FINITE ELEMENT LECTURE 01

hello good afternoon everyone my name is mohammed pervez second faculty of mechanical department is engineering college after completing my master's in mechanical engineering with specialization in computer edit design and computer added manufacturing i've been working as an assistant professor at isl engineering college for four years during my tenure i have taught many core subjects like engineering mechanics fluid mechanics strength of materials finite element analysis and hydraulic machinery and systems cad cam and in this semester i will be teaching one of the core subjects for both civil and mechanical department that is finite element analysis is the name of the subject for mechanical department f e a is for mechanical department and finite element techniques is the name of the subject for civil department however the starting two units are same so this will be a common lecture for both the departments so your respective videos will be posted in your group so that there will be no confusion so so let's get started with the subject so finite element analysis is one of the modern techniques which is employed by many departments especially in the field of mechanical civil and aerospace to analyze the different complex engineering problems so all those problems which are very difficult to solve by using the techniques of strength of materials so we can use or we can solve problems other than that also so which are not be able which cannot be solved by using the conventional formulas which we have used there so you might have studied about the different softwares like modeling software one of the commonly used modeling software is solidworks and civil engineers might have used revit and for analysis purpose you might have used ansys okay these are design softwares ansys and stat pro nissa nastran abacus so all these are analysis softwares so what runs behind these softwares is the techniques which we are going to study in this subject like what you will be doing is you will be preparing a model in solidworks and or you would be preparing a model in revit what i can do is i can export this model i can export this model to ansys or any other design softwares so what i can do in this because i would not be able to apply loads here i can only create a model for instance i prepare i prepared a cantilever beam here or a condenser here so what i would do is since i would not be able to apply loads i will export this model to other software so that is my analysis softwares so what i can do here i can apply loads and i can give boundary conditions so that which is the most critical part of that model i can analyze so why i am finding that critical part because if at all that model breaks in the future the critical part will be the starting point of failure so in order to understand this let's take one simple example which you have this problem which you might have solved in strength of materials okay let's take a bar which is subjected to a load of p is equals to some thousand newtons so let's say l is the initial length of the bar so what will happen because of this point load p okay so this is the fixed support so when i am applying the load p is equal to thousand newtons here so the length of bar will be increased so we were finding this increase in length of the bar dl by using the formula p into l by a this is what we were doing in strength of materials okay so dl so we would have got the increase in length we would also be able to find out the stresses stress is equal to load by area so why we are finding this stress value is to see whether the stress is crossing the ultimate stress of this material if it crosses the ultimate stress what will happen it will reach the breaking point so once it reaches the breaking point this will break okay so this member will fail once it reaches its breaking point so that can be analyzed by finding out these stresses so this is what we are able to do in these softwares also like we we can find out the values of stresses enhances but i can't be able to apply loads in this software solidworks so what i am doing i am only making a model because solidworks is a simple software so i can make models here and i can export this models to any other analysis software so i can apply loads i can apply loads here and i can find out the values of the displacement and stresses so what techniques these softwares are using these softwares are using finite element techniques the main idea of this technique is to divide a model into number of small and simpler elements so what a finite element software does is what a design software does is it will divide this model into number of elements okay so after applying load let's say i am applying a load of p is equals to 1000 newtons so what will happen to each and we can analyze what is what is the behavior of each and every element because of this load behavior in the sense how much this node how much this element will elongate and what are the stresses developed in each and every element of this model can be analyzed by using finite element techniques although we have softwares available for this why we are studying is to interpret the results in a better way because sometimes what might happen is you might give loads and everything and the results what you are getting might be wrong so you can find out those results you can find out you can use the calculation part mathematically and you can correlate those results so that your results are in order to find out whether the results you got in these softwares are correct or not so that's the reason why we are studying this subject okay so the simplest idea of finite element analysis is to divide a model into number of simpler elements called as finite element so this is called as discretization discretization is the first step of finite element analysis so your first step in finite element analysis is discretization discretization means the software will divide this element into number of elements that is called as discretization so that is the first step of