疲劳和断裂第七讲.ppt

单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,*,第七章 弹塑性断裂力学简介,7.1,裂纹尖端的小范围屈服,7.2,裂纹尖端张开位移,7.3 COD,测试与弹塑性断裂控制设计,返回主目录,1,用线弹性材料物理模型，按照弹性力学方法，研究含裂纹弹性体内的应力分布，给出描述裂纹尖端应力场强弱的应力强度因子,K,，,并由此建立裂纹扩展的临界条件,处理工程问题。,第七章 弹塑性断裂力学简介,线弹性断裂力学,(LEFM),线弹性断裂力学给出的裂纹尖端附近的应力趋于 无穷大。然而，事实上任何实际工程材料，都不可能承受无穷大的应力作用。因此，裂尖附近的材料必然要进入塑性，发生屈服。,2,Linear elastic fracture mechanics predicts,infinite stresses at the crack tip.In real,materials,however,stress at the crack tip are,finite because the crack tip radius must be,finite.Inelastic material deformation,such as,plasticity in metal,leads to further relaxation,of the crack tip stress.,线弹性断裂力学预测裂纹尖端应力无穷大。然而,在实际材料中，由于裂尖半径必定为有限值，故,裂尖应力也是有限的。非弹性的材料变形，如金,属的塑性，将使裂尖应力进一步松弛。,3,7.1,裂纹尖端的小范围屈服,1.,裂尖屈服区,当,r,0,时，,s,，,必然要发生屈服。,因此，有必要了解裂尖的屈服及其对,K,的影响。,无限大板中裂纹尖端附近任一点,(r,),处的正应力,x,、,y,和剪应力,xy,的线弹性解为：,s,s,x,y,2,a,dx,dy,r,q,s,y,s,x,t,x,y,s,s,q,y,a,r,=,+,2,2,1,cos,q,q,2,3,2,sin,sin,t,s,q,q,q,xy,a,r,=,2,2,2,3,2,sin,cos,cos,s,s,q,x,a,r,=,-,2,2,1,cos,q,q,2,3,2,sin,sin,(5-1),4,这里仅简单讨论沿裂纹线上屈服区域的大小。,线弹性断裂力学,裂尖附近任一点处的,x,、,y,xy,，,一点的应力状态,计算主应力,屈服准则,裂纹尖端屈服区域的形状与尺寸,s,s,q,y,a,r,=,+,2,2,1,cos,q,q,2,3,2,sin,sin,t,s,q,q,q,xy,a,r,=,2,2,2,3,2,sin,cos,cos,s,s,q,x,a,r,=,-,2,2,1,cos,q,q,2,3,2,sin,sin,(5-1)在裂纹线上,(,=0),，注意到 ，有；,a,K,p,s,=,r,K,r,a,y,x,p,s,s,s,2,2,1,=,=,=,0,=,xy,t,；,s,s,x,y,2,a,dx,dy,r,q,s,y,s,x,t,x,y,5,对于平面问题，,还,有：,yz,=,zx,=0,；,z,=0,平面应力,z,=,(,x,+,y,),平面应变,r,K,r,a,y,x,p,s,s,s,2,2,1,=,=,=,0,=,xy,t,；,则裂纹线上任一点的主应力为：,=,r,K,p,n,s,2,/,2,0,1,3,平面应力,平面应变,s,s,r,K,p,2,1,2,1,=,=,；,塑性力学中，,von Mises屈服条件为：,2,1,3,2,3,2,2,2,1,),(,),(,),(,s,s,s,s,s,2,2,ys,s,=,-,+,-,+,-,s,6,将各主应力代入,Mises,屈服条件，得到：,(,平面应力,),(,平面应变,),ys,p,r,K,s,p,=,2,/,1,ys,p,r,s,p,=,2,K,n,-,/,),2,1,(,1,式中，,ys,为材料的屈服应力，,为泊松比。,对于金属材料，,0.3,，这表明平面应变情况下裂尖塑性区比平面应力时小得多。,故塑性屈服区尺寸,r,p,为：,(,平面应力,),2,1,),(,2,1,ys,p,K,r,s,p,=,2,2,1,),2,1,(,),(,2,1,n,s,p,-,=,ys,p,K,r,(,平面应变,),(7-3),7,虚线,为弹性解，,r,0,，,y,。,由于,y,ys,，,裂尖处材料屈服，,塑性区尺寸为,r,p,。,当,=0,时,(,在,x,轴上,),，裂纹附近区域的应力分布及裂纹线上的塑性区尺寸,如图。