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异型 冲压 模具设计

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异型垫冲压模具设计,异型,冲压,模具设计

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河南科技大学毕业设计（论文）异型垫片冲压模具设计摘 要该论文分析了垫片的工艺性，介绍了其特殊复合模总体设计结构和排样方案。实践证明:该模具结构合理、可靠，并能保证产品质量，对此类零件的复合模设计有重要参考价值。采用复合模便于生产批量大，尺寸小的工件，可以提高生产率，操作安全，保证质量，但复合模工序安排较为重要，设计中应充分考虑各工序的顺序，提高工作效率，提高模具寿命和降低模具成本。主要工序包括：a落料，b冲孔。本设计分别论述了产品工艺分析，冲压方案的确定，工艺计算，模板及零件设计等问题。本设计的内容是确定复合模内型和结构形式以及工艺性，绘制模具总图和非标准件零件图。该垫片复合模总体设计结构和排样方案经分析论证和实际投产检验。实践证明:模具结构灵活、可靠，并能保证产品质量，成本低，对此类零件的级进模设计有重要参考价值。通过本论文我掌握了模具设计的基本的模具技能懂得了怎样分析零件的工艺性，怎样确定工艺方案，了解了模具的基本结构，提高了计算能力，绘图能力，熟悉了规范和标准，同时各科相关的课程都有了全面的复习，独立思考的能力也有了提高。关键词：垫片，落料, 冲孔, 模具设计 MULTI-ARM OF THE RAM DIE DESIGNABSTRACTAfter an analysis on the manufacturability of gasket, a particular progressive die, including its structure and layout design, was developed. The die was reliable in structure and good in operation. It can be important reference to the design of progressive dies for similar parts. Principal manufacturing processes, including: a punching, ,b blanking, .Respectively on the design of the product process, pressing for identification, technology, design templates and components, and other issues. The design is to determine the content of progressive die-mold-type structure and form and process of drawing the map and the non-standard mold of Parts. The structure and layout design of the gasket are as proof of actual production and testing. Practice has proved: the structure of die is flexible, reliable and can ensure product quality and low cost，it can be important reference to the design of progressive die for such parts.Through this paper we master die design of the basic skills to understand the mold of what parts of the process, how to determine Process, learn the basic structure of the mold, improved computing power, graphics capabilities, and is familiar with the norms and standards related to curriculum subjects at the same time have a comprehensive review, the ability of independent thinking has also improved.KEY WORDS: gasket， blank, punch，die design目 录摘 要.1前 言.5 第一章 设计任务书和产品图.8第二章 零件的工艺性分析.9 2.1 零件的工艺性分析.,.9 2.2 冲裁件的精度.9 2.3 确定工艺方案.9第三章 冲压模具工艺与设计计算 .,.113.1冲压模具总体结构设计.113.1.1 模具类型.113.1.2 操作与定位方式.113.1.3 卸料与出件方式.113.1.4 模架类型及精度.113.2 排样设计与计算.113.3 设计冲压力与压力中心,初选压力机. .123.4 计算凸凹模刃口尺寸及公差.13第四章 模具的总张图与零件图.,154.1 冲压模具的零件图.15 4.1.1凹模的设计.15 4.1.2凸模的设计.16 4.1.3凸凹模的设计.164.2选择坚固件及定位零件.17 4.2.1 螺钉规格的选用.174.2.2活动挡料销.184.3设计和选用卸料与出件零件.18 4.3.1 卸料板.184.4选择模架及其它模具零件.19 4,4,1 模架.19 4.4.2 模柄.20 4.4.3 垫板.21 4.4.4 凸模固定板.21 4.4.5 凸凹模固定板.22 4.5压力机的校核.23 4.6级进模具的装配图.24结 论.25参考文献.26致 谢.27外文资料.,.28 前 言模具是现代工业的重要工艺设备，随着科学技术的不断进步，它在国民经济中占有越来越重要的地位，发展前景十分广阔。模具技术水平在很大程度上决定于人才的整体水平，而模具技术水平的高低，又决定这产品的质量、效益和新产品的开发能力，因此模具技术已成为衡量一个国家产品制造水平高低的重要标志。改革开放30年来，我国的模具工业获得了飞速的发展，设计、制造加工能力和水平、都有一了很大的提高。虽然我国模具产品水平有了很大提高，但差距还很大。 近些年来，中国模具的设计和制造水平有了很大提高，CAD/CAE/CAM等计算机辅助技术、高速加工技术、热流道技术、气辅技术、逆向工程等新技术得到广泛应用。目前，国内外相继涌现出精密冲压工艺、软模成形工艺、高能高速成形工艺及无模多点成形工艺等精密、高效、经济的冲压新工艺。