2021年英文原版教材班“材料科學(xué)基礎(chǔ)”考試試題

1、學(xué)習(xí)必備歡迎下載英文原版教材班“材料科學(xué)基礎(chǔ)”考試試題試卷一examination problems of the course of“ fundament of materials science”姓名：班級：記分：1. glossary 2 points for each1) crystal structure:2) basis or motif:3) packing fractor:4) slip system:5) critical size:6) homogeneous nucleation:7) coherent precipitate:8) precipitation harde

2、ning:9) diffusion coefficient:10) uphill diffusion:2. determine the indices for the planes in the cubic unit cell shown in figure 1. 5 pointsfig. 13. determine the crystal structure for the following: a a metal with a0 = 4.9489 ., r = 1.75 . and one atom per lattice point; b a metal with a0 = 0.4290

3、6 nm, r = 0.1858 nm and one atom per lattice point. 10 points4-1. what is the characteristic of brinell hardness test, rockwell hardness test and vickers hardness test. what are the effects of strain rate and temperature on the mechanical properties of metallic materials. 15 points4-2. what are the

4、effects of cold-work on metallic materials. how to eliminate those effects. and what is micro-mechanism for the eliminating cold-work effects. 15 points5-1. based on the pb-sn-bi ternary diagram as shown in fig. 2, try to 1show the vertical section of 40wt.%sn; 4 points(2) describe the solidificatio

5、n process of the alloy 2# with very low cooling speed includingphase and microstructure changes; 4 pointso(3) plot the isothermal section at 150c. 7 pointsfig. 25-2. a 1mm sheet of fcc iron is used to contain n 2 in a heated exchanger at 1200oc. theconcentration of n at one surface is 0.04 atomic pe

6、rcent and the concentration at the second surface is 0.005 atomic percent. at 1000oc, if same n concentration is demanded at the secondosurface and the flux of n becomes to half of that at 1200c, then what is the thickness of sheet. 15 points6-1. supposed that a certain liquid metal is undercooled u

7、ntil homogeneous nucleation occurs. 15 points(1) howto calculate the criticalradius of the nucleus required. please give the deduction process.(2) for the metal ni, the freezing temperature is 1453 c, the latent heat of fusion is 2756 j/cm 3, and the solid-liquid interfacial energy is 25510-7 j/cm 2

8、. please calculate the criticalradius at 1353 c. assume that the liquid ni is not solidified.6-2. fig.3 is a portion of the mg-al phase diagram. 15 points(1) if the solidificationis too rapid, please describe the solidificationprocess of mg-10wt%al alloy.(2) please describe the equilibrium solidific

9、ation process of mg-20wt%al alloy, and calculate the amount of each phase at 300 c.fig. 37-1. figure 4 shows us the al-cubinary diagram and some microstructures found in a cooling process for an al-4%cu alloy. please answer following questions according to this figure. 20 pointsfig. 4(1) what are pr

10、ecipitate, matrix and microconstituent. please point them out in the in the figure and explain.(2) whyisneed-likeprecipitatenotgoodfordispersionstrengthening.thetypical microstructure shown in the figure is good or not. why.(3) please tell us how to obtain the ideal microstructure shown in this figu

11、re.(4) can dispersion strengthened materials be used at high temperature. please give the reasons comparing with cold working strengthening7-2. please answer followingquestions according to the time-temperature-transformationttt diagram as shown in fig. 5. 20 points1what steel is this ttt diagram fo

12、r. and what means p, b, and m in the figure. 2why dose the ttt diagram exhibits a c shape.(3) pointout what microconstituentwillbe obtained after austenite is cooled according to thecurves i, ii, iii and iv .(4) what is microstructural difference between the curve i and the curve ii. 5how to obtain

13、the steel with the structure of(a) p+b(b) p+m+a residual(c) p+b+m+a residual(d) full tempered martensiteif you can, please draw the relative cooling curve or the flow chart of heat treatment.iviiiiiifig. 5學(xué)習(xí)必備歡迎下載英文原版教材班“材料科學(xué)基礎(chǔ)”考試試題答案solutions of the course of“ fundament of materials science”1. glos

