.2
x 0.020852 59
.. 0.00417 m
=x
x = 4.17 mm 25.
A carburizing heat treatment of an FCC steel normally can be successfully completed at 1250·C in 2 h. What temperature were lowered to 1100·C? D1
-4
(0.23 x 10
) exp
time
would
be
necessary
if
the
(137,600) - (8.314)(1523)
4.3672 X 10- 10 m2/s D2
(0.23 x 10
.. D t
..2...2. D
=
t
2
-4
) exp
2
(4.3672 x 10-10 (1.3316 x .. t2 26.
(137,600) - (8.314)(1373)
= 6.56
)
(2)
10- 10 )
t
2
hours
A nitriding heat treatment of a BCC steel normally requires 2 h at 600·C. What temperature would be required to reduce the heat treatment time to 1 h? (0.0047 x 10- 4 ) exp (
D
= 1. 2316 t
10- 11 m2/s
X
=1
2 hr
t
T
=
(1.
2 -11 2.4632 x 10
=
1
:. D2
D t
76570 ) - (8.314) (873)
2
hr
2316 ~ 10
-11
(0.0047 x 10
)(2) -4
) exp
In(2.4632 x 10- 11 ) = _ 9.8564 = _ 0.0047 x 10- 4 .. T
934 K
661 0 C
60
= 2.4632
X
10-11 m2/s
(76570 ) - (8.314)T
76570 8.314(T)
27.
Suppose, during a diffusion bonding process used to join copper to nickel, that bonding was 90% complete after 1 h at 800·C. But at 900·C, only 5.9 min were required to obtain the same bonding. Assuming that the rate of bonding is related only to diffusion, calculate the activation energy for the process. Consulting Table 5-1, is it likely that bonding is controlled primarily by diffusion of Cu into the nickel or by diffusion of Ni into the copper? (Note that rate is the reciprocal of time. ) 1
Co exp [-01(8.314)(1073)]
3600
(5.9) (60)
Co exp [-Q/(8.34)(1173)
0.00027778 0.00282486
exp(- 0.000112Q) exp(-0.0001025Q)
= exp(- 0.0000095Q) 0.098334 = - 2.31939 = - 0.0000095Q
0.098334 In Q=
244,146 J/mol
This is close to the activation energy for the diffusion of nickel in Cu. 28.
Suppose the grain size of copper doubles when the metal is held at 500·C for 5 h, but doubles at 800·C in only 8 min. Estimate the activation energy for grain growth. Does this correlate with the activation energy for self-diffusion of copper? Should it? (note that rate is the reciprocal of time.) We can set up two equations: 11(5) (3600)
c
1/(8)(60)
c
0.00005563 0.0020833 0.02667 In(0.02667) :. Q
= 83,124
0 0
exp [-Q/(8.314)(773)] exp [-Q/(8.314)(1073)]
exp(-0.0001556Q) exp(-0.000112Q) exp (-0.0000436Q)
= -3.624216 = -O.0000436Q J/mol
It does not correlate. More likely, the rate of grain growth will be related to the diffusion along grain boundaries, which has a smaller activation energy.
61
29.
Suppose we would like to join
SiC
to
Si 3 N4 ,
by
a
diffusion
bonding
process. What problems might we have during the first step in the bonding process? Can you think of any way that these problems might be minimized? Would you expect that diffusion bonding of these materials would take more or less time than if two metals were joined at the same temperature? Both are brittle so the first step - localized cause fracture rather than deformation.
deformation
may
We could put a soft inter layer material between the two ceramic surfaces that would deform during the first stage and simultaneously will almost completely fill in the interface area. More time, since ceramics generally have higher activation energies. 30.
Suppose we were to produce a silver contact for an electrical relay using powder metallurgy. Sintering initially occurs very rapidly, but then slows significantly at a later time. Explain why the rate decreases during the latter stages of the 'process. Initially we would expect both surface and grain boundary to predominate. As the pores reduce in size and perhaps isolated, sintering will depend more on volume diffusion. be slower due to the higher activation energy.
31.
diffusion are even This will
At 6SO°C, the electrical conductivity of NaCl is 10-2 ohm- 1 m- 1 , at 420°C, the electrical conductivity is 10- 6 ohm- 1m- 1 . Which ion - Na or Cl - do you expect is transferring the greater portion of the charge? What is the activation energy for diffusion of this ion in NaCl? Most of the charge will be carried by sodium - it is cation expected to be smaller and more mobile. 10- 2
= exp [-Q/(S.314)(953)] = exp (-0.0001262Q)
10- 6
= exp
[-Q/(S.314)(693)]
10- 4
= exp
[(-0.0001262 + 0.0001735)Q]
= exp
and
is
(-0.0001735Q)
= exp
(0.0000473Q)
In(10 4 ) = 9.21034 = 0.0000473Q :. Q 32.
= 194,722
J/mol
A balloon filled with helium deflates more rapidly than a balloon with air. Explain. Helium atoms are smaller balloons more quickly.
62
and
can
diffuse
through
the
filled polymer
Chapter 6 MECHANICAL TESTING AND PROPERTIES
1.
A 800 N force is applied to a 2.5 mm diameter copper wire having a yield strength of 135 MPa and a tensile strength of 270 MPa. Determine (a) whether the wire will plastically deform and (b) whether the wire will experience necking.
F 800 cr = So = 4.9087
2.
N -2
mm
163 MPa
(a)
Wire will plastically deform.
(b)
Wire will not neck.
Calculate the maximum force that a 2.5 mm thick and 50 mm wide nickel strip, having a yield strength of 310 MPa and a tensile strength of 430 MPa, can withstand with no plastic deformation.
cr
-
y -
F
S-y o
310 x 125
= 387.5
kN
(ie) a force less than 387.5 kN can be deformation. 3.
= Stress = 23000/60.13 = 382.49 = 1149 Strain
no
plastic
0.5/150
0.00333
.
GPa
The modulus of elasticity of nickel is 209 GPa. Determine the length of the bar when a force of 6.82 kN is applied to a 12.5 mm x 7.5 mm bar originally 900 mm long without causing plastic deformation. 209000
= 6820/9~.75 straln
0.00033807 = strain .. 0.3133 mm
We would having a Can this force of
x 900
=x
.. length of the bar 5.
with
A titanium bar 8.75 mm in diameter will stretch from a 150 mm length to a 150.5 mm length when a force of 23 kN is exerted. Calculate the modulus of elasticity of the alloy.
E 4.
withstood
= 900.313
mm
like to plastically deform a 25 mm diameter bar of aluminum yield strength of 170 MPa and a tensile strength of 270 MPa. be accomplished using a forging machine that can exert a maximum 150 kN? F
cr = So
= 170
N mm
-2
63
s o = 490.87 :. F = 170 x
mm2
= 83.45
490.87
kN < 150 kN
Yes, the forging machine has sufficient capacity. 6.
