Mohr's Circle and the Finite Element Method Explained

Part 1

Hello everyone, welcome you all in lecture 2 of week 1. So here we are going to continue with the Mohr circles. So till now we have studied about basics of strength of material. So we have studied about the stress strain, then we have gone through the stress strain curves, Hooke's law, Poisson's ratio, and then we have gone through the Young's modulus, its significance, elasticity and plasticity. And then we have also read about the principal stress. Okay. And now we are going to have a just brief knowledge about the Mohr circle. Okay. So what is Mohr circle? So it is a representation or you can say it is a diagrammatic representation of the stress, the shear stress and the normal stress.

So in this by drawing the Mohr circle, we can actually get the stress at the different planes. Okay. So suppose this is an element on which this forces or this stress has been induced on this element. Okay. So sigma x, sigma y, yx and tau xy, tau xy, tau xy and tau yx are complement to each other. Okay. So if we'll just take a plane, okay, for this, if we'll take a plane like this, okay, and we'll see that on this, on this plane, right to this plane, we have this face and this face.

So on this face, we have a sigma x and tau xy. So this will form coordinate of x and y on the Mohr circle diagram. Okay. Because we have the Mohr circle diagram sigma normal stress on the x plane and this real stress on the y plane. Okay. So this will give you the coordinates, which we have to put it on this diagram. So suppose sigma x and tau xy is like this. So sigma x is in tensile manner. Okay. And the tau xy is producing a anticlockwise effect. Okay.

So in that scenario, if you see, we have the sigma x here and tau xy, the fourth quadrant and the third quadrant of the shear stress will give you the anticlockwise or counterclockwise effect. And the first quadrant and second quadrant will give you the clockwise effect. Okay. So similarly, if you see sigma y and tau yx. Okay. So here if you see tau yx is producing a clockwise effect. So here you can see this is fine. Okay. So it will produce, we'll write like sigma y, sigma y in the positive side only and then tau xy. Okay.

So this will give you one coordinate, one point on this tau sigma plane and one point on this one. So just we'll make a line with this two point. Okay. And with the center of this one, we'll take the center of this line and we'll draw a circle with the radius, with the radius of this one. Okay. Okay. So this entire line will act as a diameter and we'll draw a circle which will give you the maximum principle stress and minimum principle stress. This is the minimum principle stress and this is the maximum principle stress. Okay. So this has been calculated with respect to origin. Okay.

So this is the maximum principle stress. Okay. So this will give you the principal stress and then the radius of this Mohr circle, the radius of the Mohr circle will going to give you the maximum shear stress. Okay. And shear stress is having a magnitude 25 Mpa. On this phase, it is clockwise and on this phase, it is anticlockwise and the tensile stress is 50 Mpa. Okay. So as we can see, this is 80 Mpa. So it is in compressive in nature. So means that we'll just draw a graph of x-axis and y-axis we'll draw and left to the y-axis, there will be a negative sigma or negative x-axis, negative x-axis. Okay. So here it is whatever like we have the compressive nature on that phase. That is it minus 80 Mpa. So that minus 80 Mpa and so it is a clockwise shear stress.

So it will go above this. It will go above this sigma axis. Okay. So whatever above this, it is clockwise. Below this, it is counterclockwise shear. So this is counterclockwise shear stress. Okay. So this two points is going to form a coordinate system here. So at this point, it is minus 80, 25 coordinate. And at this point, it will be 50 and 25. I hope it is clear for you. So we'll just take the center of this line and taking this line as a diameter and we'll draw a circle. And wherever this circle is going to meet, this x-axis will give you the principal stress and the radius of this circle will give you the maximum shear stress. Okay. So this is the Mohr circle.

Now we are going to start with the theories of failure. Okay. So in our structural domain, in the structural engineering or the problem we solve for the, in the structural domain, we have certain theories of failure. So all the problems like we are solving are based on the assumption made by this theory of failure or based on the statement or the theory given by this principal theories. Okay. So the principal theories of the elastic failure is the maximum principal stress theory, maximum shear stress theory, distortion and energy theory, maximum strain theory, and maximum strain energy theory.

Okay. So here we are going to discuss the top three because most of the problem we are going to solve using the first three theories. Okay. So let's start with maximum principal stress theory. Okay. So what does the maximum principal stress theory state? So this theory states that the failure of a mechanical component, the theory states that the failure of the mechanical component, subjected subjected to bi-axial or tri-axial stress occurs when the maximum principal stress reaches the yield or ultimate strength of the material. Okay. So suppose the material is under, we have loaded a component. These are the principal stresses in the component.

Okay. Sigma one, sigma two, sigma three, be the three principal stress acting at a point on the component. Okay. Then the failure occurs when sigma one is going to be equal to the sigma y, that is the yield value for the ductile material and ultimate for the brittle material. Okay. Why sigma one only? Because sigma one is the maximum of all the three principal stress. Okay. That's why sigma one, if the sigma one is greater than equal to actually, it should be greater than equal to.

Okay. Then the failure occurs. Okay. So this is the limiting condition. Sigma one is equals to sigma y. Okay. And this is the boundary of the maximum principal stress under the bi-axial stress. Okay. So this is the sigma one and sigma two plane. And this is like, this is going to form a rectangular or a square section. Okay. So this boundary is rectangular type. Okay. Okay. So what is the limiting condition? Sigma two is equals to S-Y and sigma two is equals to minus S-Y-C. So this will be the in compressive and in tensile. Okay. So similarly, the sigma one and sigma one for the tensile yield value and the compressive yield value.

So when we'll plot it, a straight line will come, which is parallel to the y-axis and the x-axis, which will form a rectangular boundary for the maximum principal stress under the bi-axial stress. Okay. So coming to the next theories of failure, that is the maximum shear stress theory. The theory states that the failure of a mechanical component subjected to a bi-axial, tri-axial stress, it occurs when the maximum shear stress at any point in the component becomes equal to maximum shear stress in the standard specimen of tensile test when yielding starts.

