LECTURE 8.2LECTURE 8.2

LECTURE OUTLINELECTURE OUTLINE

Weekly ReadingWeekly ReadingLesson 08 Prototype Quiz: FeedbackLesson 08 Prototype Quiz: Feedback

CHAPTER XXVCHAPTER XXV

Chapter 25 is concerned with "macroscopic" aspects of the mechanical behavior of Chapter 25 is concerned with "macroscopic" aspects of the mechanical behavior of materials. The concepts of stress and strain (yes, they are different!) are introduced materials. The concepts of stress and strain (yes, they are different!) are introduced and it is shown that, when properly defined, stress and strain are properties of a and it is shown that, when properly defined, stress and strain are properties of a given material and independent of the geometry of that material. The stress-strain given material and independent of the geometry of that material. The stress-strain curve is introduced, as are the three stages in the deformational behavior of curve is introduced, as are the three stages in the deformational behavior of materials: the elastic regime, the plastic regime, and ultimate failure. Examples of materials: the elastic regime, the plastic regime, and ultimate failure. Examples of stress-strain curves for metals, ceramics, and polymers, are presented, and the stress-strain curves for metals, ceramics, and polymers, are presented, and the differences between the mechanical properties of the three major material types are differences between the mechanical properties of the three major material types are discussed. discussed.

The chapter shows that the mechanical behavior of materials may be almost fully The chapter shows that the mechanical behavior of materials may be almost fully described in terms ofdescribed in terms of

the Young's modulus (a measure of stiffness or rigidity).the Young's modulus (a measure of stiffness or rigidity). the yield strength (the stress at which permanent deformation occurs).the yield strength (the stress at which permanent deformation occurs). the ultimate Tensile Strength (the stress at which the material breaks).the ultimate Tensile Strength (the stress at which the material breaks). the ductility (a measure of, for example, the elongation to failure).

CHAPTER XXVICHAPTER XXVI

Chapter 26 discusses the trends observed in the previous chapter in terms of atomic processes. For example, it is shown that Young's modulus, which is a measure of stiffness, is directly related to the bond strength of the materials. The ease with which a thermoplastic polymer deforms plastically is explained in terms of the weak intermolecular bonding, and the ductility of metals is rationalized in terms of the non-directional metallic bond. Chapter 26 also describes a variety of crystalline defects, dislocation in particular. It is argued that dislocation is mainly responsible for the ease with which a metal will plastically deform. The chapter outlines a number of ways for strengthening metallic materials when armed with this information.

CHAPTER XXVIICHAPTER XXVII

A composite may be defined as "a multiphase material that exhibits a significant proportion of the properties of both constituent phases." Composites range in complexity from the simple addition of straw to bricks (a process that has been followed since well before Biblical times) to the complex multi-layered structure of many sporting goods. Chapter 27 describes a number of composites, including wood (a polymer-polymer composite), concrete (ceramic-ceramic), and fiberglass (polymer-ceramic). The reader is also introduced to the concept of "performance indices," which seek to rank or rate different materials as candidates for particular applications. The two indices chosen are specific strength (strength divided by density) and specific modulus (modulus divided by density). These performance indices are of some importance when weight savings are crucial—as in sports equipment.

INTERMEDIATE INTERMEDIATE CONSTRAINTCONSTRAINT

a

24 Mg12

23

45

67

7 Li39 Be4

23 Na1139 K19

1 H1

40 Ca2045 Sc21

48 Ti2251 V23

52 Cr2455 Mn25

56 Fe2659 Co27

59 Ni2864 Cu29

65 Zn30

4 He211 B5

12 C614 N7

16 O819 F9

20 Ne1027 Al13

28 Si1431 P15

32 S1635 Cl17

40 Ar1870 Ga31

73 Ge3275 As33

79 Se3480 Br35

84 Kr3685 Rb37

89 Y3991 Zr40

93 Nb4196 Mo42

98 Tc43101 Ru44

103 Rh45106 Pd46

108 Ag47112 Cd48

115 In49119 Sn50

122 Sb51128 Te52

127 I53131 Xe54

133 Cs55137 Ba56

139 La57178 Hf72

181 Ta73181 W74

186 Re75190 Os76

192 Ir77195 Pt78

197 Au79201 Hg80

204 Tl81207 Pb82

209 Bi83210 Po84

210 At85222 Rn86

223 Fr87226 Ra88

227 Ac89261 Ku104

262 Ha105

II IIIIVVVIVIIVIII

I

88 Sr38

Q1–3. For the following sentences, substitute the correct word/phrase for each bolded item. If you believe that the item is correct as it stands, select “a.”