A landing gear must be capable of supporting one-third of an airplane that weighs 200,000 Kg. Just to be safe, we would like the landing gear to be able to support twice its maximum load. Determine the minimum cross-sectional area of the landing gear if it is made of a heat-treated steel having a yield strength of 860 MPa and a tensile strength of 900 MPa. Load to be supported = 2 x 200,000 = 133,333 kg/gear ~
= s- = 860 F
o
3
N mm
~
= 87.69
Kgf/mm
2
133,333 87.69 S o 2 .. S = 1520 mm
o
7.
We would like to produce a copper plate that is 6.25 mm thick by a rolling process. The yield strength is 240 MPa, the tensile strength is 380 MPa, and the modulus of elasticity is 127 GPa. Calculate the separation between the rolls that we should use, assuming that the rolls do not deflect. C
=E
C
=
240 127000 = 0.001889 mmlmm
~
..
h
f
- h h
h
0
6.25 - h
.. 6.25
6.25 - h
0
h
.. 6.238 mm
0
(1 + 0.001889)
=h
o Roll separation 8.
0.001889
0
0.001889 h
0
o
0
6.238 mm
We plan to stretch a steel bar until it has a final length of 1.8 m. What length should the bar be before the forming stress is removed? The yield strength is 427 MPa, the tensile strength is 572 MPa, and the modulus of elasticity is 210 GPa. If - 10 427 I = 210000 = 0.002035 mmlmm o If - 1.8 = 1.8 x 0.002035 = 0.003664
C
=
:. If
= 1. 8037
m
64
9.
A titanium bar 9.4 mm in diameter with a gauge length of 50 mm is pulled in tension to failure. After, failure, the gauge length is 53.75 mm and the diameter is 8.8 mm. Calculate the % elongation and % reduction in area. Lu - L 3.75 % Elongation 7.50% LOx 100 = x 100 o
-so-
So - Su S o
%Reduction in area
10.
69.3977 - 60.8212 69.3977 12.4%
The % elongation of a magnesium alloy is 12.5%. If a magnesium bar 600 mm long is pulled until it breaks, and if the deformation is uniform during this process, what is the final length of the bar? If necking occurred during this process, would the final length be longer or shorter than your calculation? 12.5% of 600 mm
= 75
:. Bar's final length
mm
= 675
mm
If necking occurred the final length would be shorter. 11.
The following data were collected from a standard 12.5 mm specimen of magnesium: Load (kN)
o 5
10 15 20 22 23.9 26.4 27.2 (maximum) 26.4 (fracture)
diameter
test
Gauge Length (mm)
o
50.045 50.090 50.135 50.175 50.195 50.350 51. 250 53.250 56.375
After fracture the gauge length is 56.125 mm and mm. Plot the data and calculate (a) the 0.2% tensile strength, (c) the modulus of elasticity, (el the % reduction in area, (f) the engineering (g) the true stress at fracture.
65
the diameter is 11.54 proof stress, (b) the (d) the % elongation, stress at fracture, and
300
400
Stress I~ Po.
S tre ss .M Po
300
200 200. 100 Problem 11
0·002
0.02
0.04
0.06
0.002
Strain (mm/mm)
600
0.02
0.04
0.06
Strain(mm/mm)
300
Stress
Stress
M Po
M Po
400
200
Problem 12
100
200
100 Problem 13
0.002 0.02 0.04 Strain (mm/mm)
0.06
66
Problem 14
0.002 0.02 0.04 Strain(mm/mm)
0.06
S
0
122.72 mm
2
Stress/Nmm-2
Strain mm/mm
0 40.74 81. 49 122.23 162.97 179.27 194.75 215.13 221. 65 215.13
0 0.0009 0.0018 0.0027 0.0035 0.0039 0.007 0.025 0.065 O. 1275
(a)
0.2% offset yield strength
(b)
tensile strength
(c)
E
(d)
% elongation
= 221.65
= 191.5
N mm- 2
N mm- 2
81.49/0.0018 45272 MN m- 2 = 45.3 GPa =
56.125 - 50 x 100 50
= 12.25% (e)
% reduction in area
= rr/4(12.5)2 = i~2~~~
12.
(f)
fracture stress
(g)
true stress =
- rr/4(11.54)2
rr/4(12. 5) 2
x 100
x 100
= 114.8%
215.13 N mm- 2 26400
--~~~~
rr/4(11. 54)2
26400
-2
= 104.59 = 252 N mm
252 MPa
A standard 12.5 mm diameter tensile bar is machined from copper-nickel alloy; the results of a tensile test are described in following table. Load (kN)
Gauge Length (mm)
o
o
50.015 50.030 50.045 50.05 50.50 51. 30 57.00 68.70
5
10 15 17.6 26.4 35.2 48.4 (maximum) 39 :> (fracture)
67
a the
After fracture, the gauge length is 67.95mm and the diameter is 9.05 mm. Plot the data and calculate (a) the 0.2% proof stress, (b) the tensile strength, (c) the modulus of elasticity, (d) the % elongation, (e) the % reduction in area, (f) the engineering stress at fracture, and (g) the true stress at fracture. Stress/Nmm-2
Strain mmlmm
o
o
40.74 8~. 49 122.23 143.42 215.13 286.84 394.4 322.69
0.0003 0.0006 0.0009 0.001 0.01 0.026 0.014 0.374
(a)
0.2% proof stress
= 183
(b)
tensile strength
= 394.4
(c)
E - 0.0006 - 135816.7 MN m
(d)
% elongation
(e)
% elongation in area
_ 81. 49
N mm- 2
_
N mm- 2 -2
= 135.82 GPa
67.95 - 50 x 100 50
= nI4(12.5)2
35.9% - nI4(9.05)2
n/4( 12.5)2
47.6% (f)
fracture stress
(g)
true stress
322.69 N mm- 2
= __39_6_0_0__ n/4(9.05)2
68
615.6 N mm- 2
x 100
13.
A 25 mm diameter tensile bar is prepared from a silver alloy in a tensile test with the following results. Load
and
pulled
Gauge Length (mm)
(kN)
o
50.0000 50.0613 50. 1227 50.1848 50.50 51. 35 52.90 53.40
50,000 100,000 150,000 175,000 200,000 225,000 (maximum) 231,000 (fracture)
StresslMPa
Strain mm/mm
o
o
101. 86 203. 72 305.58 356.51 407.44 458.37 470.59
0.001226 0.002454 0.003696 0.01 0.027 0.058 0.068
After fracture, the gauge length is 53.1 mm and the diameter is 24.25 mm. Plot the data and calculate (a) the 0.2% offset yield strength, (b) the tensile strength, (c) the modulus of elasticity, (d) the % elongation, (e) the % reduction in area, (f) the engineering stress at fracture, and (g) the true stress at fracture. The load and gauge length are converted to stress and strain in table above. (a)
yield strength
320 MPa
(b)
tensile strength
= 470.59
(c)
E
(d)
%e
MPa
= 82,680 x 100 = 6.2%
305.58/0.003696 53.1 - 50.0 50.0 2
MPa
= (rr/4) (25) - (rr/4)(24.25) (rr/4) (25)2
(e)
r~
(f)
fracture stress
(g)
true stress
= 470.59
2
82.68 GPa
x 100
MPa
= 231,OOO/(rr/4) (24. 25)2
69
5.19%
500.15 MPa
the
14.