其中，精密冲裁是提高冲裁件质量的有效方法，它扩大了冲压加工范围，目前精密冲裁加工零件的厚度可达25mm，精度可达IT1617级；用液体、橡胶、聚氨酯等作柔性凸模或凹模的软模成形工艺，能加工出用普通加工方法难以加工的材料和复杂形状的零件，在特定生产条件下具有明显的经济效果；采用爆炸等高能效成形方法对于加工各种尺寸在、形状复杂、批量小、强度高和精度要求较高的板料零件，具有很重要的实用意义；利用金属材料的超塑性进行超塑成形，可以用一次成形代替多道普通的冲压成形工序，这对于加工形状复杂和大型板料零件具有突出的优越性；无模多点成形工序是用高度可调的凸模群体代替传统模具进行板料曲面成形的一种先进技术，我国已自主设计制造了具有国际领先水平的无模多点成形设备，解决了多点压机成形法，从而可随意改变变形路径与受力状态，提高了材料的成形极限，同时利用反复成形技术可消除材料内残余应力，实现无回弹成形。无模多点成形系统以CAD/CAM/CAE技术为主要手段，能快速经济地实现三维曲面的自动化成形。冲压成形是一种历史悠久的金属加工工艺方法. 本世纪前20年是我国经济社会发展的重要战略机遇期。在这样一个关键历史时期，制造业扮演着重要的角色。制造业中的冲压成形行业是发展汽车制造业、航空航天工业、金属制品业、仪器仪表电器及化工工业等等的基础，是现阶段我国最具商机，既有大好发展机遇，又面临严峻市场挑战的行业。因此，依靠技术进步，大力发展先进成形技术，振兴我国冲压行业是历史性战略任务21世纪初，制造业全球化进程的加快及信息技术、材料技术的迅猛发展，必将给冲压成形技术和行业带来一系列新的深刻的变化。冲压行业必须转变观念，改变机制，加速信息技术、数字化技术与冲压成形技术的融合，改变传统冲压行业的面貌，才能满足国民经济快速发展的需要和提升国际竞争力，把中国的冲压成形行业做大做强。冲压主要是按工艺分类，可分为分离工序和成形工序两大类。分离工序也称冲裁，其目的是使冲压件沿一定轮廓线从板料上分离，同时保证分离断面的质量要求。成形工序的目的是使板料在不破坯的条件下发生塑性变形，制成所需形状和尺寸的工件。在实际生产中，常常是多种工序综合应用于一个工件。冲裁、弯曲、剪切、拉深、胀形、旋压、矫正是几种主要的冲压工艺。 冲压用板料的表面和内在性能对冲压成品的质量影响很大，要求冲压材料厚度精确、均匀；表面光洁，无斑、无疤、无擦伤、无表面裂纹等；屈服强度均匀，无明显方向性；均匀延伸率高；屈强比低；加工硬化性低。，有6070%是板材，其中大部分是经过冲压制成成品。汽车的车身、底盘、油箱、散热器片，锅炉的汽包、容器的壳体、电机、电器的铁芯硅钢片等都是冲压加工的。仪器仪表、家用电器、自行车、办公机械、生活器皿等产品中，也有大量冲压件。全世界的钢材中冲压是高效的生产方法，采用复合模，尤其是多工位级进模，可在一台压力机上完成多道冲压工序，实现由带料开卷、矫平、冲裁到成形、精整的全自动生产。生产效率高，劳动条件好，生产成本低，一般每分钟可生产数百件。 在中国，从事模具技术研究的机构和院校已达30余家。华中科技大学模具技术国家重点实验室、上海交通大学模具CAD国家工程研究中心、北京机电研究所精冲技术国家工程研究中心和郑州工业大学橡塑模具国家工程研究中心等，在模具CADCAECAM技术、冷冲模和精冲模CAD软件、模具的电加工和数控加工技术、快速成型（PP）和快速制模技术、新型模具材料等方面都取得了显著的进步和多项成果。由此展望未来，中国模具工业将会有一个持续快速发展的机遇期。我们要很好地把握住这个机遇期，使我国模具工业有一个很大的提高。 面对这种形势，我国模具行业当前的任务是：推进改革，调整结构，开拓市场，苦练内功，提升水平，上新台阶。最终使我国成为真正的模具强国。通过四年的基础课程和专业课程的学习，我对本专业的理论知识已有了系统的掌握，为以后走上工作岗位打下了结实的基础。但在这次设计中我也发现了我的许多不足之处，并加以改正。本套模具就是在发现错误和改正错误的过程中完成的，由于知识浅薄，错误之处难免，敬请老师指正。 第一章 设计任务书和产品图 本设计零件名称为异型垫片，材料为08钢，材料厚度3mm，大批量生产，冲压件图如下图所示: 图1-1 工件图 技术要求： 1.材料：08钢； 2.料厚：3mm；第2章 零件的工艺性分析2.1 零件的工艺性分析 该零件材料为08钢结构简单,抗剪强度为300mpa形状对称,只有落料，冲孔两个工序具有良好的冲压性能，适合冲裁。工件结构相对简单，有一个20mm的孔两个8mm的孔；孔与孔、孔与边缘之间的的距离也满足要求，最小壁厚为6mm（20mm孔与35mm的孔、8mm的孔与R10mm的外圆之间的壁厚。）2.2 冲裁件的精度与粗糙度 冲裁件的经济公差等级不高于IT11级,一般落料公差等级最好低于IT10级,冲孔件公差等级最好低于IT9级,工件的尺寸全部为自由公差，可看作IT14级，尺寸精度较底，普通冲裁完全能满足要求。根据表8-14得出零件图如下： 图2-1 工件图2.3确定工艺方案 该冲裁件包括落料和冲孔两个基本工序,可采用的冲裁方案有单工序冲裁,复合冲裁和级进冲裁三种：方案一、单序模：零件属于中批量生产,因此采用单工序须要模具数量较多,生产率低,所用费用也高,不合理;方案二、：复合模：若采用复合冲,可以得出冲件的精度和平直度较好,生产率较高,凸凹模最小壁厚为3.2mm。 方案三、级进模：大批量生产、冲裁精度较高，冲裁件孔与孔、孔与边缘尺寸较小时使用，计算该件的最小壁厚为6mm。综上所述可选方案二复合模。第三章 冲压模具工艺与设计计算3.1冲压模具总体结构设计3.1.1根据零件的冲裁工艺方案,采用复合冲裁模.3.1.2导向与定位方式导向形式：滑动导柱导套导向定位方式：板料定位靠导料销和弹簧弹顶的活动挡料销完成，因为该模具采用的是条料，控制条料的送进方向采用的是导料板，无侧压装置。控制条料的送进步距采用挡料销定距。3.1.3卸料与出件方式 冲孔凸模与凸凹模冲孔，冲孔废料直接落料。利用推件块将制件顶出。3.1.4模架类型及精度该模具采用后侧导柱模架，以凹模周界尺寸为依据，选择模架规格。3.2 排样设计与计算3.2.1 根据零件形状和增大利用率而采用斜排。3.2.2 搭边值设定：查表2.5.2得a1=2.5 a= 条料宽度与导料板间距离计算 采用无侧压装置条料宽度与导料板间距离 条料的宽度B=（Dmax+2a+C）查表2.5.3得条料宽度偏差=0.9mm导料板与条料之间的最小间隙C查表2.5.5得C=0.5B=(35+2x2.8+0.5)=41.1mm,根据实际取料宽为42mm导料板间的距离:A=B+C= 42+0.5=42.5mm步距S=2X+a1如图4-1,x=35.28,S=35.28X2+2.5=73.06,图3-1 排样图3.2.4 条料的利用率=(A/BS)X100%由零件图用CAD计算得一个零件的面积为1600.77mm一个进距内的坯料面积:BXS=42X73.06=3068.52mm,因此材料利用率为:=(A/BS)X100%=(1600.77/3068.52)X100%53%3.3 设计冲压力与压力中心,初选压力机3.3.1 落料力 根据零件图,用CAD可计算出冲一次零件外周边之和L=186.6mm,冲孔内周边之和L=113.1mm,又因为=300Mpa,t=3mm,取K=1.3,则根据式,F1=KLt=1.3X186.6X3X300=218.322KN冲孔力: F2=KLt=1.3X113.1X3X300=132.327KN所以冲裁力F=F1+F2=218.322+132.327=350.649KN推件力:由表2.9.4,根据材料厚度取凹模刃口直壁高度h=9mm,故n=h/t=9/3=3,查表2.6.1,取KT=0.45则FT=n KT F冲=3X0.45X132.327 =178.641KN卸料力FX 由表2.6.1,取KX=0.04,则FX=KXXF落=0.04X218.322=8.732KN所以FZ=F+FX+FT=350.649+178.641+8.732=538.02KN公称压力F=1.3FZ=1.3X538.02=699.43KN根据公称压力选取压力机因此可选压力机型号为J23-80闭合高度为380H2903.3.2 压力中心 因为该制件是简单对称件,所以压力中心为该制件的几何中心.3.4 计算凸凹模刃口尺寸及公差3.4.1冲孔尺寸2-根据表2.3.3,由材料厚度可得Zmin=0.460mm, Zmax=0.640mm.