英文原版教材班“材料科学基础”考试试题_资格考试-教师资格考试.pdf
学习必备 欢迎下载 英文原版教材班“材料科学基础”考试试题 试卷一 Examination problems of the course of“fundament of materials science”姓名：班级：记分：1.Glossary(2 points for each)1)crystal structure:2)basis(or motif):3)packing fractor:4)slip system:5)critical size:6)homogeneous nucleation:7)coherent precipitate:8)precipitation hardening:9)diffusion coefficient:10)uphill diffusion:2.Determine the indices for the planes in the cubic unit cell shown in Figure 1.(5 points)Fig.1 3.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=0.42906 nm,r=0.1858 nm and one atom per lattice point.(10 points)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 mechanical properties of metallic materials?(15 points)4-2.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 points)5-1.Based on the Pb-Sn-Bi ternary diagram as shown in Fig.2,try to (1)Show the vertical section of 40wt.%Sn;(4 points)(2)Describe the solidification process of the alloy 2#with very low cooling speed(including 学习必备 欢迎下载 phase and microstructure changes);(4 points)(3)Plot the isothermal section at 150oC.(7 points)5-2.A 1mm sheet of FCC iron is used to contain N2 in a heated exchanger at 1200oC.The concentration of N at one surface is 0.04 atomic percent and the concentration at the second surface is 0.005 atomic percent.At 1000 oC,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 points)6-1.Supposed that a certain liquid metal is undercooled until homogeneous nucleation occurs.(15 points)(1)How to calculate the critical radius 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/cm3,and the Solid-liquid Interfacial Energy is 255 10-7 J/cm2.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)(1)If the solidification is too rapid,please describe the solidification process of Mg-10wt%Al alloy.(2)Please describe the equilibrium solidification process of Mg-20wt%Al alloy,and calculate the amount of each phase at 300 C.Fig.3 Fig.2 载学习必备欢迎下载英文原版教材班材料科学基础考试试题答案学习必备欢迎下载学习必备欢迎下载学习必备欢迎下载英文原版教材班材料科学基础考试试题试卷二姓名班级记分学习必备欢迎下载学习必备欢迎下载英文原版教材班考试试题试卷三姓名班级记分学习必备欢迎下载学习必备欢迎下载英文原版教材班材料科学基础考试试题答案学习必备欢迎下载学习必备欢迎下载学习必备欢迎下载英文原版教材班材料科学基础考试试题试卷四姓名班级记分学习必文原版教材班材料科学基础考试试题试卷五姓名班级记分学习必备欢迎下载学习必备欢迎下载英文原版教材班材料科学基础考试试题答案学习必备欢迎下载学习必备欢迎下载学习必备欢迎下载英文原版教材班材料科学基础考试试题学习必备 欢迎下载 7-1.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 questions according to this figure.(20 points)Fig.4(1)What 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)Please 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 strengthening)7-2.Please answer following questions according to the time-temperature-transformation(TTT)diagram as shown in Fig.5.(20 points)(1)What steel is this TTT diagram for?And what means P,B,and M in the figure?(2)Why dose the TTT diagram exhibits a C shape?(3)Point out what microconstituent will be obtained after austenite is cooled according to the curves I,II,III and IV.(4)What is microstructural difference between the curve I and the curve II?