A 15 mm diameter tensile bar of an aluminum alloy is pulled in a test with the following results. Load
tensile
Gauge Length (mm)
(N)
°
60.0000 60.0469 60.0938 60.1407 60.210 60.300 60.600 61. 200 63.000 63.900
10,000 20,000 30,000 35,000 37,500 40,000 42,500 45,000 (maximum) 44,200 (fracture)
Stress/MPa
Stl'ain mm/mm
o
o
56.59 113.18 169.77 198.06 212.21 226.35 240.50 254.65 250.12
0.00078 0.00130 0.00235 0.0035 0.005 0.010 0.020 0.050 0.065
After fracture, the gauge length is 63.66 mm and the diameter is 14.5 mm. Plot the data and calculate (a) the 0.2% offset yield strength, (b) the tensile strength, (c) the modulus of elasticity, (d) the % elongation, (e) the %reduction in area, (f) the stress at fracture, and (g) the true stress at fracture. The load and gauge length are converted to stress and strain in the table above. (a)
yield strength
(b)
tensile strength
(c)
E
(d)
% • e I onga ti on
(el
%RA
(f)
fracture stress
(g)
true stress
200 MPa 254.65 MPa
= 113.18/0.00130 = 87,060 = 63.6660-
MPa
87.06 GPa
60 x 100
2 = (n/4l(15) 2 - (n/4)(14.5)
6 . 1%• x 100
6.56%
(n/4) (15)2
= 250.12
MPa
= 44,200/(nI4)(14.5)2 70
267.7 MPa
15.
A three-point bend test is performed on a block of Al 20 3 that is 127
mm
long, 8.26 mm wide, and 6.35 mm tall and is resting on two supports 76.2 mm apart. A force of 1.6014 kN is required to break the ceramic. Calculate the flexural strength in MPa. 3FL/2wh2
FS
= 3(1601.4)(76.2) 2(8.26)(6.35)2
FS
16.
= 549.9
N
mm- 2
= 549.9
MPa
A three-point bend test is performed on a block of silicon nitride that is 100 mm long, 8 mm wide, and 6 mm tall and is resting on two supports 80 mm apart. The modulus of rupture is 896 MPa. Calculate the force required to break the test block in newtons. 3FL
FS=~~ 896
3.F.80
F = 2.15 kN 17.
Suppose a stainless steel is to be selected as a valve in a system designed to pump liquid helium at 4 K. A severe impact may be observed during opening and closing of the valve. Would you select a stainless steel that has an FCC or a BCC structure? Explain. FCC structure; because this crystal structure does brittle transition.
18.
not
undergo
Fiberglass is made by introducing short glass fibers into a more ductile polymer matrix. Would you expect the fiberglass to be notch sensitive or insensitive in a series of impact tests? Explain Notch insensitive: the short glass fibers already act like in the polymer matrix.
19.
a
notches
A cylindrical tool steel specimen (Figure 6-17) which is 230 mm long and 12.5 mm in diameter is to be designed so that failure never occurs. What is the maximum load than can be applied? Endurance Limit
= 413.8
MPa
95.01.t.w d3
413.8 x 1953 95.01 x 230
=w
36.98 kg
.. a load less than 36.98 kg will not cause fatigue failure.
71
20.
A cylindrical aluminum test specimen (Figure 6-17) which is 200 mm long and 18.75 mm in diameter is to be designed so that failure does not occur within ten million cycles. What is the maximum load that can be applied?
= 95.01 t w stress = 33500
IT
1 x 145
= 224.14
MPa
to cause failure in 107 cycles 224.14 X (18.75)2 90.11 x 200
= w = 82
kg
Thus a load of less than 82 kg will not cause failure in 107 cycles. 21.
A cylindrical tool steel test specimen (Figure 6-17) which is 250 mm long is to be subjected to a load of 6600 N. What is the minimum diameter of the specimen if failure is never to occur? d3= 95.01 x 6600 x 250 413.8 x 9.8306
= 38537
d = 33.78 mm 22.
A cylindrical aluminum test specimen (Figure 6-17) which is 250 mm in length is to be subjected to a load of 20,000 N. What is the minimum diameter of the specimen if failure within one million cycles is to be prevented? Stress to cause failure in 106 cycles d3 d
= 241.4
MPa
= 95.01
x 20,000 x 250 241. 4 x 9.8306
= 58.5
mm
(ie) minimum diameter to avoid failure in 106 cycles = 58.5 mm 23.
A cylindrical tool steel test specimen (Figure 6-17) which is 150·mm long and 9.4 mm in diameter is to have a fatigue life of 500,000 cycles. What is the maximum allowable load? Fatigue life
=5
x 105
Applied stress = 508.6 MPa for failure at 5 x 10 5 cycles IT
95.01
=
t w
d3
(9.4)3 x 508.6 95.01 x 150
=w
.. w = 29.64 kg (ie)
Maximum allowable load
72
29.6 kg
24.
A cylindrical aluminum test specimen (Figure 6-17) which is 175 mm long and 12.5 mm in diameter is rotating at 3000 rpm. If a load of 350 N is applied to the specimen, how many hours of operation are expected before failure? 0'
=
.. 0'
95.01. t. F d3
95.01 x 175 x 350
303 N mm- 2
02.5)3 x 9.0306
303 MPa
5
No of cycles
= 105
105 :. Time to failure = 3000 minutes 25.
0.56 hours
and 37.5 mm in diameter is rotating at 60 rpm. If a load of 11.7 applied to the specimen, how many days of operation are expected failure?
x 117000 x 250 (37.5)3 x 9.8306
= 95.01
--~----~~~~~~
536 MPa
5
kN is before
536 MPa
No of cycles
.. Time to failure
300000 60 x 60 x 24
3.47 days
To avoid failure by fatigue, the maximum force that can be applied to the end of a rotating cylindrical steel specimen which is 125 mm long and 6.25 mm in diameter is 66 N. Estimate the tensile strength of the steel. Fatigue limlt
95.01 x 125 x 66 = ~~~~~~~~
Fatlgue iimit
= 1/2
(6.25)3 x 9.8036
:. tensile strength 27.
33.33 minutes
A cylindrical tool steel test specimen (Figure 6-17) which is 250 mm long
0'
26.