finite element analysis is to divide a model into number of simpler elements so however while solving problems what we would be doing is we will divide it manually but in software we have a direct step for that that is called as in many softwares we have mesh option so when you click on mesh it will ask how many number of elements and what is the type of element you would want to divide this model into so that is called as machine for this machine also specially softwares are there like hypermesh is one of the commonly used softwares for this if you divide it into more number of elements and in a better way again there's a software for this hypermesh so what you can do is you can create a model in modeling software okay so first step is you have created a model next step is to divide that model into number of simpler elements so you can export this model from solidworks to hypermesh so what we have to do is there are many extension formats one of the extension formats so what you might be doing is you will save this file you will save this file in such a way that you can be able to open this file in any other software so that one of the extension commonly used is iges so when you save you have an option save as dot iges so that is called as initial graphic exchange specification so when you are saving this file in this format dot iges you can open this file in hypermesh so here you can discretize or you can mesh both are same both are synonyms both are same discretization or meshing anything is to divide into a number of elements so there's a software software for this hypermesh is one of the commonly employed softwares for this so next you can export this model again to one of these softwares to give the loads and to find out the values of stresses and strains or the displacements okay so that's the introduction about finite element analysis so next what we would be doing is we would be finding out the second step first step is discretization second step is to form a mathematical model of each and every element because uh we are not solving in this subject we are not solving by using solving a problem by using a software we have to do it manually in the exam so we have to find out the mathematical formulation of each and every element so those that mathematical formulation of every element first i will find out the element stiffness matrix why we are doing it in the form of a matrix that is because our system our software takes everything in terms of the matrices for example if i take it in terms of matrices what what i would be able to do is for example i am finding out the displacement of this element i would be able to find out the displacement at this point and the displacement of this point okay i am exaggerating this element i am showing here so this is the bigger view of this element so i would be able to find out the displacement at two different points so you would be able to understand once i start the problem so far for second step is to find out the element stiffness matrices next what i will do i will assemble all those elements and i will find out the stiffness of this complete model so next next step is assembly of stiffness matrices to find out global stiffness matrix the global stiffness matrix will give us the stiffness of this complete model okay so why we are finding out stiffness because what is stiffness stiffness is load per unit deflection or displacement since i am the one who is giving load and when i am able to calculate the stiffness i am giving the load being a being an engineer i am giving the load because i have to find out the design i have to find out the design so once i am able to find out the stiffness stiffness is denoted by k so i can find out the displacement how much because of this load p what is the displacement of this point or any other point inside this model so next after finding out the stiffness next i will find out the force matrix next is substituting substituting in finite element equation the finite element equation is k into u is equals to f because k is force by displacement in finite element analysis u stands for displacement f is also called as force or load so i can say that finite element equation is k into u is equals to f after this i can find out this precision strains once we are able to find out displacement we can find out stresses and strains so while solving a problem uh these five steps we will be using to solve any long answer or short answers so let's start with the simplest problems that is called as bar elements so this is a type of bar element why i am calling calling this as bar element because this is only subjected to longitudinal loads longitudinal loads because this member this member will only will only have displacements in this direction okay so if this would be if this rod would be in this way a cantilever beam why i am not calling this as beam and why i am calling this as name because beams are subjected to lateral loads beams are also beams are subjected to lateral nodes like we have beam here in this classroom that's a beam okay so beams always have lateral loads so this is a lateral load this direction along the length is called as if there is a load along the length that is called as longitudinal nodes if there is a load in the lateral direction then what will happen to this member this member will bend also okay and also there will be a displacement in this direction that is y direction in this case there is no bending at all there is no bending because of this force there is only displacement okay but in this case because of this load because this is fixed you will be having bending as well as the displacement in the downward direction that is y direction so this was called as beams because beams will have two types of displacements you will have to remember beams have two types of displacements one is movement in the long lateral direction y direction and as well as bending in this case you only have movement but there is no bending so this is called as bar and this is called as beams so we will be solving in