14、sary 2 points for each1) the arrangement of the atoms in a material into a repeatable lattice.2) a group of atoms associated with a lattice.3) the fraction of space in a unit cell occupied by atoms.4) the combination of the slip plane and the slip direction.5) the minimum size that must be formed by

15、 atoms clustering together in the liquid before the solid particle is stable and begins to grow.6) formation of a critically sized solid from the liquid by the clustering together of a large number of atoms at a high undercooling without an external interface.7) aprecipitatewhosecrystalstructureand

16、atomicarrangementhaveacontinuous relationship with matrix from which precipitate is formed.8) a strengthening mechanism that relies on a sequence of solid state phase transformations in a dispersion of ultrafine precipitates of a 2nd phase. this is same as age hardening. it is aform of dispersion st

17、rengthening.9) a temperature-dependent coefficient related to the rate at whichatom, ion,or other species diffusion. the dc depends on temperature, the composition and microstructure of the host material and also concentration of the diffusion species.10) a diffusion process in which species move fr

18、om regions of lower concentration to that of higher concentration.2. solution : a-364, b-340, c346.3. solution : afcc; b bcc.4-1. what is the characteristic of brinell hardness test, rockwell hardness test and vickers hardness test. what are the effects of strain rate and temperature on the mechanic

19、al properties of metallic materials. 15 points4-2. what are the effects of cold-work on metallic materials. how to eliminate those effects. and what is micro-mechanism for the eliminating cold-work effects. 15 points5-1. based on the pb-sn-bi ternary diagram as shown in fig. 2, try to 1show the vert

20、ical section of 40wt.%sn; 5 points(2) describe the solidificationprocess of the alloy 2# with very low coolingspeed includingphase and microstructure changes; 5 pointso(3) plot the isothermal section at 150c. 5 pointsfig. 25-2. a 1mm sheet of fcc iron is used to contain n 2 in a heated exchanger at

21、1200oc. theconcentration of n at one surface is 0.04 atomic percent and the concentration at the second surface is 0.005 atomic percent. at 1000oc, if same n concentration is demanded at the second surface and the flux of n becomes to half of that at 1200oc, then what is the thickness of sheet.15 po

22、ints6-1. supposed that a certain liquid metal is undercooled until homogeneous nucleation occurs. 15 points(3) howto calculate the criticalradius of the nucleus required. please give the deductionprocess.(4) for the metal ni, the freezing temperature is 1453 c, the latent heat of fusion is 2756 j/cm

23、 3, and the solid-liquid interfacial energy is 25510-7 j/cm 2. please calculate the critical radius at 1353 c. assume that the liquid ni is not solidified.6-2. fig.3 is a portion of the mg-al phase diagram. 15 points(3) if the solidificationis too rapid, please describe the solidificationprocess of

24、mg-10wt%al alloy.(4) please describe the equilibrium solidification process of mg-20wt%al alloy, and calculate the amount of each phase at 300 c.fig. 37-1. figure 4 shows us the al-cu binary diagram and some microstructures found in a cooling process for an al-4%cu alloy. please answer following que

25、stions according to this figure. 20 pointsfig. 4 1what are precipitate, matrix and microconstituent. please point them out in the in the figure and explain.(2) why is need-like precipitate not good for dispersion strengthening. the typical microstructure shown in the figure is good or not. why.(3) p

26、lease tell us how to obtain the ideal microstructure shown in this figure.(4) can dispersion strengthened materials be used at high temperature. please give the reasons comparing with cold working strengthening7-2. please answer followingquestions according to the time-temperature-transformationttt

27、diagram as shown in fig. 5. 20 points1what steel is this ttt diagram for. and what means p, b, and m in the figure. 2why dose the ttt diagram exhibits a c shape.(3) pointout what microconstituentwillbe obtained after austenite is cooled according to the curves i, ii, iii and iv .(4) what is microstr