查表2.4.1T=0.020 A=0.020磨损系数表可得磨损系数X1=0.5,校核: T+A=0.04Zmax-Zmin=0.18dT=(dmin+x)=(8+0.5X0.36)=8.18dA=(dT + Zmin)=(8.18+0.46)=8.643.4.2 冲孔尺寸20根据表2.3.3, 由材料厚度可得Zmin=0.460mm, Zmax=0.640mm. 查表2.4.1T=0.020 A=0.025磨损系数表可得磨损系数X1=0.5,校核: T+A=0.045Zmax-Zmin=0.18dT=(dmin+x)=(20+0.5X0.52)=20.26dA=(dT + Zmin)=( 20.26+0.46)=20.723.4.3 落料尺寸20根据表2.3.3, 由材料厚度可得Zmin=0.460mm, Zmax=0.640mm. 查表2.4.1T=0.020 A=0.025磨损系数表可得磨损系数X1=0.5,校核: T+A=0.045Zmax-Zmin=0.18DA=(Dmax-x)=(20-0.5X0.52)=19.74DT=(DA-Zmax)=(19.74-0.64) = 落料尺寸35根据表2.3.3, 由材料厚度可得Zmin=0.460mm, Zmax=0.640mm. 查表2.4.1T=0.020 A=0.030磨损系数表可得磨损系数X1=0.5,校核: T+A=0.05Zmax-Zmin=0.18DA=(Dmax-x)=(35-0.5X0.62)=34.69DT=(DA-Zmax)=(34.69-0.64) =34.053.4.5 孔心距 600.74L=L平18=6018X1.48=600.185第四章 模具的总张图与零件图4.1 冲压模具的零件图4.1.1凹模设计 各冲裁的凹模孔均采用线切割机床加工,安排凹模在模架上的位置时,要依据计算压力中心的数据将压力中心与模柄中心重合,其轮廓尺寸按公式2.9.3 2.9.4计算 用CAD计算出凹模刃口的最大尺寸b=78mm 查表2.9.5得k=0.3凹模厚度H=kb=0.3X78=23.4mm 凹模厚度取24mm凹模壁厚C=(1.5-2)H=40mm 所以凹模壁厚度为40mm 凹模长度L=l+2C=78+2X40=158mm 凹模宽度 B=l+2C=35+2X40=115mm由以上算得凹模轮廓尺寸LXBXH=158X115X24,查有关国家标准,并无厚度合适,因此我选LXB为标准尺寸,得LXB=160X125凹模材料的选用:材料选用Cr4WmoV。 图4-1 冲孔凹模4.1.2 凸模设计凸模材料：参照冲压模具设计与制造选用T10A因为所冲的孔均为圆形，而且都都不属于特别保护的小凸模，所以冲孔凸模采用阶梯式，一方面加工简单，另一方面又便于装配与更换，冲孔凸模结构如图： 图4-2 冲孔凸模凸模长度L=H固+H凹=1.8 H凹=1.8X24=43.2 H1=0.8 H凹=0.8X24= 凸凹模设计由表2.9.6得凸凹模最小壁厚=6.7 图4-3 凸凹模4.2 选择坚固件及定位零件4.2.1 螺钉规格的选用:,根据标准GB699选取材料为45钢.热处理硬度值HRC3540 卸料螺钉选用圆柱头内六角卸料螺钉,卸料板上设置4个卸料螺钉,公称直径为12mm 卸料钉尾部应留有足够的行程空间,卸料螺钉拧紧后应使卸料板超出凸凹模端面1mm,有误差时通过在螺钉与卸料板之间安装垫片来调整. 图4-4 卸料螺钉4.2.2 活动挡料销: 在卸料板上固定了3个导料销,用于条料送进的定位.在送进方向上用一个弹簧弹顶装置的挡料销, 挡料销,的尺寸如下图: 图4-5 活动档料销4.3 设计和选用卸料与出件零件卸料以卸料板卸料,出件是以凸凹模往上冲出即可,因此不用设计出件零件.固定卸料板的平面外形尺寸一般与凹模板相同,其厚度可取10，LXBXH=158X115X10材料为45号钢由以上根据凸凹模可设计出卸料板如图4-6. 图4-6 卸料板4.4 选择模架及其它模具零件4.4.1选择模架:根据GB/T 2861.5-90,由凹模周界158X115,及安装要求,选取凹模周界:LXB=160X125,闭合高度:H=170205,上模座:160X125X40下模座:160X125X50导柱25X160, 导套:25X95X38,由以上可得模架如图4-7所示. 图4-7 模架4.4.2 模柄: 由压力机的型号J23-80.可查得模柄孔的直径为60,深度为80,由装配要求,模柄与模柄孔配合为H7/m6并加销钉防转,模柄长度比模柄孔深度小510mm,由于采用固定卸料，上模座回程时受力较大，因此选用压入式模柄较合理，所以根据GB2862.1-81得图4-8所示: 图4-8 模柄 4.4.3 垫板: 垫板的作用是承受并扩散凸模传递的压力,以防止模座被挤压损伤,因此在与模座接触面之间加上一块淬硬磨平的垫板.垫板的外形尺寸与凸模固定板相同,厚度可取310mm,这里设计时,由于压力较大,根据GB2865.2-81选取规格为LXBXH=160X115X104.4.4 凸模固定板: 凸模固定板的外形尺寸与凸模的外形尺寸一致,厚度为18mm,根据核准选取板的规格为LXBXH=160X115X18; 图4-9 凸模固定板4.4.5 凸凹模固定板: 凸凹模固定板的尺寸与垫板的尺寸一致,厚度为20mm,规格为LXBXH=160X115X20; 图4-10 凸模固定板4.5 压力机的校核5.5.1 公称压力 根据公称压力的选取压力机型号为J23-80,它的压力为800699.43所以压力得以校核;5.5.2 闭合高度 由压力机型号知Hmax=380 M=90 H1=100Hmin=HmaxM=380-90=290(M为闭合高度调节量/mm,H1为垫板厚度/mm)根据GB/T 2851.3-90 得模具闭和高度为170H205,根据公称压力选用压力机闭和高度 290H380, H垫=100因为 Hmin+10-H垫HHmax-5-H垫 200H275所以H=205 选顶定压力机J23-8符合要求.H=H(上)+H(垫)+H(凸凹)+L+H(下)=40+10+43.2+H(凸凹)+50=145+ H(凸凹)145+ H(凸凹)=205 H(凸凹)=604.6模具的总张图如图4-11所示:图4-11 装配图1.模座，2.螺钉,3.