The intramolecular bonding in solid methane (CH4) is (1) hydrogen. The intermolecular bonding in methane is (2) covalent. Methane is a (3) molecular material.(1)

b) van der Waalsc) covalent

(2) b) van der Waalsc) hydrogen

(3) b) non-molecular

<F> Obtaining credit for this question requires an understanding of what constitutes a molecular material. Molecular materials are comprised of molecules. Within the molecules, the atoms are bonded covalently to each other (i.e., the intramolecular bond is covalent). However, the molecules only attract each other through much weaker secondary bonds, either van der Waals or hydrogen. This is the intermolecular bond. Almost invariably, the secondary bond is van der Waals, although in water, it is hydrogen. Molecules can be formed when non-metallic atoms bond to each other. Such is the case with methane. Thus, the answers to the above questions are:

The intramolecular bonding in solid methane is (c) covalent. The intermolecular bonding in methane is (b) van der Waals. Methane is a (a) molecular material.

Q4–6. Mark each of the bolded items as either true (a) or false (b).

The element (4) lithium is in the same group as calcium. An atom of lithium has (5) three protons, and for a neutral atom, (6) one outer-shell electron.

<F> For the answers to this question, refer to any of the periodic tables in the book. This will reveal that lithium is in Group I (which implies that it has one outer-shell electron) while calcium is in Group II. Lithium has an atomic number of three and hence three protons in its nucleus. The answers to the above questions are thus:

(4) false (5) true (6) true

Sulfur is in Group VI of the periodic table and has (7) ____ outer-shell electrons, or valence electrons. Each S atom can bond to (8)_____ other S atoms to form an eight-membered ring molecule. The bonding between adjacent rings of S atoms is (9) ____.

(7) a) two

b) eight

c) four

d) six

(8) a) two

b) eight

c) four

d) six

(9) a) covalent

b) ionic

c) metallic

d) van der Waals

<F> The answers to this question are to be found in the CRRA essay on Sulfur, or can be answered using a little intuition and reference to a periodic table. Sulfur is in Group VI of the periodic table and has six outer-shell electrons, or valence electrons. Hence, S needs to acquire two extra electrons to complete its outer shell. As a result, each S atom can form single bonds to two other S atoms. This bonding arrangement results normally in a ring of eight S atoms or, in some cases, a linear inorganic polymer molecule. The bonding between the rings of S atoms is van der Waals. The answers to the above questions are thus:

(7) six(8) two(9) van der Waals

Q10–12. The Wayward islanders have developed their own version of the periodic table and have their own names for the elements. However, they have adopted the same conventions for the naming of groups and periods as are adopted in the course textbook. One element, ralphinium (R), is known to be in Group III, Period 6. Atoms of ralphinium bond to one another to form a face-centered cubic solid. However, most ralphinium is found in a compound with oozium (Oz), which has six outer-shell electrons and is in Period 2. Oozium is a gas at room temperature. The ralphinium-oozium compound is called pentobalite.

The bonding in the pure ralphinium solid is (10) ______. It is likely that the compound pentobalite has the stoichiometry (11) ____ in which the ralphinium atoms are (12) ____ bonded to the oozium atoms.

(10) a) ionicb) covalentc) metallicd) van der Waalse) hydrogen

(11) a) ROz3

b) R3Oz2

c) R2Oz

d) R2Oz3

e) R3Oz

(12) a) ionically b) covalentlyc) metallicallyd) van der Waalse) hydrogen

<F> The answers to this question rely on the identification of ralphinium as a metal (it is in Group III, but in Period 6, and metallic behavior increases as the period number increases). The bonding in the pure ralphinium solid is likely to be metallic, a suggestion that is reinforced by the observation that ralphinium is face-centered cubic. Oozium can be identified as a non-metal, as it is in Group VI and Period 2. Hence, it can be concluded that ralphinium will bond ionically to oozium to form the compound pentobalite. Each ralphinium atom will be able to donate its three outer-shell electrons to one or more oozium atoms, whereas each oozium atom only needs two electrons to complete its outer shell. Hence, if two ralphinium atoms donate three electrons each for a total of six donated electrons, then three oozium atoms can complete their outer shells. Thus, it is likely that the compound pentobalite has the stoichiometry R2Oz3. The answers are:

(10) metallic

(11) R2Oz3

(12) ionically

THE SYMMETRY OF THE SYMMETRY OF PATTERNSPATTERNS

Q13. Figure 1 is a schematic drawing of a crystalline solid. It is a compound consisting of "A" atoms and "B" atoms. The two-dimensional lattice that best describes the material is a ______.<a> parallelogram<b+> rectangle<c> centered rectangle<d> square<e> hexagon<F> The pattern of Figure 1 is best described by a rectangle.

