FEM Spring Problems | Finite Element Analysis on Spring | Spring Analysis by FEM

[Music] hello everyone today in this lesson I am going to discuss you the finite element method on the spring elements so we can see the figure here so there are three spring elements which are connected in the series and at both the ends it is rigidly connected so the nodes which have been represented at the fixed end that is one nine four are the fixed end and the connecting from one spring to another spring there it will be unknown so this will be the node one and after that node two after the second element it will be node three and four after the third element again it would be node four so like that we have arranged in the four nodes and there are three elements three element Queens three Springs which are connected so that K 1 K 2 K 3 K 1 K 2 K 3 these are the stiffness value for the spring elements and at the node 3 there is an force which are applying towards the node 4 so that before this load is applied it means that the basic concept is that when you apply the load at the node 3 so we can find the displacement at 1 and we can obtain the displacement at two I'm at 3 also so this tree which will be under and positive deformation means it will be an tensile deformation because at the 2 it will also make the deformation in the tensile and at the 1 also it will give the tensile deformation whereas in the kk 2 well whereas in the key 3 where the deformation is under compression so this will be have a negative deformation so we will check what are the deformation which will we can obtain here now okay so now this is a question after the diagram for the spring system shown in the above now K 1 K 2 and K 3 value is mentioned where K 1 is equal to 100 Newton per millimeter the unit of the stiffness will be force per deformation that first force Newton and deformation will be millimeter and K 2 is given as 200 I kiiis 100 Newton per millimeter and the value of P P is nothing but the load that is given as 500 Newton so you need to calculate the first one the global stiffness matrix that is K case assembling of all the three elements then we obtain the global stiffness matrix again the displacement at two and three so what is the displacement at two nine three you are going to calculate so we will write here there is a displacement of to find here it will be n three and at the fixed end there is an displacement u 190 at four it will be u 4 so these are the displacement and also you need to calculate the reaction forces so reaction forces it will be obtained where we have the fixed end so at one there is a reaction forces that is we can say R 1 and reaction for set for that will be our four you need to calculate and also you need to calculate the spring in the force in the spring - so what is the force at the spring - that is also you are going to calculate so these are the question now we will calculate the stiffness matrix for each element so for the spring one so K 1 K 1 is equal to as you know a by L a by L in the bracket 1 minus 1 minus 1 1 that is for a bar element but for this spring element the K value which is directly given so we can write in the matrix form that is K 1 value is given hundred and minus hundred again - hundred and hundred so this is the stiffness equation for element 1 so after that write down the deformation for the element 1 also for the element 1 so this is 1 into 3 so for the element 1 the deformation is you want I mean you do so here it will be u 1 u 2 in the row also u 1 u 2 similarly calculate for element 2 so for the element 2 again it will be same equation same formula that is for 2 it is 200 - 200 - 200 200 again here deformation it will be changes now for the element 2 the deformation from u 2 2 u 3 so u 2 u 3 u 2 u 3 similarly for k 3 so for the k 3 the value is hundred minus hundred so minus hundred 100 so what is the deformation here u 3 u 4 again in the row u 3 u 4 so after completing the stiffness matrix for element 1 2 & 3 which are calculated now the first one you need to calculate the global stiffness matrix the global stiffness matrices is nothing but assembling of all the stiffness matrix so that is K is equal to K 1 plus K 2 plus K 3 assembling of all the ailments so now for this we have a node of we have node 4 1 2 4 so write down all the displacement that is u 1 u 2 u 3 u 4 similarly in the row u 1 u 2 u 3 u 4 now you have to look for all the three elements suppose in the first row and first column first row first column that is U 1 u 1 so u 1 u 1 the value is hundred so like that you have to calculate for on the element you want a youtube so u 1 u 2 that is minus hundred u 1 u 3 so there is no relation for u 1 u 3 so it will be 0 0 u 4 also 0 now you - you won so you - you won having - and read u2 u2 so u2 u2 this value we have 100 again we have the another value at element - also that is 200 so 200 + hundred that is 300 you can right now u2 u3 so u 2 u 3 that is minus 200 u2 u4 there is no relation so it will be 0 now you three you are no relation zero u 3 u 2 u 3 u 2 that is minus 200 u 3 u 3 so u 3 u 3 it is 200 again we have another