

January 4, 2021 at 5:53 amRameez_ul_HaqSubscriberHello there. Can anyone please guide me on how to use various options under the heading 'Integration Point Results' in Equivalent stress? What is the difference between 'Averaged'. 'Unaveraged', ..... etc in the Display Option and where and how they should be used? What does Average Across Body mean? When should we choose NO and when should we choose YES in the latter option?nThank you.n

January 4, 2021 at 1:04 pmpeteroznewmanSubscribernRead my January 3 reply in this discussion: /forum/discussion/comment/101713#Comment_101713 for an example of Averaged, Unaveraged and Nodal Difference.nIf you used shared topology to break a single part of the same material up into many bodies to help make a good mesh, you should choose YES for Average Across Bodies.nIf adjacent bodies have different materials, then choose NO for Average Across Bodies. For example, if you used shared topology to separate a thick surface coating modeled as a thin solid from the underlying solid of a different material, you want to see the stress computed in each material separately and there is a real physical discontinuity of stress at the interface that you don't want to average away.n

January 4, 2021 at 6:18 pmRameez_ul_HaqSubscriber,what I understood is that the faces of each body [both bodies having a different material] where the shared topology is created, will have equal averaged stresses if we turn ON the Average Across Bodies option, instead of turning it OFF. Is it correct?nIf it is, then consider the picture below:nwill the stresses just on the left of the shared topology (within the steel material) and just on the right of shared topology (within the Aluminum material) will also be affected by the averaged stress that we have computed on the shared topology face by turning Average Across Bodies option ON?nWhat about discontinuities occurring due to sudden geometric changes? Like having two bodies [made of same material] sharing topology, but have a sudden change in cross section from one body to the other.n

January 4, 2021 at 6:59 pmpeteroznewmanSubscribernStress is computed in each element at the integration points, then extrapolated out the the nodes of the element. If a node is connected to 8 solid hex elements, that node has 8 values of stress. If a node is connected to 4 shell quad elements, that node has 4 values of stress. You can plot the averaged value of stress at each node and the contour shows smooth transition of stress from one element to the next. Or you can plot the unaverage stress on each element. The unaveraged plot has stress values that make step changes at element boundaries. If the stress is changing slowly and the elements are small, the step is small. If the stress is changing rapidly and the elements are large, the step is large. A large step shows that the elements are not small enough to capture the stress gradient accurately. That is why you can plot the size of the step directly using Nodal Difference, to identify areas to improve the mesh.nAn example is an aluminum deposition coating a glass part to make a mirror. When the temperature changes, stress is caused by the different coefficients of thermal expansion. The stress in each material is computed by the elements. There is a layer of nodes at the interface that have a glass element on one side and an aluminum element on the other side. That node has multiple values of stress that you don't want to average across the bodies (but you do want to average within a body). One layer deeper into the mesh, there is a layer of nodes that have the same material on each side. That node also has multiple values of stress, but you do want to average those.nI think you have another discussion to talk about sudden geometric change, so let's discuss that in the other thread.n

January 5, 2021 at 11:34 amRameez_ul_HaqSubscriber. you explained so well when the 'Average across bodies' should be used and when not. And also, in which case 'Nodal Difference' option can prove to be fruitful. But I didn't understand one thing, if you may. For example, I see a huge step between few elements' boundaries using the 'Nodal Difference' option. You mentioned that the size of the mesh should be improved. But what is the goal of doing this? I mean what would I achieve in doing so?nSecondly, the option 'Nodal Difference' and 'Elemental Difference' are used side by side, or they each have a particular purpose. You described the purpose of Nodal difference, what about Elemental Difference? nThank you.n

January 5, 2021 at 12:31 pmpeteroznewmanSubscriberArraynThe purpose of mesh refinement is to improve the accuracy in the solution. There is an exact solution, which is the solution with infinitesimal element size. You can't afford to make very small elements everywhere, so looking at the maximum value of nodal difference shows you where to make the elements smaller.nI don't use Elemental Difference.n

January 5, 2021 at 1:14 pmRameez_ul_HaqSubscriberArray, I would like to go in a little bit of detail here, if you may. As you have said, for a surface body quad element, the nodal difference will show a single value. But for that node, we have 4 different stress values. What is this nodal difference value at a node showing? The maximum difference in stresses at a node or something else?nYou mentioned this, A large step shows that the elements are not small enough to capture the stress gradient accurately., even if the element size (which is quite big) is constant in a body, and some regions are showing large step while others are showing small, why would I only improve the mesh in the regions which have large stress gradient? Does the large stress gradient means the the stress values are not correct there? But the stress values will also not be quite accurate in the regions having the same element size but showing a relatively low stress gradient. nI know that you, kind of, have answered this already in your previous comment, but try to understand my point of concern here, Sir.