303 MPa
x tensile strength 2 x 95.01 x 125 x 66 = ------------(6.25)3 x 9.8036
707 MPa
Which of the following would you expect to show the temperature dependence of the creep rate (e): e - T; e - r2; e - liT; or e - exp (-ciT)? Explain your answer exp(-c/T): diffusion determines the temperature dependence of creep should diffusion.
73
rate of creep, be the same as
so that
the for
28.
The following data are the results of a creep test performed on a material. The original gauge length was 50 Mm. Calculate the creep rate in XIh. Length of Specimen
Time
(_)
(h)
50.5 50.95 51.15 51.50 52.30 53.05 53.80 54.50 55.00
5 200 500 1450 3450 5500 7500 8500 8950 II
Extension/mm
time (h)
X
0.5 0.95 1. 15 1. 50 2.3 3.05 3.8 4.5 5.0
1.0 1.9 2.3 3.0 4.6 6.1 7.6 9.0 rupture
5· 200 500 1450 3450 5550 7500 8500 8750
&1
(53.8 - 50)/50
= 0.076 mm/mm
&2
(51.50 - 50)/50
= 7.6X = 0.03 mm/mm
......,.......
.-
Ca.,
-
= 3X
:. creep rate
(7.6 - 3)/(7500 - 1450) = 7.6
29.
X
10-4 X/hour
A 50 mm diameter bar of an iron-chromium-nickel alloy is subjected to a load of 26.4 kN. and (cl 1090·C? v
How many days will i t survive at (a) 980·C, (b) 1040·C,
= 26400/(W/4)!~0)2 = 13.445 N mm
(a) (b) (c)
at 980·C rupture time = 10,000 hours = 416.7 days at 1040·C rupture time = 2,000 hours = 83.3 days at 1090·C rupture time = 300 hours = 12.5 days
74
30.
What is the maximum load that a 20 mm x 30 mm bar of an iron-chromium-nickel alloy can withstand at 870°C without falling within 10 years? Time
= (10)(365)(24) = 87,600
hours
The maximum stress at 870·C for a 10 year life is v = 20.7 MN m- 2 = 20.7 N mm- 2 F = vA = 20.7 x 20 x 30 124105 N = 12.41 kN 31.
An iron-chromium-nickel alloy is to withstand a load of 11 kN at 1090·C for 15 years. What is the minimum diameter of the bar that is necessary?
Time = (15)(365)(24) = 131,400 hours Maximum allowable stress at 1090·C is v = 2.76 MPa = 2.76 N mm- 2
A = v~
= 11000 = 3988 . 6 2.76
mm2
« = ~3988.6/(n/4) = 71.26 mm 32.
Develop an equation of the form v = ct-n that will relate the applied stress v to the rupture time for an iron-chromium-nickel alloy held at 760·C. Using this equation, calculate the maximum load that a 25 mm diameter bar of the alloy can withstand if it is to survive for 50 years. (a)
From Figure 6-24, the stress is 110 MPa at 100 hours and 55 MPa at 100,000 hours. The equatIon can be written and solved for n and c: In v
= In
In(llO) 4.7005 In(55) 4.0073
c - n In t In(c) - n In(100) In(c) 4.605 n In(c) - n In(100,OOO) In(c) - 11.5 n
(i) - (11) gives 0.6932 :. n
0.1
110 110
C(100)-o.t 0.6309 c 174.3
:. C
(b)
t
= 6.895
-
-(11 )
n
time = (50) (365) (2~J 438, 000 hou~~ = 174.3(438,000) . = 47.55 N mm F = 47.55 x (n/4) (25)2 = 23.341 kN
v
75
(i)
33.
An iron-chromium-nickel alloy is to operate for 10,000 h under a load of
154 kN. What is the maximum operating temperature if the bar is 37.5 in diameter? u
=
= 13.94
15400 1l/4(37.5)2
MPa
= 980°C
:. maximum temperature 34.
mm
Suppose a stress of 138 MPa is applied to the heat-resistant alloy shown in Figure 6-24(b). (a) Determine the equation for the Arrhenius relationship between the rupture time t f and the operating temperature and (b) determine the rupture time if the alloy operates at 500°C. (a)
The Arrhenius equation is rate = coexp(-Q/RT). Therefore we can use reciprocal time as our graph at 138 MPa we can pick off two points: t
10 4 hr l/t 10
t
= 0.0001
hr l/t = 0.1
h- 1 1/T
h- 1
l/T
rate.
= 10.4 X 10- 4 K- 1 = 7.9 x 10- 4 K- 1
From
= 961. 54 K T = 1265.82
T
0.0001 _ Co exp[-Q/(8.314)(961.54)) _ exp(-0.000125Q) - c exp[-Q/(8.314)(1265.82)) - exp(-O.00009SQ) o
~
:. 0.001
exp [( -0.000125 + 0.000095 )Q]
In(O.OOl) ~
-6.907753
=Q
230,258.5 J/mol
0.0001
Co exp[-230, 258. 5/(8.314)(961.54)]
0.0001
Co exp(-28/803)
.. c
o
3.228 x
.. t1_ - (3.228) f
(b)
= exp [-0. 00003Q]
= -0.00003Q
108 X
h- 1
10 8 ) exp
c (3.0974 x 10- 13 ) o
( _ 230,258.5) (8.314)T
773 K
at 500°C
8 ) t1f = (3 . 228 x 10 ) exp (230,258.5 - (8.314)773
t1f -_ 3.228
X
10 8 exp(-35.8283)
(3.228 x 108 )(2.754 x 10- 16 ) 8.89 x 10- 8 h- 1 .. t f = 1. 1248583
X
76
107 h
the
35.
Suppose the heat-resistant alloy shown in Figure 6-24(b) is to survive for a minimum of 20 years. Calculate the maximum temperature to which the alloy can be heated if the stress is 138 MPa. (Use the equation developed in Problem 34. ) t = (20)(365)(24)
= 175.200
h
From the equation developed in Problem 34
t1
= 1/175.200 = (3.228
8
x 10 ) exp (-27696.3/T)
1.7682 x 10- 14 = exp (-27.69S.3)/T) In(1.7682 x 10-14 ) = 31.666227 = -27.69S.3/T T = 874.6 K = 601.6°C 36.
A 12.5 mm diameter bar of tantalum alloy is originally 250 mm After operating at 131S o C for 50 h, its length must be less than mm. What is the maximum allowable load? Creep ra t e
= 12.5
mm per 250 mm 50 h
long. 262.5
0.05 mm/mm 50 h =
S~%h = 0.01% h- 1
From Figure 6-24(c). the maximum stress for this creep rate is 186 MPa (1'
= 186 N mm- 2 =
F (1l/4) (12. 5)2
:. F = 22825.6 N 37.
