 Hello everyone, my name is Saurabh Deshmukh, working as an assistant professor in Department of Mechanical Engineering, Vulture Institute of Technology, Solapur. In this video, we are going to calculate the nodal temperature using stiffness matrix, the learning outcome. At the end of this session, the learner will be able to calculate the nodal temperature using stiffness matrix. So we are going to solve this particular problem in this video. So there are three walls, first is made up of the brick, the second is made up of the insulator and third one is again made up of the brick. So their length and the temperatures are given. So we will solve this problem using stiffness matrix. So first of all, we will convert that problem into a line problem. So this is our first node, this is first wall, this is second node, this is second wall made up of the insulator and this is third node and this is third wall made up of the brick. So this is K1, K2 and this is K3. This is first node, this is second node, this is third, this is fourth. The temperature of the first node will be T1, this is at T2, this is T3 and this is T4. So we will first write the given data here. So the T1 temperature at the first node, it is given as a 200 degree Celsius. The temperature at fourth node, it is given as T4, it is 20 degree Celsius. They are also given the K of brick, it is 80 into 10 to minus 3 Watt per mm degree Celsius. Also Ki equals to 1 into 10 to minus 3 Watt per mm degree Celsius. They are also given the area of the wall that is A1 equals to A2 equals to A3 equals to 1 mm square. Also the length of the brick wall, first brick wall it is L1, it is given as 10 mm. For the insulated wall it is 5 mm and for the brick wall, third brick wall it is L3 equals to again 10 mm. So now first we will calculate the K1 that is for first brick wall. It is K1 equals to it is A1 KB upon L1 into the distributed matrix it is 1 minus 1 minus 1 1. So we will substitute the values of A1 KB and L1 here. So A1 it is 1 into KB it is 80 into 10 to minus 3 upon L1 it is 10 into 1 minus 1 minus 1 1 equals to it is 80 into 10 to minus 3 into this matrix. So we will multiplied this matrix by 8. So it will be 8 minus 8 minus 8 8 into 10 to minus 3. Now the parameters of the first brick wall and the third brick wall it these are same. So we can say that it is K1 equals to K3 equals to it is 8 minus 8 minus 8 8 into 10 to minus 3. Now similarly we will calculate the value of K for the second wall that is insulating wall. So it is K2 equals to A2 KI upon L2 into 1 minus 1 minus 1 1. So we will substitute the values of the parameters of insulating wall in this equation. So it is K2 equals to A2 is 1 into KI is 1 into 10 to minus 3 upon L2 the length of the insulating wall it is given as L2 and it is 5 mm. So divided by 5 into 1 minus 1 minus 1 1. So it is equals to it is 0.2 into 10 to minus 3 into this matrix. So we will substitute the value as 0.2 or we will multiply by 0.2 into this matrix it will be 0.2 minus 0.2 minus 0.2 and 0.2 into 10 to minus 3. Now we will find the global stiffness matrix the global matrix we denote it by K so it is K equals to what is it? It is K1 minus K1 0 0 minus K1 K1 plus K2 minus K2 0 0 minus K2 K2 plus K3 minus K3 0 0 minus K3 and K3. So we will substitute the values of K1 K2 and K3 in this global stiffness matrix so we will get it as K equals to K1 is 8 into 10 to minus 3 K2 is 0.2 into 10 to minus 3 and K3 is also same as K1 it is 8 into 10 to minus 3. So we will substitute here and we will get the global stiffness matrix. So it is 8 minus 8 0 0 minus 8 8 plus 0.2 it is 8.2 minus 0.2 0 0 minus 0.2 it is again 8.2 minus 8 into the common is 10 to minus 3. Now we know that the equation it is Q equals to the stiffness matrix that is K global stiffness matrix into the temperature. So heat flow equals to K into temperature. So we will substitute the values here so what it will be? So I will just change the page here I will substitute the Q K and T matrices Q K and T here so it will be Q1 Q2 Q3 and Q4 equals to K11 K12 K13 K14 into the temperature the nodal temperatures are T1 T2 T3 and T4. Now we will apply the boundary conditions here so applying boundary conditions. What are the boundary conditions here? Here Q2 and Q3 are the 0. So Q2 equals to Q3 equals to 0. We know the global stiffness matrix we have calculated it and we also know the temperature at node 1 and at node 4. So it is T1 equals to 200 degree Celsius and T4 equals to 20 degree Celsius. So we will substitute these values here. We do not know the value of Q1 we know Q2 Q3 and we do not know the value of Q4. We will substitute the global stiffness matrix that we have just calculated it is 8 minus 8 0 0. We will also substitute the values of T1 and T4 here that is 200 degree Celsius. We do not know the value of T2 we do not know T3 but we know the value of fourth node temperature it is 20 degree Celsius. So we can convert this matrix form into the simultaneous equations. So it will be Q1 equals to it is into 10 to 2 minus 3. So we will write it here 10 to minus 3 for the global stiffness matrix. It is 8 into 200 minus 8 into T2 into 10 to minus 3. Also for the second row it is 0 equals to it is minus 8 into 200 plus 8.2 into T2 minus 0.2 into T3 into 10 to minus 3. Similarly, it is 0 equals to minus 0 0.2 into T2 plus 8.2 into T3 and minus 8 into T4 into 10 to minus 3. Also the fourth equation is Q4 equals to it is minus 8 into T3 plus 8 into T4 it is 20. So I will write here 20 instead of T4 it is 20 into 10 to minus 3. So this is equation 1 this is equation 2, 3 and 4. These are the four simple simultaneous equations. After solving this four simultaneous equations into the calculator, we will get the values of T2 and T3 that is T2 equals to it is 195.7 degree Celsius and T3 it is 24.28 degree Celsius. These are the references. Thank you.