凸凹模固定板,4.导柱,5.弹簧,6.活动挡料销,7.导套8.垫板9.螺钉,10.连接打杆,11.推板,12.打杆,13.模柄 14.螺钉15.上模座 16.17冲孔凸模 18.凸模固定板 19.推件块 20.落料凹模 21.卸料板 22.凸凹模23.卸料螺钉 24.导料销 结 论在大学的学习过程中，毕业设计是一个重要的环节。是我们步入社会参与实际工作的一次极好的演示，我十分有幸提早把毕业设计和实际工作有机的结合起来。经过这次设计，我认识到在冲压模具设计中，主要涉及到两个方面的问题：一是模具结构的整体构思，二是技术标准的选用。在设计中，整体构思非常重要，一次完整的设计全部由整体构思所引导，若没有一个完善的整体构思，设计过程将会很难顺利进行。技术标准的选用也至关重要，选用合理的技术标准会很大程度上减小模具的设计、制造难度。在这次设计中，我综合运用本专业所学课程的理论和生产实际知识进行一次冷冲压模具设计工作的实际训练从而培养和提高学生独立工作能力，巩固与扩充了冷冲压模具设计等课程所学的内容，掌握冷冲压模具设计的方法和步骤，掌握冷冲压模具设计的基本的模具技能懂得了怎样分析零件的工艺性。但是毕业设计也暴露出自己专业基础的很多不足之处，比如缺乏综合应用专业知识的能力，对材料的不了解。这次实践是对自己大学所学的一次大的检阅，使我明白了自己知识还很浅薄。虽然马上要毕业了但是自己的求学之路还很长，以后更应在工作中学习，努力使自己成为一个对社会有所贡献的人。参考文献【1】史铁梁.冷冲压设计指导.北京：机械工业出版社，1999.【2】肖景容、姜奎华.冲压工艺学.北京：机械工业出版社，2008.【3】王爱珍.冷作成形技术手册.北京：机械工业出版社，2006.【4】涂光祺.冲模技术.北京：机械工业出版社，2002.【5】杨玉英.实用冲压工艺及模具设计手册.北京：机械工业出版社，2005.【6】李天佑.冲模图册.北京：机械工业出版社，1988.【7】谢建、杜东福.冲压工艺及模具设计技术问答.上海：上海科学技术出版社，2005.【8】何永熹、武充沛.几何精度规范学.北京：北京理工大学出版社，2006.【9】付宏生.冷冲压成形工艺与模具设计制造.北京：化学工业出版社，2006.【10】付建军.模具制造工艺.北京：机械工业出版社，2007.【11】冷冲模国家标准【12】 周玲.冲模设计实例详解.北京：化学工业出版社，2007.【13】 中国机械工业教育协会.冷冲模设计及制造.北京：机械工业出版社，2003.【14】 黄毅宏、李明辉.模具制造工艺.北京：机械工业出版社，2009.【15】 赵大兴.工程制图.北京：高等教育出版社，2003.【16】 王鹏驹、成虹.冲压模具设计师手册.北京：机械工业出版社，2009. 致 谢 通过这次毕业设计，我深切体会到团体力量的重要性。无论是参考资料的互借，还是问题的探讨，都对这次设计的顺利进行起到至关重要的作用。首先感谢洛阳理工学院，给我提供这么好的学习生活环境，在校学习和生活的日子是我一生中一段难忘的经历！本文是在金文中老师精心指导和大力支持下完成的。金老师以其严谨求实的治学态度、高度的敬业精神、兢兢业业、孜孜以求的工作作风和大胆创新的进取精神对我产生重要影响。他渊博的知识、开阔的视野和敏锐的思维给了我深深的启迪。这次毕业设计能够顺利完成，机电工程系的领导对我的大力支持和帮助，离不开同学们的热心帮助，更离不开金文中老师对我的我耐心指导和支持，借此向他们再次表示衷心的感谢。 最后，再次对关心、帮助我的老师和同学表示衷心地感谢。6 Bending of sheet6.1IntroductionBending along a straight line is the most common of all sheet forming processes; it can be done in various ways such as forming along the complete bend in a die, or by wiping, folding or flanging in special machines, or sliding the sheet over a radius in a die. A very large amount of sheet is roll formed where it is bent progressively under shaped rolls. Failure by splitting during a bending process is usually limited to high-strength, less ductile sheet and a more common cause of unsatisfactory bending is lack of dimensional control in terms of springback and thinning. If the line of bending is curved, adjacent sheet is usually deformed in the process and the sheet is either stretched, which may lead to splitting, or compressed with the possibility of buckling. There are special cases where sheet can be bent along curved lines without stretching or shrinking adjacent areas, but these require special geometric design.In this chapter, simple cases of bending along straight lines are examined for the elastic, plastic and fully plastic regimes.6.2Variables in bending a continuous sheetAs shown in Figure 6.1, we consider a unit width of a continuous sheet in which a cylindrical bent region of radius of curvature is flanked by flat sheet. The bend angle is , and a moment per unit width M, and a tension (force per unit width) Tare applied. We note that the tensionTis applied at the middle surface of the sheet. The units of Mare force length/length and of Tforce/length. Figure 6.1A unit length of a continuous strip bent