Q14. Figure 1 is a schematic drawing of a crystalline solid. It is a compound consisting of "A" atoms and "B" atoms. The rotational symmetry associated with, for example, the "B" atoms in Figure 1 is _________.<a> one-fold<b+> two-fold<c> three-fold<d> four-fold<e> six-fold<F> I can take the pattern of Figure 1 and rotate it by 180˚ and 360˚, and each time, the motif repeats itself exactly. Hence the rotational symmetry is two-fold—it repeats twice in one complete rotation or after rotations of 360/n, where n = 2 (see Figure 1).

Q15. Figure 1 is a schematic drawing of a crystalline solid. It is a compound consisting of "A" atoms and "B" atoms. The mirror symmetry associated with any one A atom on Figure 1 is given by the lines on Figure ____.<a+> 2a<b> 2b<c> 2c<d> 2d<e> 2e<F> Figure 2a is the correct answer . HINT #1: The mirror symmetry must "reflect" one half of the crystal (assumed infinite) into the other half through the mirror line.

Q16. Figure 1 is a schematic drawing of a crystalline solid. It is a compound consisting of "A" atoms and "B" atoms. It is known that atom (A) is from Group II of the periodic table. Atom B will be from group____.<a> VII<b> V<c> VIII<d> IV<e+> VI<F> The compound has the stoichiometry of AB, because both atomic species have four nearest neighbors. Hence, atom "B" must be from Group VI.

Q17. Figure 1 is a schematic drawing of a crystalline solid. It is a compound consisting of "A" atoms and "B" atoms. The bonding in the material shown in Figure 1 is _______.<a> metallic<b+> ionic<c> covalent<d> covalent plus van der Waals<F> The elements that constitute this material are from Groups II and Group VI and are thus metal and non-metal, respectively. Hence, the bonding in the material is ionic.

Q18. Figure 1 is a schematic drawing of a crystalline solid. It is a compound consisting of "A" atoms and "B" atoms. Figure U8-F1 is a schematic drawing of a ______. <a+> ceramic<b> metal<c> polymer<F> Ceramic is the correct answer. The bonding in the compound is ionic, which automatically excludes metals, because they are metallically bonded, and polymers, for which the intramolecular bonding is covalent.

Q19. Figure 1 is a schematic drawing of a crystalline solid. It is a compound consisting of "A" atoms and "B" atoms. The material shown in Figure U8-F1 displays _________.<a> short-range order (SRO)<b> long-range order (LRO)<c> orientational order<d> short-range order (SRO) + long-range order (LRO)<e+> short-range order (SRO) + long-range order (LRO)

+ orientational order<F> The material contains SRO, which states that the monomer is "ordered" (Chapter 22). The crystalline nature of the material also implies both LRO (Chapters 22 and 23) and orientational order (Materials' Essay 23B).

Q23. Strontium chloride (SrCl2) is a(n) _______.

<a> atomic solid

<b> molecular solid

<c> hydrogen-bonded solid

<d> covalent network solid

<e+> ionic solid

<F> Strontium is in Group II (hence a metal) and chlorine is in Group VII (hence a non-metal). According to the definition given in the Epilogue (see Epilogue, Table 3), SrCl2 is an ionic solid and hence, a ceramic.

Q24. Strontium chloride (SrCl2) is a(n) ______.

<a> noble gas

<b+> ceramic

<c> metal

<d> polymer

<F> Strontium is in Group II (hence a metal) and chlorine is in Group VII (hence a non-metal). According to the definition given in the Epilogue (and see Epilogue, Table 3), SrCl2 is an ionic solid and hence, a ceramic.

Q25. Solid carbon dioxide (CO2) is a(n) ________.

<a> atomic solid

<b+> molecular solid

<c> hydrogen-bonded solid

<d> covalent network solid

<e> ionic solid

<F> Carbon dioxide is molecular. It consists of a single carbon atom covalently bonded to two oxygen atoms. At “low” temperatures, the material may form a molecular solid.

Q26. Solid carbon dioxide (CO2) is a _____.

<a> noble gas

<b+> ceramic

<c> metal

<d> polymer

<F> Carbon dioxide is non-metallic and inorganic, hence, it is a ceramic.

Q27. Diamond-carbon is a(n) __________.

<a> atomic solid

<b> molecular solid

<c> hydrogen-bonded solid

<d+> covalent network solid

<e> ionic solid

<F> The bonding in diamond-carbon is covalent, but the tetrahedral bonding creates a network solid.