Okay. So what does it mean actually? So here we are doing a benchmarking. Benchmarking means comparing one problem with the other problem. So here what we are doing, we are just, we are taking an mechanical component. We are subjecting it to the bi-axial or tri-axial stress and we see where the maximum shear stress is going to occur. So at any point when the maximum shear stress is occurring at a point and when it is going to equaling, when it is getting equal to the maximum shear stress in the standard specimen of the tensile test, then the yielding starts.

Okay. So standard specimen of the tensile test is the same like, we will take a specimen of certain gauge length under the UTM. We are applying a tensile load and we are doing a test for the failure. So when the maximum shear stress in the standard specimen of the tensile test becomes. Equals to the maximum shear stress at any point in the component due to the external loading. That loading may be the biaxial or triaxial. Then the yielding will start and the component will fail. So suppose Sigma1, Sigma2, Sigma3 be the three principle.

The stress is acting at point in the component. So for the tensile test, so first taking in this benchmarking problem, we have the two tests. One is the tensile test, another one is the, our main mechanical component which we have subjected to the biaxial or the triaxial loading. So for the tensile test, we have the uniaxial loading that is Sigma2 is going to be 0, Sigma3 is going to be 0 and Tau max is going to be Sigma1 by 2. If you will ask me this formula from where this formula came, if you might have seen formula, so this you can get it from the Mohr's circle diagram also.

Okay, and secondly, you can get this from the formula of these Sigma1, 2 is going to SigmaX plus SigmaY by 2 plus minus under root of SigmaX minus SigmaY by 2 whole square plus Tau XY square. So that one very long formula is there. So you can use that to get this formula also or you can use the Mohr's circle diagram. So if you see the Mohr's circle diagram, what we have solved here, so if you see, we know that this maximum shear stress is going to be the radius of the circle.

What will be the radius of the circle? Sigma1 minus Sigma2 by 2, right? That is going to be the radius of the circle, correct? Okay, and if you see here, when the Sigma2 or Sigma3 is going to be 0, then Tau max will be Sigma1 by 2 only. Okay, and when the specimen start yielding, Tau max is equals to SYT by 2. Okay, so that Sigma1 will get replaced by the yield value of the stress. Okay, so now for the component, for the mechanical component in which we have subjected to a triaxial loading, we have a component of the shear stress on three planes like Tau12, Tau23 and Tau31.

Okay, so what will be the Tau12? That is the maximum shear stress acting on the plane 12. So that is going to be the Sigma1 minus Sigma2 by 2. Okay, using the Mohr's circle concept or the Lindyar formula, you can use the either way, you will get this formula. Tau23 is equals to Sigma2 minus Sigma3 by 2. Tau31 is equals to Sigma3 minus Sigma1 by 2. Okay, so suppose we have, as we know that Sigma1 is greater than Sigma2 is greater than Sigma3. Suppose Tau12 is the maximum of all this.

So all the three maximum shear stress. Okay, then Tau12 is maximum this. Then what will be the formula? Tau12 should be equal to Syt by 2. Okay, Tau12 should be equal to Syt by 2. And if you will put this formula Sigma1 minus Sigma2 by 2 will be equal to, you just put this formula, equate this two equation, you are going to get Sigma1 minus Sigma2 by 2 is equals to Syt by 2. And we will take a factor of safety. Because we generally take a factor of safety to, for the design consideration for make our design under the yielding.

So it will not get yield under the biaxel or the triaxel loading. Okay, so this is the boundary for the maximum shear stress under the biaxel stress. Okay, so if you see Sigma1 minus Sigma2 is equals to minus Syt here. And actually, this is the equation of the straight line. Okay, so we have the two straight line equation here. Okay, and Tau max is equals to Syt by 2. So this is the, this equation. Okay. Sigma1 is equals to Syt by 2 and Sigma2 is equals to Syt by 2.

So here, the Sigma1, this is the Sigma1 value. We just plot it, it equals to 0.5 Syt. Okay. And here, then you will get a hexagonal boundary for the maximum shear stress. Okay, so now maximum distortion energy theory. This theory states that failure of a mechanical component subjected to biaxel or triaxel stress occurs when the strain energy of distortion per unit volume at any point in the component becomes equal to the strain energy per unit volume of the standard specimen in tensile test when yielding starts.

We are just benchmarking the strain energy per unit volume of the tensile test with the mechanical component which is subjected to the biaxel or the triaxel stress. Okay, so Sigma1 is greater than Sigma2 is greater than Sigma3. Suppose we have this three principal stress with this relation. Sigma1 is greater than Sigma2 is greater than Sigma3. Okay, this is the shear strength. The shear strength is equal to the 0.577 times the yield strength. Okay, so the yield strength in the shear is 0.577 times the yield strength in the tension.

Okay, so this is the von Mises stress equation is given by this. Okay, Sigma1 square minus Sigma1 Sigma2 plus Sigma2 square under root and Syt. And this is an elliptical equation. So we will have an ellipse here. Okay, and this is the boundary for the distortion energy theory under biaxel stress. Okay, so this is the very important point here. The yield strength in shear is 0.577 times the yield strength in tension. Okay, so we are going to select the theory for ductile material because if you are going for most of the structure engineering projects in the industry deals with the ductile material.

Whether it is automotive, heavy engineering or railway transportation in which domain you are going to get the job you are going to work for. Mainly it is going to deal with the ductile material. So, we know that the ductile material are weak on shear plane and brittle material are weak on normal plane. The Maximum Principle Stress Theory is suitable for the brittle material and Maximum Sheer Stress Theory and the von Mises theory that is the distortion energy theory is suitable for the ductile material. So, these are the theories of failure we have read.

So, we have read the Mohr Circle, we have read the theories of failure and we have read the comparison. Okay, this is what we have to learn for the strength of material perspective. And we can now start the other part.

