复合型裂纹的断裂准则
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第六章 二维脆性断裂§6.1 引 言破裂判据是断裂力学的核心问题, 这需要从微观、亚微观、宏观三个层次进行研究。
所谓微观就是涉及物体的终极结构单元发生相对运动时其间内聚力的破坏。
亚微观涉及颗粒及粒间界面这一水平上的破坏。
宏观涉及肉眼可以看得见的破坏。
破裂判据是针对某一特定尺度、特定层次提出的, 做为一个完整的破裂判据,至少应该能够回答两个问题: ① 破裂在什么可测条件下起始或继续? ② 破裂向什么方向扩展? 岩石微观、亚微观破裂机制与宏观不同, 因而破裂判据也不同。
实际上, 迄今为止并不存在一种万能判据, 能够同时包括这三个层次。
为有所区分, 本文仍沿袭惯例, 对于微观、亚微观的 Griffth 裂纹, 按照Ⅰ型、Ⅱ型、Ⅲ型命名, 对于宏观断层模型化的裂纹,按照张破裂、平面内剪切裂纹、反平面剪切裂纹命名。
一些共用名词, 例如内聚力, 内聚区等, 在宏观中的含义也与微观不同。
对于岩石中的断裂机理的研究, 最早可以追溯到Griffith(1921)提出的脆性破坏理论, 该理论认为, 当裂纹端部扩展一小段长度时, 弹性势能的释放率如果大于或等于表面能的增加率时, 裂纹才能持续扩展。
在这之后, 发展了两种受压闭合裂纹模式, 即扁椭圆裂纹模式和Griffith 裂纹模式(Jaeger 和Cook, 1979)。
扁椭圆裂纹模式在第四章中已经介绍。
这里讨论的是Griffith 裂纹模式(也叫做数学裂缝), 是在Irwin(1957)引入应力强度因子的概念之后发展起来的, 它以断裂韧性作为材料抗脆断能力的指标, 也叫做K 判据。
断裂力学的其它模式和判据都是在这个模式的基础上加以修正或发展起来的, 也是断裂动力学的基础模式。
K 判据不能回答破裂方向问题, 特别是复合型裂纹问题, 因此产生了一系列脆性断裂理论。
线弹性断裂力学中关于脆性断裂的理论可分为两类:一类是应力场参数法,以应力场的某一特征量为参数。
复合材料失效准则复合材料是由两种或多种不同材料的复合结合而成的材料,具有较好的强度、刚度和重量比等优点。
然而,复合材料在使用过程中也会出现失效现象,为了保证复合材料的可靠性和安全性,需要建立相应的失效准则来预测和评估其失效情况。
复合材料的失效准则主要有以下几种:强度失效准则、疲劳失效准则和断裂失效准则。
强度失效准则是指在复合材料承受外载荷作用下,由于应力达到或超过了材料本身的强度极限而导致失效。
强度失效准则主要分为静力失效准则和动力失效准则两种。
静力失效准则是指在静态载荷作用下,当复合材料中的应力达到或超过其本身的强度极限时,会引发失效现象。
常见的静力失效准则有最大应力准则、最大应变准则和von Mises准则等。
最大应力准则是利用材料本身的强度敏感参数(如材料的最大拉伸强度、最大压缩强度等)来判断材料的失效情况,当应力超过这些强度参数时,材料发生失效。
最大应变准则是通过根据材料的最大应变来判断失效情况,当应变超过材料的最大应变极限时,材料发生失效。
von Mises准则是利用材料的等效应力与材料的强度参数进行比较,当等效应力超过强度参数时,材料发生失效。
动力失效准则是指在动态载荷作用下,由于应力的急剧变化或加载速率的变化而导致的失效。
动力失效准则主要有Johnson-Cook准则、Cockcroft-Latham准则和Chaboche准则等。
这些准则是根据材料的动态力学性能参数来判断材料的失效情况,如动态增强系数、应变硬化指数等。
这些参数是在试验或数值模拟中得到的,通过与材料的动态力学性能进行比较,可以判断材料的失效情况。
疲劳失效准则是指在复合材料中,由于长时间作用的循环载荷引起的失效现象。
疲劳失效准则主要有S-N曲线准则和e-N曲线准则等。
S-N曲线准则是指在不同应力水平下,循环载荷下发生失效的循环次数与应力的关系,通过试验得到的S-N曲线可以用来预测材料的疲劳寿命。
e-N曲线准则是指在不同应变水平下,循环载荷下发生失效的循环次数与应变的关系。
复合型裂纹准则实际的裂纹往往是张开型和滑移型(I、II)并存或张开型和撕开型(I、III)并存。
Irwin断裂准则不能简单地用于复合型裂纹问题(Irwin的K准则理论假定裂纹按原方向开裂)1)I、III型裂纹一般按原方向开裂2)II型裂纹一般不按原方向开裂3)复合型裂纹一般不按原方向开裂复合型裂纹要解决的问题1)裂纹沿什么方向开裂2)裂纹在什么条件下开裂1.最大切向应力准则(Erdogan and Sih, 1963)I、II复合型裂纹尖端应力场o - 3 cos^ . sm 幺in 为X、而2〔 2 2 )ttuwxwu 一』 sin 与 + cos i cos当回 21 2 2 )o 二上cos i[1 + sin幺in当y ?冗r2( 2 2 )K .00 30+ 开 sin — cos — cos —J2兀r222rrrrmTmnK 0 . 0 30T = i cos — sin - cos 一冲,:2 兀 r 2 2 2K + iiA ro r cos— 1 . sm 0sin30)转化为极坐标形式o = o cos 20 +a sin 20 + 2T sin0cos0。
0 =。
sin 20+o cos 20 -2T sin0 cos0T 0 = -(o -o )sin0 cos0 +T (cos 2 0 -sin 2 0)u = u cos 0 + v sin 0u 0 = - u sin 0 + v cos 01 0 1 得: o = ------ ----- K (3 — cos 0) cos + ----- ---- K 22r i2 2V2兀rii(3cos 0 - 1)sin : 1 八 0 o = ------- ----- [K (1 + cos 0) — 3K sin 0 ]cos - 0 2“ 2兀 r i ii 2 T r 0 12V 2K 7 [K sin 0+ K (3cos 0 - 1)]cosii最大切向应力准则的基本假设1)裂纹沿最大切向应力。
