bj0244-車床尾座體機(jī)械加工工藝與工裝設(shè)計外文翻譯

1、沖壓變形沖壓變形工藝可完成多種工序，其基本工序可分為分離工序和變形工序兩大類。分離工序是使坯料的一部分與另一部分相互分離的工藝方法，主要有落料、沖孔、切邊、剖切、修整等。其中有以沖孔、落料應(yīng)用最廣。變形工序是使坯料的一部分相對另一部分產(chǎn)生位移而不破裂的工藝方法，主要有拉深、彎曲、局部成形、脹形、翻邊、縮徑、校形、旋壓等。從本質(zhì)上看，沖壓成形就是毛坯的變形區(qū)在外力的作用下產(chǎn)生相應(yīng)的塑性變形，所以變形區(qū)的應(yīng)力狀態(tài)和變形性質(zhì)是決定沖壓成形性質(zhì)的基本。因此，根據(jù)變形區(qū)應(yīng)力狀態(tài)和變形特點進(jìn)行的沖壓成形分類，可以把成形性質(zhì)相同的成形方法概括成同一個類型并進(jìn)行系統(tǒng)化的。絕大多數(shù)沖壓成形時毛坯變形區(qū)均處于平面

2、應(yīng)力狀態(tài)。通常認(rèn)為在板材表面上不受外力的作用，即使有外力作用，其數(shù)值也是較小的，所以可以認(rèn)為垂直于板面方向的應(yīng)力為零，使板材毛坯產(chǎn)生塑性變形的是作用于板面方向上相互垂直的兩個主應(yīng)力。由于板厚較小，通常都近似地認(rèn)為這兩個主應(yīng)力在厚度方向上是均勻分布的?；谶@樣的分析，可以把各種形式?jīng)_壓成形中的毛坯變形區(qū)的受力狀態(tài)與變形特點，在平面應(yīng)力的應(yīng)力坐標(biāo)系中(沖壓應(yīng)力圖)與相應(yīng)的兩向應(yīng)變坐標(biāo)系中(沖壓應(yīng)變圖)以應(yīng)力與應(yīng)變坐標(biāo)決定的位置來表示。也就是說，沖壓應(yīng)力圖與沖壓應(yīng)變圖中的不同位置都代表著不同的受力情況與變形特點(1)沖壓毛坯變形區(qū)受兩向拉應(yīng)力作用時，可以分為兩種情況：即 0t=0和 0， t=0。再

3、這兩種情況下，絕對值最大的應(yīng)力都是拉應(yīng)力。以下對這兩種情況進(jìn)行分析。1)當(dāng) 0 且t=0 時，安全量理論可以寫出如下應(yīng)力與應(yīng)變的關(guān)系式：(1-1) 式中 /（ - m）= /（ - m）= t/（ t - m）=k ， ， t分別是軸對稱沖壓成形時的徑向主應(yīng)變、切向主應(yīng)變和厚度方向上的主應(yīng)變； ， ， t分別是軸對稱沖壓成形時的徑向主應(yīng)力、切向主應(yīng)力和厚度方向上的主應(yīng)力； m平均應(yīng)力， m=（ + + t）/3；k常數(shù)。在平面應(yīng)力狀態(tài)，式（11）具有如下形式：3 /（2 - ）=3 /（2 - t）=3 t/-（ t+ ）=k（12）因為 0，所以必定有 2 - 0 與 0。這個結(jié)果表明：在兩

4、向1拉應(yīng)力的平面應(yīng)力狀態(tài)時，如果絕對值最大拉應(yīng)力是應(yīng)變一定是正應(yīng)變，即是伸長變形。 ，則在這個方向上的主又因為 0，所以必定有-（t+ ）0與t2 時， 0；當(dāng) 0。在雙向等拉力狀態(tài)時，的變化范圍是 = =0 = ，有式（12）得式（22） = 0及 0 且 t=0 時，有式（12）可知：因為 0，所以1）定有 2態(tài)，當(dāng) 0 與 0。這個結(jié)果表明：對于兩向拉應(yīng)力的平面應(yīng)力狀 的絕對值最大時，則在這個方向上的應(yīng)變一定的，即一定是伸長變形。又因為 0，所以必定有-（t+ ）0與t ， 0；當(dāng) 0。 的變化范圍是 = =0。當(dāng) = 時， = 0，也就是在雙向等拉力狀態(tài)下，在兩個拉應(yīng)力方向上產(chǎn)生數(shù)值相

5、同的伸長變形；在受單向拉應(yīng)力狀態(tài)時，當(dāng) =0 時， =- /2，也就是說，在受單向拉應(yīng)力狀態(tài)下其變形性質(zhì)與一般的簡單拉伸是完全一樣的。這種變形與受力情況，處于沖壓應(yīng)變圖中的 AOC 范圍內(nèi)（見圖 11）；而在沖壓應(yīng)力圖中則處于AOH 范圍內(nèi)（見圖 12）。上述兩種沖壓情況，僅在最大應(yīng)力的方向上不同，而兩個應(yīng)力的性質(zhì)以及它們引起的變形都是一樣的。因此，對于各向同性的均質(zhì)材料，這兩種變形是完全相同的。(1)沖壓毛坯變形區(qū)受兩向壓應(yīng)力的作用，這種變形也分兩種情況分析，即 t=0 和 0， t=0。1）當(dāng) - 0 0 且t=0 時，有式（12）可知：因為 0，一定有2與 0。這個結(jié)果表明：在兩向壓應(yīng)力

