下载文档原格式
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。
二 K i',=dxJ(a 2-x 2)10分一、 简答题（本大题共5小题，每小题6分，总计30分）1、 （1）数学分析法：复变函数法、积分变换；（2）近似计算法：边界配置法、 有限元法；（3）实验应力分析法：光弹性法.（4）实验标定法：柔度标定法；2、 假定：（1）裂纹初始扩展沿着周向正应力；一、为最大的方向；（2）当这个方向上的周向正应力的最大值（；=）max 达到临界时,裂纹开始扩展•S3、 应变能密度:W，其中S 为应变能密度因子，表示裂纹尖端附近应力场r密度切的强弱程度。
4、 当应力强度因子幅值小于某值时，裂纹不扩展，该值称为门槛值。
5、 表观启裂韧度，条件启裂韧度，启裂韧度。
二、 推导题（本大题10分）D-B 模型为弹性化模型，带状塑性区为广大弹性区所包围，满足积分守恒的 诸条件。
积分路径：塑性区边界。
AB 上：平行于％，有dx 2 r O’ds r d %兀》s BD 上：平行于 ％，有 dx 2 = 0 , ds = d% , T 2 - sJ(WdX 2 -T 凹 ds) T 2 竺 dX !X-IAB rBDA ；「s VB =：；S (V A ' V D )三、计算题（本大题共3小题，每小题20分，总计60分）1、利用叠加原理：微段一集中力qdx — dKi = 2q ；a 2 dx 业（a-x 2）2007断裂力学考试试题 B 卷答案T 2 土 dx ,BD 2 :x,1SvZ 二.—(sin 2b -sin ( a) 2b 二（a ））2兀a 2 -(sin 2b )31 uJ-L u,cos = 12b2b JE JEJE it二 sin ——cos 一a cos 一 sin — a2b2b2bTt .. Tt二——cos ——a sin 2b 2b■ .2'- 22二[sin (a)] = () cos a 2b2b 2b—0 时，sin 2b sin =( a)二2bn a2b 仝 2b 2b - nn IT 2cos ——a sin ——a (sin — a)b 2b 2bb.在所有 裂纹 内部 应力 为零.y =0, -a ::: x ::: a, -a _ 2b ：：： x ：：： a _ 2b 在区间内C.所有裂纹前端；「y •匚 单个裂纹时Z - —^Z —Jz 2—a 2又Z 应为2b 的周期函数二 Z 二J 兀z 2 兀a 2 、(sin —)2- (sin —)2Y 2b 2b采用新坐标：『：=z - a令 x=acosv= \ a -x = acosv, dx 二 acosrdr 匚 K “ 2q. a ：n1(a1a )咤 d 一Yu '0 a cos 日当整个表面受均布载荷时，耳-；a. K i = 2q J^s in10分2、 边界条件是周期的：a. Z 、，二y 7 一；「.兀z 二sin b10分sin A (a /a)10分当V -0时,第3页 共3页一、简答题（80分）1•断裂力学中，按裂纹受力情况，裂纹可以分为几种类型？请画出这些-： - 2 ■ ■ 2=[sin (a)] -(sin a) 2 cos asin a2b2b 2b 2b 2bZ -0 =.na二 sin 2b2“'： ：■. a 二acos ——sin ,2b 2b 2b二 sin -2b K I 二 lim 、尹Z =-=口0 Ji na 兀 a in ———cos 2b 2b 2b ■: a2b =匚二a 、，—tan —10分 3、当复杂应力状态下的形状改变能密度等于单向拉伸屈服时的形 状改变能密度，材料屈服，即：注 意 行 为 规 范2 2 2 2（匚1-匚2）（二2-匚3）（匚3-匚1）=2j对于I 型裂纹的应力公式：cr +cr J cr -cr nX丫 * xy二亠cos 邛一沐］2 2-2遵 守 考 场 纪律二3 =0（平面应力,薄板或厚板表面）r =cos 2[1 _3si n 2』]2 210分--平面应力下，I 型裂纹前端屈服区域的边界方 管导核字主领审签类型裂纹的受力示意图。