,r,p,a,x,y,ys,A,B,D,o,H,K,与原线弹性解,(,虚线,HK),相比较，少了,HB,部分大于,ys,的应力。,假定材料为弹性,-,理想塑性，屈服区内应力恒为,ys,，,应力分布应由实线,AB,与虚线,BK,表示。,8,The simple analysis as above is,not strictly correct because it was,based on an elastic crack tip,solution.When yielding occurs,stress must redistribute in order,to satisfy equilibrium.,r,p,a,x,y,ys,A,B,D,o,H,K,上述简单分析是以裂纹尖端弹性解为基础的，故,并非严格正确的。屈服发生后，应力必需重分布，,以满足平衡条件。,The region ABH represents forces that would be,present in an elastic material but cannot be carried,in the elastic-plastic material because the stress,cannot exceed yield.The plastic zone must increase,in size in order to carry these forces.,ABH,区域表示弹性材料中存在,的力，但因为应力不能超过屈,服，在弹塑性材料中却不能承,受。为了承受这些力，塑性区,尺寸必需增大。,9,为满足静力平衡条件，由于,AB,部分材料屈服而少承担的应力需转移到附近的弹性材料部分，其结果将使更多材料进入屈服。因此，塑性区尺寸需要修正。设修正后的屈服区尺寸为,R,；,假定线弹性解答在屈服区外仍然适用，,BK,平移至,CD,，,为满足静力平衡条件，修正后,ABCD,曲线下的面积应与线弹性解,HBK,曲线下的面积相等。,由于曲线,CD,与,BK,下的面积是相等的，故只须,AC,下的面积等于曲线,HB,下的面积即可。,r,p,a,x,y,ys,A,B,o,H,K,R,C,D,10,于是得到：,积分后得到，平面应力情况下裂尖的塑性区尺寸,R,为：,ys,K,R,p,r,2,),(,1,2,1,=,=,s,p,注意到式中：,y,=,平面应力时：,K,r,p,2,/,1,2,1,),(,2,1,ys,p,K,r,s,p,=,=,p,ry,ys,dx,x,R,0,),(,s,s,r,p,R,a,x,y,ys,A,B,C,D,o,H,K,11,依据上述分析，并考虑到平面应变时三轴应力作用的影响，,Irwin,给出的塑性区尺寸,R,为：,上式指出：,裂纹尖端的塑性区尺寸,R,与,(K,1,/,ys,),成正比；,平面应变时的裂尖塑性区尺寸约为平面应力情况的,1/3,。,2,1,),(,1,2,ys,p,K,r,R,s,ap,=,=,=,2,2,1,a,(,平面应力,),(,平面应变,),(7-4),12,Most of the classical solution in fracture,mechanics reduce the problem to two dimensions.,That is at least one of the principal stresses or,strains is assumed to equal zero(plane stress and,plane strain respectively).,断裂力学中的大部分经典解都将问题减化为,二维的。即主应力或主应变中至少有一个被假设,为零，分别为平面应力或平面应变。,In general,the conditions ahead of a crack are,neither plane stress nor plane strainbut are,three-dimensional.There are,however,limiting,cases where a two dimensional assumption is valid,or at least provides a good approximation.,一般地说，裂纹前的条件既不是平面,应力，也不是平面应变，而是三维的。然,而，在极限情况下，二维假设是正确的，,或者至少提供了一个很好的近似。,13,2.,考虑裂尖屈服后的应力强度因子,曲线,CD,与线弹性解,BK,相同。,假想裂纹尺寸由,a,增大到,a,+r,p,则裂纹尖端的线弹性解恰好就是曲线,CD,。r,p,r,p,a,x,y,ys,A,B,C,D,o,H,K,r,r,o,对于理想塑性材料，考虑裂纹尖端的屈服后，裂尖附近的应力分布应为图中,ACD,曲线。