6、的平面應(yīng)力狀態(tài)時，如果2絕對值最大拉應(yīng)力是縮變形。 0，則在這個方向上的主應(yīng)變一定是負(fù)應(yīng)變，即是壓又因為 0與t0，即在板料厚度方向上的應(yīng)變是正的，板料增厚。在 方向上的變形取決于 與 的數(shù)值：當(dāng) =2 時， =0；當(dāng) 2 時， 0；當(dāng) 0。這時的變化范圍是 與 0 之間。當(dāng) = 時，是雙向等壓力狀態(tài)時，故有 = 0；當(dāng) =0 時，是受單向壓應(yīng)力狀態(tài)，所以 =- /2。這種變形情況處于沖壓應(yīng)變圖中的EOG 范圍內(nèi)（見圖 11）；而在沖壓應(yīng)力圖中則處于 COD 范圍內(nèi)（見圖 12）。2) 當(dāng)一定有 2 0 且 t=0 時，有式（12）可知：因為 0，所以0 與 0。這個結(jié)果表明：對于兩向壓應(yīng)力的

7、平面應(yīng)力狀 ，則在這個方向上的應(yīng)變一定時負(fù)的，即一定是壓態(tài)，如果絕對值最大是縮變形。又因為 0與t0，即在板料厚度方向上的應(yīng)變是正的，即為壓縮變形，板厚增大。在 方向上的變形取決于 與 的數(shù)值：當(dāng) =2 時， =0；當(dāng) 2 ， 0；當(dāng) 0。這時， 的數(shù)值只能在 = =0之間變化。當(dāng) = 時，是雙向等壓力狀態(tài)，所以 = 0。這種變形與受力情況，處于沖壓應(yīng)變圖中的GOL 范圍內(nèi)（見圖 11）；而在沖壓應(yīng)力圖中則處于 DOE 范圍內(nèi)（見圖 12）。(1)沖壓毛坯變形區(qū)受兩個異號應(yīng)力的作用，而且拉應(yīng)力的絕對值大于壓應(yīng)力的絕對值。這種變形共有兩種情況，分別作如下分析。1）當(dāng) 0， | |時，由式（12）

8、可知：因為 0， | |，所以一定有 - 02及 0。這個結(jié)果表明：在異號的平面應(yīng)力狀態(tài)時，如果絕對值最大應(yīng)力是拉應(yīng)力，則在這個絕對值最大的拉應(yīng)力方向上應(yīng)變一定是正應(yīng)變，即是伸長變形。又因為 0， | |，所以必定有 0 0， 0， | |時，由式（12）可知：用與前項相同的方法分析 0。即在異號應(yīng)力作用的平面應(yīng)力狀態(tài)下，如果絕 ，則在這個方向上的應(yīng)變是正的，是伸長變形；而在對值最大應(yīng)力是拉應(yīng)力壓應(yīng)力 方向上的應(yīng)變是負(fù)的（ 0， 的變化范圍只能在 =- 與 =0 的范圍內(nèi)。當(dāng) =- 時，0， 0， | |時，由式（12）可知：因為 0， | |，所以一定有 - 02及 0， 0，必定有 - 0

9、，即在拉應(yīng)力方向上的應(yīng)變是正的，是伸長變形。這時 的變化范圍只能在 =- 與 =0的范圍內(nèi) 。當(dāng) =- 時， 0 0， 0， | |時，由式（12）可知：用與前項相同的方法分析 0， 0， 0, 0 AONGOH+伸長類 AOCAOH+伸長類雙向受壓 0, 0 EOGCOD壓縮類 0, | |MONFOG+伸長類| | |LOMEOF壓縮類異號應(yīng)力 0, | |COB+伸長類| | | |DOEBOC壓縮類+ + / /+- - + - - 圖 13 沖壓應(yīng)變圖7擴(kuò)口變形區(qū)質(zhì)量問題的表現(xiàn)形式變形程度過大引起變形區(qū)產(chǎn)生破裂現(xiàn)象壓力作用下失穩(wěn)起皺成形極限主要取決于板材的塑性，與厚度無關(guān)可用伸長率及

10、成形極限 DLF 判斷主要取決于傳力區(qū)的承載能力取決于抗失穩(wěn)能力與板厚有關(guān)變形區(qū)板厚的變化減薄增厚提高成形極限的方法改善板材塑性使變形均勻化，降低局部變形程度工序間熱處理采用多道工序成形改變傳力區(qū)與變形區(qū)的力學(xué)關(guān)系采用防起皺措施化學(xué)成分組織塑 性變形條件硬化性能伸長類抗縮頸能 力應(yīng)力狀態(tài)變形應(yīng)變梯度硬化性能變形均化與擴(kuò)展能力模具狀態(tài)變形區(qū)的成形極限力學(xué)性能抗起皺值與值能力相對厚度化學(xué)成分沖壓成形極限壓縮類塑性組織變形變形條件強(qiáng)度變形抗力傳動區(qū)的成形極限變形力及其 變 化硬化性能抗拉與抗壓縮失衡能力各向異性值圖 13體系化方法舉例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 與 =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