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毕业论文（设计）外文翻译题 目： 三维接触问题的有限元分析 2015 年 3 月 9 二维有限元接触解决方案的验证2.1赫兹接触对于不同材质的经典三维接触解决方案的验证，正确的地方是先从一个简单的二维模型开始。然后通过这个实验获取准确的结果，再由二维模型转移到三维维问题。目前关于两个气缸之间的赫兹接触分析,它们之间的接触压力是由赫兹发现表面的弹性变形与牛顿光学干涉条纹之间的差距玻璃镜片产生的。他假设一般接触面积是椭圆的干涉条纹。由于先进的方法可用于弹性 半空间的边值问题,然后他简化和假定每个相互接触是一个弹性半无限。这使得它可以计算在接触变形。为了满足上述条件的要求是:接触面必须是小的尺寸相比,每个物体和接触面积必须小于相对的表面的曲率半径。第一个要求是必要的,为了不影响高强度的边界接触地区。第二个要求是:以确保外表面接触是大致平面。另一个要求是应变接触区必须小。最后但并非最不重要的假设是假定为无摩擦接触的表面,所以没有切向牵引力在接触区域中的赫兹问题。 赫兹接触假设的摘要：1.每个个体在接触中被认为是弹性半空间应力计算。2.接触被假定为无摩擦。3.这两个机构之间的接触是相容的接触。4.接触区域附近的应变非常小。5.相比接触体的尺寸的接触面积比较小。二维平面应变分析 在一些工程问题，例如在内部压力的管道，一坝进行水负荷或圆柱滚子强行压缩，如图2.1，有显著应变仅在一个平面上;应变在一个方向上比在另外两个方向所述应变小得多。在这种情况下，更小的应变会被忽略，工程问题就解决了作为二维平面应变问题。图2.1：二维平面应变问题的说明 应变垂直于XY平面z作用示于图2.2和剪切应变XZ和YZ被假定为零。假设为平面应变是沿Z方向具有作用于仅在X或Y的负载均匀的横截面方向的，不以Z方向变化。对于本图进行正常负载假设为2D平面应变问题，并且解决了如图2.1厚度有限元模型。图2.2:应变在x - y平面两个气缸之间的联系本数值本案涉及两个气缸中接触，其轴线平行于彼此。力共同沿着接触的方向上施加。气缸因接触面变形2A的负载。图2.3：在赫兹接触变形的机构 如果具有曲率半径R1和R2的两个气缸等于作为曲率半径R的气缸. 两个气缸的材料特性是杨氏模量为E1和E2，泊松比为1分别2 来自Johnson理论结果20对于上述分析是半接触长度和最大接触压力。半接触长度A由下式给出： 用于在接触区域的压力分布的如下： 接触压力是最大的触点的中心，并且由给定的方程在接触的边缘的压力等零。2.1.1有限元关于无摩擦的解决方案图2.4：有限元二维接触模型 使用商业有限元软件ANSYS21进行了上述数值案例的分析。这个问题的建模涉及每个气缸二维分析th和运用对称边界条件建模。 PLANE42元素被用来啮合二维维度圆柱体。PLANE42是二维四边形元件具有4个节点，每个节点具有两个自由度的，如在一个平面上的两个方向的平移。关于啮合，该接触区域附近的啮合精细，具有最小元素大约0.26毫米和网孔是粗如移离具有最大元件尺寸0.92毫米接触区域。正常负载沿着共同接触点施加。两个圆柱体和接触模型的载荷数的材料性能示于表2.1表2.1：无摩擦有限元二维接触模型数据材料特性尺寸数据气缸1（B1）1.弹性模量 ( E1 )=30000 X 106 Pa2.泊松比 ( 1 )=0.25半径 ( R1 )=10 mm气缸2（B2）1.弹性模量( E 2 )=29120 X 106 Pa2.泊松比( 2 )=0.3半径( R2 )=13 mm负载数据： 负载（P）=4300N用于啮合的接触区域，从21是CONTA172和TANGE169，并且在21接触的向导选项在这里用来创建这些接触上的接触区域的元件。 CONTA172是用3个节点和2度的二维元件在一个平面上，自由在每个节点和用于模型表面到表面的接触。TARGE169是用于表示与所述接触相关目标表面的二维元件元素。在半接触长度和最大接触被压问题解决之后，接触长度是使用节点的解决方案，从一般处理程序菜单中选择。接触区是认可的色差，接触的状态从列表中每个节点可以读取结果，选择一般处理程序。接触长度测量接触边界的节点的距离。有限之间的百分比差异元件解决的结果和分析解决方案的结果是1,它证明了有限元素的解决方案的结果与赫兹分析结果一致。比较了有限元解结果和赫兹解析解结果所示表2.2。表2.2：对于分析结果的有限元分析与赫兹接触模型的比较分析结果解析结果差异的半接触长度（mm）1.18721.11971.04峰值接触长度（mm）1711.21698.10.72.2斯彭斯解决方案 在上一节中有一些假设的赫兹接触分析，如摩擦接触，小应变的接触等。本节讨论的非赫兹接触问题的分析。在非赫兹接触问题，我们已经通过假设放宽了引入摩擦接触之间的摩擦的情况下，赫兹摩擦接触气缸之间的效果将存在两个气缸，具有用于正常的不同的材料性质接触问题。由于机构之间接触不同与切线位移，滑移将发生与总接触区域，成为组合中央区包围滑移区。 机构滑移的数量取决于摩擦系数之间的联系等,如果摩擦的数量较多,滑移会减少,如果摩擦的数量少,那么滑移将更多。还可以增加摩擦系数的数量,使得滑移是零。存在一个切线牵引力，由于摩擦滑动区这些切线牵引力在相反的方向，两个接触面的接触区域和由下式给出： （2.5） 因为正常负荷增加，接触尺寸也增加。在任意两点在两个接触面接触区进行不同的切向位移。一旦他们进入区之间没有相对切向位移点。应力和应变的大小会增加接触长度的比例。 斯宾塞10溶液指出，对于相同的材料特性，即，杨氏模量，泊松比和摩擦系数滑移区域的量是相同的。所有的物体所具有的形状轮廓Z = AX n和等于平放穿孔。他还得出的结论是在单调载荷下，坚持区域大小的比率和接触区域大小只取决于材料性能。例如摩擦和系数泊松比。他还介绍了不同数值的不同泊松比范围从0到0.5的求解，并观察到更准确的结果接近0.5区段尺寸c给出由下式： （2.6）是第一类的完整椭圆积分和，解决了使用命令ELLIPKE从数学规划工具Matlab22。第一类椭圆积分可配制成其中是等于：参数的数值用来衡量不同的弹性常数不同的材料。 的值等于零时两种材料在接触是类似的材料，为零时两个材料具有泊松比等于0.5，即对于不可压缩的材料，对于极端值0.5时，一个主体和其它具有零泊松比的材料是不可压缩的。2.2.1有限元接触摩擦的解决方案图2.5：有限元二维模型接触摩擦非赫兹接触问题有限元的解决方案也进行了同样的方式，赫兹接触除了ANSYS改变一些选项。系统的建模使用气缸进行分析。 PLANE42二维4联接的QUAD元件被用来接触两个网络体，元件CONTA172和TARGE169被用于在接触区域靠近啮合。附近的接触网状区优良具有最小单元尺寸4.5e-2毫米。粗网格距离具有最大元件尺寸0.95毫米接触。两者之间的摩擦接触是在ANSYS材料模型选项指定。斯宾塞9提出了独特的接触问题的解决方案和独立的加载路径。后来斯宾塞10为幂律硬度的计算提出了解决方案，材料性能，摩擦系数和载荷，独立的加载路径。因此，我们没有考虑负载介入这个摩擦接触问题的有限元解。表2.3:有限元二维接触与摩擦模型数据材料特性维度数据柱面1（B1）1. 弹性模量 ( E1 )=30000 X 106 Pa2. 泊松比( 1 )=0.25半径 ( R1 )=10 mm柱面2（B2）1. 弹性模量( E 2 )=29120 X 106 P2. 泊松比( 2 )=0.3半径 ( R2 )=13 mm负载数据： 负载 (P)=3200 N摩擦： 摩擦系数()=0.01-0.05 解决问题之后,结果是使用节点从通过处理程序菜单中选择的区域大小和接触大小获得的。接触的区域,区域被认可的颜色的区别和联系每个节点的状态也可以从中读出，从结果列表中选择后处理。接触长度和根区大小测量之间的距离分别为接触和根区边界。接触模型解决了不同摩擦系数。表中给出的数值结果李2和坎波斯23表明,摩擦系数的增加会导致增加区域大小和导致减少接触的大小，观察到这些结果，摩擦系数是导致了粘区的大小增加。有限元件的比较结果与分析结果绘制在图2.6。该有限元结果与所述彭斯解析协议结过相同，根据有限元计算结果和斯宾塞之间的最大百分比差异分析的结果，观察到的摩擦值系数低。与这个幅度百分比差19.23，更多的错误（19.23）相比于在该元件的尺寸，是由于更小的区域大小和区域位置。这种错误可能是由附近的接触网格细化减少。在这项工作中网格细化由软件版本的限制。有限元计算结果会更准确的在更高的系数值。