期末作业讲评.ppt

1、第 四 章作 业 讲 评,第九次作业,4-1： 铝的弹性模量为70GPa,泊松比为0.34，在 83MPa的静水压时，此单位晶胞的体积是多少？ 解：由E=3K(1-2) 得 K=E/3(1-2) =70Gpa/3(1-2*0.34)=72.9Gpa V/V=/K=83Mpa/72.9GPa=1.14 V=4.0496310-30(1-1.14)=66.310-30 (m3) 答：此单位晶胞的体积是66.310-30 (m3),P176 表39 点阵常数单位,4-14.有哪些途径可以提高高分子材料的刚性？ 结晶 交联 提高分子量 热处理 复合材料,7.22 Cite the primary di

2、fferences between elastic, anelastic, and plastic deformation behaviors. 概念、 原子论角度、 施加应力后的应变、 材料的差别（或对应的材料）,4-12热处理（退火）的实质是什么？它对材 料的拉伸强度、硬度、尺寸稳定性、 冲击强度和断裂伸长率有什么影响？ 结构弛豫 ，非晶至结晶转变 P112-113 拉伸强度、硬度、尺寸稳定性 冲击强度和断裂伸长率,第十次作业,4-13 有哪些方法可以改善材料的韧性，试举例 说明。 晶粒细化（晶格类型）； 成分，高分子共混橡胶，金属种杂质； 热处理； 高分子中的银纹。,4-24 按照粘附摩

3、擦的机理，说明为什么极性高 聚物与金属材料表面间的摩擦系数较大， 而非极性高聚物则较小。 摩擦系数：=S/Pm S为剪切强度；Pm为抗压强度，极性聚合 物作用力大，S大。,第十一次作业,4-29试从金属、陶瓷和高聚物材料的结构差别 解释它们在热容、热膨胀系数和热导率等 性能方面的差别。 原子键合方式不同； 结构不同，包括结晶。,热导率 金属无机非金属高分子 导热的微观机制不同 热容 高分子无机非金属金属 热运动基团单元大小不同 热膨胀系数 金属无机非金属高分子 化学键的键能不同,自由电子,晶格振动,分子传导,1mol材料升高 1所需的能量,原子在热能增加 时的平均振幅,17.6 (a) Bri

4、efly explain why Cv rises with increasing temperature at temperatures near 0 K. (b) Briefly explain why Cv becomes virtually independent of temperature at temperatures far removed from 0 K. (a) The Cv is zero at 0 K, but it rises rapidly with temperature; this corresponds to an increased ability of

5、the lattice waves to enhance their average energy with ascending temperature. (b) Above what is called the Debye temperature D, Cv levels off and becomes essentially independent of temperature at a value of approximately 3R, R being the gas constant. Thus even though the total energy of the material

6、 is increasing with temperature, the quantity of energy required to produce a one-degree temperature change is constant.,17.23 For each of the following pairs of materials, decide which has the larger thermal conductivity. Justify your choices. (a) Pure silver; sterling silver (92.5 wt%Ag 7.5 wt% Cu

7、). (b) Fused silica; polycrystalline silica. (c) Linear polyethylene (Mn=450,000g/mol); lightly branched polyethylene (Mn=650,000g/mol). (d) Atactic polypropylene (Mw=106g/mol); isotactic polypropylene (Mw = 5105g/mol). （a）纯银 合金的杂质会妨碍自由电子的运动，减少传导作用 （b）熔融SiO2 熔融SiO2有对流现象 （c）LLPE 结晶度更大 （d）isotactic PP

8、 isotactic PP 结晶,全同立构,无规,4-30 何为半分解温度？它与高分子化学键 之间有什么关系？ 半分解温度为高聚物在真空中加热30min后质量损失一半所需要的温度 K=Ae- E/RT， 键能越大，半分解温度越高 杂环或环状结构、元素高分子,4-35增加高分子材料的阻燃性一般有哪些方法？ 结构和组成 提高热稳定性 引入卤族、磷、氮等元素 添加阻燃剂和无机填料 吸收热量 降低温度 隔离氧,4-55、将2400A.m-1的磁场作用到相对磁导率 为5000的材料上，计算磁感强度和磁 化强度。,解：磁化强度： M=(r-1)H=(5000-1)2400A.m-1=1.2107A.m-1

9、 磁感强度： B= 0 (H+M)=410-7(H.m-1) (2400A.m-1 +1.2107A.m-1) =15.1Wb.m-2=15.1tesla 或B= r 0H=5000 410-7H/m 2400A/m=15.1tesla,第十三次作业,真空磁导率,18.7 The magnetization within a bar of some metal alloy is 3.2 105 A/m at an H field of 50 A/m. Compute the following: (a) the magnetic susceptibility 磁化率 (b) the perme

