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。