January 5, 2021 at 2:25 pmpeteroznewmanSubscribernHere is a page in the ANSYS Help where I learned the definition of ELEMENTAL DIFFERENCE.nAveraged vs. Unaveraged Contour Results (ansys.com)nNodal Difference: Computes the maximum difference between the unaveraged computed result (for example, equivalent stress) for all elements that share a particular node.nElemental Difference: Computes the maximum difference between the unaveraged computed result (for example, equivalent stress) for all nodes in an element, including midside nodes.nThe reason you are plotting stress is to find out how close to yield the material has come. Often, the area of high stress is also the area of high stress gradients. The goal is to get an accurate estimate of the highest stress in the model. Large elements can have a much larger error to the true solution than small elements in the area around the maximum stress where the stress gradient is also high. Large elements in an area of low stress gradient will have smaller errors, and low stress gradient is often in low stress areas which are of no concern, so it makes sense to spend resources (add elements) where the error is large and there is a high interest in accuracy due to it being the maximum stress in the model.n

January 7, 2021 at 12:16 pmRameez_ul_HaqSubscriber,I just want to confirm this statement what I am about to make.nThe displacements are computed for the nodes during the solution, by solving the stiffness matrix. These displacements of the nodes are used to compute the displacements within each element using the shape functions. Points within an element are chosen where the stresses are computed using those points' displacements. These points are known as integration points. After computing the stresses at the integration points, those are interpolated onto the element's nodes, again using shape functions. This means that each node will have an interpolated stress coming from each element it is a part of. An 'averaged stress' will mean that the all stresses at a node is averaged out with all of its values at that node, while an 'unaveraged stress' doesn't average out the stresses at a node. Afterwards, if you have chosen any of the options i.e. 'Averaged' or 'Unaveraged', the stresses within the element are again computed using shape functions, by making use of the stress already calculated at the node.nn

January 7, 2021 at 12:52 pmpeteroznewmanSubscribernExactly.n

January 8, 2021 at 7:42 amRameez_ul_HaqSubscriber,can we see a singularity stress if we use 'Unaveraged Stress'?nWhat I think is that we cannot, since each of the stress at a node is independent, so the shape functions used to calculate the average stress at a node no longer interacts, so there is no chance for the discontinuity to happen at the node within the shape functions, hence no singular stress at the node. This would also mean that the unaveraged stress that we see within an element cannot be affected by the singular stress at a node, so basically it is gives us more dependable results as compare to an averaged stress. Am I right?n

January 8, 2021 at 1:01 pmpeteroznewmanSubscribernThe stress singularity exists in the geometry (or boundary condition on the geometry). There is a mathematical exact solution such as that known at a crack tip is that the stress is inversely proportional to the distance squared from the crack tip, where the stress at the tip is infinite. Finite Element solutions estimate this exact solution, where stress is derived from strain which is derived from difference in nodal displacements. If there was just one integration point, it would be at the center of the element which is about half an edge length away from the corner with the mathematical infinite value, so the stress at the integration point gives a first value. Now reduce the element size by a factor of 2, the distance of the integration point to the infinite point is reduced by a factor of 2, so the stress goes up. Repeat and the stress goes up again. Repeat over and over and the stress goes up each time. Observation that the stress goes up without limit as the element gets smaller and smaller is proof that there is a singularity.nJust looking at unaveraged stress doesn't tell you that there is a singularity, but just looking at a sharp interior corner on the geometry does tell you that there is a singularity.n

January 8, 2021 at 1:22 pmRameez_ul_HaqSubscriber,so you mean to say that even if we are using 'unaveraged stress', a corner 90 deg node will still have a singular stress there. This is bound to happen because of the mathematics behind it. Interpolating each stress from each set of integration points (from every element it is a part of) to this node will still give a singular stresses at that node. This means that all these elements will have an effect of this singular stress (occurring on this node) on the stresses within each of the element. But this effect will only remain within the elements this node is a part of. But beyond these elements, all the elements will have nodes where the singular stress is not happening, and hence these elements won't experience any effect of that singular node.n

January 8, 2021 at 4:36 pmpeteroznewmanSubscriberArraynThere is no such thing as singular stress on an element.nThere is geometry that has a vertex where a stress singularity exists, mathematically. It is only at the vertexwhere the stress is infinite. Any small distance away from the vertex, the stress is not infinite.nElements have a finite dimension, so the integration point is always a small distance away from the vertex.nWe say there is a stress singularity at the corner because the stress value at that corner from averaged or unaveraged nodal values keeps getting larger as the elements get smaller and the distance from the integration point to the vertex gets smaller.nA stress singularity doesn't happen when there is a fillet radius in the corner. The stress will go up as you make the elements smaller, but the curve will flatten out as the elements get smaller. You can extrapolate the last few element sizes on that curve to a true value of stress at zero element size.n

January 8, 2021 at 8:59 pmRameez_ul_HaqSubscriber,I completely understand your explanation, Sir. I am just wondering that since the shape functions are used to calculate the stress at the nodes from the integration points, it means that the stresses on the nodes depend on the stresses at the integration point, isn't it? But at the same time, the stress at that vertex have to be a singularity, even though the integration points are not showing any signs of the singularity. I know there is some mathematics behind this dilemma, which I think I need to research extensively to find out. Because as you said, the stress is supposed to be infinite at that geometric singularity, but not at the integration points. If I can find out the basic mathematics behind this, it would be easier me to judge if that geometric feature or a boundary condition will result in a singularity or not, even before conducting the analysis.nIntegration points are used for every nodes to calculate their stresses, but only the singularity nodes are rising in the stress to an infinite value but not other nodes. You gave an analogy to the crack tip, that the stress is supposed to be infinite there. And its exact solution exists. This implies that ANSYS is using an exact solution on some nodes directly (singularity nodes) and on others, it is not. I need to research quite deeply about this, I guess.