Chapter Five Fracture criterion for mixed mode crackIn the material mechanics, for the multiaxial stress state, four strength theories have been developed. In the fracture mechanics for the mixed mode crack problem, we need to develop the fracture theory accordingly. Many fracture theories have been developed. Two key questions must be answered.(1) What direction does a crack propagate along?(2) What is the critical case?In what follows, five theories will be introduced.§5-1 Maximum normal stress criterionMaximum stress criterion can be applied to the mixed mode crack of mode I and mode II.The asymptotic stress solution is)23cos 2cos 2(2sin 2)23sin 2sin 1(2cos 2θθ+θπ-θθ-θπ=σrK r K II I xx 23cos 2cos 2sin 2)23sin 2sin 1(2cos 2θθθπ+θθ+θπ=σr K r K II Iyy )23sin 2sin 1(2cos 223cos 2sin 2cos 2θθ-θπ+θθθπ=σrK r K II I xy By application of the coordinate transformation formulas, we can obtain the expressions of three stress components in the polar coordinates (r , θ). The circumferential normal stress is]sin 3)cos 1([2cos 2121θ-θ+θπ=σθθII I K K r The circumferential normal stress intensity factor θθK is defined as]sin 3)cos 1([2cos 212lim 0θ-θ+θ=σπ=θϑ→θθII I r K K r KHence, θθσ can be written asr K π=σθθθθ2Assumptions:(1) Crack initiation direction is the direction of the maximum θθK ;(2) When θθK reaches its critical value c K θθ, break occurs. c K θθ is a material constant.The crack initiation angle 0θ can be determined from0=θ∂∂θθK , 022<θ∂∂θθK The result is0)1cos 3(sin 00=-θ+θII I K K0)5cos 9(2sin )cos 31(2cos 0000<+θθ+θ-θII I K K The critical condition is c II I K K K K θθθ=θ-θθ=)sin 232cos (2cos0020maxDetermination of c K θθ:For mode I crack, 0≠I K , 0=II K , 00=θ, the critical condition reduces to c Ic K K K θθθ==maxNote that c K θθ is a material constant. When, 0≠I K and 0≠II K , there still prevails Ic c K K =θθ. The maximum stress criterion is expressed asIc II I K K K ≤θ-θθ)sin 232cos (2cos 0020Application to mode II crack:For a mode II crack, 0=I K and 0≠II K . The crack initiation angle can be solved,o 5.700-=θ. FromIc c II I K K K K K ==θ-θθ=θθθ)sin 232cos (2cos 0020max one can obtain thatIc IIc K K =149.1, Ic IIc K K 87.0=,The fracture criterion for Mode II crack can be derived from the maximum stress criterion thatIIc II K K ≤It is convenient for the engineering application. However, there is no difference between the plane stress and plane strain.§5-2 Maximum normal strain criterionNear the crack tip, the circumferential normal strain is]}2sin )1cos 3(2cos sin 3[2cos )]cos 3()cos 1[({2121)(111111θ-θν+θθ-θθ-ν-θ+π=σν-σ=εθθθθII I rr K K E r E E E =1, νν=1, for plane stress; 211ν-=E E , ννν-=11, for plane strain. The circumferential normal strain intensity factor *θθK is defined as]}2sin )1cos 3(2cos sin 3[2cos )]cos 3()cos 1[({212lim 1110*θ-θν+θθ-θθ-ν-θ+=επ=θθ→θθII I r K K E r K Then,r K π=εθθθθ2*Assumptions:(1) Crack initiation direction is the direction of the maximum *θθK ;(2) When *θθK reaches its critical value *c K θθ, break occurs. *c K θθ is a materialconstant.The cracking angle 0θ satisfies0*=θ∂∂θθK , 02*2<θ∂∂θθK The critical value *c K θθ can be determined by Ic K . For Mode I, 0=II K , 00=θ. It can be obtained thatIc c K E K 11*1ν-=θθ The maximum normal strain criterion isIc II I K K K ≤θ-θν+θθ-θθ-ν-θ+ν-}2sin )]1cos 3(2cos sin 3[2cos )]cos 3()cos 1[({)1(210010000101Now the plane stress and plane strain can be distinguished.