the first unit we will be solving the simplest problems that is called as problems on bar elements okay so these are called as one dimensional bar elements the reason why we are calling it as one dimensional bar elements because there is displacement only in one direction okay there is displacement only in one direction whereas for beams there are two two types of displacements okay so let's start the problems on bar elements i will directly get started with the problems because theoretical part has already been explained and i will give you the notes also in order to understand the theoretical part so these are the five steps we will be using so i will write down the formulas which will be using for solving bar element problems bar problems so the first formula is stiffness matrix so first step is to find out the stiffness matrix of course we will be discretizing we will be discretizing manually but there is a formula for stiffness so stiffness matrix for bar so stiffness is denoted by k is equals to a into e by l 1 minus 1 minus 1 1 for now you remember it as formula i will be explaining why this formula is in this way while we are solving problems okay so k is called as stiffness k is stiffness a is area of cross section is called as young's modulus also called as modulus of elasticity l is length of the bar or length of the element also length okay so these are the terms k is stiffness a is area of cross section is called as young's modulus or modulus of elasticity and l is length of the element so that's the first formula for solving problems on bar element so first we have to use this formula for finding out stiffness of each and every element okay depending on the number of it depends on the number of elements we are dividing that model into so that's the first formula you'll have to remember this formula okay so next second step what we will be doing is we will be forming all these stiffness matrixes then we will be assembling all the stiffness matrix that's a method next we have force matrices so what are the different types of forces we have i will erase this part so on a member or on a beam what are the different types of force you can have okay forces so you can have a point force nexter is self weight of the member self weight of the member because weight is also a force because of gravity so there is also a weight force it is because of gravity of course whatever the analysis we are doing we are doing it on earth so there will be gravity so you have to take the weight force also into consideration but sometimes the weight is negligible so you can understand it by seeing the terms what has been given in the question i will help you understand how to take this weight into consideration or when to take this weight into consideration and when not to take this weight into consideration next is point force self-rate force temperature force why i am calling the temperature as force temperature load also we can say force or load in a better way okay so what will happen because of temperature load what will happen because of temperature for example i am giving i am heating something i am hitting let us say i am heating a fixed beam okay so what will happen to this fixed beam so when i'm heating this we know that solids expand on heating okay this is a solid member and when i'm hitting it solids expand on heating so since you have fixed supports on both the sides you have fixed supports on both the sides the solid which i am hitting wants to expand but what the supports are doing the supports are stopping this from expanding the supports are stopping the supports are not allowing it to expand so this solid wants to expand in every direction it will expand in this direction also it will expand in this direction also it will expand in z direction also but along this direction of length what the supports are doing the supports are not allowing it to expand so this supports are preventing it from expansion by giving some force and because of this force the stresses will develop whenever i am trying to stop a member from expanding by giving supports in the lateral direction what will happen the stresses will be developed so this supports are not allowing this member to expand so the stresses will develop if if this would have been if there would be no supports what will happen every member has a support on this earth but if there are no supports then this is a case of free expansion in case of free expansion no stresses will be developed in case of free expansion because i am not stopping it from anything i am not stopping it it can expand in this direction it can expand in this direction this member can expand in all the directions since i am not stopping this from anything so that's a case of in this case no stresses are developed in this case you have stresses because the supports are preventing it from expanding okay whenever like if you want to do something and someone is preventing you to do from that thing so what will happen in that case stresses will develop in your mind so think in that way so that it would be easier for you to understand so so the supports are stopping this member from preventing so stopping this from expanding so in this case you have stresses developed so there can be stresses developed because of temperature loads there can be stresses which can be developed because of temperature load so you have point force self weight force temperature force or temperature load next is udl and uvl we all know what is udl uniformly distributed load and uniformly varying loads okay so what are the formulas for these types of floats we have point load is simply a value like we took one example there it it's directly acting at one point so we will take that p okay we will add that wherever we have point load we will add that p in terms of matter because we are solving in terms of matrices we will add this in terms in this way next is self weight the self weight force is given by rho g a l by 2 minus 1 minus 1 this is the formula for self weight this row is called as density of that