January 8, 2021 at 10:28 pmpeteroznewmanSubscribernYou have to separate in your mind two things: the theoretical value of stress at a corner vertex and the FE calculation of stress at a corner node. The node and shape function have no knowledge of what the theoretical value of stress is at the corner vertex. All the element knows is what the deformations were at the nodes connected to it. From that it can calculate stress at the integration point and using the shape function, the stress at every point in the element is known, including at the corner node. The stress at the vertex is theoretically infinite, but all you can do is use smaller and smaller elements to get higher and higher nodal stress in the corner. The stress got higher in the corner because the stress got higher at the integration point.nIntegration points are used for every nodes to calculate their stresses, but only the singularity nodes are rising in the stress to an infinite value but not other nodes.nNot true. Suppose you mesh a solid body with 4 elements along a 1 inch edge that ends at an interior corner. The node at the singularity shows a stress of 100 MPa while the node 1/4 inch away shows 50 MPa. Remesh with 8 elements. The stress at the singularity goes up to 200 MPa, and the next node, 1/8 of an inch away, shows 100 MPa, while the node at 1/4 inch is now 55 MPa. Remesh with 16 elements. The stress at the singularity goes up to 400 MPa, and the next node, 1/16 of an inch away, shows 200 MPa while the node at 1/4 inch shows 56 MPa. These numbers are just made up to illustrate the point.nn

January 9, 2021 at 7:51 amRameez_ul_HaqSubscriberArray, well that explanation was very helpful. Let me check out the video that you also shared.nAny small distance away from the vertex, the stress is not infinite., you mentioned this in one of your comments. After how many elements from the Stress singularity geometric feature, we can safely depend on the values of the stress? Some articles say after like 2 or 3 elements, if the mesh is fairly refined, you can believe that the stress is accurate.n

January 9, 2021 at 1:30 pmpeteroznewmanSubscribernTony says something about that in the video. I have seen one or two elements ignored around a stress singularity.nBut if the high stress is due to a stress concentration in the design, then no elements can be ignored, because that is likely to be a point at which failure occurs.n

January 9, 2021 at 10:00 pmRameez_ul_HaqSubscriberDoes a change in loading condition change the location of the singularity regions (I mean like moving from one corner to the other), or no matter what the loading condition is, the singularity is supposed occur on some specific geometric features always, and not others? n

January 9, 2021 at 10:18 pmpeteroznewmanSubscribernThe singularity is a feature of the geometry or boundary condition. The size of the stress computed at that singularity will change as the load changes. There can be multiple singularities in the structure, so the highest stress, which should be ignored, could move around as the load changes.nA good practice is to split the structure some distance away from the location of a singularity, which may have knowingly been created by geometry simplification to reduce mesh density in that area, so that a refined mesh can be used in an area of concern. That way, you request stress from the body which has no singularity, and the maximum value can be used directly.n

January 21, 2021 at 11:57 amRameez_ul_HaqSubscriber,kindly observe the pictures below.nnBelow are the results for 0.5 mm of element size (Averaged):nnBelow are the results of 0.125 mm of element size (Averaged):nnnI believe the maximum stress location is a singularity. This is because the stress at that location (or node) keeps on soaring up as the mesh is refined. As you have already mentioned, at the same distance of 0.5 mm from the singular node, the mesh refinement makes the stress gradient to soar up from 431.7 MPa to 1043.4 MPa.nIt is also known that the integration points within that element (of which the singular node is a part of) also have drastic increase in the stresses as singular node has to have an infinite stress theoretically. These integration points will also cause the stresses within the other nodes of the same element to rise up because of shape functions. nMy question is for how many more elements can we expect to see the effect of this singularity (for both, the 0.5 mm element size and 0.125 mm element size? I mean after how many elements will the stress results can be dependable. For the element of which this singular node is a part of, it makes sense to see high stresses within it. But beyond this element, should we expect the integration points of the other elements to also have very high stresses (which in turn will cause the stresses within these elements to soar up)? nI know that you recommended to watch Tony Abbey's video to see how is the effect of singularity on the other elements, but I was not able to grasp the concept properly. nAlso, why is this singularity happening? Is it happening because of the sharp 90 deg transition from the top surface to the hole's side surface? Or is it happening because the stiffness is suddenly changed from an infinite value to a finite value? Either if one or both of these reasons are causing the singularity to occur at that node, then why only that node? I mean there are nodes like this on the entire top edge of the hole, then why only one node is a singularity?.

