 Deshmukh, working as an assistant professor in Department of Mechanical Engineering, Wollchen Institute of Technology, Solapur. In this video, we are going to find the shape function for three-noded 1D element, the learning outcome. At the end of this session, the learner will be able to find the shape functions for three-noded 1D element. So, we will find the shape functions for three-noded 1D element. We consider a truss element here. This is the first node of that truss element. This will be our second node and this will be our third node of that particular truss element. Let us assume the displacement at first node will be u1, the displacement at second node u2 and displacement at third node as u3. Now, we will take a reference plane at first node. So, this will be our reference plane. So, for the first element with the reference plane, the Cartesian coordinate will become 0. So, at first node, the x1 equals to 0. Assume the length of this element as l. So, total length of this element will be l. So, the Cartesian coordinate of the second node will be l by 2. That is, x2 equals to l by 2. And from the reference plane, the Cartesian coordinate of the third point that is x3 equals to l because the length of that element is l. Now, for 1D truss element, the displacement at any point within the element can be expressed by the quadratic expression. So, the quadratic equation becomes u equals to a1 plus a2x plus a3x square. This will be our equation 1, where u is displacement or the field variable at any point within the element a1, a2, a3 are the constants and x is the Cartesian coordinate of that particular point. So, now we will apply the boundary conditions here. Now, at x equals to x1 or at x equals to x1 means it is 0. What is the value of u here? It is u1. So, u equals to u1. Therefore, from equation 1, what we will get? u1 equals to a1 and these x are 0. So, these terms will be eliminated. So, this is the value of our first constant that is a1. Now, at x equals to x2 which is l by 2. So, at l by 2, what is the displacement? It is u2. So, u equals to u1. So, the equation 1 becomes so, u2 equals to what is the value of a1? It is u1 plus a2 into x it is l by 2 plus a3 it is l square by 4. This will be our equation 2. Also, at x equals to x3 equals to l, u equals to it is u3. Therefore, u3 equals to u1 plus a2 into value of x it is l plus a3 l square. This is our equation 3. Now, we need to calculate the values of constants that is a2 and a3. So, we need to find the value of a2 and a3. So, we need to solve these two equations that is equation 2 and equation 3. Therefore, from equation 2, we get if you multiplied equation 2 by 2, we will get 2 u2 equals to 2 u1 plus if you multiplied this by 2, 2 will get cancelled. So, it is a2l plus a3 l square by 2. This is our equation 4. Now, we need to subtract equation 4 from equation 3 to calculate the values of a2 and a3. So, equation 3 minus equation 4. So, what we will get? It is u3 minus 2 u2. u3 minus 2 u2 equals to u1 minus 2 u2 it is minus u1. a2l will get cancelled and for a3 it will be plus l square by 2 a3. So, we will get the value of a3 here. So, what will be a3 here? a3 equals to it will be 2 u3 plus 2 u1 minus 4 u2 upon l square. This is value of a3. Now, we need to calculate the value of a2. So, we will substitute this value of a3 in equation 3. So, from equation 3 we will get from equation 3 it is u3 equals to u1 plus a2l plus a3. The value of a3 is 2 u3 plus 2 u1 minus 4 u2 by l square. So, we will substitute it here 2 u3 plus 2 u1 minus 4 u2 upon l square multiplied by l square this l square. So, l square square we will get cancelled. Therefore, u3 equals to u1 plus a2l plus 2 u3 plus 2 u1 minus 4 u2. So, from this equation we will get the value of a2. It will be a2 equals to 4 u2 minus 3 u1 minus u3 upon l. If we simplify this equation we will get the value of a2. Now, we need to substitute this value of a1 a2 and a3 in equation 1. So, from equation 1 it is u equals to what is the value of a1 it is u1. So, it is u1 plus what is the value of a2 it is 4 u2 minus 3 u1 minus u3 upon l into x plus we need to substitute the value of a3 also. So, what is the value of a3? It is 2 u3 plus 2 u1 minus 4 u2 upon l square. So, we will substitute the value of a3 here it will be 2 u3 plus 2 u1 minus 4 u2 upon l square into x square. So, we need to further simplify this equation to get the shape functions of three noted one detrus element. So, we will solve this equation will be u equals to u1 plus multiply this x into this bracket and multiply this bracket by x square. So, what we will get? 4 u2 x by l minus 3 u1 x by l minus u3 x by l plus 2 u3 x square by l square plus 2 u1 x square by l square plus sorry this is minus 4 u2 by l square into x square. After simplification we will get u equals to we will collect the terms of u1. So, this is u1 another u1 is here and the third u1 is here. So, we will detect common u1. So, it will be 1 plus this will be minus actually minus 3 x by l plus 2 x square by l square u1 plus. Now, we will collect the terms with u2 and take u2 as a common. So, this is u2 and the second u2 is here. So, we will take common as u2 it will be 4 x by l minus this is 4 x square by l square u2 plus u3 it will be minus x by l plus another term is here it will be 2 x square by l square u3. Therefore, u equals to n1 this bracket we will consider as n1 u1 plus n2 u2 plus n3 u3 where n1 equals to 1 minus 3 x by l plus 2 x square by l square. The value of n2 is 4 x by l minus 4 x square by l square and the value of n3 that is third shape functions will be minus x by l plus 2 x square by l square. These are the shape functions for 3 nodded 1 d truss l. These are the references. Thank you.