along a line.Figure 6.2Deformation of longitudinal fibres in bending and tension.6.2.1Geometry and strain in bendingIn bending a thin sheet to a bend radius more than three or four times the sheet thickness, it may be assumed that a plane normal section in the sheet will remain plane and normal and converge on the centre of curvature as shown in Figure 6.2.In general, a line CD0 at the middle surface may change its length to CD if, for example, the sheet is stretched during bending; i.e. the original length l0 becomesLs = A line AB0at a distance y from the middle surface will deform to a lengthThe axial strain of the fibre AB isWhere a is the strain at the middle surface or the membrane strain and b is the bending strain. Where the radius of curvature is large compared with the thickness, the bending strain can be approximated as,The strain distribution is approximately linear as illustrated in Figure .2Plane strain bendingIf the flat sheet on either side of the bend in Figure 6.1 is not deforming it will constrain the material in the bend to deform in plane strain; i.e. the strain parallel to the bend will be zero. In this work, plane strain conditions will be assumed, unless stated otherwise.The deformation process in bending an isotropic sheet is thereforeBending of sheet83Figure 6.3Assumed strain distribution in bending.Following Equations 2.18(b) and 2.19(c), for, = 0, = 1/2, we obtainWhere S is the plane strain flow stress. (Equation 6.6 assumes the von Mises yield condition. If a Tresca yield criterion is assumed,1= f= S.) The stresses on a section along the bend axis are illustrated in Figure 6.4. Clearly, at the edge of the sheet, the stress along the bend axis will be zero at the free surface and plane strain will not exist. It is usually observed that the edge of the sheet will curl as illustrated. This happens because the stress state is approximately uniaxial tension near the edges of the sheet; the minor strain will be negative near the outer surface and positive near the inner surface giving rise to the anticlasticcurvature as shown. Within the bulk of the sheet, however, plane strain deformation is assumed with the minor strain along the axis of the bend equal to zero.Free edgeFigure 6.4Stress state on a section through the sheet in plane strain bending.6.3Equilibrium conditionsWe consider a general stress distribution on a normal section through a unit width of sheet in bending, as shown in Figure 6.5. The force acting on a strip of thickness dy across the unit section is1 dy 1. The tensionTon the section is in equilibrium with the integral of this force element, i.e.84 Mechanics of Sheet Metal Forming85Figure 6.5Equilibrium diagram (a) for a section through a unit width of sheet and (b) at y cal stress distribution.Integrating the moment of the force element, we obtain(We note too that there is a third equilibrium equation for forces in the radial direction arising from the tension T . This is given in Section 4.2.5 by Equation 4.11.)6.4Choice of material modelFor the strain distribution given