图2.6.FEA和斯彭斯的图像比较结果本文摘译自美国辛辛那提大学3D接触问题的有限元分析理学硕士2. Validation of 2D Finite Element Contact Solution2.1 Hertz Contact For validation of the classic 3D contact solution with different materials, the right place to start with a simple 2D model. Then use this experience gained in obtaining accurate results for 2D model to move to the 3D problem. The present section deals with the analysis of the Hertz contact between two cylinders. Hertz found the elastic deformation of the surfaces from Newtons optical interference fringes in the gap between the glass lenses due to the contact pressure between them. He made the hypothesis that in general contact area is elliptical from the observations of the interference fringes. Due to well developed methods available for the boundary value problem for the elastic half-space, he then simplified and assumed that each body in contact as an elastic half-space. This makes it possible to calculate the deformation at the contact. In order to satisfy the above conditions the necessary requirements are: contact area must be small compared to the dimension of each body, and contact area must be small compared to the relative radii of the curvature of the surfaces. The first requirement is necessary in order to not to influence the boundaries from the highly stressed contact region. The second requirement is necessary to make sure that the surface outside the contact region is approximately plane. Another requirement is that strains in the contact region must be small. The last but not least assumption is the contacting surfaces are assumed to be frictionless, so there are no tangential tractions atthe contact region in the Hertz problem.SUMMARY OF THE ASSUMPTIONS FOR HERTZ CONTACT1. Each body in contact is considered as elastic half-space for stress calculations.2. The contact is assumed to be frictionless.3. The contact between the two bodies is non-conforming contact.4. The strains near the contact region are assumed to be small.5. The contact area is small compared to the dimensions of the contacting bodies.2D Plane Strain AnalysisIn some engineering problems, such as a pipe under internal pressure, a dam subjected to water loading, or cylindrical roller compressed by force as in Figure 2.1, have significant strain only in one plane; strain in one direction is much less than the strains in other two directions. In such cases smaller strain is ignored and the engineering problem is solved as a 2D plane strain problem. Strain normal to the X-Y plane as shown in Figure 2.2 and shear strains and are assumed to be zero. The assumptions for plane strain are, long bodies along the Z-direction having uniform cross section subjected to loads that act only in X and/or Y directions and do not vary in Z-direction. For the present case cylinders in contact subjected to normal load is assumed as 2D plane strain problem and solved the unit thickness finite element model as shown in Figure 2.1. Contact between Two Cylinders The present numerical case involves the two cylinders in contact, whose axes are parallel to each other. Force is applied in the direction of common normal of contact along the axis of the cylinder. Cylinders deformed at the contact surfaces of size 2a due to the Load. If the two cylinders having the radii of curvature as R 1 and R 2 then relative radius of curvature is equal to R. (2.1)Material properties of the two cylinders are, Youngs modulus as E 1 and E 2 , Poissons ratio as and respectively. (2.2)Theoretical results from Johnson 20 for the above analysis are semi contact length and the peak contact pressure. Semi contact length a is given by: (2.3)The