= 22.826
kN
A 30 mm diameter bar of tantalum alloy is originally 600 mm long. It operates at a load of 198 kN at 131S o C. Assuming it does not fail. what is the length after 8 h? 198.000
= 280 MPa
(1l/4) (30)2
Creep rate = l%/hour In 8 hours. the total creep will be 8% If - 600 600 x 100 .. If 38.
=8
600 + 48 = 648 mm
A ductile cast iron bar is to operate at a stress of 28 MPa for without failing. what is the maximum allowable temperature?
2500
For a stress of 28 MPa. the Larson-Miller parameter is about 36 . .. 36 .• 36 .. T
(T/I000)[36 + 0.78 In(2S00)] (T/I000) (42.1) 8SSK=S82°C
77
h
39.
A ductile cast iron bar is to operate at 600°C for 10,000 h. If the has a diameter of 18.75 mm what is the maximum allowable load? (873/1000)[36 + 0.78 1n(10,OOO)) (873/1000)(43.184) = 37.7
Larson Miller parameter (1'
F
40.
= 172
MPa
= (172.367)(n/4)(18.75)2 = 47593.5 N = 47.594 kN
A 3000 kg load is applied to a 10mm diameter
indentor, producing an impression having a diameter of 2.2 mm on a steel plate. Calculate the Brinell hardness number of the steel, then estimate the tensile strength and fatigue limit of the steel. 3000
BHN = _ _ _ _3_0_0_0-;:=:::;:::=:==;(n/2)(10)(10 -1(10)2 - (2.2)2) tensile strength
5n(10 - "95.16)
= (3. 45)BHN = (3.45)(780)
endurance limit
41.
bar
= 2691 MPa
(1/2) tensile strength (0.5)(2691) 1345 MPa
A 500 kg load is applied to a
10 mm diameter indentor, impression on a steel plate having a tensile strength Estimate the diameter of the impression. BHN
780
producing an of 520 MPa.
520 150.7 = 3.45 =
150; 7 = _ _ _--'5:....:0...::..0_--;::.:=:=~ (nl2) (10) (10 - /100 - n2 )
+ Aoo - n2
.. 10
:. Aoo - n2 100 -
n2
= 5001 (1l12) (10) (150.7)
0.2112
= 9.7888
= 95.82
D = 2.044 mm 42.
Estimate the Brinel1 hardness number of a steel having a tensile strength of 937 MPa. . 937 BHN ~3.45
43.
= 271.6
The fracture toughness of a materici1 is 45,000 psi inl12 • . fracture toughness expressed in MPa m1/ 2? K Ic
= (45,000
psi in
1/2
What
' . 1/2 )(0.006895 MPa/psi )(0; 0254 mlin)
= 49.4496 MPa. m1/2
78
is
the
44.
A Ni-Cr steel with a yield strength of 1640 MPa (Table 6-8) contains internal flaws that may be as long as 0.025 mm. What is the maximum allowable applied stress if these flaws are not to propagate? How does this compare with the yield strength? K = fcrV'all
= 50.33
MPa ml12 2.5 x 10- 5 m a = 1. 25 X 10-5 m
where 2a
~
= 50.33/~.25
X
10- 5 n
= 8031
MPa
Because the allowable stress is much strength, the flaws will not propagate. 45.
greater
than
yield
An aluminum-copper alloy with a yield strength of 325 MPa (Table 6-8) contains surface flaws that are 0.0075 mm deep. What is the maximum allowable applied stress if these flaws are not to propagate? this compare with the yield strength? K ~
46.
the
How does
h.
36.26 MFa. mIn = (J" 5 X 10-6 n 7470 MPa > yield strength and flaw will not propagate.
A ceramic has a fracture toughness of 2.75 MPa m1 / 2 and must experience an applied stress of 100 MPa. Calculate the size of the maximum allowable (a) internal flaw and (bl surface flaw. 2.75 = 100 v'ai 0.0007562 = an 2.4 x 10-4 m = a if a surface flaw and 4.8 x 10- 4 m
47.
if an internal flaw
A polymer material has a fracture toughness of 1.65 MPa m1 / 2 experience an applied stress of 30 MFa. Calculate the maximum allowable (a) internal flaw and (b) surface flaw. 1. 65 = 30 v'ai 0.003025 = an 9.63 x 10- 4 m
1.926 48.
= 2a
X
=a
10- 3 m
and must size of the
if an internal flaw
= 2a
if a surface flaw
A titanium alloy (Table 6-8) contains internal flaws that
are 5 mm in length. Will these flaws propagate if a stress of one-half the yield strength is applied? Consider both a high strength and a low strength alloy.
K crV'aIl = (0.5)(YS) vO.005 n = 0.06267(YS) For high strength Ti, YS = 896 MPa
K = (0.06267)(896)
=
1/2
56.15 MPa m
79
> KIC
= 55
iO~ ~~~0:~~;~:~2~i~ ~~ ~:6~lW~
K IC = 98.9 MPa mi /2
The cracks may propagate in the high strength alloy but not low strength alloy. 49.
in
A sIlicon nitride ceramic (Table 6-8) contains surface flaws that are 0.02 mm in length. Will these flaws propagate if a stress of one-half
the yield strength is applied? K
=
~ x
(0.5)(551)
"'>
2. 186 MP
mi /2
5 MPa. mi
K
10-5 W
thus the flaws will not propagate.
80
the
Chapter 7 DEFORMATION, STRAIN HARDENING AND ANNEALING
1.
A 12.5 mm diameter bar with a gauge length of 50 mm is subjected to a
tensile test. When a force of 52.8 kN is applied, the specimen has a diameter of 12.30 mm and a gauge length of 50.115 mm. When a force of 100 kN is applied, the diameter is 10.54 mm and the gauge length is 58.85 mm. (a) Determine the strain-hardening coefficient. (b) Would you expect the metal to have the FCC structure? Explain. (a)
= 52.8
~T1
kN/(n/4) (12.30 mm)2
= In(58.85150) = 0.16291
~T2
In(0.444) In(1. 146)
InK + nln(0.003494) InK + nln(0.16291)
0.948 = 3.843 n
2.
kN/mm2
100 kN/(n/4)(10.54 mm)2 = 1.146 kN/mm2
~T2
(b)
= 0.444
In(50.175150) = 0.003494
£T1
n
11.033 11.991
InK - 5.651 n InK = 1.841 n
= 0.25
the strain hardening coefficient of 0.25 is a little low for the FCC metals listed in Table (1-1).
A copper-nickel alloy tensile bar, originally having a 12.5
and a 50 mm gauge length, has a strain-hardening The bar falls at an engineering stress of 675 MPa; the time of failure is 56.5 mm and its diameter is occurred. Calculate the true stress on the bar strain was 0.05.