§5-3 Strain energy density factor theoryStrain energy density factor theory was proposed by Prof. G . C. Sih that can be applied to the three dimensional problem.When, 0≠I K , 0≠II K , 0≠III K , the asymptotic stress solution is)23cos 2cos 2(2sin 2)23sin 2sin 1(2cos 2θθ+θπ-θθ-θπ=σrK r K II I xx 23cos 2cos 2sin 2)23sin 2sin 1(2cos 2θθθπ+θθ+θπ=σr K r K II Iyy )23sin 2sin 1(2cos 223cos 2sin 2cos 2θθ-θπ+θθθπ=σrK r K II I xy 2sin 222cos 22θπν-θπν=σr K r K II I zz 2cos 2θπ=σrK III yz , 2sin 2θπ-=σr K III zx The strain energy density w is)(21)()(21222222zx yz xy xx zz zz yy yy xx zz yy xx E E w σ+σ+σμ+σσ+σσ+σσν-σ+σ+σ= The strain energy density w can be expressed in the form ofrS w = where233222122112IIIII II I I K a K a K K a K a S +++=, strain energy density factor )cos )(cos 1(16111θ-κθ+πμ=a )1cos 2(sin 16112+κ-θθπμ=a )]1cos 3)(cos 1()cos 1)(1[(16122-θθ++θ-+κπμ=a πμ=4133a Assumptions: it is physics, not mathematics.(1) Crack initiation direction is the direction of the minimum S ;(2) When S reaches its critical value c S , break occurs. c S is a material constant.The cracking angle 0θ can be solved from0=θ∂∂S , 022>θ∂∂S The critical condition isc S S S =θ=)(0minDetermination of S c :For mode I, it can be derived that2421Ic c K S πμν-= The minimum strain energy density factor criterion can be expressed asS ≤S c , i.e.,223322212211]2[214Ic III II II I I K K a K a K K a K a ≤+++ν-πμ.Mode II crack: 0==III I K K , )321arccos(0ν-=θ, Ic IIc K K 2)1(2)21(3ν-ν-ν-= Take 31=ν. There is 7383o 0'-=θ, Ic IIc K K 9.0=Recall that for the maximum normal stress criterion, there iso 05.70-=θ, Ic IIc K K 87.0=Two results have little difference.§5-4 Modified maximum normal stress criterionSometime the maximum normal stress criterion is not so good. A modified maximum normal stress criterion has been proposed.It has been known that in view ofrS w = a strain energy density factor S is defined. For the mixed mode of mode I and II, S canbe written as222122112IIII I I K a K K a K a S ++= Let constant ==C w .)(]2[122212211θ=++===F K a K K a K a CC S w S r II II I I For different values of C , we can obtain a group of curves called as isolines of strain energy density.The circumferential normal stress is]sin 3)cos 1([2cos 2121θ-θ+θπ=σθθII I K K r Let )cos 1(2cos 21)(θ+θ=θI f , θθ-=θsin 2cos 23)(II f . )]()([21θ+θπ=σθθII II I I f K f K rLet )]()([21)(θ+θπ=θII II I I f K f K S f . This gives )(θ=σθθf rS On the isolines of the strain energy density, C S r =, the circumferential normal stress is)(θ=σθθf CThe circumferential normal stress intensity factor θθK is identical with §5-1. )()(2lim 0θ+θ=σπ=θθ→θθII II I I r f K f K r K Assumptions:(1) Crack initiation direction is the direction of the maximum θθK on the isoline of the strain energy density. The crack initiation angle 0θ can be determined from0=θ∂∂θθK , 022<θ∂∂θθK(2) When θθK reaches its critical value