value at Hellman 3 that is hundred so 200 a + hundred so it will be 300 again u 3 u 4 u 3 u 4 so 100 now u 1 u 4 u 1 0 u 4u to 0 u 4u 3 so you for u 3 that is minus hundred now you for you for we got as hundred okay so here u 3 u 4 so u 3 u 4 that is - and ready out right so this is global supplies matter once again I will write in the need form so here it will be or we can write here K is equal to 100 minus hundred 0 0 so minus hundred three hundred again - 200 0 0 minus 200 300 minus hundred 0 0 - cent 800 so these are the displacement now to write the displacement also u1 u2 u3 you for again u 1 u 2 u 3 u 4 ok so this is nothing but the global stiffness matrix so the first option is completed here we have calculated this global stiffness matrix so here we can see the one this ailment this matrix is M a symmetric matrix so here you can see it is an asymmetry okay so it is minus under it it's entirely symmetry about the axis now after that the equilibrium equation you've read the equilibrium equation for the whole system whole system means for the entire spring system you have to write where the equation for the stiffness matrix that is K u is equal to F so as we have calculated the difference matrix K same thing you have to write once again - hundred three hundred minus two hundred zero zero minus two hundred three hundred zero zero minus hundred hundred now after that the second one is the displacement so we have total for displacement that is u1 u2 u3 9004 is equal to the force vector so here we can denote the for sector at one there will be a fun at two there will be able to add three we have the P and at four it will be f4 now you can say there is no force at one kind for I add two also so there is a force at three only okay so same write down your fun I know for set F 2 that we can write 0 and for set 3 that is P 4 set for F 4 so this will become 0 now one and therefore by applying the boundary condition the next step is to apply the applying the boundary condition so what are the boundary conditions here the boundary conditions we have to find out now so whenever there is an fixed end when this one I'm for is a fixed end so it means that we can directly take the value of displacement it will be 0 so u 1 and u 4 that should be 0 so this is nothing but the boundary condition when it is a fixed end and also the load at F 3 we have applied load of P that is 500 Newton so this is also another boundary condition now here in the next step apply this boundary condition in the equation so again you have to write the same equation and apply the boundary condition so u 1 is equal to 0 u 4 that is again 0 and the P has 500 Newton so I made changes in the same equation so now by applying the elimination method so the displacement is zero so their first row and their first column it will get canceled and similarly the third row sorry the fourth row and their fourth column it will get cancelled so what remains here now so this is the remaining value so 300 - 200 - 200 300 so it is in the matrix now the remaining of tree elimination u2 and u3 is the remaining is equal to zero 500 so this same equation number 1 we can write so after that so after that we are again going to calculate the displacements at 2 & 3 so making the equation where 300 u2 minus 200 u 3 is equal to 0 the second equation so this into you how to multiple this one now so this column is going to multiple with the row so 200 u2 plus 300 u 3 is equal to 500 so this 2 equation which are formed after that so now just anything is possible to get cancel so we you cannot cancel because it's 300 and 200 if you make the multiply sine of 2 to the equation 1 and 3 to the equation 2 it will change us now so it will be 600 you - - 400 u 3 is equal to 0 so you have to multiple entire equation after that 3 into this equation so again minus 600 u 2 plus 900 u 3 is equal to 3 into 5 it is 1500 so this becomes the equation now so this 2 will get cancelled 600 600 plus 600 and this is minus 600 so after that 900 minus 5 400 so you'll get 500 you 3 is equal to 1500 so now you 3 is equal to 3 you will get so the displacement it will be in millimeter so this same deformation which are calculated for 3 node 3 and also for the 2 you can calculate substitute this value in any one of the equation you will get the you to value so 600 300 u2 minus 200 u 3 is equal to 0 so this is better because it's in short equation so here it will be 300 u 2 minus 200 into u3 value that is 3 is equal to 0 now u 2 is equal to it will be 600 in the RHS divided by 300 so you will get 2 mm the displacement at node 2 is 2 millimeter so the second option which are calculated now the displacement at two so after that we are going to calculate the reaction forces so reaction forces at 194 you are going to calculate please subscribe the channel

Channel: Mahesh Gadwantikar

Views: 45,219

Keywords: Analysis of Spring, Spring Problems in FEM Finite Elements methods Mechanical Engineering Finite Element Methods Lecture, Spring problems in mechanics, Spring problems in physics

Id: dxtmAOz1DFM

Length: 16min 28sec (988 seconds)

Published: Sun Apr 07 2019