= vAo = (615
F
x 106 Pa)(n/4) (0.0125 m)2 = 82800 N
= F/A = (82800
VT
mm diameter coefficient of 0.48. its gauge length at 11.76 mm; no necking when the engineering
N)/(n/4) (0.01115 m)2
= 188.6 MPa
= In(lllo) = In(56.5150) = 0.1222
£ ~
T
= 1(£
0.48
T
188.6 = 1(0.1222)°·48 = 0.3646 I(
I(
= 2163 MPa
If the engineering strain is 0.05, then If = 52.5 mm £
= In(lll
)
= In(52.5150) = 0.04879
T 0 v t = 2163(0.04819)°·48 = 507.5 MPa
81
3.
The Frank-Read source in Figure 7-3(e) has created four dislocation loops from the original dislocation line. Estimate the total dislocation line present in the photograph and determine the percent increase in the length of dislocations produced by the deformation. If the length of the original dislocation line is 1 mm on the photograph, then we can estimate the circumference of the dislocation loops. The loops are not perfect circles, so we might measure the smallest and largest diameters, then use the average. first loop:
D = 10 mm; Dlarge small circumference = 12.0'lr
14 mm D average
second loop:
D = 20 mm; Dlarge small circumference = 19.0'lr
18 mm D average
19.0 mm
third loop:
Dsmall = 28 mm; D large circumference = 29.0'lr
30 mm D average
29.0 mm
fourth loop:
Dsmall = -42 mm; Dlarge circumference = 43.5'1r
45 mm D average
43.5 mm
= 12.0
mm
Therefore in the photograph itself: total length
=1
= 362.2
+ (12.0 + 19.0 + 29.0 + 43.5)'Ir
The magnification in the photograph is 30,000. total length (actual)
= 362.2
/ 30,000
%increase 4.
5.
= (0.0109
Therefore
= 0.0109
The original dislocation line is 1 mm / 30,000 mm / 0.0000333 mm) x 100
mm
mm
= 0.0000333
mm
32,700%
Suppose we begin with a copper plate 75 mm thick. Calculate the total percent cold work i f (a) we reduce the plate-to 12.5 mm thick and (b) we reduce the plate first to 25 mm, then later to 12.5 mm.
= 6~55
(a)
%CW
75 ;5 12 .5 x 100
(b)
%CW
83.3% (the intermediate step does not affect total CW).
x 100
83.3%
Suppose we begin with an aluminum bar 75 mm thick. Calculate the total percent cold work if (a) we reduce the bar to a 12.5 mm diameter and (b) we reduce the bar first to a 25 mm diameter, then later to a 12.5 mm diameter. (75)2 - 02.5)2 5625 - 156 (a) %CW x 100 97.2% X 100 5625 (75)2 (b)
%CW
97.2% (the intermediate step does not affect total CW).
82
6.
Calculate (a) the percent cold work and (b) the final properties i f we reduce a 3105 aluminum plate from an original thickness of 50 mm to a final thickness of 6.25 mm (See Figure 7-22). (a)
50 - 6.25 x 100 50
%CW
87.5% 7.
tensile strength yield strength %elongation
667 MPa 450 MPa 0%
properties of 25 mm
if
to
we a
(Assuming no fracture. )
Calculate (a) the total percent cold work and (b) the final properties if we reduce a copper plate from 37.5 mm to 25 mm to 18.75 mm to 15 mm in three passes through a rolling mill. (See Figure 7-6. ) %CW
= 37.5
- 15 x 100 37.5
(b) tensile strength yield strength %elongati6n
60% 9.
tensile strength 210 MPa yield strength 200 MPa %elongation 2.5% (Assuming no fracture during deformation. )
Calculate (a) the percent cold work and (b) the final reduce a C-30% Zn brass bar from an original diameter final diameter of 7 mm (See Figure 7-24). (25)2 _ (7)2 (a) %CW = x 100 = 92.2% (25)2 (b)
8.
(b)
560 MPa 550 MPa 2%
We wish to produce a 3105 aluminum rod having a final yield strength of at least 170 MPa and a final diameter of 5.625 mm. Using Figure 7.23, calculate the minimum original diameter required. To obtain the yield strength, we need at least 50% cold work. d 2 _ (5.625)2
%CW
= _0"--_--::-_----,-_ d 2
°
0.5d 2 o 10.
31. 64
x 100 = 50 d
°
= 7.95
mm
We wish to produce a Cu-30% Zn brass plate having a final % elongation of at least 20% and a final thickness of 9.375 mm. Using Figure 7.24, calculate the maximum original thickness required. We must use no more than 25% cold work. t
%CW = 0.75t
0
- 9.375 t
0
x 100
°
9.375
t
25 0
= 12.5 mm
83
11.
We would like to produce a copper plate 6.25 mm thick having at least 10% elongation and a 310 MPa yield strength. Is this possible? If so, calculate (a) the required percent cold work and (b) the original thickness of the plate. (See Figure 7-6. ) (a)
possible if the percent cold work is between 20 and 30 percent.
(b)
%CW
minimum
=
t t
%CWmaximum
0
0
- 6.25 t
x 100
= 20%
t
x 100
= 30%
t
0
- 6.25 t
0
0
= 7.81
mm
= 8.93
mm
Any original thickness between 7.81 and 8.93 mm will work. 12.
We would like to produce a Cu-30% Zn wire 2.5 mm in diameter with at least 20% elongation and a 400 MPa tensile strength. Is this possible? If so, calculate (a) the required percent cold work and (b) the original diameter of the wire. (See Figure 7-24. ) (a)
(b)
possible if the percent cold work is between 18 and 23 percent. d 2 _ (2.5)2 %CW = __o__~_____ x 100 = 18% do = 2.76 mm minimum d 2 o
d = 2.85 mm o
d 2 o
Any diameter between 2.76 and 0.285 mm will work. 13.
We would like to produce a 3105 aluminum wire 1.25 mm in diameter with at least 10% elongation and a 170 MPa tensile strength. Is this possible? If so, calculate (a) the required percent cold work and (b) the original diameter of the wire. (See Figure 7-23). (a)
(b) 14.
To obtain at least 10% elongation, the cold work should be less than 12%; however, to obtain at least 170 MPa tensile strength, the cold work should be more than 33%. Consequently it is not possible to do this. Does not apply.
