优质材料科学与工程基础英文版第五版课后习题

1、Homework 11.1 What are materials? List eight commonly encountered engineering materials.Answer1.1: Materials are substances of which something is composed or made. Steels, aluminum alloys, concrete, wood, glass, plastics, ceramics and electronic materials.1.2 What are the main classes of engineering

2、 materials?Answer1.2: Metallic, polymeric, ceramic, composite, and electronic materials are the five main classes.1.3 What are some of the important properties of each of the five main classes of engineering materials?Answer1.3:Metallic Materials? many are relatively strong and ductile at room tempe

3、rature? some have good strength at high temperature? most have relatively high electrical and thermal conductivitiesPolymeric Materials? generally are poor electrical and thermal conductors? most have low to medium strengths? most have low densities? most are relatively easy to process into final sh

4、ape? some are transparentCeramic Materials? generally have high hardness and are mechanically brittle ? some have useful high temperature strength? most have poor electrical and thermal conductivitiesComposite Materials? have a wide range of strength from low to very high? some have very high streng

5、th-to-weight ratios (e.g. carbon-fiber epoxy materials)? some have medium strength and are able to be cast or formed into a variety of sha (e.g. fiberglass-polyester materials)? some have useable strengths at very low cost (e.g. wood and concrete)Electronic Materials? able to detect, amplify and tra

6、nsmit electrical signals in a complex manner? are light weight, compact and energy efficient1.8 What are nanomaterials? What are some proposed advantages of using nanomaterials over their conventional counterparts?Answer1.8: Are defined as materials with a characteristic length scale smaller than 10

7、0 nm. The length scale could be particle diameter, grain size in a material, layer thickness in a sensor, etc. These materials have properties different than that at bulk scale or at the molecular scale. These materials have often enhanced properties and characteristics because of their nano-feature

8、s in comparis on to their micro-featured coun terparts. The structural, chemical, electr on ic, and thermal properties (am ong other characteristics) are ofte n enhanced at the nano-scale.Homework 2Chapter 3, Problem 4What are the three most com mon metal crystal structures? List five metals that ha

9、ve each of these crystal structures.Chapter 3, Solution 4The three most com mon crystal structures found in metals are: body-ce ntered cubic (BCC), face-centered cubic (FCC), and hexagonal close-packed (HCP). Examples of metals having these structures in clude the followi ng.BCC: - -iron, vanadium,

10、tungsten, niobium, and chromium.FCC: copper, alum in um, lead, ni ckel, and silver.HCP: magnesium, - - titanium, zinc, beryllium, and cadmium.Chapter 3, Problem 5For a BCC un it cell, (a) how many atoms are there in side the un it cell, (b)what is the coord in ati onn umber for the atoms, (c) what i

11、s the relati on ship betwee n the len gth of the sidea of the BCC un itcell and the radius of its atoms, and (d) APF = 0.68 or 68%Chapter 3, Solution 5(a) A BCC crystal structure has two atoms in each unit cell. (b) A BCC crystal structure has a coord in ati on n umber of eight. (c) In a BCC un it c

12、ell, one complete atom and two atom eighths touch each other along the cube diag on al. This geometry tran slates into the relati on ship 、3a 二 4R.Chapter 3, Problem 6For an FCC un it cell, (a) how many atoms are there in side the un it cell, (b) What is the coord in ati on4Rn umber for the atoms, (

13、c)a =, and (d) what is the atomic pack ing factor?Chapter 3, Solution 6(a) Each unit cell of the FCC crystal structure contains four atoms. (b) The FCC crystal structure has acoord in ati on n umber of twelve. (d) By defi niti on, the atomic pack ing factor is give n as:volume of atoms in FCC un it

14、cell Atomic pack ing factorvolume of the FCC un it cellThese volumes, associated with the four-atom FCC unit cell, areVatoms=44_3兀163and Vunit cell = awhere a represe nts the lattice con sta nt.Substituti ng4R2Vunit cell = a64R32,2The atomic pack ing factor the n becomes,APF (FCC un it cell)f316JiR3

15、(1展、咒逅(32R3 丿6=0.74Chapter 3, Problem 7For an HCP un it cell (c on sider the primitive cell), (a) how many atoms are there in side the un it cell, (b) What is the coord in ati on n umber for the atoms, (c) what is the atomic pack ing factor, (d) what is the ideal c/a ratio for HCP metals, and (e) re

16、peat a through c considering thelarger ” cell.Chapter 3, Solution 7The primitive cell has (a) two atoms/u nit cell; (b) The coord in ati on n umber associated with the HCPcrystal structure is twelve. (c)the APF is 0.74 or 74%; (d) The ideal c/a ratio for HCP metals is 1.633 ; (e) all answers remain

17、the same except for (a) where the new answer is 6.Homework 3Chapter 3, Problem 25Lithium at 20 C is BCC and has a lattice constant of 0.35092 nm. Calculate a value for the atomic radius of a lithium atom in nano meters.Chapter 3, Solution 25For the lithium BCC structure, which has a lattice con sta

18、nt ofa = 0.35092 nm, the atomic radius is,4(0.35092 nm)=0.152 nmChapter 3, Problem 27a inPalladium is FCC and has an atomic radius of 0.137 nm. Calculate a value for its lattice constant nano meters.Chapter 3, Solution 27Letting a represent the FCC unit cell edge length and R the palladium atomic ra

19、dius,.2a =4R or(0.137 nm)=0.387 nmMultiplyi ng by 2 to clear the fractio n, the Miller in dices are(6 3 4).Chapter 3, Problem 31 Draw the following directions in a BCC unit cell and list the position coord in ates of the atoms whose cen ters are in tersected by the directi on vector:(a) 100(b) 110(c) 111Chapter 3, Solution 31错误！未找到引用源。(a) Positi on Coord in ates:(b) Positi on Coord in ates:(c) Positi on Coordin ates:(0, 0, 0), (1, 0, 0)(0, 0, 0), (1, 1,0)(0, 0, 0), (1, 1, 1)Chapter 3, Problem 32Draw directi on vectors in un it cells for the followi ng cubic dir