(5)How 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 martensite If you can,please draw the relative cooling curve or the flow chart of heat treatment.Fig.5 I II III IV 载学习必备欢迎下载英文原版教材班材料科学基础考试试题答案学习必备欢迎下载学习必备欢迎下载学习必备欢迎下载英文原版教材班材料科学基础考试试题试卷二姓名班级记分学习必备欢迎下载学习必备欢迎下载英文原版教材班考试试题试卷三姓名班级记分学习必备欢迎下载学习必备欢迎下载英文原版教材班材料科学基础考试试题答案学习必备欢迎下载学习必备欢迎下载学习必备欢迎下载英文原版教材班材料科学基础考试试题试卷四姓名班级记分学习必文原版教材班材料科学基础考试试题试卷五姓名班级记分学习必备欢迎下载学习必备欢迎下载英文原版教材班材料科学基础考试试题答案学习必备欢迎下载学习必备欢迎下载学习必备欢迎下载英文原版教材班材料科学基础考试试题学习必备 欢迎下载 英文原版教材班“材料科学基础”考试试题答案 Solution s of the course of“fundament of materials science”1.Glossary(2 points for each)1)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 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)A precipitate whose crystal structure and atomic arrangement have a continuous 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 a form of dispersion strengthening.9)A temperature-dependent coefficient related to the rate at which atom,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 from regions of lower concentration to that of higher concentration.2.Solution:A(-364),B(-340),C(346).3.Solution:(a)fcc;(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 mechanical properties of metallic materials?(15 points)4-2.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 points)5-1.Based on the Pb-Sn-Bi ternary diagram as shown in Fig.2,try to (1)Show the vertical section of 40wt.%Sn;(5 points)(2)Describe the solidification process of the alloy 2#with very low cooling speed(including phase and microstructure changes);(5 points)(3)Plot the isothermal section at 150oC.(5 points)载学习必备欢迎下载英文原版教材班材料科学基础考试试题答案学习必备欢迎下载学习必备欢迎下载学习必备欢迎下载英文原版教材班材料科学基础考试试题试卷二姓名班级记分学习必备欢迎下载学习必备欢迎下载英文原版教材班考试试题试卷三姓名班级记分学习必备欢迎下载学习必备欢迎下载英文原版教材班材料科学基础考试试题答案学习必备欢迎下载学习必备欢迎下载学习必备欢迎下载英文原版教材班材料科学基础考试试题试卷四姓名班级记分学习必文原版教材班材料科学基础考试试题试卷五姓名班级记分学习必备欢迎下载学习必备欢迎下载英文原版教材班材料科学基础考试试题答案学习必备欢迎下载学习必备欢迎下载学习必备欢迎下载英文原版教材班材料科学基础考试试题学习必备 欢迎下载 5-2.A 1mm sheet of FCC iron is used to contain N2 in a heated exchanger at 1200oC.The concentration of N at one surface is 0.04 atomic percent and the concentration at the second surface is 0.005 atomic percent.At 1000 oC,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 points)6-1.Supposed that a certain liquid metal is undercooled until homogeneous nucleation occurs.(15 points)(3)How to calculate the critical radius of the nucleus required?Please give the deduction process.(4)For the Metal Ni,the Freezing Temperature is 1453 C,the Latent Heat of Fusion is 2756 J/cm3,and the Solid-liquid Interfacial Energy is 255 10-7 J/cm2.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 solidification is too rapid,please describe the solidification process of 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.3 7-1.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 questions according to this figure.