ANSYS热应力分析实例教学教材

第四讲 热分析上机指导书CAD/CAM 实验室，USTC实验要求：1、通过对冷却栅管的热分析练习，熟悉用ANSYS 进行稳态热分析的基本过程，熟悉用直接耦合法、间接耦合法进行热应力分析的基本过程。

2、通过对铜块和铁块的水冷分析，熟悉用ANSYS 进行瞬态热分析的基本过程。

内容1：冷却栅管问题问题描述：本实例确定一个冷却栅管（图a ）的温度场分布及位移和应力分布。

一个轴对称的冷却栅结构管内为热流体，管外流体为空气。

冷却栅材料为不锈钢，特性如下： 导热系数：25.96 W/m ℃弹性模量：1.93×109 MPa热膨胀系数：1.62×10-5 /℃泊松比：0.3边界条件：（1）管内：压力：6.89 MPa流体温度：250 ℃对流系数249.23 W/m 2℃（2）管外：空气温度39℃对流系数：62.3 W/m 2℃假定冷却栅管无限长，根据冷却栅结构的对称性特点可以构造出的有限元模型如图b 。

其上下边界承受边界约束，管内部承受均布压力。

练习1－1：冷却栅管的稳态热分析步骤：1. 定义工作文件名及工作标题1) 定义工作文件名：GUI: Utility Menu> File> Change Jobname ，在弹出的【Change Jobname 】对话框中输入文件名Pipe_Thermal ，单击OK 按钮。

2) 定义工作标题：GUI: Utility Menu> File> Change Title ，在弹出的【Change Title 】对话框中2D Axisymmetrical Pipe Thermal Analysis ，单击OK 按钮。

3) 关闭坐标符号的显示：GUI: Utility Menu> PlotCtrls> Window Control> Window Options ，在弹出的【Window Options 】对话框的Location of triad 下拉列表框中选择No Shown 选项，单击OK 按钮。

Temperature distribution in a CylinderWe wish to compute the temperature distribution in a long steel cylinder with inner radius 5 inches and outer radius 10 inches. The interior of the cylinder is kept at 75 deg F, and heatis lost on the exterior by convection to a fluid whose temperature is 40 deg F. The convection coefficient is 0.56 BTU/hr-sq.in-F and the thermal conductivity for steel is 0.69 BTU/hr-in-F.1. Start ANSYS and assign a job name to the project. Run Interactive -> set working directory and jobname.2. Preferences -> Thermal will show -> OK3. Recognize symmetry of the problem, and a quadrant of a section through the cylinder is created using ANSYS area creation tools. Preprocessor -> Modeling -> Create -> Areas -> Circle -> Partial annulusThe following geometry is created.4. Preprocessor -> Element Type -> Add/Edit/Delete -> Add -> Thermal Solid -> Solid 8 node 77 -> OK -> Close5. Preprocessor -> Material Props -> Isotropic -> Material Number 1 -> OKEX = 3.E7 (psi)DENS = 7.36E-4 (lb sec^2/in^4)ALPHAX = 6.5E-6PRXY = 0.3KXX = 0.69 (BTU/hr-in-F)6. Mesh the area and refine using methods discussed in previous examples.7. Preprocessor -> Loads -> Apply -> Temperatures -> NodesSelect the nodes on the interior and set the temperature to 75.8. Preprocessor -> Loads -> Apply -> Convection -> LinesSelect the lines defining the outer surface and set the convection coefficient to 0.56 and the fluid temp to 40.9. Preprocessor -> Loads -> Apply -> Heat Flux -> LinesTo account for symmetry, select the vertical and horizontal lines of symmetry and set the heat flux to zero.10. Solution -> Solve current LS11. General Postprocessor -> Plot Results -> Nodal Solution -> TemperaturesThe temperature on the interior is 75 F and on the outside wall it is found to be 45. These results can be checked using results from heat transfer theory.BackThermal Stress of a Cylinder using Axisymmetric ElementsA steel cylinder with inner radius 5 inches and outer radius 10 inches is 40 inches long and has spherical end caps. The interior of the cylinder is kept at 75 deg F, and heat is lost on the exterior by convection to a fluid whose temperature is 40 deg F. The convection coefficient is 0.56 BTU/hr-sq.in-F. Calculate the stresses in the cylinder caused by the temperature distribution.The problem is solved in two steps. First, the geometry is created, the preference set to'thermal', and the heat transfer problem is modeled and solved. The results of the heat transfer analysis are saved in a file 'jobname.RTH' (Results THermal analysis) when you issue a save jobname.db command.Next the heat transfer boundary conditions and loads are removed from the mesh, the preference is changed to 'structural', the element type is changed from 'thermal' to 'structural', and the temperatures saved in 'jobname.RTH' are recalled and applied as loads.1. Start ANSYS and assign a job name to the project. Run Interactive -> set working directory and jobname.2. Preferences -> Thermal will show -> OK3. A quadrant of a section through the cylinder is created using ANSYS area creation tools.4. Preprocessor -> Element Type -> Add/Edit/Delete -> Add -> Solid 8 node 77 -> OK ->Options -> K3 Axisymmetric -> OK5. Preprocessor -> Material Props -> Isotropic -> Material Number 1 -> OKEX = 3.E7 (psi)DENS = 7.36E-4 (lb sec^2/in^4)ALPHAX = 6.5E-6PRXY = 0.3KXX = 0.69 (BTU/hr-in-F)6. Mesh the area using methods discussed in previous examples.7. Preprocessor -> Loads -> Apply -> Temperatures -> NodesSelect the nodes on the interior and set the temperature to 75.8. Preprocessor -> Loads -> Apply -> Convection -> LinesSelect the lines defining the outer surface and set the coefficient to 0.56 and the fluid temp to 40.9. Preprocessor -> Loads -> Apply -> Heat Flux -> LinesSelect the vertical and horizontal lines of symmetry and set the heat flux to zero.10. Solution -> Solve current LS11. General Postprocessor -> Plot Results -> Nodal Solution -> TemperatureThe temperature on the interior is 75 F and on the outside wall it is found to be 43.12. File -> Save Jobname.db13. Preprocessor -> Loads -> Delete -> Delete All -> Delete All Opts.14. Preferences -> Structural will show, Thermal will NOT show.15. Preprocessor -> Element Type -> Switch Element Type -> OK (This changes the element to structural)16. Preprocessor -> Loads -> Apply -> Displacements -> Nodes(Fix nodes on vertical and horizontal lines of symmetry from crossing the lines of symmetry.)17. Preprocessor -> Loads -> Apply -> Temperature -> From Thermal AnalysisSelect Jobname.RTH (If it isn't present, look for the default 'file.RTH' in the root directory)18. Solution -> Solve Current LS19. General Postprocessor -> Plot Results -> Element Solution - von Mises StressThe von Mises stress is seen to be a maximum in the end cap on the interior of the cylinder and would govern a yield-based design decision.Back。