January 21, 2021 at 12:19 pmpeteroznewmanSubscribernThis singularity is because of the Fixed Support. It is on the top edge because that is where there is an outside surface, the top flat face, which is not fixed except for the edge of the hole. That free surface allows the node on the element opposite the fixed node to deform more than an equivalent element anywhere deeper in the hole.nYou can expect the stress in the part to be almost unchanged by mesh density about 1 or 2 diameters from the edge of the hole. What is the hole diameter, 5 mm? Put a coordinate system at the center of the hole on the top face. Rotate it so that X points toward the end with the force. Insert a Path starting at the edge of the hole, X = 2.5 and ending at X = 12.5, which is two diameters away from the edge of the hole. Plot the stress on the Path for the two meshes you used. You could use a Mesh sizing Sphere of Influence with a Radius of 15 on that new Coordinate System to put a high density of elements near the edge of the hole without filling the whole solid with small elements. That would make it faster to create a few more element sizes to plot.nn

January 21, 2021 at 12:52 pmRameez_ul_HaqSubscriber,so you mean to say that the singularity is because of the sudden change in the stiffness right?nOkay I made a path for for each mesh size (the hole diameter is 4 mm btw), only the region near the singularity has high stresses for both paths (obviously the mesh size of 0.125 mm has higher stresses than 0.5 mm) and the stresses at 10 mm are almost the same for both mesh sizes. But sir what is the point of this? I am concerned about the region in the vicinity of the singularity since I am seeing quite high stresses there, even after the element (of which the singular node is a part of) for like 6 or 7 more elements for a mesh size of 0.125 mm and 1 or 2 more elements for a mesh size of 0.5 mm as shown in the pictures below.n

January 21, 2021 at 3:27 pmRameez_ul_HaqSubscriber,consider another example given below.nnAVERAGED stresses for mesh size 0.5 mm:nnAVERAGED stresses for mesh size 0.125 mm:nnMakes me believe that the location of maximum stress here is again singularity, for the same reasons as for the previous example. Can you tell me now what is the reason behind this singularity?nMoreover, there is a stress concentration also which you can see near the stress singularity [like the teal and aqua regions]. This stress concentration was also present in the previous example. I just want to know should I believe this stress concentration? Like would it be happening in the reality as well [assuming the fixed support region is very very rigid in reality, and there is no fillet radius in reality as well from the top surface to the extended hole outer surface].nAfter how many elements will the stresses be dependable for each of the mesh sizes?n

January 21, 2021 at 8:06 pmpeteroznewmanSubscribernPrevious example was proximity to a Fixed Support. This example is an interior corner on the geometry, which is a well known singularity. The corrective action to eliminate the singularity is to add a fillet radius on the interior corner. nIf you leave the singularity in the model, values of stress one diameter away from that corner will not be sensitive to element size. n

January 21, 2021 at 8:36 pmRameez_ul_HaqSubscriber,what I concluded from your comment is that we can be sure that atleast after 1 (or 2) diameters from the hole, we know that the stresses there are going to be accurate and we can depend on them, rather it be for a smaller element size or larger element size. But less than a diameter away from the singularity region, we cannot give a verdict if they will occur in the reality as well or not, which means we cannot be sure and depend on these stress results. The stress singularity might be having an effect on the stresses within the region less than a diameter away, or it might not. The best way to be sure in this case (i mean for the regions less than 1 diameter away) is to add a fillet radius (with sufficiently refined mesh locally there) and then we can 100% be sure and depend on the results. Since those stresses may now become stress concentrations.n

January 22, 2021 at 4:14 ampeteroznewmanSubscribernYes, you understand it correctly.n

January 22, 2021 at 3:16 pmRameez_ul_HaqSubscribertake a look at the geometric feature here please.n[The face is flat on the other side]nnMesh quality:nnResults:nnWould you classify the red marked region as singularity? The reason being? After how much distance can I depend on the result (since you said that for a singularity occurring near a hole in the previous examples, I can depend on them beyond 1 or 2 diamaters). How am I supposed to put a fillet radius in this region?.

January 24, 2021 at 9:56 amRameez_ul_HaqSubscriber,can I have your opinion on this one please?n

January 24, 2021 at 12:49 pmpeteroznewmanSubscribernYes, it's a singularity because there is a sharp interior corner along the edge to the left of that point.nThe legend tells me that somewhere on this part, the stress has reached 284 MPa. The stress at this point is less than 100 MPa. I would not spend any time thinking about this hot spot. I would spend my time thinking about what is going on in the area of the maximum stress.nTo fillet, just select the three edges and Pull.n

January 24, 2021 at 1:29 pmRameez_ul_HaqSubscriber,if i donot want to put a fillet, then after how much distance can I depend on the results? You already have helped me with the hole singularity [like 1 or 2 diameters away from the hole], what about this kind of singularity?n

January 24, 2021 at 7:47 pmpeteroznewmanSubscribernStress values about 2 elements away from the edge won't change much as the element size gets smaller.nWhat is the yield strength of this material? If it is 160 MPa, then a stress of less than 100 MPa at this location is unremarkable. The attention should be at the location of the 284 MPa stress.nIf this location was the highest stress in the model and the yield strength was 80 MPa, that would require a finer mesh to improve the accuracy of the mesh and if the singularity was a problem, then you would need to add the fillet.n