by Equation 6.3, the stress distribution on a section can be determined if a stress strain law is available. In general, the material will have an elastic, plastic strain-hardening behaviour as shown in Figure 6.6(a). In many cases, it is useful to approximate this by a simple law and several examples will be given. The choice of material model will depend on the magnitude of the strain in the process. The strain will depend mainly on thebend ratio, which is defined as the ratio of the radius of curvature to sheet thickness, /t.6.4.1Elastic, perfectly plastic modelIf the bend ratio is not less than about 50, strain-hardening may not be so important and the material model can be that shown in Figure 6.6(b). This has two parts, i.e. if the stress Figure 6.6Material models for bending. (a) An actual stress strain curve. (b) An lastic, perfectly plastic model. (c) A rigid, perfectly plastic model. (d) A train-hardening plastic model. is less than the plane strain yield stress, SWhere the modulus of elasticity in plane strain is slightly different from the uniaxial Youngs modulus , E; i.e.Where is Poissons ratio.For strains greater than the yield strain,Where Sis constant. In isotropic materials,Sis related to the uniaxial flow stress byEquation 6.6 for the von Mises yield condition.6.4.2Rigid, perfectly plastic modelFor smaller radius bends, and where we are not concerned with elastic springback, it may be sufficient to neglect both elastic strains and strain-hardening. A rigid, perfectly plastic model is shown in Figure 6.6(c), where and Sis a value averaged over the strain range as indicated in Section .4.3Strain-hardening modelWhere the strains are large, the elastic strains may be neglected and the power law strain-hardening model used, whereFor a material having a known effective stress strain curve of the formThe strength coefficient Kcan be calculated using Equations 6.6. This model is illustrated in Figure 6.6(d).6.5Bending without tensionWhere sheet is bent by a pure moment without any tension being applied, the neutral axis will be at the mid-thickness. This kind of bending is examined here for several types of material behaviour . In these cases, a linear strain distribution as illustrated in Figure 6.3 is assumed and the equilibrium equations, Equations 6.7 and 6.8, will apply.Figure 6.7 Linear elastic bending of sheet showing the material model (a), the strain distribution (b), and the stress distribution (c).6.5.1Elastic bendingThe material model is illustrated in Figure 6.7(a) where the yield stress isS. The stress strain relation is given by Equation 6.9 and for the strain distribution shown in Figure 6.7(b), the stress distribution in Figure 6.7(c) will be obtained. The stress at a distance y from the neutral axis, from Equations 6.4 and 6.13, isThe moment at the section, from Equation 6.8, isFrom Equation 6.15, we have and Equation 6.16 can be written in the familiar form for elastic bending.CurvatureFigure 6.8Moment curvature diagram for elastic bending. Where I = t3/12 is the second moment of area