c K θθ, break occurs.c II II I I K f K f K K θθθθ=θ+θ=)()(00maxIt can be derived from Mode I problem thatIc c K K =θθThe fracture criterion isIc II II I I K f K f K ≤θ+θ)()(00§5-5 Energy release rate theoryNear the crack tip, the stresses in the polar coordinates are]sin 3)cos 1([2cos 2121θ-θ+θπ=σθθII I K K r )]1cos 3(sin [2cos 2121-θ+θθπ=σθII I r K K r Let]sin 3)cos 1([2cos 21θ-θ+θ=θII I I K K K )]1cos 3(sin [2cos 21-θ+θθ=θII I II K K K There resultsr K I π=σθθθ2, r K II r π=σθθ2Energy release rate θG along the angle θ:G denotes the energy release rate along the direction θ=0. Now we need to know the energy release rate θG along the direction θ.It is known that002lim =θ→σπ=yy r I r K , 002lim =θ→σπ=yx r II r K , )(8122II I K K G ++=μκRecall the definitions of θI K and θII K . It is known thatθθ→θσπ=r K r I 2lim 0, θ→θσπ=r r II r K 2lim 0Comparing two cases, we know that θI K and θII K are the stress intensity factors of the virtual crack. The stress fields for two cracks are completely same. The conclusion is that the energy release rate θG along angle θ for the real crack is equal to the energy release rate G along its own direction for the virtual crack. Hence, we have )(8122θθθ+μ+κ=II I K K GAssumptions:(1) Crack initiation direction is the direction of the maximum θG . The crack initiation angle 0θ can be determined from0=θ∂∂θG , 022<θ∂∂θG (2) When θG reaches its critical value c G θ, the break occurs.In a same way, it is obtained that281Ic c K G μ+κ=θ The cracking angle 0θ satisfies the equation0)cos 31(sin )cos cos (sin 2)cos 1(sin 00200202002=θ-θ+θ-θ-θ-θ+θII II I I K K K K The fracture criterion is2020020)]cos 35(sin 4)cos 1()[cos 1(41Ic II II I I K K K K K ≤θ-+θ-θ+θ+§5-6 Fatigue crack propagation problemFatigue process:(1) Fatigue crack initiation period: empirical formula (Miner ’s liner damage accumulation theory) or damage mechanics;(2) Fatigue crack propagation period: fracture mechanics.max σ, maximum stress; min σ, minimum stress; )(21min max σ+σ=σm , mean stress; min max σ-σ=σ∆, stress amplitude; maxmin σσ=R , cyclic stress ratio. In a fatigue process, the stress intensity factor )(t K I also varies with time t. a K K K I I I πσ∆=-=∆min maxThe fatigue crack propagation rate dN da / depends on the amplitude of SIF. )(I K f dNda ∆=Experimental result:Fracture and Damage Mechanics Chapter Five Fracture criterion for mixed mode crack 77Region I: small crack, microscopic effect is important.Region II: crack stable propagation.Region III: crack instable propagation to failure.Paris equation: 1960s, Lehigh University, USA For the region II, the relation can be given byn I K C dNda )(∆= )log(log )log(I K n C dNda ∆+=, straight line Parameters C and n can be determined by the experimental data, which depend on the stress ratio R , material property, temperature and so on.The fatigue crack growth life can be calculated by using the Paris equation. There are many improvements for Paris equation.第五章完。