bj0244-车床尾座体机械加工工艺与工装设计外文翻译

1、冲压变形冲压变形工艺可完成多种工序，其基本工序可分为分离工序和变形工序两大类。分离工序是使坯料的一部分与另一部分相互分离的工艺方法，主要有落料、冲孔、切边、剖切、修整等。其中有以冲孔、落料应用最广。变形工序是使坯料的一部分相对另一部分产生位移而不破裂的工艺方法，主要有拉深、弯曲、局部成形、胀形、翻边、缩径、校形、旋压等。从本质上看，冲压成形就是毛坯的变形区在外力的作用下产生相应的塑性变形，所以变形区的应力状态和变形性质是决定冲压成形性质的基本。因此，根据变形区应力状态和变形特点进行的冲压成形分类，可以把成形性质相同的成形方法概括成同一个类型并进行系统化的。绝大多数冲压成形时毛坯变形区均处于平面

2、应力状态。通常认为在板材表面上不受外力的作用，即使有外力作用，其数值也是较小的，所以可以认为垂直于板面方向的应力为零，使板材毛坯产生塑性变形的是作用于板面方向上相互垂直的两个主应力。由于板厚较小，通常都近似地认为这两个主应力在厚度方向上是均匀分布的。基于这样的分析，可以把各种形式冲压成形中的毛坯变形区的受力状态与变形特点，在平面应力的应力坐标系中(冲压应力图)与相应的两向应变坐标系中(冲压应变图)以应力与应变坐标决定的位置来表示。也就是说，冲压应力图与冲压应变图中的不同位置都代表着不同的受力情况与变形特点(1)冲压毛坯变形区受两向拉应力作用时，可以分为两种情况：即 0t=0和 0， t=0。再

3、这两种情况下，绝对值最大的应力都是拉应力。以下对这两种情况进行分析。1)当 0 且t=0 时，安全量理论可以写出如下应力与应变的关系式：(1-1) 式中 /（ - m）= /（ - m）= t/（ t - m）=k ， ， t分别是轴对称冲压成形时的径向主应变、切向主应变和厚度方向上的主应变； ， ， t分别是轴对称冲压成形时的径向主应力、切向主应力和厚度方向上的主应力； m平均应力， m=（ + + t）/3；k常数。在平面应力状态，式（11）具有如下形式：3 /（2 - ）=3 /（2 - t）=3 t/-（ t+ ）=k（12）因为 0，所以必定有 2 - 0 与 0。这个结果表明：在两

4、向1拉应力的平面应力状态时，如果绝对值最大拉应力是应变一定是正应变，即是伸长变形。 ，则在这个方向上的主又因为 0，所以必定有-（t+ ）0与t2 时， 0；当 0。在双向等拉力状态时，的变化范围是 = =0 = ，有式（12）得式（22） = 0及 0 且 t=0 时，有式（12）可知：因为 0，所以1）定有 2态，当 0 与 0。这个结果表明：对于两向拉应力的平面应力状 的绝对值最大时，则在这个方向上的应变一定的，即一定是伸长变形。又因为 0，所以必定有-（t+ ）0与t ， 0；当 0。 的变化范围是 = =0。当 = 时， = 0，也就是在双向等拉力状态下，在两个拉应力方向上产生数值相

5、同的伸长变形；在受单向拉应力状态时，当 =0 时， =- /2，也就是说，在受单向拉应力状态下其变形性质与一般的简单拉伸是完全一样的。这种变形与受力情况，处于冲压应变图中的 AOC 范围内（见图 11）；而在冲压应力图中则处于AOH 范围内（见图 12）。上述两种冲压情况，仅在最大应力的方向上不同，而两个应力的性质以及它们引起的变形都是一样的。因此，对于各向同性的均质材料，这两种变形是完全相同的。(1)冲压毛坯变形区受两向压应力的作用，这种变形也分两种情况分析，即 t=0 和 0， t=0。1）当 - 0 0 且t=0 时，有式（12）可知：因为 0，一定有2与 0。这个结果表明：在两向压应力

6、的平面应力状态时，如果2绝对值最大拉应力是缩变形。 0，则在这个方向上的主应变一定是负应变，即是压又因为 0与t0，即在板料厚度方向上的应变是正的，板料增厚。在 方向上的变形取决于 与 的数值：当 =2 时， =0；当 2 时， 0；当 0。这时的变化范围是 与 0 之间。当 = 时，是双向等压力状态时，故有 = 0；当 =0 时，是受单向压应力状态，所以 =- /2。这种变形情况处于冲压应变图中的EOG 范围内（见图 11）；而在冲压应力图中则处于 COD 范围内（见图 12）。2) 当一定有 2 0 且 t=0 时，有式（12）可知：因为 0，所以0 与 0。这个结果表明：对于两向压应力的