January 26, 2021 at 1:58 pm

January 26, 2021 at 5:52 pmpeteroznewmanSubscribernIf there are no constraints such as a Fixed Support, sharp angle A is a singularity and sharp angle B is not.n

January 26, 2021 at 7:34 pmRameez_ul_HaqSubscriber,whats the difference?n

January 26, 2021 at 9:44 pmpeteroznewmanSubscribernWell, if you make the bottom face a Fixed support, then you just created a singularity at A.n

January 26, 2021 at 9:49 pmRameez_ul_HaqSubscriberIf there are no constraints such as a Fixed Support, sharp angle A is a singularity and sharp angle B is not.Constraints like fixed support where? At location of A? Because in the last comment you said, Well, if you make the bottom face a Fixed support, then you just created a singularity at AnArray, why won't sharp angle at B might result in a singularity for below condition? I mean both of the corners, A and B, have a sharp angle. What is the difference between them?nn

January 27, 2021 at 2:24 pmRameez_ul_HaqSubscriber,a reply from you would be extremely helpful.nn

January 27, 2021 at 5:28 pm

January 27, 2021 at 6:08 pmpeteroznewmanSubscribernPoint A is an interior corner while point B is an exterior corner.nUnder the support and load shown, the theoretical stress at point A is infinity and at point B is zero.nAs you make smaller elements, the FE stress result will get closer to the theoretical values.nA singularity is defined as stress increasing without limit as the element size is reduced.nNo one cares if the nodal stress of the corner element keeps getting closer to zero as the element size is reduced.n

January 27, 2021 at 9:31 pmRameez_ul_HaqSubscriber@peteroznewman, , I admire it so much that you can actually just by looking at the geometry see what the theoretical stress should be there, according to the loading condition. That is simply amazing. But since I am a fresh engineer, I guess it would take time for me to actually reach to that level nNevertheless, can any different type of loading condition cause the corner B to become a singularity? I mean to say that (and you have also already mentioned in one of your comment somewhere) there exists alot of theoretical singularities within some regions of a model, and for some specific loading conditions, the relevant singularities would be vivid and some other loading conditions, other singularities would be vivid. nFor this loading condition, corner A will result in a singularity. But is B a singularity as well? Since it is also a sharp exterior corner? Maybe not for this loading condition, but for some other. I just want to ensure this. I have read many articles, but for exterior corners, none of them have any mention.n

January 27, 2021 at 10:52 pmpeteroznewmanSubscribernI wrote my reply to the earlier question and posted it before I saw your new Example 1 and 2.nFor both of those, all the interior corners are singularities, all the external corners are not.nInterior = EX1 edge 1, EX2 edge 1 and edge 2.n

January 28, 2021 at 9:13 amRameez_ul_HaqSubscriber,but sir if I make the the exterior corners more and more sharp, wont they become something like this which you have already said is a singularity in your previous comments?nYou in that comment said that this is a sharp interior corner along the edge to the left of that point, then I ask what is the difference between this above corner and the sharp exterior corners I shared in my previous comment which you said wont have a singularity.nI hope you got it what I am trying to put here.n

January 28, 2021 at 11:18 amRameez_ul_HaqSubscriber,and also the picture you shared, to make the sharp interior corner as shown in the above comment go away, you used the pull command and it resulted in a feature like this below. I know this is still a problem for the mesh since it may result in a very bad quality mesh, or the mesh solver might even fail, which could result in very unreasonable stresses, but will the geometric feature, as marked below with blue, is also a geometric singularity right?nMoreover, this statement I made earlier is also true, right?n''there exists alot of theoretical singularities within some regions of a model, and for some specific loading conditions, some singularities would be much more vivid and some other loading conditions, other singularities would be much more vivid.''n

January 31, 2021 at 6:36 pmRameez_ul_HaqSubscriber,an answer from you would be so much appreciated.n

January 31, 2021 at 7:45 pmpeteroznewmanSubscribernThe blend marked with blue, is not a geometric singularity because it is a blend. All surfaces are tangent. As the elements get smaller, the element quality will improve.n''there exists a lot of theoretical singularities within some regions of a model, and for some specific loading conditions, some singularities would be much more vivid and some other loading conditions, other singularities would be much more vivid.''nThat is true.n

January 31, 2021 at 9:06 pmRameez_ul_HaqSubscriber,if we apply a fixed support for example to a sharp corner, then the singularity that is supposed to be seen due to the geometry will no longer be seen in the results, right? I mean we still might be seeing a singularity because of the sudden change from the fixed support (having infinite stiffness) to the normal stiffness of the structure, but the reason for the singularity will no longer be sharp corner, right?n

January 31, 2021 at 10:58 pmpeteroznewmanSubscribernIf you had a sharp interior corner that was on two free faces of a structure, and you put a fixed support on those two faces, the singularity at the corner would disappear, but a singularity might appear where the fixed support ends.n

February 4, 2021 at 8:19 amRameez_ul_HaqSubscriber,if you may, let me ask some basic questions.nI want to know the reason for each singularity as marked in the above pictures. The basic reason, in my knowledge, for a singularity to occur is that the force is essentially passing through a single node, but what is happening in the above regions where we are seeing a singularity? A sudden change in a stiffness matrix maybe? But that must cause a convergence problem, why is that a singularity? n

