ME 5510/6510 INTRODUCTION TO FINITE ELEMENTS AUTUMN 2005 Solve the following heat transfer problem (steady-state conduction). Use Link32 (2-D Conduction Bar) elements for the structure. Step1: Set Preferences To Include Thermal Analysis PreferencesÆ(window) select structural and thermal buttonsÆOK. Step 2: Define Nodal Locations PreprocessorÆModelingÆCreateÆNodesÆIn Active CSÆ(window) define node 1 with label and XYZ coordinates, hit Apply and repeat for nodes 2-4Æwhen finished, select OK to exit window. Step 3: Select Element Type PreprocessorÆElement TypeÆAdd/Edit/DeleteÆ(window) Add…Æ(window) highlight Themal Mass-Link and 2D Conduction 32,ÆOKÆCLOSE. Step 4: Define Material Properties PreprocessorÆMaterial PropertiesÆMaterial ModelsÆ(window) double click Thermal/Conductivity/IsotropicÆ(window) input value for thermal conductivity of material 1ÆOKÆrepeat for materials 2 and 3. Step 5: Define Real Constant PreprocessorÆRealConstantsÆAdd/Edit/Delete Æ(window) Add…Æ(window with element type 1 {Link 32} highlighted) OKÆ(window) input Cross-sectional area of 1*ÆOKÆCLOSE. *Note: since the material is approximated as infinite in the y-direction, we really have 1-d steady state conduction, which is the same for any arbitrary constant cross section==>use 1m2 area for all elements.Step 6: Build Elements Between the Nodes PreprocessorÆModelingÆCreateÆElementsÆUser NumberedÆThru NodesÆ(window) assign element as No. 1, select OKÆ(selection window) pick nodes 1 and 2ÆOK (creates element 1). Now change the default material to mat2 and create element 2; then change to mat3 and create element 3. (Remember you can change material type using PreprocessorÆModelingÆCreateÆElementsÆ Element Attributes). Step 7: Apply Boundary Conditions (Temperature Loads) PreprocessorÆLoadsÆDefine LoadsÆApplyÆThermalÆTemperatureÆOn NodesÆ(window) pick node 1 then select OKÆ(window) highlight TEMP only; make sure it shows Apply As: Constant Value; enter value as 500, select OK. Repeat for applying temperature at node 4. Step 8: Solve SolutionÆSolveÆCurrent LSÆ(asks you to review summary info) select OKÆANSYS will begin solving the problem and will post a message “Solution is done!” when it has finished. Close message windows and go to next step. Step 9: View Results Temperatures are analogous to displacements in a structural analysis—they are the dof allowed at the nodes. Therefore, we can list the nodal temperatures similar to how we listed nodal displacements: List Nodal Temperature: General PostprocÆList ResultsÆ Nodal SolutionÆ(window) highlight DOF solution and Temperature; select OK. You should get the following: NODE TEMP (°C) 1 500.00 2 420.27 3 121.26 4 100.00 Heat flux is analogous to stress in a structural analysis. Recall for these 1-d type elements, we need to use the element table to access element output: List Heat Flux: This needs to be done using the command input window. Type the following in the window: etable,flux,smisc,4 (enter) pretab,flux (enter) You should get the following: Element flux (W/m2) 1 3986.7 2 3986.7 3 3986.7 Note that these could have been calculated easily from the nodal temperatures using the 1-d form of Fourier’s Heat Conduction Law: which is approximated as: So, for example, the heat flux in element 2 could be calculated using know distances and temperatures as: dxdTKqxxx−=xTKqxxxΔΔ−=228.3986)060.0()27.42026.121()8.0(mWmCCmWq =−−=oo/com, Structural /com, Thermal /prep7 n,1,0,0,0 n,2,0.1,0,0 n,3,0.16,0,0 n,4,0.24,0,0 et,1,32 mp,kxx,1,5 mp,kxx,2,0.8 mp,kxx,3,15 R,1,1 mat,1 e,1,2 mat,2 e,2,3 mat,3 e,3,4 d,1,temp,500 d,4,temp,100 fini /solu solve fini /post1 etable,flux,smisc,4 /output,heat,out prnsol pretab,flux /output,, fini !set preference to include structural and thermal analysis !node 1 !node 2 !node 3 !node 4 !element type = link32 (2D conduction) !conductivity of material 1 !conductivity of material 2 !conductivity of material 3 !cross-sectional area = 1 !use material 1 !create element between node 1 and 2 !use material 2 !create element between node 2 and 3 !use material 3 !create element between node 3 and 4 !apply temperature of 500C to node 1 !apply temperature of 100C to node 4 !create result table containing heat flux !switch output to file heat.out !list nodal solution !list result of heat flux !switch output back to screen Results from heat.out: PRINT DOF NODAL SOLUTION PER NODE ***** POST1 NODAL DEGREE OF FREEDOM LISTING ***** LOAD STEP= 1 SUBSTEP= 1 TIME= 1.0000 LOAD CASE= 0 NODE TEMP 1 500.00 2 420.27 3 121.26 4 100.00 MAXIMUM ABSOLUTE VALUES NODE 1 VALUE 500.00 PRINT ELEMENT TABLE ITEMS PER ELEMENT ***** POST1 ELEMENT TABLE LISTING ***** STAT CURRENT ELEM FLUX 1 3986.7 2 3986.7 3 3986.7 MINIMUM VALUES ELEM 3 VALUE 3986.7 MAXIMUM VALUES ELEM 1 VALUE