(20 points)Fig.2 载学习必备欢迎下载英文原版教材班材料科学基础考试试题答案学习必备欢迎下载学习必备欢迎下载学习必备欢迎下载英文原版教材班材料科学基础考试试题试卷二姓名班级记分学习必备欢迎下载学习必备欢迎下载英文原版教材班考试试题试卷三姓名班级记分学习必备欢迎下载学习必备欢迎下载英文原版教材班材料科学基础考试试题答案学习必备欢迎下载学习必备欢迎下载学习必备欢迎下载英文原版教材班材料科学基础考试试题试卷四姓名班级记分学习必文原版教材班材料科学基础考试试题试卷五姓名班级记分学习必备欢迎下载学习必备欢迎下载英文原版教材班材料科学基础考试试题答案学习必备欢迎下载学习必备欢迎下载学习必备欢迎下载英文原版教材班材料科学基础考试试题学习必备 欢迎下载 Fig.4(1)What 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)Please 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 strengthening)7-2.Please answer following questions according to the time-temperature-transformation(TTT)diagram as shown in Fig.5.(20 points)(1)What steel is this TTT diagram for?And what means P,B,and M in the figure?(2)Why dose the TTT diagram exhibits a C shape?(3)Point out what microconstituent will be obtained after austenite is cooled according to the curves I,II,III and IV.(4)What is microstructural difference between the curve I and the curve II?(5)How 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 martensite If you can,please draw the relative cooling curve or the flow chart of heat treatment.Fig.5 I II III IV 载学习必备欢迎下载英文原版教材班材料科学基础考试试题答案学习必备欢迎下载学习必备欢迎下载学习必备欢迎下载英文原版教材班材料科学基础考试试题试卷二姓名班级记分学习必备欢迎下载学习必备欢迎下载英文原版教材班考试试题试卷三姓名班级记分学习必备欢迎下载学习必备欢迎下载英文原版教材班材料科学基础考试试题答案学习必备欢迎下载学习必备欢迎下载学习必备欢迎下载英文原版教材班材料科学基础考试试题试卷四姓名班级记分学习必文原版教材班材料科学基础考试试题试卷五姓名班级记分学习必备欢迎下载学习必备欢迎下载英文原版教材班材料科学基础考试试题答案学习必备欢迎下载学习必备欢迎下载学习必备欢迎下载英文原版教材班材料科学基础考试试题学习必备 欢迎下载 英文原版教材班“材料科学基础”考试试题 试卷二 Examination problems of the course of“fundament of materials science”姓名：班级：记分：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 one another.(5 points)2.Plot the melting temperature of the elements in the 1A column of the periodic table versus atomic number(i.e.,plot melting temperatures of Li through Cs).Discuss this relationship,based on atomic bonding and binding energy.(10 points)3.Above 882,titanium has a BCC crystal structure,with a=0.332 nm.Below this temperature,titanium has a HCP structure,with a=0.2978 nm and c=0.4735 nm.Determine the percent volume change when BCC titanium transforms to HCP titanium.Is this a contraction or expansion?(10 points)4.The density of BCC iron is 7.882 g/cm3 and the lattice parameter is 0.2866 nm when hydrogen atoms are introduced at interstitial positions.Calculate(a)the atomic fraction of hydrogen atoms and(b)the number of unit cells required on average to contain one hydrogen atom.(15 points)5.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.Calculate the carbon content at 0.01 cm,0.05 cm,and 0.10 cm beneath the surface after 1 h.(15 points)载学习必备欢迎下载英文原版教材班材料科学基础考试试题答案学习必备欢迎下载学习必备欢迎下载学习必备欢迎下载英文原版教材班材料科学基础考试试题试卷二姓名班级记分学习必备欢迎下载学习必备欢迎下载英文原版教材班考试试题试卷三姓名班级记分学习必备欢迎下载学习必备欢迎下载英文原版教材班材料科学基础考试试题答案学习必备欢迎下载学习必备欢迎下载学习必备欢迎下载英文原版教材班材料科学基础考试试题试卷四姓名班级记分学习必文原版教材班材料科学基础考试试题试卷五姓名班级记分学习必备欢迎下载学习必备欢迎下载英文原版教材班材料科学基础考试试题答案学习必备欢迎下载学习必备欢迎下载学习必备欢迎下载英文原版教材班材料科学基础考试试题学习必备 欢迎下载 6.The following data were collected from a standard 0.505-in.-diameter test specimen of a copper alloy(initial length(to)=2.0 in.):Load Gage Length Stress Strain(lb)(in.)