expression for pressure distribution in the contact area is given by: (2.4) Contact pressure is maximum at the center of contact and is given by the equation. At the edge of contact the contact pressure equals to zero.2.1.1 Finite Element Contact Solution without FrictionThe analysis of the above numerical case was performed using the commercial finite element software ANSYS 21. Modeling of this problem involves modeling of th of each cylinder for the present 2D analysis and applying symmetry boundary conditions. PLANE42 elements were used to mesh the two quarter cylindrical bodies. PLANE42 is 2D quad element with 4 nodes and each node has 2 degrees of freedom as translations in two directions in a plane. Regarding meshing, the mesh near the contact region is fine, having minimum element size 0.26 mm and the mesh is coarse as moved away from the contact region having maximum element size 0.92 mm. Normal load is applied along the common normal at the point of contact. Material properties of the two cylindrical bodies and load data of the contact model are shown in Table 2.1.Table 2.1: Finite element 2D Contact model data without frictionMaterial PropertiesDimensional DataCylinder 1 (B1)1. Modulus of Elasticity(E )=30000X 10Pa2. Poissons ratio ( )=0.25Radius(R )=10mmCylinder 2 (B2)1. Modulus of Elasticity(E )=2912010 Pa2. Poissons ratio ( )=0.3Radius()=13 mmLoad Data: Load (P)=3200 NElements used to mesh the contact region, from 21 are CONTA172 and TANGE169, and the contact wizard option in 21 is used here to create these contact elements on the contact region. CONTA172 is a 2D element with 3 nodes and 2 degrees of freedom in a plane at each node and is used to model surface to surface contacts.TARGE169 is a 2D element used to represent target surfaces associated with the contact elements.After solving the problem, the results semi contact length and the peak contact pressure were postprocessed. Contact length was postprocessed using the nodal solution option from the general postprocessor menu. The contact region was recognized by the color difference and also contact status of the each node can be read from the list results option in the general postprocessor. Contact length was measured as distance between the nodes that are in contact boundary. The percentage difference between the finite element solution results and the analytical solution results is 1% and it proves that finite element solution results are in agreement with Hertz analytical results. Comparison for finite element solution results and Hertz analytical solution results are shown in the able 2.2.Table 2.2: Comparison of finite element analysis results with analytical results for Hertz contact modelANSYS RESULTSANALYTICALRESULTS% DIFFERENCESEMI CONTACTLENGTH (mm)1.18721.1997 1.04PEAK CONTACTPRESSURE (XPa) 1711.2 1698.1 0.72.2 Spence SolutionIn the previous section there are some assumptions for the Hertz contact analysis such as frictionless contact, small strains at the contact, etc. The present section deals with the analysis of non-Hertzian contact problems. In the non-Hertzian contact problem, we have relaxed the frictionless contact assumption by introducing the friction between the contacts for the Hertz case. The effect of friction between the contacting cylinders will exist only if the two cylinders have the different material properties for the normal contact problem. Due to different tangential displacements between the contacting bodies, slip will take place and the total contact region becomes the combination of central stick region surrounded the slip region.The amount of slip depends