目录目录 (I)第1章绪论 (1)1.1 数学模型概述 (1)1.2 机械设计 (1)1.3 机械优化设计 (2)第二章数学模型在机械设计中的应用 (3)2.1数学模型在机械故障诊断技术中的应用 (3)2.1.1 故障诊断技术中的数学模型 (3)2.1.2 故障诊断技术中傅里叶变换的使用 (3)2.2基于数学模型的圆柱齿轮减速机的设计 (4)2.2.1 基于数学模型的圆柱齿轮减速器的设计的优势 (4)2.2.2 减速器数学模型设计研究 (5)2.3数学模型在可靠性设计中的应用 (6)2.3.1 可靠性设计的数学模型 (6)2.3.2 可靠性数学模型分析 (6)第3章总结与展望 (8)参考文献 (9)第1章绪论1.1 数学模型概述数学模型就是针对或参照某种问题的特征和数量相依关系，采用形式化语言，概括或近似地表达出来的一种数学结构。
数学模型因问题不同而异，建立数学模型也没有固定的格式和标准，甚至对同一个问题，从不同角度、不同要求出发，可以建立起不同的数学模型。
数学模型与我们学习的数学课程有一些区别，它需要熟练的数学技巧、丰富的想象力和敏锐的洞察力，需要大量阅读、思考别人的模型，尤其要自己动手，亲身体验。
建立数学模型的一般分为以下几步：确定问题系统及变量关系；确定最佳的试验方案和方法；确定合理的模型结构；确定模型中的最佳参数；检验修改模型。
首先，基于一系列基本的简化假设，把实际问题中的数学描绘明确地表述出来，也就是说，通过对实际问题的分析、归纳、简化，给出用以描述该问题的数学提法；然后采用数学的理论和方法进行求解，得出结论；最后再返回去阐释所研究的实际问题，总结一般规律。
详细来讲，就是在对目标系统分析的基础上，确定描述问题的变量及相互关系以及问题所属系统，模型大概的类型，提出有关假说。
在进行试验时，必须配置性能稳定，要严格保持试验条件稳定，精心操作，详细记录，对数据进行正确的判断、筛选和分析。
模型结构反映了实际过程的内在规律，对试验数据的拟合精度有着本质的影响。
岩石裂纹扩展-破断规律及流变特征曹平;曹日红;赵延林;张科;蒲成志;范文臣【摘要】讨论岩石断裂力学研究近年来的若干进展,主要内容包括扩展机理、断裂准则、实验加载方式与裂纹定位方法、数值计算方法在岩石断裂力学研究中的应用.基于室内实验研究单轴加载下预制裂纹间的贯通模式与多裂纹试样的破坏模式、压剪复合作用下混合裂纹间的贯通类型与破碎规律.与此同时,针对岩石亚临界裂纹扩展问题进行相关讨论并给与实例分析.结果表明:处于同一应力水平时,水岩化学作用能加速亚临界裂纹扩展;水化学腐蚀后岩石的断裂韧度均小于其在空气中的断裂韧度.此外,还对岩石流变断裂模式及考虑原生裂隙的非线性流变模型进行分析:结合岩石断裂力学与流变力学推导出压剪应力环境下裂纹流变断裂判据与理论模型;引入损伤因子和裂隙塑性体构建了能描述原生节理影响的岩体非线性蠕变模型.最后,展望岩石断裂力学未来的发展前景,并就岩石裂纹萌生与扩展的研究阐述几点认识.【期刊名称】《中国有色金属学报》【年(卷),期】2016(026)008【总页数】26页(P1737-1762)【关键词】裂纹扩展;贯通模式;亚临界扩展;流变断裂;理论模型【作者】曹平;曹日红;赵延林;张科;蒲成志;范文臣【作者单位】中南大学资源与安全工程学院,长沙410083;中南大学资源与安全工程学院,长沙410083;中南大学资源与安全工程学院,长沙410083;中南大学资源与安全工程学院,长沙410083;中南大学资源与安全工程学院,长沙410083;中南大学资源与安全工程学院,长沙410083【正文语种】中文【中图分类】TU52在经历了亿万年的地质作用后，自然岩体中广泛存在着不同尺度、不同赋存状态的原生不连续面，它们包括裂隙、节理、弱面以及断层等，这些不连续面对岩体的稳定性造成了显著的影响[1−2]。
岩体节理的变形和破坏规律研究也是深入研究重大岩石工程破坏和稳定性的基础，具有重大的工程应用背景，很多大型岩体工程中的重大地压灾害都是由于岩体中的节理扩展和相互连通诱发的。
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。