February 4, 2021 at 11:55 ampeteroznewmanSubscribernThe stickfigure illustrations might lead some readers to think you are talking about beam elements. Beam elements do not have singularities. Solid elements most easily reveal the presence of a singularity.nThe basic reason for a singularity is that the theoretical stress is infinite. That happens when a force is applied to a point on solid geometry meshed with solid elements. Stress is Force/Area. The area of a point is zero, therefore stress is infinite. n

February 4, 2021 at 2:00 pmRameez_ul_HaqSubscriberArray, That happens when a force is applied to a point on solid geometry meshed with solid elements, but the singularity regions I have already shown in my previous comments, thats also going to occur if that were shell elements, right? Also the sharp interior corner singularity is going to occur on a surface body with shell elements, right?nAssume I am scoping a face of a solid body to a remote point, and then I apply a constraint equation to that remote point. Should I expect a singularity in the node/element just right adjacent to that scoped face? Since there is a sudden change in the constraints between that face and the elements just right adjacent to it.n

February 4, 2021 at 7:01 pmRameez_ul_HaqSubscriberAssume the remote point is scoped with a DEFORMABLE behavior.n

February 4, 2021 at 11:59 pmpeteroznewmanSubscribernWhat do you mean by a sharp interior corner using shell elements? There are two configurations. A planar configuration has all shell elements in one plane and the shape of the boundary in that plane includes a sharp interior corner. A right angle configuration has shell elements in two orthogonal planes and the corner is created at the line of intersection of the two planes. I recommend you build each type and perform a mesh refinement study to find out if either one has a singularity.nWhat behavior have you set on the remote point? Is it Rigid or Deformable. You will get different results from each of those. What is the constraint equation you imaging layering on top of the remote point? Take a specific example and perform a mesh refinement study to find out if you have a singularity or not.n

February 5, 2021 at 7:48 amRameez_ul_HaqSubscriberArray, I understand the right angle configuration, but can you provide an image for the planar configuration? I didn't understand what you meant by shape of the boundary in that plane includes a sharp interior corner. If you can provide an image for this condition that would be helpful.nAnd I have already mentioned in previous comment that the remote point is set to DEFORMABLE behavior. For the RIGID behavior, it is understandable that there definitely would be a singularity due to the abrupt change from the infinite stiffness to the normal stiffness of the structure. Take a deformable fixed joint, for instance. I scope a fixed joint between two faces of two separate bodies. I just want to ask that at the transition of those scoped faces to their adjacent faces (which doesn't have a fixed joint applied to it i.e. no constraint equations scoped to it), should I expect a singularity or a stress concentration at that transition? I mean I can create a model and check for myself, and I have done it previously numerous times, but having an idea beforehand can make the judgements a little bit more easier.n

February 6, 2021 at 3:15 am

February 6, 2021 at 9:44 amRameez_ul_HaqSubscriber,I am going to address an important aspect of FEA and I would like to know your thoughts on this one.nIf you observe the pictures I shared in my previous comments regarding the singularity, can you tell a way to overcome these singularities? For the sharp corner, we already know a fillet will work. But now consider the singularities in the first picture. There exists a singularity right next to the fixed support (for solid elements). In reality, the stiffness is not going to be infinite there. But there would definitely be an elevated stiffness there (as compare to the rest of the model) and there would also exist a reaction force there. Now, if I try to make a convergence study for this in FEA, I would see that the stress is rising and diverging just at the transition from the fixed support to the regular stiffness of the structure. Maybe 2 or 3 elements away, the stress is going to be okay to depend upon. But can we conclude that even though the transition is showing a singularity, but in reality the stress at that location is going to be somewhat close to the stresses which we are seeing 2 or 3 elements away from there? I mean is this a sensible conclusion? Or the stress at that transition is not going to be close to the stresses 2 or 3 elements away from there? nThe same goes for the singularity at the location where there is a transition between the force area and the area right next to it where there is no force (in the first picture). That is a singularity, meaning the stress would be diverging. But is there a way to know the actual stress at that transition by overcoming the singularity? What should we expect at that location in the reality?nAnd also (as shown in the second picture) can you tell what happens to the stresses when there is a abrupt transition from one material to the other in reality? In FEA, it might show up as a singularity, but what about in reality? How can we overcome this singularity so that we can draw some reasonable stress results.n

February 7, 2021 at 1:40 ampeteroznewmanSubscribernStress analysis is used to evaluate if the structure will survive the applied loads without failure. This involves simplifying the geometry, connections and boundary conditions in order to get a first solution quickly. This first solution reveals where the highest stress is in the structure. The simplified geometry and boundary conditions might have introduced many singularities, but they are of no interest if they are not at the location where the highest stress was found.nIf there is a singularity at the location of highest stress, then making a change to remove it is warranted. Adding a blend radius, or replacing a simple Fixed Support with a Bolt Preload and Contact.nIt's not useful to say that the stress 2 or 3 elements away is correct. We want to know the correct maximum value of stress and you can't know that without looking at the element with the maximum value of stress.nDid you complete the study to see if there is in fact a singularity at the edge of a face that has an applied force on a solid element mesh?nWhen two dissimilar materials are bonded together, such as one material deposited onto the surface of another material to form a coating, there is a step change in stress at the boundary of two materials with different values of Young's Modulus. Since the strain at the interface must be equal, the stress has to be different. But don't confuse a step change in stress at a boundary with a singularity. You can have a finite step change in stress at the boundary of dissimilar materials that does not increase without limit as the elements get smaller.n