Part 2

Till now, we have learned about the stress-strain curve, stress-strain, linear elasticity, Hooke's Law, Poisson's Ratio, Young's modulus, elasticity, and plasticity, and theories of failure. Okay, now we are going to learn about the finite element methods, how we are going to use it to solve the industrial problem, and how we are going to, these are the concepts which we are generally using the software's. So, software solves these concepts, basically, in the background, and gives us the result. Okay, so, what are the learning objectives? So, after going through this presentation, the students will be able to understand the concept of the different methods to solve the engineering problems, numerical methods, understanding basics of the FEM and FEA, whatever.

People, some people called FEM and some people called FEA. So, FEA is basically more relevant to the industry, and FEM is more relevant to the academics. Okay, so, then degree of freedom, type of element, and types of analysis, yeah. So, it is going to be a more detailed kind of discussion. So, let's start. So, method to solve any engineering problem. So, what are the methods to solve any engineering problem? So, we have Q methods. So, what methods we have? We have analytical method, numerical method, and experimental method.

Okay. So, what is analytical method? It is a classical method. It is 100% accurate, and they have the limitation, but can be used for the simple problem. So, this is the classic method. This is like hand calculation result. Okay. So, we can give the solution using this method for the simple problems. Now, what is the numerical methods? So, this is the mathematical representation of the problem, and it can solve the complicated problems also. As it gives the approximate result, physical testing needs to be done. Okay.

So, numerical method, this is the area which we are going to focus on because the FEA or whatever the computational structural problem, computational structure problem, use this method to solve the problem. Okay. And it will give the approximate result. Okay. It does not give the actual 100% practical problem result, but nowadays the softwares are too competitive and too, it is giving very realistic kind of behavior for the physical problems also, which is simulated in the virtual using the FEA. Okay. Then we have the experimental method.

So, it is the actual measurement. It consumes time and money. It requires a physical prototype. So, these are the examples, strain gauge and photoelasticity are two of the examples of the experimental method. So, how does the numerical method solve a problem? So, the numerical method is generally based on the discretization of the integral form of the equation. It solves the problem only at a finite number of points and interpolates the result for the entire domain, that is the surface and volume. Okay. Before getting at the solution, a few assumptions are made based on the condition, like certain kind of assumptions, what are the type of element to discretize, linear or parabolic values, variation of the displacement over the domain.

Okay. So, these are the few assumptions which we are going to make before getting the, before solving the problem. Okay. So, the numerical method is the approximate method and needs correlation and needs correlation with the tested data to accept the result. So, in industry, we do the FEA. We solve the problem in our computer interface using this high-end software and we generally correlate this result by doing the testing of the actual component in the real-life scenario. Okay. So, when there is a match of the result approximately above 85%, then it is said to be okay that we can accept the finite element method result using the computer-aided engineering, which has been solved using the computer-aid.

Okay. So, what is FEM or FEA? So, it is a mathematical representation of an actual problem. Okay. And it is solved based on the numerical analysis. Okay. So, let's go more into deeper of the finite element method. So, it is a three-word analysis. Okay. So, this is the actually a general definition. And let's have a three-word analysis of this finite element method. Okay. So, first word is finite. Think on it. What it can be. What does it mean finite? Okay. So, every object. Suppose we are surrounded by the n number of objects.

Thousands on lakhs of objects we are surrounded with. Okay. Have you ever think of like we are applying a certain load on a certain object or leave a word that loading part or this part. Okay. How these objects are made. How you think about this? So, these objects are made by the integrity of infinite number of points. Am I right? It is made based on the infinite number of points or element. It depends how finely you are discretizing that object. Discretizing means making that object into smaller pieces.

Suppose you have a mobile phone. In hand, the mobile phone looks like a single object. Okay. Our delete uses. Okay. So, this every object that is to be analyzed have infinite degree of freedom. So, what is that infinite degree of freedom? That is the number of points, number of nodes you can say in the terminology of the finite element method. Number of nodes or number of points that object can be discretized to. Okay. In the order to solve it, we should reduce the infinite degree of freedom to finite degree of freedom with the help of discretization.

Okay. So, now your mobile phone is made up of the infinite number of points. So, those infinite number of points is infinite. It's uncountable. Okay. So, we need to make it countable. So, in the order to do that, we'll discretize it with the known number of elements. So, that will get a known number of points. Okay. So, this is actually turning our infinite degree of freedom object to finite degree of freedom object. Okay. With the help of discretization. Now, coming to the second word, element. All calculations are made at a limited number of points that is known as nodes.

Okay. At the entity joining the nodes forming a specific shape is known as element. Suppose this is an object. Okay. Now, it is having infinite number of points. So, we cannot solve those infinite number of points. Okay. So, what we'll do? We'll just make it finite number of points. Okay. Suppose these elements we are having. Okay. So, this finite number of here also everywhere it is. We have discretized it too with finite number of element. And with finite number of nodes. Okay. Now, if you can see the entity joining the nodes.

One, two, three, four. Okay. So, I'll just make one element here. So, this is node. One node. Two node. Three node. Four node. Okay. So, this four node form a specific shape that is known as element. Okay. So, we came to know what is finite. What is element. We have already read about numerical method. Which we have discussed just now. Now, degree of freedom. Okay. So, suppose we have one object. We have an object which is fixed at the bottom. And an inclined force acts on it.

Which result in the deformation of the object. Okay. The minimum number of parameter required to define the position of any entity that is completely in space is known as degree of freedom. Okay. So, here if you see the minimum number of parameter required to define the position of any entity that is completely in space is known as degree of freedom. Now, try to think on this statement. We are going to come back to this after knowing this definition. Okay. So, when an object has infinite point, it results in infinite degree of parameter.

parameter that is required to attain the solution. Suppose the number of nodes of your mobile phone is infinite and the degree of freedom per node is 6. What are these degrees of freedom per node? It is 3 translation, 3 rotation. OK, so the total number of equation because we have to solve it at a node. We have to solve our problem at node. So the total number of equation is infinity. 6 multiplied by infinity is infinity. OK, now we have discretized our object, our mobile phone such that the number of node is 21.