We would like to produce a Cu-30% Zn wire 5 mm in diameter having at least 10% elongation and a 480 MPa tensile strength. The original diameter of the bar Is 100 mm. The maximum allowable cold work is -75%. Describe the steps required, including percent cold work and intermediate thicknesses. (See Figure 7-24. ) In order to obtain the desired properties, we need between 25 and 35 percent cold work. Therefore the intermediate diameter should be d 2 _ (5)2
minimum: __i _______ x 100 d 2
= 25%
di
= 5.77
mm
1
d 2 _ (5)2
maximum: ....;::.i_-:-_ x 100 d 2 i
84
35%
d i .. 6.20 mm
Method 1: Cold work 75% from 100 mm to 50 mm; anneal; cold work 75% from 50 to 25 mm; anneal; cold work 75% from 25 to 12.5 mm; anneal; cold work 75% from 12.5 to 6.25 mm; anneal; cold work from 6.25 to between 5.77 and 6.20 mm; anneal; cold work the final 25 to 35 percent to 5 mm. Method 2: Hot work from 100 mm to between 0.577 and 6.20 work the final 25 to 35 percent to 5 mm. 15.
mm;
cold
We would like to produce a copper sheet 3.75 mm thick from an original plate that is 45 mm thick. The final required properties include at least a 345 MPa yield strength and 5% elongation. The maximum allowable cold work per pass is 80%. Describe the steps required, including percent cold work and intermediate thicknesses. (See Figure 7-6. ) In order to obtain the desired properties, we need between 25 and 40 percent cold work. Therefore the intermediate thickness should be minimum:
maximum:
t.
1
-
3.75
t.1 t.
1
-
3.75
t.
x 100
= 25%
t.
=5
x 100
= 40%
t.1
= 6.25
1
1
rom
mm
Method 1: Cold work 80% from 45 to 9 mm; anneal; cold work mm to between 5 and 6.25 mm; anneal; cold work the final 25 to 3.75 mm. Method 2: Hot work from 45 mm to between 5 and 6.25 mm; the final 25 to 40% to 3.75 mm.
16.
from 9 to 40%
cold
work
A 5 rom titanium wire is passed through a 4.5 mm diameter die in a wire-drawing process, producing a wire with a 400 MPa yield strength and a 580 MPa tensile strength. If the modulus of elasticity of the titanium is 112 GPa, calculate the diameter of the final wire product.
c
= u/E = 400
x 106 /112
X
109
= 0.00357
The stress is compressive as the wire passes through the 45 mm die; the diameter of the wire increases elastically after deformation. Therefore the sign of the strain should be negative. -0.00357 4.5 ,.. d f 4.5
-0.00357
4.516 mm
85
17.
A 1.5 mm magnesium wire with a yield strength of 170 MPa is to be produced by a wire-drawing process. If the modulus of elasticity of the magnesium is 45.5 GPa calculate the necessary diameter of the opening in the die.
= ~IE = 170
E
= 0.003736
x 10 6 /45.5 x 109
The stress is compressive as the wire passes through the die; the diameter of the wire increases elastically after deformation. Therefore the sign of the strain should be negative. - 1. 5
d
= -o-d.---- = -0.003736
E
d
18.
o
- 1.5
o
= -0.003846do
d
o
= 1. 4944
mm
We plan to draw a 6.25 mm diameter Cu-30% Zn wire having a yield strength of 70 MPa into a 5 mm diameter wire. (a) Using Figure 7-24, calculate the draw force, assuming no friction. (b) Will the drawn force cause the drawn wire to break? (Prove by calculating the maximum force that the drawn wire can withstand. ) (6.25)2 - (5)2 CW = x 100 36% (6.25)2 Final yield strength (a)
(b)
Fdraw F ~
= ~yAo
= 400
MPa
(70 MPa) (n/4) (6.25/1000 m)2
2147 N
= 400 = F/(n/4) (5/1000)2
F
= 7854 N
The wire will not break because the draw force is less than the maximum force the drawn wire can withstand. 19.
A 3105 aluminum wire 2.5 mm in diameter is to be made having a tensile strength of 170 MPa. Using Figure 7-23, determine (a) the original diameter of the wire, (b) the required draw force, and (c) whether the as-drawn wire will survive the drawing process. (a)
We need 30% cold work to obtain the desired tensile strength d 2 30%
(2.5)2
o
d 2
x 100
o
(b)
o
The yield strength before cold working is 56 MPa. F
(c)
2.99 mm
d
=
(56 MPa) (n/4) (2. 99 x 10- 3 m)2
= 372
N
The yield strength after cold working 30% is 147 MPa. F
= (20,000) (n/4) (2. 5/1000)2 = 720
N
Because the draw force is less than the maximum force that wire can withstand, wire drawing is possible. 86
the
20.
Successful wire drawing requires that strain hardening deformation; what other process(es) in Figure 7-4 might strain hardening?
occur during also require
Deep drawing. 21.
(a) From the data below, estimate recovery, recrystallisation, and grain growth temperatures. (b) Recommend a suitable temperature for stress relief heat treatment. (c) Recommend a suitable temperature for a hot-working process. (d) Estimate the melting temperature of the alloy.
Annealing Temperature (C) 200 400 600 800 1000 1200 1400 1600 2000
(a)
Electrical Conductivity (x 10 7 Q- 1 m- )
Yield Strength (MPa)
0.85 0.86 1. 08 1. 24 1. 25 1. 25 1. 26 1. 26 1. 26
Grain Size (mm) 0.1 0.1 0.1 0.1 0.0375 0.0375 0.0625 0.1625 0.375
830 830 830 830 620 580 565 550 545
Electrical conductivity begins to increase between 400 600°C; yield strength and grain size both decrease between and 1000°C; grain size begins to increase between 1200 and 1400 °C. Therefore recovery temperature recrystallization temperature grain growth temperature
and 800
500°C 900°C 1300°C
(b)
Stress relieve above the recovery temperature but below the recrystallization temperature. A temperature of about 700°C might be appropriate.
(c)
hot-work above the recrystallization temperature but below the grain growth temperature. A temperature of about 1100°C might be appropriate.
(d)
The recrystallization temperature is approximately the absolute melting temperature of the metal: T
r
Tmp
= 0.4Tmp (abs) = 900
= 2933
K = 2660°C
87
+ 273
=
1173 K
0.4
times
22.
(a) From the data below, estimate the recovery, recrystallization, and grain growth temperatures. (b) Recommend a suitable temperature for a stress relief heat treatment. (c) Recommend a suitable temperature for a hot-working process. (d) Estimate the melting temperature of the alloy.
Annealing Temperature (oC)
Residual Stresses (MPa)
Yield Strength (MPa)
Grain Size (mm)
100 200
310 310 310 0 0 0 0 0
550 550 550 550 310 260 255 255
0.20 0.20 0.20 0.20 0.06 0.06 0.08 0.18
~O
400 500 600 700 800 (a)
Residual stresses are eliminated between 300 and 400°C; yield strength and grain size decrease between 400 and 500°C; grain size increases between 600 and 700°C. therefore recovery temperature recrystallization temperature grain growth temperature
(b)
Stress relieve above the recovery but below temperature. A stress relief temperature of appropriate.
(c)
Hot-work above the recrystallization but below the irain growth temperature. A hot-working temperature of about 550 C might be appropriate.
(d)
Recrystallization occurs at about 0.4 temperature of the metal. Tr Tmp
23.