,a,+r,p,称为有效裂纹长度，用,a,+r,p,代替,a,，,由原来的,线弹性断裂力学结果可直接给出考虑,Irwin,塑性修,正的解答。即有：,),(,1,p,r,a,K,+,=,p,s,(7-5),14,r,p,r,p,a,x,y,ys,A,B,C,D,o,H,K,r,r,o,考虑,Irwin,塑性修正后，裂尖应力强度因子,K,为：,),(,1,p,r,a,K,+,=,p,s,(7-5),裂纹线上,(,=0),的应力,y,为：,ysy,s,s,=,2,1,y,r,K,=,p,s,r,2r,p,；,r,2r,p,；,),(,2,1,p,r,r,K,-,=,p,15,例,7.1,无限宽中心裂纹板，受远场拉应力,作用，,试讨论塑性修正对应力强度因子的影响。,解,：由线弹性断裂力学给出无限宽中心裂纹板的,应力强度因子为：,K,a,p,s,=,1,考虑塑性修正时，由,(7-5),式有：,),(,1,p,r,a,K,+,=,p,s,将,(7-4),式给出的,r,p,代入上式，得到：,2,/,1,2,),(,2,1,ys,a,a,s,p,s,ap,p,s,+,=,1,K,2,/,1,2,),(,2,1,1,ys,a,s,s,a,p,s,+,=,或写为：,2,/,1,2,),(,2,1,1,ys,s,s,a,l,+,=,1,K,l,=,1,K,；,16,对于平面应力情况，,=1,；若,(,/,ys,)=0.2,，,=1%,；,若,(,/,ys,)=0.5,，,=6%,；,当,(,/,ys,)=0.8,时，,达,15%,。,对于平面应变情况，,3,，二者相差要小一些。,考虑塑性修正后有：,2,/,1,2,),(,2,1,1,ys,s,s,a,l,+,=,1,K,l,=,1,K,；,1,，故考虑塑性修正后应力强度因子增大。,二者的相对误差为：,=,=,1,-,l,1,1,1,-,K,K,K,可见，,(,/,ys,),越大，裂尖塑性区尺寸越大，,线弹性分析给出的应力强度因子误差越大。,17,3.,小范围屈服时表面裂纹的,K,修正,前表面修正系数通常取为,M,f,=1.1,；,E(k),是第二类完全椭圆积分。,考虑裂尖屈服，按,Irwin,塑性修正，,用,a,+r,p,代替原裂纹尺寸,a,，,故有：,),(,),(,1,.,1,1,k,E,r,a,K,p,+,=,p,s,无限大体中半椭圆表面裂纹最深处处于平面应变状,态，故由,(7-4),式知：,2,1,),(,2,4,1,ys,p,K,r,s,p,=,无限大体中半椭圆表面裂纹最深处的应力强度因子为：,),(,1,k,E,a,M,K,f,p,s,=,18,可见，小范围屈服时，表面裂纹的,K,计算只须用,形状参数,Q,代替第二类完全椭圆积分,E(k),即可。,Q,a,K,p,s,1,.,1,1,=,2,/,1,2,2,),(,212,.,0,),(,ys,k,E,Q,s,s,-,=,(7-8),代入整理后即得：,形状参数,利用,E(k),的近似表达，,Q,可写为：,2,/,1,2,64,.,1,),/,(,212,.,0,),/,(,47,.,1,1,ys,c,a,Q,s,s,-,+,=,越大，,Q,越小，,K,越大，裂尖屈服区越大。,/,ys,s,s,利用,E(k,),式的近似表达，可将形状参数,Q,写为：,19,例,7.2,某大尺寸厚板含一表面裂纹，受远场拉应力,作用。材料的屈服应力为,ys,=600MPa,断裂韧,性,K,1c,=50MPam,1/2,，试估计：,1),=500MPa,时的临界裂纹深,a,c,。,(,设,a,/c=0.5),2),a,/c=0.1,，,a,=5mm,时的临界断裂应力,c,；,解,：,1),无限大体中半椭圆表面裂纹最深处的,K,最大，,考虑小范围屈服，,在发生断裂的临界状态有：,c,c,K,Q,a,K,1,1,1,.,1,=,=,p,s,p,s,2,2,1,2,21,.,1,c,c,K,Q,a,=,32,.,1,),600,/,500,(,212,.,0,),5,.,0,(,47,.,1,1,2,64,.,1,2Q,；,20,故得到：,mm,m,K,Q,a,c,c,47,.,3,00347,.,0,14,.,3,500,21,.,1,50,32,.,1,21,.,1,2,2,2,2,1,2,=,=,=,=,p,s,2),断裂临界状态有：,c,c,K,Q,a,K,1,1,1,.,1,=,=,p,s,Q,是,c,的函数：,),600,/,(,212,.,0,),1,.,0,(,47,.,1,1,2,64,.,1,2,-,+,=,Q,s,c,=,),600,/,(