 In previous video, we have studied how to generate the global matrix. Now in this video, we will substitute the local stiffness matrix into the global stiffness matrix and we will calculate the deformation stress strain of the system. Learning Outcomes At the end of this session, the learner will be able to solve uniaxial element problem analytically. I will substitute the values of k 1, k 2 and k 3 in the global stiffness matrix. So, I will formulate it here k equals to k 1 value of k 1 is 10 raise to 5 is common. So, I will write it 10 raise to 5 here. So, now just add the stiffness element stiffness matrix of first element this to this. So, it will be 3 minus 3 minus 3 3. So, now just look at here we will add k 2. So, k 2 the value is plus 1 here then minus 1 minus 1 1 then I will add k 3. So, k 3 plus 0.6 minus 0.6 minus 0.6 and 0.6 other elements are 0. So, this is global stiffness matrix. Now, the formula is to calculate the we will find the deformation at the 4 nodes. So, the formula is vector matrix of the force that is F equals to the global stiffness matrix that is k into vector matrix of deformation that is u. So, for first node the deformation is u 1 for second node it is u 2 similarly u 3 and u 4 for third and fourth node respectively. So, we will write the we will expand this matrix matrices here. So, it will be F 1, F 2, F 3 and F 4 F 1, F 2, F 3, F 4 are nothing but the forces at first node, second node, third node and fourth node ok. So, equals to the global stiffness matrix is 3 minus 3 0 0 minus 3 4 and this will be minus 1 0 0 minus 1 1.6 minus 0.6 ok. So, 0 0 minus 0.6 and 0.6 into the deformation at first node that is u 1, u 2, u 3 and u 4. So, now we will substitute the force values of forces and the deformation that we know ok. So, as you will see here we have constrained or fixed supported this end of the beam. So, at this node, this node the reaction will be 0 and all the reaction will be applied at this particular node ok or key point and the deformation at this node is 0 ok. The load is 50 kilo Newton, but the deformation is 0 here. So, we will just substitute the values here. So, this will be 50 kilo Newton. So, I will write only 50 0 0 and the force at fourth node is also 50 into 10 dash to 3 equals to we have forgot the 10 dash to 5 here of stiffness matrix 10 dash to 5. So, we will write it again 3 minus 3 0 0 minus 3 4 minus 1 0 0 minus 1 1.6 into now the reaction or deformation sorry the deformation at first node is 0 and we will calculate it for u 2, u 3 and u 4. Just formulate the equations now just multiplied by the matrix multiplication we need to know here. So, we will calculate we will form the simultaneous equations just multiplied by multiplied this matrix by this matrix ok. So, we will calculate here this will be 0 and then minus 3 u 2 equals to or into 10 dash to 5 remaining as usually I forgot here equals to the 50 into 10 dash to 3. This will be first equation our then for the second equation will be this 0 4 into u 3 u 2 minus u 3 into 10 dash to 5 equals to 0. This will be second equation then 0 u 2 minus u 2 here. Then plus 1.6 u 3 minus 0.6 u 4 equals to 0 this will be equation third and the fourth equation is minus 0.6 u 3 plus 0.6 u 4 equals to 50 this will be equation 4. Now we can solve the simultaneous equation on the calculator and we will get the values here. Obviously, the u 1 is 0 you will get u 2 equals to 0.16 mm u 3 equals to 0.66 mm and u 4 equals to 1.5 mm. You can now calculate the strain here the formula for this is deformation now the strain the formula is change in length upon original length. So, u 2 minus u 1 upon l 1 for first element. So, it will be 0.16 minus 0 upon l 1 it is l 1 is 200 mm. So, it will be 8 into 10 dash to minus 4. Similarly, for second element it will be 250 sorry the strain it is we will write formula here u 3 minus u 2 upon l 2. So, after substituting you will get the value 1.25 into 10 dash to minus 4 and for third element it will be 2.8 into 10 dash to minus 4. And then from this strain and deformation we can calculate the stress developed. So, stress will be sigma 1 equals to Young's modulus into strain. So, that will be 2 into 10 dash to 5 into 8 into 10 dash to minus 4. So, it will be 167 Newton per mm square for sigma 2 it will be 250 Newton per mm square and for sigma 3 for the third element it will be 555.53 Newton per mm square. So, in next video we will solve the similar problem in Epidel and Wackbench. So, these are the references. Thank you.