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1、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,第十次作业 中文,4-20玻璃的理论强度超过7000MPa。一块平板玻璃在60MPa弯曲张力下破坏。我们假定裂纹尖端为氧离子尺寸（即裂纹尖端曲率半径为氧离子半径，RO2-=0.14nm），问对应这种低应力断裂，相应的裂纹深度为多大？ max=01+2（a/）0.5 = 2f（a/）0.5 7000=601+2（a/0.14）0.5 a=468nm，2a=936.5nm 括号中1省略则a=476nm,4-21某钢材的屈服强度为1100MPa,抗拉强度为1200M
2、Pa，断裂韧性（KIC）为90MPam1/2。（a）在一钢板上有2mm的边裂，在他产生屈服之前是否会先断裂？（b）在屈服发生之前，不产生断裂的可容许断裂缝的最大深度是多少？（假设几何因子Y等于1.1，试样的拉应力与边裂纹垂直） = KIC/Y（a）0.5 =1032Mpa，先断裂 a=（KIC/Y）2/ =1.5mm 1100 1.76 1200 1.5,7.29 A cylindrical specimen of aluminum having a diameter of 12.8 mm and a gauge length of 50.800 mm is pulled in tension
3、. Use the loadelongation characteristics tabulated below to complete problems a through f. (a) Plot the data as engineering stress versus engineering strain. (b) Compute the modulus of elasticity. (c) Determine the yield strength at a strain offset of 0.002. (d) Determine the tensile strength of thi
4、s alloy. (e) What is the approximate ductility, in percent elongation? (f ) Compute the modulus of resilience.,(a) A0=d02/4 =3.14*12.82 mm2/4 = 128.6mm2 =F/A0 = l /l0 , l0 =50.8mm (b) E=slope=/=(2-1)/(2-1) =(57.0-0)MPa/(0.001-0)=57.0GPa 117.4MPa (d) 369.4MPa (e) %EL=(l f - l 0)/ l 0 *100=(59.182 50.
5、8)/50.8 *100 =16.5 (f) U r =1/2 *y*y=0.5*117.4*106*0.002J/m3 =117400 J/m3,7.40 For some metal alloy, a true stress of 415 MPa produces a plastic true strain of 0.475. How much will a specimen of this material elongate when a true stress of 325 MPa is applied if the original length is 300 mm ? Assume
6、 a value of 0.25 for the strain-hardening exponent n. T =K Tn , K= T / Tn =415MPa/(0.4750.25)=500MPa T =( T/K)1/n=(325MPa/500MPa) 1/0.25=0.179 T =ln(l i / l 0), l i = l 0 * eT =300mm*e0.179=300mm*1.196=358.8mm,9.17 Some aircraft component is fabricated from an aluminum alloy that has a plane strain
7、fracture toughness of 35 MPa.m1/2. It has been determined that fracture results at a stress of 250 MPa when the maximum (or critical) internal crack length is 2.0 mm. For this same component and alloy, will fracture occur at a stress level of 325 Mpa when the maximum internal crack length is 1.0 mm?
8、 Why or why not? KIC=Y(a)1/2, Y= KIC /(a)1/2=(35 Mpa m 1/2 ) / (250 Mpa) (3.14*0.002/2m)1/2=2.50 另外受力： = KIC /(a)1/2 Y=(35 Mpa m 1/2 )/ (3.14*0.001/2m)1/2 *2.50=353.3MPa, 因此不断裂。,9.18 Suppose that a wing component on an aircraft is fabricated from an aluminum alloy that has a plane strain fracture to
9、ughness of 40 Mpa m1/2 . It has been determined that fracture results at a stress of 365 MPa when the maximum internal crack length is 2.5 mm. For this same component and alloy, compute the stress level at which fracture will occur for a critical internal crack length of 4.0 mm .,KIC=Y(a)1/2, Y= KIC
10、 /(a)1/2=(40 Mpa m 1/2 ) / (365Mpa)(3.14*0.0025/2m)1/2=1.75 另外受力： = KIC /(a)1/2 Y=(40 Mpa m 1/2 )/ (3.14*0.004/2m)1/2 1.75=288.4MPa,9.32 A 12.5mm diameter cylindrical rod fabricated from a 2014-T6 alloy (Figure 9.46) is subjected to a repeated tension-compression load cycling along its axis. Compute
11、 the maximum and minimum loads that will be applied to yield a fatigue life of 1.0107cycles. Assume that the stress plotted on the vertical axis is stress amplitude, and data were taken for a mean stress of 50 MPa.,At a fatigue life of 1.0107 cycles, the stress amplitude a is about 175(160)MPa. a =(
12、max - min )/2, 英文书P257 m =(max + min )/2, 英文书P258 So max = m +a , min =m - a , F max /A= m +a F max =(m +a )A=27598N (25758) F min =(m -a )A=-15332N (-13492),思考题 4-7 一条长212cm的铜线，直径为0.76mm。当外加载荷为8.7kg时开始产生塑性变形(a)此作用力是多少牛顿？(b)外加载荷为15.2kg时，此线的应变为0.011，则去除载荷后，铜线的长度为多少？ 此铜线的屈服强度是多少？ 解：（a）F1=mg=8.7*9.8=85
13、.3N （b）=F2 /S= F2 /(d2/4) =15.2*9.8 /(* 0.762/4)=329(MPa) =/E=329MPa/110.3Gpa=3 L=L0（1+）=212*（1+0.011-0.003）=213.7cm （3）= F1 /S= 85.3N / (* 0.762/4) =187.9Mpa,4-9 从拉伸试验如何获得常用的力学性能数据？ 拉伸强度、屈服强度、断裂强度、断裂伸长率、弹性模量等，公式可计算。,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。,4-19某钢板的屈服强度为69
14、0MPa，KIC值为70MPam1/2，如果可容许最大裂缝是2.5mm，且不许发生塑性变形，则此钢的设计极性强度是多少？ KIC=c (a)1/2, c = KIC (a)-1/2 =70Mpa.m1/2 (3.14*1.25mm)-1/2 =1117MPa,9.6 Briefly explain (a) why there may be significant scatter in the fracture strength for some given ceramic material, and (b) why fracture strength increases with decrea
15、sing specimen size. 陶瓷材料的结构，晶相、玻璃相和气相;裂纹等缺陷。,9.28 Briefly explain why BCC and HCP metal alloys may experience a ductile-to-brittle transition with decreasing temperature, whereas FCC alloys do not experience such a transition. FCC 韧性，甚至在低温。P254 跟裂纹有关。Cracks in ductile materials are said to be stable
16、；For brittle fracture, cracks are unstable, and the fracture surface is relatively flat and perpendicular to the direction of the applied tensile load.,4-24 按照粘附摩擦的机理，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 中文,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