upon the coefficient of friction between the contacting bodies such that if the amount of friction is more, the slip will be less and if the amount of friction is less, then the slip will be more. It is also possible to increase the coefficient of friction to an amount such that slip will be zero. There exists a tangential traction due to the friction in the slip zone. These tangential tractions act in opposite directions on the two contacting surfaces at the contact region and is given by: (2.5)As the normal load increases, the contact size also increases. Any two points in the contact region on two contact surfaces undergo different tangential displacements. Once they come into the stick zone there is no relative tangential displacement between those points. The magnitude of stress and strain grows in proportion to the contact length.The Spence 10 solution states that for the same material properties, i.e., Youngs modulus, Poissons ratio and coefficient of friction the amount of slip region is same for all the bodies those have the shape profile and is equal to the flat punch. He also concluded that under monotonic loading, the ratio of stick zone size to contact zone size depends only on the material properties, such as coefficient of friction and Poissons ratio . He was also presented the numerical solution with varying values of Poissons ratio ranging from 0 to 0.5 and observed more accurate results as approaches to 0.5The stick zone size c is given by the following equation: (2.6)is the complete elliptic integral of first kind and.Here kwas solved using the command ELLIPKE from mathematical programming tool Matlab 22. The elliptic integral of first kind can be formulated as: (2.7)where is equal to the: (2.8)The parameter is the measure of the differences in elastic constants of the two different materials. The value of is equal to zero when both the materials in contact are similar materials and also zero when both the materials have the Poissons ratio equal to 0.5, i.e., for incompressible materials. The extreme values for are 0.5 when one body is incompressible and other has zero Poissons ratio.2.2.1 Finite Element Contact Solution with FrictionThe finite element solution of non-Hertzian contact problem is also carried out in the same way as Hertzian contact except with changing some options in ANSYS. The system was modeled using of the cylinders for analysis. PLANE42 2D four noded QUAD elements were used to mesh the two contacting bodies. Elements CONTA172 and TARGE169 were used to mesh near the contact region. The mesh near the contact region is fine having the minimum element size 4.5e-2 mm and coarse mesh away from the contact having the maximum element size 0.95 mm. Friction between the two contacting bodies was specified in the material models option in ANSYS. Spence 9 presented unique solution for the contact problem with fiction independent of load path. Later Spence 10 presented solution for power law indenter that depends on geometry, material properties, friction coefficient and loading, independent of load path. We therefore have not considered load stepping in the finite element solution of this frictional contact problem.Table 2.3: Finite element 2D Contact model data with frictionMaterial PropertiesDimensional DataCylinder 1 (B1)1. Modulus of Elasticity()=30000XPa2. Poissons ratio ( )=0.25Radius ( )=10 mmCylinder 2 (B2)1. Modulus of Elasticity()=29120 X Pa2. Poissons ratio ( )=0.3Radius ( )=10 mmLoad Data:Friction:Load (P)=3200 NCoefficient of friction ()=0.01-0.05After solving the problem, t
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