material what will the density indicate the density will indicate the heaviness or lightness of any body mass per unit volume is called as density so how heavy a body is or how what is the weight of that body can be determined by using this term density so that's the reason we have density in this formula g is acceleration due to gravity so of course weight is because of gravity and that value of acceleration due to gravity is 9.81 meters per second square a is area of the member and l is length of the member and these are constant values okay so that's the formula for self weight okay this is point force it's a direct value we don't have any formula for that so self weight formula is rho g a l by 2 minus 1 and minus 1 in the matrix okay this is also a matrix here in stiffness matrix and we have a matrix here i will explain how we are using this matrices okay so one thing uh many students what they make mistake in the final exam is when they are substituting rho in terms of what they do is they substitute area in terms of millimeters millimeter square length in terms of millimeters but what they do is they substitute g as 9.81 but g is 9.81 meters per second square so there you should be very careful so while substituting this g value so g is 9.81 it's not just 9.81 constant value you have units also okay so that small g value is 9.81 meter per second square so if you write down 9.81 here this is in terms of meters meters per second square so you should be substituting area also in terms of meters and length also in terms of meters okay got it i think so if your area is in terms of millimeters and everything is in terms of millimeters then you can convert this also in terms of millimeters so you have to convert this meters in terms of millimeters so you can convert you have conversion okay so one meter is okay milli meter when i say milli means that is 10 power minus 3 meters when i say 70 so value of centi is 10 power minus 2 meters if you remember these two it will help you in many ways to do the conversions because the first step in solving a problem in finite element analysis is to convert the units in one form so i want you all to remember these two techniques one is for me when i say milli it is it means 10 power minus 3. so 10 power minus 3 meters when i say 70 centi means 10 power minus 2 so centimeter is 10 power minus 2 meters apart from this we have other values also kilo when i say kilo that means thousand like we say kilogram that means 1000 grams like sometimes we say 2019 as 2k 19. so k is thousand there so value of k kilo is 1000 10 power 3 sometimes we have mega mega is 10 power 6 next is giga giga is 10 power 9 okay so i want you all to remember these five things so these five things are sufficient for you all to do number of conversions okay for instance we have these two are in terms of millimeters and i want to convert this in terms of millimeters okay but what i know millimeter is 10 power minus 3 meters so what is meter if i send this 10 power 3 to the other side this will become 10 power 3 millimeter so 1 meter is 10 power 3 millimeter so 9.81 so in place of meters i will be writing 10 power 3 millimeter per second square okay so what i have done in place of meter i have written 10 power 3 millimeters because if i send this 10 power 3 to the other side so it will become meter is equal to 10 power 3 millimeter so meter is equals to 10 power 3 millimeter so in place of meter i can substitute 10 power 3 millimeter per second square so value of g is eight one into ten power three millimeter per second square you should be very careful while substituting these formulas when solving problems okay so next is so this was about the self weight force self weight force formula you have to remember rho g a l by 2 minus 1 minus 1 okay next is the for formula of temperature force so temperature force is given by e a alpha delta t minus 1 1 e a alpha delta t minus 1 1 where e is called hence modulus okay we have already written here young's modulus or modulus of elasticity a is area of cross section alpha is called as coefficient of thermal expansion alpha is called as coefficient of thermal expansion you might have come across across this term while you were solving problems of strength of materials coefficient of thermal expansion this value will determine because of temperature how much my member will expand for instance in order to i will give you a brief overview of coefficient of thermal expansion so that it will help you to solve the problems so let us say you have two members let us say one member is a steel okay and you have member same length member is there let us say copper so this coefficient of thermal expansion is a material dependent value like the coefficient of thermal expansion of steel is 12.1 into 10 power minus 6 per degree centigrade and coefficient of thermal expansion of copper is 18 into 10 power minus 6 per degree centigrade approximately so what will happen is because of this values if i am increasing this member if i am heating this member by one degree centigrade the strain the strain in this member will be 12.1 into 10 power minus 6 that will be the strain in this member if i am heating copper by 1 degree centigrade the strain the strain in copper will be 18 into 10 power minus 6 that means strain what is strain strain is the change in dimension compared to its original dimension change in length by original length or change in diameter by original diameter so for so what will happen so my steel steel will elongate less because the strain is less the strain is less strain is changing dimension so for every one degree centigrade the change in dimension of steel is a bit lesser as compared to the change in dimension of copper copper along it more because copper has more coefficient of thermal expansion steel has less coefficient of thermal expansion so copper will elongate more and still will elongate less

Channel: Mohammed Parvez

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Id: OZVhNAbv9kY

Length: 31min 0sec (1860 seconds)

Published: Sat Sep 19 2020