If you see here, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21. So here we have the total number of nodes, 21, and the degree of freedom per node is 6. How many equations it is required to solve this problem? It requires 21 multiplied by 6, that is 126 number of equation. OK, so what are the minimum parameters we required here? What we required? We are applying a force and we are fixing the bottoms. So when we are fixing the bottoms, we are constraining all the degree of freedom at the bottom.

And here with this node, whatever the node, which are constrained, it will participate in the calculation of the deformation. OK, so suppose we have the 21 number of free node, which is going to deform by this external load. OK, and that 21 multiplied by 6, OK, is going to give 126 equation. After solving it using the software, we are going to get the desired result, that is the stress, strain, deformation. OK, and then we can optimize our component based on the result. Types of element. There are basically three types of elements, 1D, 2D and 3D.

So this is 1D element. So this is 1D element. Here, one of the dimension is very large and compared to the other two. This element is represented by a line, cross section and the length need to be defined by the user to complete the 1D geometry configuration. The element type are beam, tube, pipe, rod. It depends on the software. Different software are different terminology. OK, and the practical component, which we can represent using this 1D element are bolt, shaft, joints, etc. OK, 2D elements. So the two dimension are very large in compare with the third one.

So here, if you see in this picture, X and Z are very large in compare to this Y. OK, this is the Y is along the thickness and the thickness need to be defined by the user to complete the geometric configuration. OK, generally the meshing is carried out by extracting the mid surface, which is shown in the pink here, if you can see, on the mid plane. So element type is basically the quadrilateral and triangular. OK, so this is for the 2D elements and this is 3D element.

So in this, all three dimensions are comparable with each other. So X, Y, Z are comparable with each other. Generally, we have the hexahedron, tetrahedron, penta and pyramid. These are the four types of elements we generally use in the industry to represent the 3D geometry or 3D element. Can you use the other shape of element also? Why we have restricted our element configuration to hexahedron, tetrahedron, penta elements, 1D? Why we can't have the hexagonal shape, octa shape? OK, penta 2D shape. Obviously, already we have the pentagonal element we have, but not as a pentagonal shape 2D element we don't have.

Why? Actually, we can use any shape of the element. We should know the stiffness matrix of the elements. Practically, most applications can be solved using the discussed element. OK, so we can use any configuration. We just need to know the what is the stiffness matrix of those elements. OK, why the number of tria elements are restricted to 5% and avoided in the critical areas? What are the critical areas? So, the critical areas of the component are basically the fillet region, the washer region, the hole region, and it depends on the types of analysis and the requirement.

OK, if we are focusing on particular area to study, we have to restrict the tria elements to 5% on the overall mesh of the component. So, quad element, generally it is more accurate than the triangular element due to a better interpolation function known as the shape function. OK, so interpolation function or shape function establishes the relationship between the displacement at any point in the element with the modal displacement of the element. The interpolation function is of polynomial type. OK, it is telling, it is more accurate than the triangular element.

Why? In the lame language, the quad element is less stiffer than the tria element. OK, so when you use a lot of tria element, it will show the lesser stress and the displacement when using the critical location. OK, so in the practical scenario, suppose we have a component and we have measured it as a tria element. OK, and it is showing 5 mm of deformation. But actually, 5 mm of deformation and 100 MPA of stress. OK, 5 mm and 100 mm of, 100 MPA of stress and 5 mm of deformation.

OK, now in the practical scenario, when that object is getting loaded in the practical operation, that component under that loading is showing 7 mm or 8 mm of displacement and 150 or 180 MPA of stress. OK, suppose you have 150 MPA of yield value or the allowable value. With the factor of safety, we have the allowable stress of 150 MPA. So, when you do the FEA in this software using the tria element, you are going to conclude that this component is not going to fail in the practical application when this kind of loading it is subjected to.

But in the practical scenario, we are getting a stress of 180 MPA, which is actually exceeding the allowable stress and in the other language, it is failing. So, what does it mean that we have made our model way too much stiffer in the order to carry out the finite element analysis in the software using computer? OK, so, we should not do that. If we make it stiffer, it will show lesser stress and lesser deformation. But in actual scenario, it is failing. But our FEA result is telling it is not failing.

So, here it is going to be a very disastrous condition when the industry will fully rely on the FEA and it will tell that, OK, it is not going to fail in the practical scenario. But if it is going to fail and by accidentally, if someone get injury due to that, then it is going to be a total stake on an FEA engineer. So, while choosing the element, you have to make sure that you are not going to make your model too stiff. It is going to deviate from the practical result.

OK, so that's why the tria elements are restricted to 5% and avoided in the critical areas. OK, so this is the strain displacement matrix for the tria and the quad element. So, displacement is approximated by U is equals to capital ND, OK, and strain is, we know that du by dx and du by d epsilon, d epsilon by dx. We have written in the partial differential form, OK. OK, b is the element strain displacement matrix, OK, and here if you see the approximation of the strain.

So, epsilon x, epsilon y and gamma xy, this is the shear strain. OK, when you will solve this, you are going to get constant strain triangle. If you see here, this is a constant strain. This is a constant strain triangle, OK, and here for the quad element, it is not constant. If you see here, the stress displacement matrix B is not constant. It is a function of the variable x and y, OK. So, don't bother about this formula too much. If you want to really know, you can Google the strain displacement relationship of the quad element and the tria element.

This is just to make you understand that why we call it as a constant strain triangular element and we won't call it the constant strain quad element. So, because the strain... Quad element are not constant within that element, but the strain in the trial element is constant. If you see the strain displacement matrix, you will come to know that. Okay. So here we have covered till this. So we are going to cover the fully integrated versus under integrated element. Then hexahedral versus tetrahedral element. Then we'll solve one cantilever beam problem.

Okay. Using the FEM method and the analytical method, then we are going to study about the different kinds of analysis, contacts, everything. Okay. So that we can continue in the next class. Okay. So we have to continue from the fully integrated versus under integrated elements. Thank you.