,212,.,0,2,-,s,c,1.034,将断裂判据式二边平方，再将,Q,2,代入，得：,c,c,K,a,1,21,.,1,p,s,2,2,s,),/,(,212,.,0,2,-,s,c,1.034,ys,21,12622.8,),/,(,212,.,0,21,.,1,034,.,1,2,1,2,1,2,=,+,=,s,p,s,ys,c,c,c,K,a,K,=,s,112.4,MPa,c,即有：,),(,1,k,E,a,M,K,f,p,s,=,讨论：若不考虑屈服，有：,c,13600,21,.,1,034,.,1,2,1,2,=,=,p,s,c,a,K,=,s,116.6,MPa,c,则：,c,=,21,.,1,034,.,1,250,p,0.005,不考虑屈服，将给出偏危险的预测。,22,一般地说，只要裂尖塑性区尺寸,r,p,与裂纹尺寸,a,相比,是很小的,(,a,/r,p,=20-50),，,即可认为满足小范围屈服条,件，线弹性断裂力学就可以得到有效的应用。,对于一些高强度材料；,对于处于平面应变状态,(,厚度大,),的构件；,对于断裂时的应力远小于屈服应力的情况；,小范围屈服条件通常是满足的。,23,Plasticity correcting can extend LEFM beyond,its normal validity limits.One must remember,however,that Irwin correction are only rough,approximate of elastic-plastic behavior.When,nonlinear material behavior becomes significant,one should discard stress intensity and adopt a,crack tip parameter(such as the crack tip,opening displacement,CTOD)that takes the,material behavior into account.,塑性修正可将,LEFM,延用至超过其原正确性限制。,但必需记住,Irwin,修正只是弹塑性行为的粗略近似。,当非线性材料行为为主时，应抛弃应力强度因子,而采用如,CTOD,的裂尖参数考虑材料的行为。,24,When Wells attempted to measure K,1c,value in a,number of structural steels,he found that these,materials were too tough to be characterized by,LEFM.This discovery brought both good news and,bad news:high toughness is obviously desirable to,designers and fabricators,but Wells experiments,indicated that existing fracture mechanics theory,was not applicable to an important class of materials.,Wells,试图测量结构钢材的,K,1c,时，发现这些材料韧,性太大而不能用,LEFM,描述。这一发现带来的既有,好消息也有坏消息：高韧性显然是设计及制造者所,希望的，但,Wells,的试验指出现有的断裂力学理论不,能用于这类重要的材料。,25,While examining fractured test specimens,Wells,notice that the crack faces had moved apart prior to,fracture;plastic deformation blunted an initially,sharp crack.The degree of crack blunting increased,in proportion to the toughness of material.This,observation led Wells to propose the opening at the,crack tip as a measure of fracture toughness.Today,this parameter is known as the crack tip opening,displacement.,检查已断的试件，,Wells,注意到断裂前裂纹面已分开；塑性,变形使原尖锐的裂纹钝化。