February 8, 2021 at 5:30 pmRameez_ul_HaqSubscriberArray, I conducted a mesh convergence study to see if there would be a singularity at the edge i.e. at the transition from the area, at which the force is applied, to the area next to which, at which there is no force.nThe element size are as follows: 2 mm, 1 mm, 0.5 mm, 0.25 mm.n

February 8, 2021 at 9:35 pmpeteroznewmanSubscribernA shell model is not a good place to look for singularities, especially when applying a face load. Shell elements have zero stress normal to the shell face.nUse a solid model to look for singularities at the edge of a load.nYou can have a structure that has many singularities, but because of the loading, the stress in the elements around the singularity is not the highest in the model. If the loading was changed, then maybe the highest stress can be at the singularity. Most models have many load cases that must be studied.nn

February 12, 2021 at 4:10 pmRameez_ul_HaqSubscriber,I tried building a solid model (exactly the same as the previously shared shell model) and tried to observe the singularities at the transition from the force face to adjacent faces. However, I was still not able to find any singularity there. But, as also mentioned by Tony Abbey, they should exist there. Maybe the singularity is not apparent in the region of interest for this loading, but might be apparent in a different loading case. What would you like to say about this? Like sometimes the singularities are vivid on the transition from the force face to the adajcent faces, and sometimes they are not? What do you think?nAlso, can you kindly tell me if we can observe any singularity effects on contact regions, like bonded (linear) or frictionless (NonLinear)? An example would be so much appreciated.nn

February 13, 2021 at 1:15 pmpeteroznewmanSubscribernListen to what Tony Abbey says at time stamp 14:00. Switching an applied force from a point load to an area makes the singularity go away.nWhere did you get the idea that the boundary of an area that has an applied force creates a singularity?nBonded contact can easily create a singularity. One common example is a part with a hole in it and a shaft in the hole with bonded contact between the hole and shaft. Each solid mesh on its own has no singularity, but the bonded contact creates an interior edge and a singularity. This is essentially the same singularity you would get if you united the solid bodies in CAD, creating an interior sharp corner.n

February 13, 2021 at 1:53 pmRameez_ul_HaqSubscriber,I mean on the first slide of his episode, it is literally written on the second bullet 'Creating abrupt local transitions in local loading', and he even explains it there. Please have a look.nregarding the singularities seen from the contacts, I was actually concerned about the maybe seeing a singularity at the transition from the face which is bonded (or having a frictionless connection), to its adjacent face which is not in contact with any other face.n

February 13, 2021 at 2:05 pmpeteroznewmanSubscriber

March 22, 2021 at 11:35 amRameez_ul_HaqSubscriberArray, if you recall, I posted a comment like this which you affirmed.nThe displacements are computed for the nodes during the solution, by solving the stiffness matrix. These displacements of the nodes are used to compute the displacements within each element using the shape functions. Points within an element are chosen where the stresses are computed using those points' displacements. These points are known as integration points. After computing the stresses at the integration points, those are interpolated onto the element's nodes, again using shape functions.nI want to ask that does those Guass point (onto which the displacements are computed coming from the nodal displacements of an element) interpolate the displacements within the whole of element by using their own displacements [after which strains and then stresses are calculated by using hooke's law within the whole of element], or they just interpolate the strains and the stresses within the whole of element by using their own strains and stresses?nAnd also, for example, when the displacements of the nodes are already known, then it means that we can also calculate the strains and stresses at all the nodes by using elasticity equations/hooke's law. I don't understand why it is not done in this manner. Why are Guass points so important? Additionally, when we interpolate the either the displacements or strain/stresses coming from the Guass points onto the nodes, don't they need to be equal to the ones already found at the nodes?nBasically what I am trying to say is I don't understand why does the stresses at the nodes coming from the elements it is a part of, is different, although there is a single displacement, strain and hence stress associated with it.nn

March 22, 2021 at 12:34 pmpeteroznewmanSubscribernHere is a reference that describes Finite Element theory. nhttp://solidmechanics.org/Text/Chapter8_1/Chapter8_1.phpnThe stiffness matrix has integral equations that must be numerically integrated. Gauss quadrature is the numerical integration method used. Gauss points are the points in the element where the displacement values are used to compute the integral. The displacement values at the Gauss points are computed by the shape functions which describe the displacement at any point in the element given the displacements of the nodes. nWhen the displacement at a node is already known, such as at a Fixed Support, those nodes are eliminated from the stiffness matrix prior to solving for the unknown nodal displacements.n