Part 3

All in the week two lecture, we'll continue the week two lecture. Okay. And till now we have studied about the basics of the finite element methods and we have studied about the finite element method definition. What is this? And then we have studied about the degrees of freedom, types of element, what is discretization and everything. Okay. Okay. Few observations and types of element and why we use a particular kind of element for critical areas. So this things we have studied, then we have covered the strain displacement matrix. Okay.

And for the tri and the quad element and we came to conclusion that why it is constant strain triangle and why it is not constant strain quad element. Okay. Now we'll start from this topic, fully integrated versus under integrated elements. So in finite element method, we have basically two types of elements. So this fully integrated and under integrated, actually there are two rules, Simpson and the Gaussian quadrature rules are there to solve the numerical integration for the stiffness matrix. Okay. Okay. Because we have the known force and the stiffness matrix we have to calculate. So these are the two rules which we use to solve the stiffness matrix.

So the Simpson rules is a well-known method for numerical integration and can be adapted to two dimensional integration. Okay. However, Simpson's rules uses fixed intervals. Okay. But the Gaussian quadrature, it is accurate. Okay. But it requires the less function because it is not restricted to the fixed intervals. Okay. Okay. So this method is used in finite element calculation. Okay. So reduced integration uses a lesser number of Gaussian coordinates when solving the integral. Okay. So for the under integrated element. So this is the reduced integration uses a lesser number of Gaussian coordinates when solving the integral.

So the more Gaussian coordinates you have for each element, the more accurate your answer will be. But this will be weighted up against the cost of the computation time. Okay. So one of the area where the local stress concentration are considerable, we use as fully integrated elements. Okay. So for example, I will show you a few fully integrated elements. So this is fully integrated element. This is reduced integrated or under integrated element. So if you see these are the Gauss points 1, 2, 3, 4. So this is for the quad element. We have the four Gauss points. It is fully integrated.

Suppose this is the linear element. When it will be quadratic element. It will have 9. Correct. It will have 1, 2, 3, 4, 5, 6, 7, 8, 9 Gauss elements. So it is going to be fully integrated quad element. Okay. But the nodes are 8. In some software, you can choose the number of integration points also. So in this it is 9 integration points. This is the CPS 8 element. This is the CPS 8 element. This is the Abacus software element. Okay. CPS 4 and CPS 8. Similarly, we have this reduced integrated element. This is 1 here and this is 1, 2, 3, 4. Okay. So this is all about the fully integrated and under integrated element. Okay.

And now we are going to move to hexahedral versus tetrahedral elements. Okay. So tetrahedral can fit better in complex geometry. Okay. But however, when you integrate the save function, it is less accurate than the hexahedral elements. Okay. One of the factors that determines the quality of the mesh is distortion of the element. Okay. When we are going to use the hexahedral and tetrahedral element. Okay. When the geometry is simple, our best option would be to mesh it with the hexahedral elements. But if it is not possible when you have the curved geometries and acute angles or little bit complex, then we are going to use the tetrahedral elements.

But when you have to probe the stress for the tetrahedral elements. Okay. So it is always preferable that you should use the second order tetrahedral elements. Okay. So this is the hexahedral versus tetrahedral element. So I have a problem for you. If you have studied the finite element method in college, then you can solve this problem or you have brief knowledge about it already. So we have a cantilever problem. Okay. So we have a cantilever beam problem. Okay. One meter length and load is 100 Newton and EI that is the modulus of multiplication of the modulus of elasticity and the moment of inertia is 1000 Newton meter square.

Now we have to find the deflection. Okay. So I will give you some 2 minutes to solve it. Okay. I will just go to the next slide after 1 minute. Okay. So please solve this problem by your analytical method. Then we are going to solve it using FEM method and we will do the benchmarking. So let's solve it. So I hope you have solved. So what will be the formula delta is equal to deflection is equal to PLQ by 3EI for the cantilever beam configuration. Okay.

So here if you will put the formulae on this equation, if you put the data in this equation, you should have a unit consistency in this equation. Okay. So if 1000 load is 100 Newton, then we have length is 1000 millimeter. So 1000 to the power 3 is this and EI is Newton millimeter square this much. Okay. So when you put it in this formula, we will get the deflection of 33.33 mm. Okay. Now let's solve this problem using this F is equals to KX by using the stiffness matrix.

I am asking you to solve this problem because it will give you a feel how the background software will do the work for this small kind of problem. For the complex problem anyway, we cannot do the hand calculation for each of the stiffness matrix. We cannot solve. But for this simple problem, we should get an idea how we calculate the stiffness matrix and all everything. Okay. So if you don't know the formula, this is the formula of this stiffness matrix for the beam element. Okay. K is equals to EI by LQ. This 12, 6 minus 12, 6 L is multiplied at this second and the fourth column.

Similarly here L square is here 4 is minus 6 L, 2 L square minus 12 minus 6 L and it is a symmetric matrix. About this diagonal. Okay. So this term and this terms are same. Okay. So let's solve this. That's why I have given you the simple problem. L is equals to 1. So you don't have to do anything. You just have to put the L is equals to 1 and your matrix will reduce to K is equals to EI by 1 and this is minus 12 everything is.

So here you shouldn't better. You do not act smartly. Don't change the unit. Keep it in meters. Okay. And now you should put the equation F is equals to K X. Now we know what is the F. It is 100 Newton. So if you see the equation, the exact equation will be like this force, moment, force, moment. Here is the deflection. This is the angular movement. Okay. You can say slope and this is again deflection. This is slope. So suppose one is fixed end and two is free end. So this new one and theta one will be zero because there will be no deflection.

Slope also will be zero and new deflection also will be zero. So this will be zero. This will be zero. Okay. And if you apply the formula for this matrix, how to solve the matrix, you will reduce to this. So this will get crossed F1 will be zero. It will get zero zero. If you will multiply it by this one. Okay. And you will see only F2 and M2 will left. F2 is 100 and M2 is zero. So EI is 1000. This entire matrix reduced to two cross two. What it is going to be?

