材料科学与工程基础作业讲评-7

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。
第十次作业 中文

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


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？
a=（KIC/Yσ ）2/π 1100 1200 =1.5mm 1.76 1.5


Hale Waihona Puke Baidu
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. Use the load–elongation 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 this alloy. (e) What is the approximate ductility, in percent elongation? (f ) Compute the modulus of resilience.
400 350 300
Stress (MPa)
250 200 150 100 50 0 0 0.05 0.1 Strain 0.15 0.2
(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)
T
9.17 Some aircraft component is fabricated from an aluminum alloy that has a plane strain 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? Why or why not?
第十一次 中文
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。
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.
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690MPa，KIC值为 70MPa· 1/2，如果可容许最大裂缝是2.5mm， m 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

σmax=σ0[1+2（a/ρ）0.5]
= 2σ f（a/ρ ）0.5
7000=60[1+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1200MPa， 断裂韧性（KIC ）为90MPa· 1/2 。（a）在一钢板上有 m 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） σ = KIC/Y（π a）0.5 =1032Mpa，先断裂
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=85.3N （b）σ =F2 /S= F2 /(π d2/4) =15.2*9.8 /(π * 0.762/4)=329(MPa) ε =σ /E=329MPa/110.3Gpa=3‰ L=L0（1+Δ ε ）=212*（1+0.011-0.003）=213.7cm （3）σ = F1 /S= 85.3N / (π * 0.762/4) =187.9Mpa
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 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 * e =300mm*e0.179=300mm*1.196=358.8mm
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 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.
=(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 – 50.8)/50.8 ]*100
=16.5
(f) U r =1/2 *y*y=0.5*117.4*106*0.002J/m3 =117400 J/m3
FCC 韧性，甚至在低温。P254 跟裂纹有关。Cracks in ductile materials are said to be stable；For brittle fracture, cracks are unstable, and the fracture surface is relatively flat and perpendicular to the direction of the applied tensile load.
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 decreasing specimen size.
At a fatigue life of 1.0×107 cycles, the stress amplitude a is about 175(160)MPa. a =(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)
9.18 Suppose that a wing component on an aircraft is fabricated from an aluminum alloy that has a plane strain fracture toughness 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 /[(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,