7、平面应力状 ，则在这个方向上的应变一定时负的，即一定是压态，如果绝对值最大是缩变形。又因为 0与t0，即在板料厚度方向上的应变是正的，即为压缩变形，板厚增大。在 方向上的变形取决于 与 的数值：当 =2 时， =0；当 2 ， 0；当 0。这时， 的数值只能在 = =0之间变化。当 = 时，是双向等压力状态，所以 = 0。这种变形与受力情况，处于冲压应变图中的GOL 范围内（见图 11）；而在冲压应力图中则处于 DOE 范围内（见图 12）。(1)冲压毛坯变形区受两个异号应力的作用，而且拉应力的绝对值大于压应力的绝对值。这种变形共有两种情况，分别作如下分析。1）当 0， | |时，由式（12）

8、可知：因为 0， | |，所以一定有 - 02及 0。这个结果表明：在异号的平面应力状态时，如果绝对值最大应力是拉应力，则在这个绝对值最大的拉应力方向上应变一定是正应变，即是伸长变形。又因为 0， | |，所以必定有 0 0， 0， | |时，由式（12）可知：用与前项相同的方法分析 0。即在异号应力作用的平面应力状态下，如果绝 ，则在这个方向上的应变是正的，是伸长变形；而在对值最大应力是拉应力压应力 方向上的应变是负的（ 0， 的变化范围只能在 =- 与 =0 的范围内。当 =- 时，0， 0， | |时，由式（12）可知：因为 0， | |，所以一定有 - 02及 0， 0，必定有 - 0

9、，即在拉应力方向上的应变是正的，是伸长变形。这时 的变化范围只能在 =- 与 =0的范围内 。当 =- 时， 0 0， 0， | |时，由式（12）可知：用与前项相同的方法分析 0， 0， 0, 0 AONGOH+伸长类 AOCAOH+伸长类双向受压 0, 0 EOGCOD压缩类 0, | |MONFOG+伸长类| | |LOMEOF压缩类异号应力 0, | |COB+伸长类| | | |DOEBOC压缩类+ + / /+- - + - - 图 13 冲压应变图7扩口变形区质量问题的表现形式变形程度过大引起变形区产生破裂现象压力作用下失稳起皱成形极限主要取决于板材的塑性，与厚度无关可用伸长率及

10、成形极限 DLF 判断主要取决于传力区的承载能力取决于抗失稳能力与板厚有关变形区板厚的变化减薄增厚提高成形极限的方法改善板材塑性使变形均匀化，降低局部变形程度工序间热处理采用多道工序成形改变传力区与变形区的力学关系采用防起皱措施化学成分组织塑 性变形条件硬化性能伸长类抗缩颈能 力应力状态变形应变梯度硬化性能变形均化与扩展能力模具状态变形区的成形极限力学性能抗起皱值与值能力相对厚度化学成分冲压成形极限压缩类塑性组织变形变形条件强度变形抗力传动区的成形极限变形力及其 变 化硬化性能抗拉与抗压缩失衡能力各向异性值图 13体系化方法举例8Categories of stamformingMany de

11、formation proses can be done by stam, the basic proses ofthe stamcan be dividedo two kinds: cutting and forming.Cutting is a shearing prost one part of the blis cut form the other .Itmainly includes bling, punching, trimming, parting and shaving, where punchingand blbling are the most widely used. F

12、orming is a prost one part of thehas some displacement form the other. It mainly includes deep drawing,bending, local forming, bulging, flanging, necking, sizing and spinning.In substance, stamforming icht the plastic deformation occurs inthe deformation zone of the stambcaused by the external force

13、. The stressse and deformation characteristic of the deformation zone are the basic factors todecide the properties of the stamforming. Based on the stress se anddeformation characteristics of the deformation zone, the forming methods can bedividedo several categories with the same forming propertie

14、s and to be studiedsystematically.The deformation zone in almost all types of stam stress se. Usually there is no force or only small force appforming isd on the blhe planesurface.When it is amedt the stress pendicular to the blsurface equal to zero,two principal stresses pendicular to each other an

15、d act on the blsurfaceproduce the plastic deformation of the material. Due to the small thickness of thebl, it is amed approximayt the two principal stresses distributeuniformly along the thickness direction. Based on thisysis, the stress se and9the deformation characteristics of the deformation zon

16、e in all kind of stamforming can be denoted by the poin the coordinates of the plane princ ipalstress(diagram of the stamstress) and the coordinates of the correspondingplane principal stains (diagram of the stamstrain). The different poshefigures of the stamstress and strainsess different stress se

17、anddeformation characteristics.(1)When the deformation zone of the stamblibjected toplanetensilestresses,it can be dividedo two cases,t is 0,t=0and 0,t=0.Inboth cases, the stress with theum absolute value is always a tensile stress.These two cases are 2)In the caserelationships betyzed respectively

18、as follows.t 0andt=0, according to n stresses and strains are:theegral theory, the/（-m）=/（-m）=t/（t -m）=kwhere, ，t are the principal strains of the radial,1.1tangential and thicknessdirections of the axial symmetrical stamforming; ，and tare the principalstresses of the radial, tangential and thicknes