钝化的程度随材料的韧性而增,加。这一观察使,Wells,提出,用裂尖的张开作为断裂韧性的度,量。此参数即现在的裂纹尖端张开位移。,26,7.2,裂纹尖端张开位移,(CTOD-Crack Tip Opening Displacement),2,a,W,s,s,屈服区,则塑性区将扩展至整个截面，造成全面屈服，,小范围屈服将不再适用。,如果作用应力,大到使裂纹所在截面上的净截面应力,净,=,W/(W-2,a,),ys,中低强度材料,ys,低,K,1c,高,断裂,c,大,裂尖,r,p,大,27,2,a,COD,x,y,o,显然，,COD,是坐标,x,的函,数，且裂纹尺寸,a,越大，,COD,越大。,裂尖张开位移,(CTOD),是,在,x=a,处的裂纹张开位移。,裂尖端屈服范围大,CTOD,LEFM,Irwen,修正不再适用,断裂与裂纹张开尺寸相关,裂纹张开位移,(COD),c,大，,r,p,大，,裂纹越来,越张开。,可用于,建立适于大范围屈服的弹塑性断裂判据。,28,Dugdale,设想有一虚拟裂,纹长,a,eff,=,a,+r,p,在虚拟裂纹,上、下裂纹面上加上,=,ys,的应力作用而使裂纹闭合，,然后进行准弹性分析。,平面应力条件下，在全面屈服之前,净,/,ys,1,，,Dugdale,给出裂尖张开位移,与,间的关系为：,(7-10),),2,lnsec(,8,ys,ys,E,a,s,ps,p,s,d,=,2,a,COD,x,y,o,2,a,eff,=2,a,+2r,p,CTOD,ys,29,如果,/,ys,1,，,则可将上式中,sec,项展开后略去高次项，得到：,1,2,2,2,8,1,ln,ys,s,s,p,),2,lnsec(,ys,s,ps,-,=,-,Dugdale,解：,(7-10),),2,lnsec(,8,ys,ys,E,a,s,ps,p,s,d,=,2,2,2,2,2,2,8,),8,(,1,ln,ys,ys,s,s,p,s,s,p,=,+,=,),2,lnsec(,ys,s,ps,得到：,注意到当,x,t,，,可忽略筒体曲率的影响。,视为无限大中心裂纹板，且为平面应力,.,),2,lnsec(,8,ys,ys,E,a,s,ps,p,s,d,=,),1200,800,2,lnsec(,10,200,14,.,3,1200,8,3,=,p,a,=0.0106,a,由,(7-10),式有：,s,s,在临界状态下有：,=0.0106,a,c,c,得到：,a,c,0.05/0.0106=4.71mm,故可以容许的缺陷总长度为,2,a,=9.42mm,。,48,讨论,：,假设按小范围屈服计算，由,(7-11),式有：,或写为,E,K,ys,s,d,2,1,=,E,a,ys,s,p,s,d,2,=,可容许的缺陷总长度为,2,a,=11.94mm,。,故,当,/,ys,较大时，小范围屈服假设将引入较大的,误差，且结果偏危险。,对于本题则断裂判据写为：,c,ys,c,E,a,d,s,p,s,d,=,2,即,:,14,.,3,800,800,10,200,1200,05,.,0,3,2,=,p,s,s,d,E,a,ys,c,c,=5.97mm,49,1),线弹性断裂力学给出裂尖应力趋于无穷大，,故裂尖附近的材料必然要发生屈服。,小 结：,2)Irwin,给出的塑性区尺寸,R,为：,2,1,),(,1,2,ys,p,K,r,R,s,ap,=,=,=,2,2,1,a,(,平面应力,),(,平面应变,),(7-4),平面应变情况下塑性区尺寸约为平面应力的,1/3,。,3),小范围屈服时，,Irwin,给出修正后的,K,为：,1,K,),(,p,r,a,+,=,p,s,用 代替,a,),(,p,r,a,+,50,无限大体中半椭圆表面裂纹最深处，有：,Q,a,K,p,s,1,.,1,1,=,用形状参数,Q,代替椭圆积分,E(k),4)Dugdale,给出裂纹尖端张开位移,(CTOD),与作用,应力,间的关系为：,),2,lnsec(,8,ys,ys,E,a,s,ps,p,s,d,=,(7-10),小范围屈服时有：,(7-12),=1,，平面应力；,=(1-,2,)/2,，平面应变。,E,K,ys,s,b,d,2,1,=,51,6),裂纹尖端张开位移,可以通过实验测定。,7),临界,CTOD,值,(,c,),可作为弹塑性断裂材料参数。,以,CTOD,为控制参量的弹塑性断裂判据为：,c,d,d,8),以裂纹尖端张开位移为基础，已经发展了一些,用于弹塑性断裂控制和缺陷评估的方法。,如中国“压力容器缺陷评定规范”中的,CVDA,安全,设计曲线、英国方法、日本规范等等。,弹塑性断裂问题复杂，仍在进一步研究。,52,