 My name is Saurabh Deshmukh. I am working as an assistant professor in Department of Mechanical Engineering, Vulture Institute of Technology, Solaapur. In this video, we are going to learn how to perform thermal analysis using APDL. So the learning outcomes. At the end of this session, the learner will be able to perform steady state thermal analysis in APDL using link 33 element. We will solve this simple problem in APDL. So we are considering here three walls of length 10 mm, 5 mm and 10 mm. Okay. The inner temperature of the wall is T1 that is 200 degree Celsius and the outer temperature of the wall is T4 that is 20 degree Celsius. The area we are considering as a uniform that is 1 mm square and other thermal properties are given here. So we will solve this problem in APDL. So we have opened the APDL tab here. So now we need to first choose the preferences. So click on the preferences. We are going to perform the thermal analysis. So click on the thermal here and click okay. So after that we need to do preprocessors. Click on the preprocessor then element type. Add LD, delete element. Click on the add. So we are going to choose the link element here. Just click on the link and 3D conduction 33. So this is the link 33 element. So just click on okay here. Here the link 33 element is selected for the analysis purpose. Close this window. After that we need to give the real constants here. Okay so click on the real constant. Add LD, delete. Click on add. The first we have selected here the link 33 element. Click okay. So just we need to give here the real constant set number. We are going to apply the uniform cross sectional area of 1 mm square. So there is no need to give the real constant set number. Okay so it will be 1 only. Now we need to give the cross sectional area of 1 mm square. Click on okay. Let's close this. Now we need to go to the material properties. We need to change the units here. Okay so click on the temperature units. You'll find Kelvin ranking, Celsius and Fahrenheit. We are going to select the Celsius here. Click on Celsius and click on okay. Then go to the modeling. Create key points in active coordinate system. We are going to apply the key points in active coordinate system. So click on the active coordinate system. As I have told in previous video, there is no need to give the key point number. The insist APDL take it by default. So just we will give here the locations in active coordinate system for x, y and z axis. Okay so first the key point will be at origin only. So click on the apply. The second will be at 10 mm away from the origin. So x axis is 10, y and z is 0. Click on apply. The third element will be at 15 mm. Apply and the fourth element will be at 25 mm from the origin. Okay so it will be 25 and click on okay. So we'll find here the first element is created at the origin. The second is at 10 mm away from the origin. Third is at 15 mm and fourth is at 25 mm away from the origin. So now we will create the lines here. We'll join these key points. Click on the lines, lines, straight line. Select first and second key point. Second and third, third and fourth. Just click on okay here. Now we need to mesh these. Okay so before that we have not given the material properties. The first line represents the brick wall of the thermal conductivity 80 into 10 to minus 3. The second one is of insulating wall of the thermal conductivity 1 into 10 to minus 3. And the third line is also of the brick wall with the same thermal conductivity 80 into 10 to minus 3. So we need to apply the material properties here. So go to the material properties, then material models, the thermal conductivity problem and isotropic. Okay the first one is brick element of thermal conductivity 80 into 10 to minus 3 or e minus 3. Just click on the okay. So the material property of the brick wall is we have either assigned here. Now we need to assign the another element for the insulating wall. So go to the materials, new model that will be defined by the id2. So click on the okay. So you will find it here the first material number is one and the second number is two. It is the number two is of insulating wall and number one is of the brick wall which is assigned for the first line and third line. Okay so we will apply for the material properties for the insulating material. So it will be same thermal conductivity and isotropic. Just click on isotropic and it will be one into 10 to minus 3 that is e minus 3. Okay click on okay. So you will find here the thermal conductivity of isotropic material. Okay so click on close button. Now we will go for the machine. So before machine we need to apply the materials here. Okay so click on the mesh attributes. Pick lines. We will select the first and third line. Okay the first and third line of the brick material that is they are made of a brick wall. So click on both lines and click on okay. So you will find the material numbers here. You will find two numbers here. One is of brick wall and second is for insulating wall. So the first and third wall are of the brick. So click on the one and just apply it. Now we need to change the material of this middle wall that is second wall that is insulating wall here. So click on that line and then click on okay. And just change it the material from one to two. That is we have assigned here the properties of insulating wall here. Just click on okay. Now we need to go for meshing here. Go to the mesh tool. We are going to discretize these lines into one element. So click on the lines here. Select all three lines. Click on okay. And then set here the number of element division as one. Okay and then mesh it. Just click on the mesh tool again. You'll find here mesh option. Click on the mesh. Select all three lines and then mesh it. We have meshed all the three lines. So now we need to apply the load conditions. Okay boundary conditions. So click on the loads. Then define loads. Then apply and obviously this thermal analysis. So we are going to click on the thermal and temperature here. Either you can apply it on the key point or node because we have discretized the lines into one element. So the key points are also we are considering here as nodes. So either you can click on the key points or either on the nodes. There is no problem. So we'll click here on the key points. Select first element. Just click on the okay here. So we are going to consider here the temperature. Okay here all degrees of freedom. There is no problem. We are considering only temperature here. So at this key point the temperature is about 200 degree Celsius. Okay we have already changed the unit of the temperature which is we have changed from Kelvin to degree Celsius. So directly apply here the temperature value of 200 degree Celsius. Just click on okay. Then again select the key point. We are applying the temperature at this node also which is of 20 degree Celsius. So click on the okay. Just click here and put the value of 20 degree Celsius. Just click on the okay. We have applied the boundary conditions that is the temperature at this end of the wall is 200 degree Celsius and temperature at the other end of the wall is 20 degree Celsius. And we are going to calculate the temperatures at the second and third node. Okay so we'll solve this problem now. Just go to the solution. Solve current load step here. Click on the okay. The solution is done. Now we'll see the values which are occurring at the second and third node. So go to the general processing. Then list results. We are going to find the values at the node. So go to the nodal solution here. Then degrees of freedom solution. Then the nodal temperature. Click on okay. You'll get here the temperatures at the first node it is of 200 degree Celsius. For the second node it is 195.71 degree Celsius. For the third node it is 24.28 degree Celsius. And at the fourth node that is we applied it is 20 degree Celsius. You can request you guys to solve this problem with the hand calculation by using matrix method. That is stiffness it is similar to the structural problem also. So just watch our video. This solution of a simple uniaxial element using stiffness matrix. We have solved it there. For the structural you can use the same formulas for the thermal analysis for the thermal calculations. So I request you guys to solve this problem with the hand calculations. Cross check your all the nodal solutions here. So first to fourth node you will get the temperatures here. And the maximum temperature is obviously occurring at the first node that is 200 degree Celsius. So I request you guys to solve this same problem with the hand calculations. And just cross check your answers. These are the references. Thank you.