for a unit width of sheet and 1/is the curvature.The limit of elastic bending is when the outer fibre at y = t /2 reaches the plane strain yield stress S. The limiting elastic moment is given byAnd the curvature at this moment isWithin this elastic range, the moment, curvature diagram is linear as shown in Figure 6.8, i.e.The bending stiffness of unit width of the sheet is E t3/Rigid, perfectly plastic bendingIf the curvature is greater than about five times the limiting elastic curvature, a rigid, perfectly plastic model, Equation 6.12, as shown in Figure 6.6(c), may be appropriate, although this will not give information on springback. The stress distribution will be as shown in Figure 6.9. In Equation 6.8, the stress is constant and integrating, we obtain the so-called fully plastic moment Mp asThe moment will remain constant during bending and is illustrated in Figure 6.10.Figure 6.9Stress distribution for a rigid, perfectly plastic material bent without tension.Figure 6.10The moment curvature diagram for a rigid, perfectly plastic sheet bent without tension.Figure 6.11Stress distribution for an elastic, perfectly plastic sheet bent without tension.6.5.3Elastic, perfectly plastic bendingFor curvatures beyond the limiting elastic curvature(1/)e and below that where the moment reaches the fully plastic momentMp, an elastic, perfectly plastic model, as inSection 6.4.1, is often used. The model is illustrated in Figure 6.6(b); the flow stress is constant and for plane strain, The stress distribution is illustrated in Figure 6.11; fory ye, the material is plastic with a flow stress S . As the curvature increases, ye decreases and at any instant is given byFrom Equation 6.19,The equilibrium equation, from Equation 6.16, isThe moment, curvature characteristic is shown in Figure 6.12 and it may be seen that this is tangent to the elastic curve at the one end and to the fully plastic curve at the other. It may be seen that with this non-strain-hardening model, the moment still increases beyond the limiting elastic moment and reaches 1.5Me before becoming constant. For this reason, elastic plastic bending is usually a stable process in which the curvature increases uniformly in the sheet without kinking. It may be shown that for materials that do not at this elastic, perfectly plastic model, for example aged sheet having a stress strain curve as shown in Figure 1.4, the moment characteristic is different and kinking may occur. In real materials it is very difficult to predict precisely the moment curvature character-istic in the region covered by the bold curve in Figure 6.12 from tensile data. The moment characteristic is extremely sensitive to material properties at very small strain and these properties often are not determined accurately in a tension test.Figure 6.12Moment curvature diagram for an elastic, perfectly plastic sheet bent without tension.6.5.4Bending of a strain-hardening sheetIf a power law strain-hardening model of the kind shown in Section 6.4.3 and Figure 6.6(d) is used, the stress distribution