= 350°C 450°C = 650°C
= 0.4Tmp (abs) = 450 = 1808
+ 273
times
the
recrystallization about 400°C is
absolute
melting
723 K
K = 1535°C
Aluminum is often added to liquid steel, causing deoxidation by producing tiny Al a0 3 inclusions which are dispersed uniformly throughout the steel after solidification. What effects will the alumina particles have on the recrystallized grain size and the temperature at which grain growth will become a problem? The aluminum oxide particles will produce a smaller grain size and will increase the temperature required for grain growth to occur.
88
24.
From the photomicrographs in Figure 7-16, estimate the ASTM grain size numbers and plot the grain size number versus the annealing temperature. The approximate number of photomicrograph at 75x is: 400·C: N
=
26/in 2 @l 75x
per
14.6 = 2 n- 1 (n - 1)1n(2) = (n - 1) (0. 693)
3/in2 @l 75x
(3)(75/100)2
(3) (75/100)2
1. 7
inch
each
n
= 4.9
n
= 1.8
@l100x
(n - 1)1n(2) (n - 1)(0.693)
0.7/in 2 @l 75x
(0.7)(75/100)2
(0.7) (75/100)2
0.4 = 2 n- 1 (n - 1)ln(2) (n - 1) (0. 693)
In(0.4) (-0.92)
@l 100x
n
= -0.3
Using the data in Table 7-4, plot the recrystallization temperature versus the melting temperature of each metal, using absolute temperatures. Measure the slope and compare with the expected relationship between these two temperatures. Is our approximation a good one? converting to absolute temperature and plotting, we approximation is accurate. T
T r
~
Al Mg Ag Au Cu Fe Pt Ni Mo Ta W
26.
in
= -f1- 1
In( 1. 7) 0.53 800·C: N
square
(26) (75/100)2 @l100x
(26) (75/100)2 In(14.6) 2.683 650·C: N
25.
grains
933 923 1235 1337 1358 1811 2042 1726 2883 3269 3683
K K K K K K K K K K K
423 K 473 K 473 K 473 K 473 K 723K 723K 873 K 1173 K 1273 K 1473 K
find
that
the
-
2000
:i'
...= 1000
Melting Temperature (K)
When copper is joined to nickel by a roll bonding process, a minimum of 85% deformation is required. Measure the thickness of a quarter, then estimate the minimum combined original thickness of the copper and nickel alloy sheets from which the quarter was produced. The thickness of a quarter is approximately 0.062 in t - 0.062 = 0.413 in -O--t,------- x 100 = 85 t %CW o
o
89
27.
Two sheets of annealed aluminum alloy 3105, each 2.5 mm thick, are to be joined by cold indentation welding. A total deformation of 67% is required at the joint to obtain bonding. (a) What is the final thickness of the joint? (b) What is the final yield strength of the material in the joint? (c) Estimate the force that each individual sheet could withstand and the force that the joint could withstand if the force was applied parallel to the surface of the sheets. (Refer to Figure 7-23. ) 1. 65 mm
(a)
%CW
(b)
177 MPa
(c)
Each individual sheet, whose cross-sectional area is 2.5 w, where w is the width in mm, has a yield strength of 56 MPa. The joint, which has a cross-sectional area 1.65 w, has a yield strength of 177 MPa. Therefore
67
sheet: F
(56 x 106 )(2.5w/10 6 )
29.
N/mm length
292 N/mm length
joint: F 28.
= 140
We plan to produce a cold indentation weld in annealed copper alloy (Figure 7-6) sheet material. Each sheet is 5 mm thick. the final thickness of the joint is 3.75 mm. (a) Estimate the percent cold work done in the joining process. (b) Estimate the yield strength in the original material and in the final joint.
= 10
~03.76
%CW
(b)
The yield strength of the original material yield strength of the joint is 480 MPa.
x
100
= 62.5%
(a)
is
150
MPa;
the
Suppose two sheets of aluminum alloy 3105 were to be jOined by cold indentation welding. What effect, if any, would there be on the joining process if the aluminum sheet had been cold worked 50% prior to JOInIng compared with the originally annealed sheet? (Refer to Figure 7-23) Less deformation could be done without embrittling the joint. If insufficient deformation could be accomplished due to this embrittling potential, then no bonding would occur.
30.
If you Annealing of a nickel alloy requires 1 h at 620°C. complete the annealing process in 15 min, what temperature recommend?
wanted to would you
The time for annealing approximately doubles for every decrease of 10°C, or it would be cut in half for each increase of 10°C. If it takes 60 minutes at 620°C, then it would take 30 minutes at 630°C, and would take 15 minutes at 640°C. Therefore we recommend an annealing temperature of 640°C.
90
31.
32.
You would like to produce the following.products. For each one, tell whether you would recommend hot working, cold working. or a combination of cold working and annealing; tell which, if any, of the processes shown in Figure 7-4 would be suitable; and explain your choice. (a)
Paper clips: cold working would be appropriate - the parts are small so little force is required, processing would be very rapid, and little or no finishing to remove oxidation products would be necessary.
(b)
I-beams that will be welded to produce a portion of a bridge: hot working would be more appropriate = the parts are rather large, so less force would be required for deformation, a continuous hot forming process could be developed, and surface finish is not critical.
(c)
Copper tubing that will connect a water tap to the main copper plumbing: a cold work and annealing cycle would be desired the cold working would permit a good surface finish to be obtained, while annealing would then soften the tubing so it could be deformed during installation and would also improve the corrosion resistance.
(d)
The steel tape in a tape measure: cold working would be desired to produce a good surface finish, to give rapid production rates, and to provide good stiffness to the tape.
(e)
A head for a carpenter's hammer formed from a round rod: hot working would be desired to assure adequate ductility and metal flow during a forging process, giving the detail that is necessary. The hammer head would later be heat treated to obtain the desired strength, hardness, and toughness.
What deformation bonding process would you following? Explain your choices.
recommend
for
each
of
the
(a)
Joining a 100 mm diameter steel shaft to a 100 stainless steel shaft: friction welding would be since the shafts are both symmetrical.
(b)
Sealing an electronic device in a protective box made from mm thick stainless steel: ultrasonic bonding would appropriate since the material is rather thin.
(c)
Joining a 25 mm thick steel plate to a 6.25 mm thick titanium plate, where the length and width of the plates are each 1.8 m: roll bonding or explosive bonding would both be appropriate for producing complete bonding in material of this size.
91
mm diameter appropriate, 0.5 be
Chapter 8
SOLIDIFICATION AND GRAIN SIZE STRENGTHENING 1.
Plot the maximum observed undercooling versus the freezing temperature for the metals listed in Table 8-1. Do the data in the table confirm the relationship expressed in Equation 8-5? Metal Ga Bi Pb Ag
Cu Ni Fe
T (K) m
l1T(K) 76 90 80 250 236 480 420
30 271 327 962 1085 1453 1538
+ + + + + + +
273 273 273 273 273 273 273
500 400 303 544 300 600 1235 g200 H 1358