28、uctural difference between the curve i and the curve ii. 5how to obtain the steel with the structure of(a) p+b(b) p+m+a residual(c) p+b+m+a residual(d) full tempered martensiteif you can, please draw the relative cooling curve or the flow chart of heat treatment.iviiiiiifig. 5英文原版教材班“材料科學(xué)基礎(chǔ)”考試試題試卷二e

29、xamination problems of the course of“ fundament of materials science”姓名：班級：記分：1. you would like to be able to physically separate different materials in a scrap recyclingplant. describe some possible methods that might be used to separate materialssuch as polymers, aluminum alloys, and steels from o

30、ne another.5 points2. plot the melting temperature of the elements in the 1a column of the periodic tableversus atomic number i.e., plot melting temperatures of li through cs. discuss this relationship,based on atomic bonding and binding energy.10 points3. above882 ,titaniumhasabcccrystalstructure,w

31、itha=0.332nm.belowthis temperature,titaniumhasa hcpstructure,witha= 0.2978nmandc = 0.4735nm.determine the percent volume change when bcc titanium transforms to hcp titanium. is this a contraction or expansion.10 points34. thedensityof bccironis 7.882g/cmandthelatticeparameteris 0.2866nmwhenhydrogen

32、atoms are introduced at interstitial positions. calculate a the atomic fraction ofhydrogenatomsandbthenumberofunitcellsrequiredonaveragetocontainone hydrogen atom.15 points5. a carburizing process is carried out on a 0.10% c steel by introducing 1.0% c at thesurface at 980 , where the iron is fcc. c

33、alculate the carbon content at 0.01 cm,0.05 cm, and 0.10 cm beneath the surface after 1 h.15 points6. the following data were collected from a standard 0.505-in.-diameter test specimen of acopper alloy initial length tloado = 2.0 in.:gage lengthstressstrainlbin.psiin/in.02.0000000.03,0002.0016715,00

34、00.0008356,0002.0033330,0000.0016657,5002.0041737,5000.0020859,0002.009045,0000.004510,5002.04052,5000.0212,0002.2660,0000.1312,4002.50 max load62,0000.2511,4003.02 fracture57,0000.51after fracture, the gage length is 3.014 in. and the diameter is 0.374 in. plot the data and calculate a the 0.2% off

35、set yield strength, b the tensile strength, c the modulus ofelasticity,d the %elongation,e the %reductionin area,f the engineeringstressat fracture, g the true stress at fracture, and h the modulus of resilience.15 points7. a 1.5-em-diameter metal bar with a 3-cm gage length is subjected to a tensil

36、e test. the following measurements are made.change in force ngage length cmdiameter cm16,2400.66421.202819,0661.47541.088419,2732.46630.9848determine the strain hardening coefficient for the metal. is the metal most likely to be fcc, bcc, or hcp. explain.15 points8. based on hume-rothery s condition

37、s, which of the following systems would beexpected to display unlimited solid solubility. explain.15 points(a) au-agb al-cuc al-audu-we mo-taf nb-wg mg-znh mg-cd英文原版教材班“材料科學(xué)基礎(chǔ)”考試試題答案solutions of the course of“ fundament of msactienricaels”1. steels can be magnetically separated from the other materi

38、als; steel or carbon-containing iron alloys are ferromagnetic and will be attracted by magnets. density differences could be used polymers have a density near that of water; the specific gravity of aluminum alloys is around 2.7;that of steels isbetween 7.5 and 8. electrical conductivity measurements

39、 could be used polymers are insulators, aluminum has a particularly high electrical conductivity.5 points2. t ocl i 180.7 na97.8 k 63.2 rb 38.9as the atomic number increases, the melting temperature decreases,10 points3. we can find the volume of each unit cell. two atoms are present in both bcc and

40、 hcp titanium unit cells, so the volumes of the unit cells can be directly compared.v bcc= 0.332 nm 3 = 0.03659 nm3v hcp= 0.2978 nm 20.4735 nmcos30 = 0.03637 nm 3 v=x 100 =×100= -0.6%therefore titanium contracts 0.6% during cooling.10 points4. a7.882 g/cm3 =x = 0.0081 h atoms/cellthe total atom