文档收集于互联网，已重新整理排版.word版本可编辑.欢迎下载支持. 6-1•本章练习稳态热分析的模拟，包括：A. 几何模型B. 组件-实体接触C. 热载荷D. 求解选项E. 结果和后处理F. 作业6.1• 本节描述的应用一般都能在ANSYS DesignSpace Entra或更高版本中使用，除了ANSYS Structural• 提示：在ANSYS 热分析的培训中包含了包括热瞬态分析的高级分析K T T= Q T –在稳态分析中不考虑瞬态影响–[K] 可以是一个常量或是温度的函数–{Q}可以是一个常量或是温度的函数•上述方程基于傅里叶定律：• 固体内部的热流（Fourier’s Law）是[K]的基础；• 热通量、热流率、以及对流在{Q} 为边界条件；•对流被处理成边界条件，虽然对流换热系数可能与温度相关•在模拟时，记住这些假设对热分析是很重要的。

•热分析里所有实体类都被约束：–体、面、线•线实体的截面和轴向在D esignModeler中定义• 热分析里不可以使用点质量（Point Mass）的特性•壳体和线体假设：–壳体：没有厚度方向上的温度梯度–线体：没有厚度变化，假设在截面上是一个常量温度• 但在线实体的轴向仍有温度变化•唯一需要的材料特性是导热性（Thermal Conductivity）•Thermal Conductivity在Engineering Data 中输入•温度相关的导热性以表格形式输入若存在任何的温度相关的材料特性，就将导致非线性求解。

•对于结构分析，接触域是自动生成的，用于激活各部件间的热传导–如果部件间初始就已经接触，那么就会出现热传导。

–如果部件间初始就没有接触，那么就不会发生热传导（见下面对pinball的解释）。

–总结：–Pinball区域决定了什么时候发生接触，并且是自动定义的，同时还给了一个相对较小的值来适应模型里的小间距。

•如果接触是Bonded（绑定的）或no separation（无分离的），那么当面出现在pinball radius内时就会发生热传导（绿色实线表示）。

热流体在代有冷却栅的管道里流动，如图为其轴对称截面图。

管道及冷却栅的材料均为不锈钢，导热系数为1.25Btu/hr-in-oF，弹性模量为28E6lb/in2泊松比为0.3。

管内压力为1000 lb/in2，管内流体温度为450 oF，对流系数为1 Btu/hr-in2-oF,外界流体温度为70 oF，对流系数为0.25 Btu/hr-in2-oF。

求温度及应力分布。

7.3.2菜单操作过程7.3.2.1设置分析标题1、选择“Utility Menu>File>Change Title”，输入Indirect thermal-stress Analysis of a cooling fin。

2、选择“Utility Menu>File>Change Filename”，输入PIPE_FIN。

7.3.2.2进入热分析，定义热单元和热材料属性1、选择“Main Menu>Preprocessor>Element Type>Add/Edit/Delete”，选择PLANE55，设定单元选项为轴对称。

2、设定导热系数：选择“Main Menu>Preprocessor>Material Porps>Ma terial Models”，点击Thermal，Conductivity，Isotropic，输入1.25。

7.3.2.3创建模型1、创建八个关键点，选择“Main Menu>Preprocessor>Creat>Keypoints>On Active CS”，关键点的坐标如下：3、设定单元尺寸，并划分网格：“Main Menu>Preprocessor>Meshtool”，设定global size为0.125，选择AREA，Mapped，Mesh，点击Pick all。

7.3.2.4施加荷载1、选择“Utility Menu>Select>Entities>Nodes>By location>X coordinates，From Full”，输入5，点击OK，选择管内壁节点；2、在管内壁节点上施加对流边界条件：选择“MainMenu>Solution>Apply>Convection>On nodes”，点击Pick，all，输入对流换热系数1，流体环境温度450。