12 new two minus six theta two minus six new two and four theta two. So you have to multiply like this and this will be the equation. You will get the two equation. You have to solve the two equation, two unknowns. You solve those equation and you will get the new two. That is one by 30 meter. If you multiply it by 1000, you will get 33.33 millimeter. Okay. Hope everything is clear here. Just have a recap again. So first we have solved the analytical problem by delta source to PLQ by 3i.

And then we have solved using the FEM method. So we came across, okay, this is the stiffness matrix and we have solved the stiffness matrix. And then we have put it in this general formula. I have to suppose to KX. Okay. And there is a force or the application of load is a function of the force and the moments. Moment is zero. And we have calculated like this. Yeah. F1 and M1 is zero because it is at the fixed end and F2 is 100, M2 is zero because we are not applying any external moment.

Okay. And this is new one and theta one is zero because at the fixed end we know for the cantilever room it will be zero. So based on that, we will get the two equation, two unknowns and we'll solve the problem. Now we are going to come, arrive at the very interesting part, types of analysis in FEA. So the term FEA, finite element analysis, with the support of computer, so we'll call it in some industry, it will be termed as a computer-aided engineering CAE or FEA.

But CAE is a general term. Thank you. Thank you. Design engineer also will come into that, but FEA is a more specific term, only the simulation engineer or FEA engineer will come into that. So we FEA as a engineer, if we know FEA, we should know these types of analysis. So these are the few types of analysis I have mentioned here, linear static analysis, non-linear analysis, dynamic analysis, buckling analysis, thermal analysis, fatigue analysis, optimization, CFD analysis, crash analysis, NVH analysis, okay. So optimization, it is a part of analysis because when until unless we are not going to analyze the structural part, we cannot optimize it.

And if we cannot optimize it, we cannot make a cost material relation, we cannot fulfill it. Okay, so this whole FEA, this analysis we do to like, obviously we have a very huge commercial profit after when the industry started using the FEA. Okay, and they have introduced this kind of analysis in the FEA and so we'll go through each of the analysis and okay, so linear is static analysis. So we'll start with this. So what is linear is static analysis? So it is having three terms, okay.

So this is also three term analysis. What is linear? Anything is linear, okay. Sometimes you people also think about, okay, my life is going very monotonous. It is going very linear, okay. So nothing is happening. Okay, so linear means basically a straight line in our analysis, okay. So in the context of FEA, we define linear as a straight line. So what is straight line here? That will come to know. Static. What is static? Static means it is with respect to time, it is zero, okay. It may be anything, okay.

So in a structural analysis, what parameter with respect to time is zero? Force, right. Mainly the force. It is not going to change with respect to time. Okay, so that is the meaning of static, okay. The general meaning of the static is, it is non-transient activity. Whatever you are doing, it is not with respect to time. Okay, that's why people will tell you, okay, which is constant with respect to time. Okay, and we have already read, we have already read in the stress strain curve that sigma is equals to epsilon, okay, and that is straight line is analogous to y equals to mx, and it will start from the origin, okay.

So sigma equals to epsilon, it is plot on the sigma epsilon curve. It starts from the origin, zero, zero, okay. And now what we are observing here, why this analysis has come into the picture? So if you see here, if you see here, this is the green line, if you can see here, this green line, okay. So what it will do, we all already know this stress strain curve, the strain hardening, the proportionality limit, ultimate point, necking, everything we know. So I am not going to repeat it again, okay.

So, but here we have a straight line equation, zero to suppose this is a here. O to A is a line that is following the Hooke's law, and it is, this stress is proportional to strain, and it is a straight line. Okay, so what will happen when we'll solve a problem of structure analysis in software domain? Okay, how it is going to compute the relation, the answer, okay. So what it will do, like when you are solving linear static problem, okay, so the component will not show that it has failed, it will not show necking, it will just show that this part is in red zone or like whatever the color you choose.

So, okay, so after crossing a certain limit, that limit is defined by U. Suppose I have a yield value of 300 MPa, okay, and I have a factor of safety of 2, okay. So what should be the R allowable stress? It should be 300 divided by 2, that is 150 MPa, okay. So that much stress we have to check for, based on the application of the load on a certain component, whether any part of the component is crossing to 150 MPa or not, okay. Suppose we have 150 at this point, okay.

This is 1 and this is, suppose this is 400, okay. I write 0.2 is 400 and 1 is 150, okay. So what it will do that when your stress is within 150, obviously the stress strength curve, it will follow the straight line, but when it will cross the allowable limit or the yield limit of 300, whatever, the behavior will be straight only. It is not going to follow this curve, okay. It is not going to show you, okay, it's a, but behavior, you can see that bending behavior and all those, the elongation behavior, you can see those behavior, but the curve it will follow is still this point because we have not for that particular material.

We are defining in the linear static analysis is the linear material. So it will solve the problem in the linear zone. It is not going to take up any curve. You can give the curve, then that problem will come to the nonlinear zone, okay. So that we are going to study in the later part of this course. When you are solving the linear static analysis and you have not defined any curve for the material which you have defined for the, for a structure, then automatically it is going to follow a linear path like this.

It is going to interpolate the result on this linear path only. That's it. Okay. So a single unbroken part with the location of the failure in the red color zone will be shown on your computer screen when you solve the problem. So that red color is by default, you can change that color also, okay. It will be a single unbroken part. It has crossed the ultimate point. So let's, that material should break from there and all. Okay. It will not do that like that. Okay. So this is the meaning of the linear static analysis.

Static means the change of parameter that is force with respect to time is zero. Okay. And the force and moment should be balanced. That is a compatibility equation for the linear static analysis. This should be balanced with each other. This is the summation of the forces and summation of the moment should be zero. Okay. This is the, all about the linear static analysis. Now we'll study about the non-linear analysis. Okay. So non-linear analysis has been categorized into three parts mainly. Geometric non-unity, contact non-unity and material non-unity.