41、s per cell include 2 fe atoms and 0.0081 h atoms. thus:10 pointsbsince there is 0.0081 h/cell, then the number of cells containing h atoms is: cells = 1/0.0081 = 123.5or1 h in 123.5 cells 5 points5. d = 0.23 exp-32,900/1.9871253 = 4210-8 cm2/s×c x= 0.87% cc x= 0.43% cc x= 0.18% c15 points26.=fi

42、 /40.505= f/0.2= l- 2 / 2(a) 0.2% offset yield strength = 45,000 psi(b) tensile strength = 62,000 psi6c e = 30,000 - 0 / 0.001665 - 0 = 18 x 10psi(d) %elongation =(e) %reduction in area =(f) engineering stress at fracture = 57,000 psi(g) true stress at fracture = 11,400 lb / tc/40.374 2 = 103,770 ps

43、i h from the graph, yielding begins at about 37,500 psi. thus:15 points7.forcelbgage lengthin.diameterin.true stresspsi true strainpsi16,2403.664212.0281430.20019,0664.475410.8842050.40019,2735.46639.8482490.6002t =k torln143=ln k + n ln0.2ln 249 = ln k + nln 0.6 n = 0.51a strain hardening coefficie

44、nt of 0.51 is typical of fcc metals.15 points8.the au ag, mo ta, and mg cd systems have the required radius ratio, the same crystal structures, and the same valences. each of these might be expected to display complete solid solubility.the auag and mot ad ohave isomorphous phase diagrams. in additio

45、n, themg cdalloysall solidifylikeisomorphousalloys;howeveranumberof solidstatephasetransformations complicate the diagram.15 points英文原版教材班“材料科學(xué)基礎(chǔ)”考試試題試卷三examination problems of the course of“ fundament of materials science”姓名：班級：記分：1. you would like to be able to identify different materials without

46、 resorting to chemicalanalysisor lengthytesting procedures. describe some possible testing and sortingtechniquesyou might be able to use based on the physical properties of materials.5 points2. plot the meltingtemperaturesof elementsin the4a to8-10 columns of theperiodictable versus atomic number i.

47、e., plot melting temperatures of ti through ni, zr through pd, and hfthrough pt. discuss these relationships, based on atomic bonding and binding energy, a as the atomic number increases in each row of the periodic table and b as the atomic number increases in each column of the periodic table.10 po

48、ints3. beryllium has a hexagonal crystal structure, withao= 0.22858 nm andco= 0.35842 nm. the3atomic radius is 0.1143 nm, the density is 1.848 g/cm, and the atomic weight is 9.01 g/mol.determine a the number of atoms in each unit cell and b the packing factor in the unit cell.10 points4. suppose we

49、introduce one carbon atom for every 100 iron atoms in an interstitial position in bcc iron, giving a lattice parameter of 0.2867 nm. for the fe-c alloy, find a the density and b the packing factor.15 points5. iron containing 0.05% c is heated to 912oc in an atmosphere that produces 1.20%c at the sur

50、face and is held for 24 h. calculate the carbon content at 0.05 cmbeneath the surface if(a) the iron is bcc and b the iron is fcc. explain the difference.15 points6. the following data were collected from a 0.4-in. diameter test specimen of poly vinyl chlorideafter fracture, the gage length is 2.09

51、in. and the diameter is 0.393 in. plotcalculateelasticity,a the0.2%offsetyieldstrength,the data andb thetensilestrength,c the modulusofd the %elongation,e the %reductionin area,f the engineeringstressatfracture, g the true stress at fracture, and h the modulus of resilience.15 points7.a titanium all

52、oy contains a very fine dispersion of tiny er203 particles. what will be the effectof these particles on the grain growth temperature and the size of the grains at any particularannealing temperature. explain.15 points8.suppose 1 at% of the following elements is added to copper forming a separatealloywitheach element without exceeding the solubility limit. which one would be expected to givethe higher strength alloy. is any of the alloying elements expected to have unlimited solidsolubility in copper.a aub mnc srd sie co 15 pointsl 0= 2.0 in:loadlbgage