 Hello friends, 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 see the direct formulation of uniaxial element using matrix method, learning outcomes. At the end of this session, the learner will able to solve uniaxial element problem analytically using matrix method. So we are going to solve a simple step bar problem. You will find here three step bars. One is of area 300 mm square, second is of area 200 mm square and third is of cross sectional area 90 mm square. And the length of first bar is 200 mm, the length of second bar is 400 mm and the length of third bar is 300 mm. The one end of the bar is fixed supported and from other end we are applying tensile force of 50 kilo Newton. So we will write first the given data. The area of first bar suppose we will call it as A1. So A1 equals to it is 300 mm square. The area of second bar suppose it is A2 equals to it is 200 mm square and A3 it is 90 mm square. The length of first bar is L1 equals to 200 mm, L2 it is 400 mm and L3 it is 300 mm. We will take a material as a simple structural steel of Young's modulus E equals to 200 gigapascal. We will convert it into Newton per mm square. So it will be 2 into 10 dash to 5 Newton per mm square. And the applied force is of course here we have written it F equals to 50 kilo Newton. So first of all we will convert this bar into a single line bar line problem. So this is the fixed support and the first bar will present by first line. Here is this end point of this bar then we will start the second line for to represent the second bar and then third. So we will apply here force F equals to 50 kilo Newton. So we have converted it into a line problem. So this is the first node, this is second node or we will call it as a key point here. This is third key point and this is fourth key point. So we will now calculate the element stiffness matrix. We denote it by letter K. So for first bar we will say it is K1, for second bar K2 and third bar it is K3. So we will calculate it K1 equals to its formula is A1 E1 by L1 into the matrix 1 minus 1 1. So we will substitute here the values, respective values that is 300 into E is 2 into 10 plus to 5 upon L1 it is 200. This matrix is called as distribution matrix 1 minus 1 minus 1 1. So we will calculate it, it will get cancelled, 0 will get cancelled. So 3 into 10 plus to 5 that is 1 minus 1 minus 1 1. So we will just insert these values into the matrix, so 10 plus to 5 outside the matrix only, just sake of calculation convenience. So 10 plus to 5 into 3 minus 3 minus 3 3. We have calculated this for first element K1. Similarly we will go for the second element that is K2. So I will just shift here. Now we will calculate it for K2 equals to, similarly we will get the values for the second bar. So A2 E2 upon L2 into the distribution matrix 1 minus 1 minus 1 1. So we will substitute the values here, A2 equals to 200 into E2 the Young's modulus of second bar is we are selecting the same element, so same material here. So that will be 2 into 10 plus to 5 upon L2 it is 400 into 1 minus 1 minus 1 1. Equals to it will get cancelled to 2 0 also, so it will be 1 into 10 plus to 5 into minus 1 sorry just here 1 minus 1 minus 1 1. So just it will be same matrix as it is because we are multiplying by 1. So for K3 now equals to A3 E3 by L3 into the distribution matrix 1 minus 1 minus 1 1 equals to we will substitute the values of third bar here. So that will be 90 into it is 2 into 10 plus to 5 upon 300. So into the stiffness matrix 1 minus 1. So it will be it will become 0 will get cancelled it is 0.6 into 10 plus to 5 1 minus 1 minus 1 1 multiplied 0.6 into the matrix so 10 plus to 5 at it is 0.6 minus 0.6 minus 0.6 and 0.6. So we have calculated the stiffness element matrix for each element. So the first element, the second element and the third element here. So now we will formulate the global stiffness matrix. So for the global stiffness matrix it is shown by K I will write it here global stiffness matrix K equals to it is it is very simple to formulate the global stiffness matrix from the element stiffness matrix just there are 4 nodes 4 key points. So the global stiffness matrix is of order 4 by 4. So just see here how we formulate the global stiffness matrix first write K1 then minus K1 minus K1 K1. For the first row second row and first column and second column we have written the stiffness the element stiffness matrix of first element. So just we have to continue it for the second and third element. So I will add it here K2 same here what is here? So just 1 minus 1 minus 1 1. So K2 minus K2 now this is also minus K2 and K2. And just add here in this K2 this stiffness element stiffness matrix of third element. So I will just add here plus K3 minus K3 minus K3 and K3 just complete this matrix here and the remaining parts or the element of the matrix will be 0 just write it 0 0 0 0 0 and 0 this is real simple to formulate the global stiffness matrix. In this video we have studied how to form a global stiffness matrix. Now in next video we will learn how to substitute the local stiffness values into global stiffness matrix and how to find the stresses the strain and the deformation. So these are the references thank you.