Okay. Now material non-unity, till now we have studied about the linear material, linear static analysis. Now we'll go through the material non-unity, contact non-unity and geometric non-unity. So what is geometric non-unity? This happens because of the large deformation. Okay. Even though the material is within the elastic limit, the large length and a small force cause a very large deformation. Okay. So this you can see when you have a very long cantilever kind of beam. Okay. Or a very long column in the buckling analysis. Okay. So there you can see that it is deflecting in the lateral direction at a very large extent.

Okay. So in that scenario, the geometric non-unity will come into the picture and there are some parameter in the analysis setting you have to switch on to see the behavior of the geometric non-unity. So that's all about the geometric non-unity. What is contact non-unity? So contact non-unity is due to the contact simulation. We'll study about this in more detail. Okay. Here we have the contact non-unity. So we'll study about this in more detail. So let's see about the material non-unity. So material non-unity, it happens when the material cross the elastic limit.

Okay. Or when we have the non-metal material like plastic or rubber or the creep deformation when material is at certain creep. It is undergoing in a certain creep. So what is creep? Creep is a deformation at the constant stress for prolonged time and at elevated temperature. Okay. So we'll focus on the first one as of now. Material cross the elastic limit. Okay. Because the other part for the plastic and rubber, rubber we don't have a significant elastic zone. So all the non-metal material is non-linear by default.

Okay. So material non-linear data. So this is in the ANSES, we will feed the, we have to feed the non-linear stress strain curve while solving the non-linear material problem. Okay. So while solving the material non-linearity problem, we provide a stress strain curve for the non-linear region after the elastic limit. Okay. So in the CAE software, either we have. The bilinear curve or the multilinear curve. In the both scenario, in using this both option, we can provide the stress-strain curve for the nonlinear material. So this we are going to use it in the explicit analysis. We are going to study about the impact test and crash test.

So at that period of time, we are going to study about this one in detail with the application. Now coming to the contact. So, contact. So, why contact nonlinear is required? What is actually contact? So, when you do Namaste, one palm, right hand palm is touching to the left hand palm or vice versa. So, these both are in contact. Okay. So, and when you are pressing your each hand. Okay. So, there is an application. You are applying a contact force from one of the hand. So, that the other hand is also resisting the same contact force to the other hand.

The action and reaction is there. Okay. So, when you are doing Namaste, it is in contact. When you are pressing from right hand to the left hand doing Namaste Mudra. Okay. Then, what is going to happen? Suppose you are pressing the right hand to the left hand. Then, to be in that same position, the left hand apply an equal and opposite force on the right hand. So, that it will be in the equilibrium position. So, that is the contact. So, this is the contact problem. The problem is categorized as the boundary nonlinearity. So, what is the boundary? Our one palm touching to the other palm.

So, between that one boundary has been created and that is why our contact problem is categorized as the boundary nonlinearity. So, why it is nonlinear? Because contact boundary and the contact stresses both are unknown and there can be abrupt changes in the contact force. So, this is why it is nonlinear. So, if you see here body 1 and body 2, it cannot penetrate each other but it can apply a force on each other. So, this is the contact force. So, you can use your both hand doing the Namaste and you can feel this force.

Okay. That if you apply from one hand or left hand applying a load force to the left palm. So, the left will also provide a contact stress to the right hand. Okay. So, but both cannot penetrate each other. So, this application is enormous. So, in every industry in all application, the contact problem has been used. Okay. So, nowhere in the FEA. This is a very frequent application you are going to use in the FEA. Whether it is industry like automobile, heavy machinery, anywhere. You are going to use this contact problem.

Okay. So, there are different kinds of types of contacts are there. So, based on the surface, it is frictional contact, freeze contact, frictionless contact. Okay. So, frictional contact is applied between the two surface with friction coefficient having the relative motion. Freeze contact, bonded contact, tie contact is applied between the two surface having no relative motion. Okay. And the frictional contact, the normal pressure is equals to zero. Okay. If separation occurs and just gap can form in a model between the body depending on the loading. Okay. Okay. And the zero coefficient of friction is assumed here and thus allowing the freeze sliding.

Okay. But it cannot move in the normal direction. But it can slide. Okay. Here is the rough contact. It is also like the frictional less setting. But the settings, the setting models have perfectly rough frictional contact. Where there is no sliding and it has infinite coefficient of the friction. So, these types of contact you are going to see in the near softwares. So, based on the contact area, we have edge to surface, edge to edge and surface to surface. The difference of software is having different compatibility with this type of contact.

Okay. So, yeah. This is based on the area of contact. Okay. So, this is the Newton method and this is we use to solve the non-linear quasi-static process. Basically, the contact problem. Okay. So, why we use this one? Okay. So, first of all, the quasi-static implies that the process occurring in the real time is being simulated infinitely slowly. Okay. So, this allows the equilibrium equation to be satisfied as each step and the inertia moment effect are ignored. Therefore, the possibly dynamic problem can be solved as a static problem.

Okay. So, what we are doing here? We are basically applying our, suppose we have F Newton we have to apply. So, we are not applying at the whole Newton, the whole load, certain point of time, the full load we are not going to apply. We are just loading our structure in steps. Okay. So, when we are loading our structure in our, in the steps, then we are solving this F is equals to KX problem. Okay. At each step and if you see here, okay. Okay. So, what it says, non-linear problem are generally history dependent.

In the order to achieve a certain level of accuracy, the solution must be obtained in a series of small increment. For this purpose, you need to solve the equilibrium equation at each increment and corresponding increment size is selected. Okay. So, that increment size need to be selected. So, suppose you can divide the total load in the two sub-step, five sub-steps. Okay. So, suppose you have a hundred Newton of load you need to apply. Then, if you want to choose, like you can divide it to the five sub-step.

So, at each step, it is going to apply the 20 Newton. But, this is also driven if the software feels that the load is converging properly, then it can internally, it can tweak those data to converge the problem in a faster way. Okay. So, suppose in the first step, it applies a 20 Newton and it's all very pretty good. There are, the convergence factor is very good. But, it takes automatically a decision that in next sub-step, instead of 40, so means 20 after that, it will be 40.