will be as shown in Figure 6.13. The whole section is assumed to be deforming plastically and the stress at some distance,y, from the middle surface is (6.24)Figure 6.13Stress distribution for a power-law-hardening sheet bent without tension.The equilibrium equation can be written as These equations can be combined to give a set of equations for bending a non-linear material, i.e. whereThe moment, curvature diagram is shown in Figure 6.14.Figure 6.14Moment curvature diagram for a strain-hardening sheet bent without tension.The power law equation is not a good fit for most materials at very small strains so that Equation 6.26 will not predict the moment curvature characteristic near the elastic, plastic transitionEquations 6.26 reduces to some familiar relations for special cases. For the linear elastic model,n = 1, K= E , and we obtain Equations 6.17. For the rigid, perfectly plastic model, n = 0, K= S, and Equation 6.26 reduces to Equation .5(Worked example) momentsA hard temper aluminium sheet, 2 mm thick, has a flow stress of 120 MPa, that is approx-imately constant. Determine the moment per unit width to bend the sheet to the limiting elastic state. What is the radius of curvature at this moment Determine the fully plastic moment if the sheet is bent further. The elastic modulus is 70 GPa and Poissons ratio is 0.3.SolutionFrom Equation 6.6, the plane strain plastic bending stress isFrom Equation 6.10, the elastic modulus in plane strain isFrom Equation 6.18, the limiting elastic bending moment isThe radius of curvature, from Equation 6.19, isThe fully plastic moment from Equation 6.21 is外文资料译文6板料弯曲6.1 介绍 沿着直线弯曲是所有板料成形过程最常见的; 它可以做以各种各样的方式实现，例如：在模具里形成完全弯曲，或者通过蚀刻工艺的弯曲，用专用机械进行的折叠或者翻边。还有很多板料是加载的逐渐增加的辊轮压力下成型的。 弯曲的失败通常是因为高强度，低韧性的板料和不好的弯曲条件或是缺乏尺寸控制和没有跟据正确的回弹和变薄而造成的。 如果是在线弯曲中，相邻的板料在变形过程中通常被变形，并且板料如果是拉伸，也会导致增加拉裂，或者压缩的时候出现褶皱的情况。 有板料可沿着弯曲的线弯曲的特殊情况，无需相邻区域的拉伸或收缩，但是这些是要求有特别的几何设计。在本章当中，例子是简单的沿着直线弯曲，变形分别有弹性、塑性和完全塑性的情形。 6.2 在弯曲连续的板料的变化因素 如图6.1所显示的，我们考虑曲率半径为的一个圆柱形弯的区域连续变形的板料的一个单位宽度 由平板侧相接的区域。弯曲角度是和力矩每单位宽度M和拉伸力（每个单位宽度) T。 我们注意到，拉伸 T 是应用的在板料的中间表面。 M单位 是力量 长度/长度和的T 力量/长度。 图形 6.1 连续的小条沿着直线的单位长度弯曲。图形 6.2 弯曲和拉伸的纵向纤维的变形。6.2.1 在弯曲中的几何分析和张力分析在弯曲过程当中，薄板料的到弯曲半径超过三或四倍板料厚度，也许假设如图6.2所显示，在板料的平面正常部分在曲率中心将依然是平面和正常的并且聚合。在弯曲期间，如果，例如，板料被拉伸成一条线CD0在中间表面，也许一般来说，改变它的长度到CD; 即原始的长度l0成为Ls = A线从中间表面的一个距离y将变形长度的AB0at纤维AB的这个轴向张力是当 a是张力在中间表面或膜张力和b是弯曲应变。 那里曲率半径是比厚度大的多的，弯曲应变大概接近，这个张力分布是大约线性的如图6.3所示。6.2.2 平面变形弯曲如果在弯曲变形的任一边平板在表6.1不变形它的在平面变形将压抑在弯的材料变形; 即张力将是零。 在这工作，平面变形情况是假设的，要不然是不会有这样的情况的。这个在各向同性弯曲的板料的变形过程是图形 6.3 在弯曲的假设的张力分布。下面的式2.18 (b)和2.19 (c)，为， = 0， = 1/2，我们获得当 S是平面变形流程应力。 (式6.6采用的Mises屈服准则情况。 如果Tresca屈服准则标准假设， 1= f= S.)在一个部分的应力沿弯轴在表6.4清楚地说明。 在板料的边缘，在自由表面沿弯轴的应力将是零，并且平面变形不会存在。 通常被观察板料的边缘将卷曲如被说明的情况。 因为应力状态是大约单轴的拉伸在板料的边缘附近这发生; 较小张力将是消极近的外表面和正面近提升相对的内在表面 曲度如显示。 在板料的大多数之内，然而，平面变形变形假设与沿弯的轴的较小张力相等到零。活动的边缘情况。 如果Tresca屈服准则标准假设， 1= f= S.)在一个部分的应力沿弯轴在表6.4被说明。 清楚地，在板料的边缘，沿弯轴的应力将是零在自由表面，并且平面变形不会存在。 通常被观察板料的边缘将卷曲如被说明。 因为应力状态是大约单轴的拉伸在板料的边缘附近，这发生; 较小张力将是消极近的外表面和正面近提升相对的内在表面 曲度如显示。 在板料的大多数之内，然而，平面变形变形假设与沿弯的轴的较小张力相等到零。活动的边缘图形 6.4 通过在平面变形弯曲的板料强调在部分的状态。6.3 平衡情况We通过板料的单位宽度在弯曲的