(psi)(in/in.)0 2.00000 0 0.0 3,000 2.00167 15,000 0.000835 6,000 2.00333 30,000 0.001665 7,500 2.00417 37,500 0.002085 9,000 2.0090 45,000 0.0045 10,500 2.040 52,500 0.02 12,000 2.26 60,000 0.13 12,400 2.50(max load)62,000 0.25 11,400 3.02(fracture)57,000 0.51 After fracture,the gage length is 3.014 in.and the diameter is 0.374 in.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,(g)the true stress at fracture,and(h)the modulus of resilience.(15 points)7.A 1.5-em-diameter metal bar with a 3-cm gage length is subjected to a tensile test.The following measurements are made.Change in Force(N)Gage length(cm)Diameter(cm)16,240 0.6642 1.2028 19,066 1.4754 1.0884 19,273 2.4663 0.9848 Determine the strain hardening coefficient for the metal.Is the metal most likely to be FCC,BCC,or HCP?Explain.(15 points)8.Based on Hume-Rotherys conditions,which of the following systems would be expected to display unlimited solid solubility?Explain.(15 points)(a)Au-Ag (b)Al-Cu (c)Al-Au (d)U-W (e)Mo-Ta (f)Nb-W (g)Mg-Zn (h)Mg-Cd载学习必备欢迎下载英文原版教材班材料科学基础考试试题答案学习必备欢迎下载学习必备欢迎下载学习必备欢迎下载英文原版教材班材料科学基础考试试题试卷二姓名班级记分学习必备欢迎下载学习必备欢迎下载英文原版教材班考试试题试卷三姓名班级记分学习必备欢迎下载学习必备欢迎下载英文原版教材班材料科学基础考试试题答案学习必备欢迎下载学习必备欢迎下载学习必备欢迎下载英文原版教材班材料科学基础考试试题试卷四姓名班级记分学习必文原版教材班材料科学基础考试试题试卷五姓名班级记分学习必备欢迎下载学习必备欢迎下载英文原版教材班材料科学基础考试试题答案学习必备欢迎下载学习必备欢迎下载学习必备欢迎下载英文原版教材班材料科学基础考试试题学习必备 欢迎下载 英文原版教材班“材料科学基础”考试试题答案 Solutions of the course of“fundament of materials science”1.Steels can be magnetically separated from the other materials;steel(or carbon-containing iron alloys)are ferromagnetic and will be attracted by magnets.Density differences could be usedpolymers have a density near that of water;the specific gravity of aluminum alloys is around 2.7;that of steels is between 7.5 and 8.Electrical conductivity measurements could be used polymers are insulators,aluminum has a particularly high electrical conductivity.(5 points)2.T(oC)Li 180.7 Na 97.8 K 63.2 Rb 38.9 As the atomic number increases,the melting temperature decreases,(10 points)3.We can find the volume of each unit cell.Two atoms are present in both BCC and HCP titanium unit cells,so the volumes of the unit cells can be directly compared.VBCC=(0.332 nm)3=0.03659 nm3 V HCP=(0.2978 nm)2(0.4735 nm)cos30=0.03637 nm3 V=x 100=100=-0.6%Therefore titanium contracts 0.6%during cooling.(10 points)4.(a)7.882 g/cm3=x=0.0081 H atoms/cell The total atoms per cell include 2 Fe atoms and 0.0081 H atoms.Thus:(10 points)(b)Since there is 0.0081 H/cell,then the number of cells containing H atoms is:cells=1/0.0081=123.5 or 1 H in 123.5 cells(5 points)5.D=0.23 exp-32,900/(1.987)(1253)=42 10-8 cm2/s 载学习必备欢迎下载英文原版教材班材料科学基础考试试题答案学习必备欢迎下载学习必备欢迎下载学习必备欢迎下载英文原版教材班材料科学基础考试试题试卷二姓名班级记分学习必备欢迎下载学习必备欢迎下载英文原版教材班考试试题试卷三姓名班级记分学习必备欢迎下载学习必备欢迎下载英文原版教材班材料科学基础考试试题答案学习必备欢迎下载学习必备欢迎下载学习必备欢迎下载英文原版教材班材料科学基础考试试题试卷四姓名班级记分学习必文原版教材班材料科学基础考试试题试卷五姓名班级记分学习必备欢迎下载学习必备欢迎下载英文原版教材班材料科学基础考试试题答案学习必备欢迎下载学习必备欢迎下载学习必备欢迎下载英文原版教材班材料科学基础考试试题学习必备 欢迎下载 C x=0.87%C C x=0.43%C C x=0.18%C(15 points)6.=FI(/4)(0.505)2=F/0.2 =(l-2)/2 (a)0.2%offset yield strength=45,000 psi(b)tensile strength=62,000 psi(c)E=(30,000-0)/(0.001665-0)=18 x 106 psi(d)%Elongation=(