So, I am going to apply another 25 to 30 Newton extra. So, it will be 20 plus 25 or 30. Okay. So, like that, it can also select those numbers. So, that can be program control also. And, you can also define the initial sub-step at least. So, here, if you can see in this F versus U, force versus displacement diagram, F is the global load vector and LU is the non-linear response of the system. Okay. That is the nodal reaction, you can say. Okay. So, for a linear problem, LU would be simply KU.

So, that is, LU will be a KU is the displacement. It is similar to the F equals to KX. Okay. So, but if we are applying this Newton's method, it will solve in the iterative process. Okay. And, the final residual, if you see the residual at the beginning of the solution, at the beginning of the solution, it is R0. It will converge when the residual is very, it is very negligible and our load and the residual, the difference between the our load and the residual is very close to 0 or 0.

Okay. So, here, you can see here, we have divided the load into the n number of sub-step, F1, F2, F3, like this. And, we have applied the, we are solving it, the F equals to KX equation at each load step. And, this is cumulative process and we are doing this problem as an iterative process. Okay. This is all about the Newton's method. So, this much mathematical equation is okay, fine, you can, it's always good to know those mathematical equations. But, the concept should be clear that in the Newton's method, basically, we will use to solve the contact-norrative problem and, we'll solve it at the iterative process.

Intermediate loads will apply and we'll solve it in the iterations. Okay. So, another types of analysis is the dynamic types of analysis. So, in the dynamic analysis, there is time dependency. What matters is the intensity of the load, the distribution of the load and the distribution of the load. It has been categorized into two parts. First one is the free vibration and the force vibration. So, in the free vibration and object is allowed to vibrate after disturbance without an external force. It happens when an body is disturbed from the equilibrium position and released.

Okay. Forced vibration, external force, we have to, if a structure has been excited by the external force and it is in periodic in nature, then it is termed as the forced vibration. So, forced vibration is further sub-categorized into this 3A analysis, frequency response analysis, transient response and random vibration analysis. Okay. And this is the general equation for the dynamic analysis. M A double dot C A dot plus K A equals to P T. P is the load, external excitation. Okay. Which is a function of time in the general equation.

M is the mass. A is the acceleration. C is the damping coefficient. A double dot is the acceleration. A dot is the velocity. A is the displacement and K is the stiffness. So, dynamic analysis also like this. We modal analysis and frequency response analysis which we are going to study in the upcoming lecture in this course. So modal analysis calculates the natural frequency of the system alone and based on the external excitation that is F as a function of time, FRA will give the correlation whether it matches to the natural frequency obtained from the modal analysis or not.

Okay, so modal analysis it will just tell you okay these are the natural frequency of this structure but the frequency response analysis will tell you whether it is going to resonate on with this frequency or not. So this everything we are going to cover in the in detail in the upcoming lecture this dynamic analysis. Crash analysis we are going to study this one in detail in the crash analysis class. Okay, it is a highly non-linear transient dynamic event. Okay, so crash force simulation is one of the typical area application of the finite FEA in which non-linear FE simulation are effective.

Okay, so what is basically crash analysis? So we provide an input velocity at which the vehicle is going to impact at a certain kinetic energy with a barrier and later it converts to the different forms of energy. So all those different forms of energy we are going to study in the crash analysis topic explicit dynamics. Okay, so if you see in this course there are n number of energies you can see. So internal energy, kinetic energy, total energy, hourglass and contact energy. So this is a basic explanation I have given here that is total energy is the kinetic energy plus internal energy plus interface plus hourglass energy.

Internal energy is computed based on the six components of the stress and strain. Okay, the calculation is done incrementally of each element as follows. Internal energy new is equal to internal energy old plus the sum of all six direction. That is the stress multiplied by delta that is incremental strain and the volume. Okay, hourglass energy or hourglass modes are non-physical zero energy modes or deformation that produces zero strain and no stress. Hourglass modes occur only in the under integrated solid shell and the thick shell elements.

Okay, if you see unnecessary distortion you can see in the elements with the hourglassing. Okay, and with the no hourglassing it is smooth. So these are the buckling analysis. We know here that the buckling analysis we apply a compressive load on a elongated or column and we'll analyze it whether it can sustain that buckling load or not or if it will lead to any lateral deflection or not. So this is a linear kind of analysis. What we generally do like we have a certain formulation for the buckling analysis which I have not mentioned here I think so.

Okay, so in that you might have also read about the buckling analysis that is a critical load. We have a formulation for the critical load. F critical is equals to pi e square i by l square. Okay, for l will be different supporting condition pin fixed free and free free like that. Different loading conditions are there not free free I think. It is the four basic conditions are there. That is a pin pin condition, free fixed condition, fixed condition and fixed and pin connection. So based on that analytical result or if you have the simple problem you can use that formula but if you have a complex problem you need to use a formula which is actually using the industry design handbook.

So there are other formulas also for the buckling calculation and so the process is to solve this problem is you have to solve the problem analytically with the hand calculation and you have to solve it in the software. And you have to see whether it is correlating or not your hand calculation buckling factor buckling load factor and the buckling load factor we are getting from the this one. Okay, so that is the buckling analysis. So this optimization why we do the optimization I told you in the beginning of this lecture.

So optimization is basically the we can do optimization topology topography and material. Okay, we can optimize the material also we can optimize the geometry also. So this part is more like once we have done one set of analysis. Okay, and then we are observing a part is either over constrained or like some extra material is there or we need to add some material. So like this we take a decision either based on the experience we can suggest to the designer that we need to add some or reduce some stiffness from the spark or we can optimize the product using the software and number of software we are for the optimization.

Okay, so this optimization we are not going to study here but this is also type of analysis which comes under the umbrella of the types of analysis. Okay, thank you. So now in the next we are going to revise next lecture and then we are going to start the workbench. We are going to start the ANSYS workbench and see you in the next lecture everyone. Bye.