10、ability 磁导率 (c) the magnetic flux density within this material. 磁感强度 (d) What type(s) of magnetism would you suggest as being displayed by this material? Why?,（a）m=M/H=（3.2105A/m）/（50A/m）=6400 （b）r=m+1=6401 =r0=2.5610-3H/m (c) B=H=2.5610-3H/m(50A/m)=0.402tesla （d）铁磁体 m为103数量级,18.9 Confirm that there

11、 are 2.2 Bohr magnetons associated with each iron atom, given that the saturation magnetization is 1.70 106 A/m, that iron has a BCC crystal structure, and that the unit cell edge length is 0.2866 nm.,将number of Bohr magnetons 代入公式计算 Ms=nBB/a3 =22.29.2710-24/(0.286610-9m)3 =1.733106A/m,18.25 Figure

12、18.25 shows the B-versus-H curve for a steel alloy (a) What is the saturation flux density?(b) What is the saturation magnetization?(c) What is the remanence?(d) What is the coercivity?(e) On the basis of the data in Tables 18.5 and 18.6, would you classify this material as a soft or hard magnetic m

13、aterial? Why?,(a) The saturation flux density,(c) What is the remanence?,Bs=1.4 tesla,Ms=Bs/ 0=1.4/(410-7H/m)=1.11 106A/m,(b) Bs= 0(Hs+Ms) 0Ms,Br=0.8 tesla,(d) What is the coercivity?,(e) On the basis of the data in Tables 18.5 and 18.6, would you classify this material as a soft or hard magnetic ma

14、terial? Why?,Hc=75 A/m,4-62 已知碲化锌的Eg=2.26eV，它 对哪一部分可见光透明？,第十四次作业,=hc/Eg=0.55m h=6.6210-34J.s/1.610-19 =4.1310-15eV.s c =3108m.s-1 对波长0.550.7 m的可见光透明,19.22 Briefly explain what determines the characteristic color of (a) a metal and (b) a transparent nonmetal.,金属：吸收和再发射 透明非金属：透射、部分吸收（选择性吸 收）。,19.25 Br

15、iefly explain why amorphous polymers are transparent, while predominantly crystalline polymers appear opaque or, at best, translucent.,半晶聚合物，结晶与非晶区密度差异造成折射指数差异，当晶区尺寸大于可见光波长，造成光散射，造成不透明或半透明。,4-74 在一块单向玻纤增强不饱和聚酯单面板中，若纤 维直径是20m，纤维体积分数是0.45，Ef=4Gp， Em=80MPa。求（a）单位体积界面的面积； （b）当该复合材料沿纤维方向受到200MPa拉应力 时，玻纤和

16、基体各承受的拉应力是多少？,解（a）设该单向板为 1cm1cm1cm = 1cm3正立方体 单根纤维的体积V1 = （0.002cm / 2）21cm = 3.1410-6cm3 单根纤维的界面面积S1=0.002cm1cm = 0.00628 cm2 则该1cm3正立方体单向板复合材料界面面积： Sc = (1 cm3 0.45S1)/V1 = 900cm2,b） Ff/Fm = VfEf/(1-Vf)Em = 0.454000MPa / (0.5580MPa) = 40.90 = F/A （假定Ac=1m2) F=200 106N Fc = Ff + Fm = 40.90 Fm + Fm

17、= 200 106N Fm = 4.77 106N Ff =200-4.77=195.23 106N m=4.77 106N /0.55m2=8.68MPa f=195.23 106N /0.45m2=433MPa,4-78 一边长为25mm的立方体用薄铝板和硫化橡胶（厚 度分别为0.5mm和0.75mm）交替层压制成。问该 复合板的热导率为多少？ （a）平行板方向 （b）垂直板的方向 （Al=0.22 R=0.00012 单位Js-1m-1K-1）,解（a）model: 铝板和橡胶板分别都为25mm /（1.25mm）= 20层,VA1=10/25=0.4 VR=15/25=0.6 l =V

18、 A1 A1 +VRR =0.4*0.22+0.6*0.00012 =0.088+0.000072 =0.088 ( Js-1m-1K-1 ),(b) Solution: 因为： 1/t=V A1 /A1 +Vr/r =0.4/0.22 +0.6/0.0012=1.82+5000=5001.82 t=1/5001.82=2.0*10-4 ( Js-1m-1K-1 ),15.11 A continuous and aligned fiber-reinforced composite is to be produced consisting of 30 vol% aramid fibers and 70 vol% of a polycarbonate matrix; mechanical characteristics of these two materials are as follows: Also, the stress on the polycarbonate matrix when the aramid fibers fail i