19、s directions of the axial symmetricalstamforming;m is the average stress,m=（+t）/3; k is a constant.In plane stress se, Equation 1.13/（2-）=3/（2-t）=3t/-（t+）=k1.2Since 0,so 2-0 and 0.It indicatestwo axial tensile stresses, if the tensile stress with thet in plane stress se with um absolute value is ,t

20、is, the deformation belongsthe principal strainhis direction must beitive,10to tensile forming.In addition, because 0，therefore -（t+）0 and t2,0；and when 0.The range of is=0 .he equibiaxial tensile stress se = ，according to Equation 1.2,=0 andt 0 and t=0, according to Equation 1.2 , 2 0 and0,This res

21、ult showst for the plane stress se with two tensile stresses, whenthe absoluste value of is the strainhis direction must beitive,t is, itmust behe se of tensile forming.Also because0，therefore -（t+）0 and t,0;and when 0.11The range of is = =0 .When =,=0,t is, in equibiaxialtensile stress se, the tens

22、ile deformation with the same values occurs in the twotensile stress directions; when =0, =- /2,t is, in uniaxial tensile stress se,the deformation characteristichis case is the same ast of the ordinary uniaxialtensile.This kind of deformation ishe region AON of the diagram of the stamstrain (see Fi

23、g.1.1), and in the region GOH of the diagram of the stam(see Fig.1.2).stressBetn above two cases of stamdeformation, the properties ofand,and the deformation caused by them are the same, only the direction of theum stress is different. These two deformationshomogeneous material.are same for isotropi

24、c(1)When the deformation zone of stamblis subjected to twocompressive stressesand(t=0), it c0,t=0 and 0,t=0.so be dividedo two cases, which are1）When 0 and t=0, according to Equation 1.2,2-0 与 =0.Thisresult showsthe plane stress se with two compressive stresses, if the stresswith theum absolute valu

25、e is 0, the strain in this direction must behe se of compressive forming.negative,t is,Also because 0 and t0.The strainhe thicknessdirection of the blt isitive, and the thickness increases.The deformation condition in the tangential direction depends on the values12of and .When =2,=0;when 2,0;and wh

26、en 0.The range of is 0.When =,it is in equibiaxial tensile stress se,hence=0; when =0,it is in uniaxial tensile stress se, hence =-/2.Thiskind of deformation condition ishe region EOG of the diagram of the stamstrain (see Fig.1.1), andhe region COD of the diagram of the stamstress (seeFig.1.2).2）Whe

27、n 0and t=0, according to Equation 1.2,2- 0 and 0. Thisresult showsthe plane stress se with two compressive stresses, if the stresswith theum absolute value is , the straine of compressive forming.his direction must be negative,t is,Alsohe sbecause 0 and t0.The strain in thethickness direction of the

28、 blt isitive, and the thickness increases.The deformation conditionhe radial direction depends on the values ofand . When =2, =0; when 2,0; and when 0.The range of is = =0 . When = , it is in equibiaxial tensile stress se, hence =0.This kind of deformation ishe region GOL of the diagram of the stams

29、train (see Fig.1.1), and in the region DOE of the diagram of the stamstress(see Fig.1.2).(3) The deformation zone of the stamblibjected to two stresseswith opite signs, and the absolute value of the tensile stress is largernt ofthe compressive stress. There exist two cases to beyzed as follow:131)Wh

30、en 0, |, according to Equation 1.2, 2-0 and0.This result showsthe plane stress se with opite signs, if the stresswith theum absolutevalue is tensile, the strain in theum stressdirection isitive,t is,he se of tensile forming.Also because 0, |, therefore =-. When =-, then 0,0,0, |, according to Equati

31、on 1.2, bystress (seemeans of the sameysis mentioned above, 0,t is, the deformation zone ishe plane stress se with opite signs. If the stress with theum absolutevalue is tensile stress , the strainhis direction isitive,t is,he se oftensile forming. The strainse of compressive forming.he radial direc

32、tion is negative （=-. When =-, then 0, 0, 0,|, according to Equation 1.2, 2- 0 and0 and 0, therefore 2- 0. The strainhetensile stress direction isitive, orhe se of tensile forming.The range of is 0=-.When =-, then 0,0,0, |, according to Equation 1.2 and by means ofthe sameysis mentioned above,=-.Whe

33、n =-, then 0, 0, 0,0 AONGOH+TensileAOCAOH+TensileBiaxial compressive stress se 0,0 EOGCODCompressive0,|MONFOG+Tensile|LOMEOFCompressiveSeof stresswith opite signs 0,|COB+Tensile| |DOEBOCCompress iveTable 1.2 Comparison betn tensile and compressive forming20ItemTensile formingCompressive formingRepre

34、senion of thequality problem in the deformation zoneFracturehe deformationzoneduetoexsive deformationInstability wrinkle caused by compressive stressForming limitMainly depends on the plasticity of the material, and isirrelevant to the thicknessCan be estimated byextensibility or the forming limit DLFMainly depends on the loading capability in the force tr