March 22, 2021 at 1:21 pmRameez_ul_HaqSubscriberArray, The displacement values at the Gauss points are computed by the shape functions which describe the displacement at any point in the element given the displacements of the nodes. I still didn't get what I was looking for. So after the displacements are computed at each point within an element using shape functions and displacements at the guass points, does it mean that these displacements calculated at any point within the element is used to calculate the stress and strain at that point [using elasticity equations/hooke's law]? Or the stress and strain is first computed at the gauss points and then these values are interpolated at any point within an element?nFor the former case, if it is done so, then the displacements at the nodes coming from the interpolation from the gauss points must be equal to the ones already determined at the nodes. This means a single displacement exist at the node, meaning single strain and hence stress value. (No matter that node is connected to how many elements).nIf latter is the case, then I can believe that the stresses at any of the node might have different values coming from different elements since the shape functions are going to be different for each element and hence the interpolation might give different stresses at the connected node of various elements.n

March 22, 2021 at 2:43 pmpeteroznewmanSubscriberKeep in mind that when the stiffness matrix is being assembled, it is for the Integral form of the governing equations of linear elasticity. Those equations include: nthe straindisplacement equation,nthe elastic stressstrain law (Hooke's law), nthe equation of static equilibrium for stresses (which requires an integral and uses Gauss points)nthe boundary conditions on displacement and applied forces.nThey are all rolled up together into a set of simultaneous equations that allow the solution of the unknown nodal displacements. After the solution of nodal displacements, post processing allows element stress and strain to be computed from the solved displacements.nThe shape function allows interpolation of displacement, strain and stress throughout the element. Each element has its own nodal stress value so if one node is connected to 8 hex elements, there will be 8 values of stress at that node, one for each element. That is why the nodal stresses are averaged.n

March 24, 2021 at 10:51 amRameez_ul_HaqSubscriber,I have just taken the basic finite element theory course, and therefore I am not aware of the indepth assembling of the stiffness matrix for a large global model. I guess I definitely need to read and research more about this, or take any advanced finite element courses to illuminate my mind on how does the complicated finite element works.nHowever, I just have a simple, naive question and I would be glad if you could clarify it here. I mean when the displacements are interpolated from the guass/integration points to the rest of the element, this would mean that a node connected to 8 hex elements would have 8 different displacements coming from each element it is a part of. Is this true? If it is, then does it mean that the nodal displacements that we see in our FEA results are basically also averaged out ones?nBut what I think is that the shape function behaves and works in such a manner that when it tries to interpolate the displacements back onto the nodes, it has to give the value of the displacement that was calculated by solving the simultaneous equations within the stiffness matrix. This means that even if a node is connected to 8 hex elements, all the shape functions of all the elements must eventually give the same displacement value at that node which is found by solving the global stiffness matrix. If this is how FEA works, then why can't we have the same thing happening for the strains and stresses as well for a node which a part of many elements?n

March 24, 2021 at 11:25 ampeteroznewmanSubscriberArray na node connected to 8 hex elements would have 8 different displacements coming from each elementnThat is not the right way to think about it. Instead, think that displacement of a node connected to 8 elements is unknown, but the stiffness of those 8 elements will be used to compute the displacement of that node in a big stiffness matrix that has all the other nodes in it as well. nEach linear hex element has 8 nodes, but only one node on each element is shared by the 8 elements. The other 7 nodes on each element have different displacements, so each element has its own unique strain, and therefore stress.nThe nodal displacements are the solution. That is the starting point for all postprocessing. The shape function is used to interpolate displacement locally within the element.n

March 24, 2021 at 12:30 pmRameez_ul_HaqSubscriberArrayInstead, think that displacement of a node connected to 8 elements is unknown, but the stiffness of those 8 elements will be used to compute the displacement of that node in a big stiffness matrix that has all the other nodes in it as well. Why can't this be further extended to calculate the strain as well as as stress at that node then?n

March 25, 2021 at 12:25 pmpeteroznewmanSubscribernThe system of linear equations for statics is [K]{x} = {F} where x is the vector of unknown displacements.nThe stiffness matrix [K] includes the straindisplacement equation, the elastic stressstrain equation, and the equation of static equilibrium.nAfter solving the system of equations, the displacements become known and the straindisplacement equation is used to recover the strains and the stressstrain equation is used to recover the stress.n

March 25, 2021 at 1:19 pmRameez_ul_HaqSubscriberArray, After solving the system of equations, the displacements become known and the straindisplacement equation is used to recover the strains and the stressstrain equation is used to recover the stress. strains and stresses of what? Of the same nodes within the model whose displacements were first found out? If yes, then why there is a need for shape functions to find stresses at the nodes (using guass points) when we can already find the strains of all the nodes using straindisplacement equations and stresses of all the nodes using stressstrain equation?n

March 25, 2021 at 2:50 pmRameez_ul_HaqSubscriberNow, after finding the strains and stresses at the nodes of an element from the straindisplacement equations and stressstrain relations, we can interpolate them directly within the element using shape functions, without having a need of guass points.n

March 25, 2021 at 7:15 pmpeteroznewmanSubscribernThe nodal displacement solution allows the post processor to compute the element stresses and strains. The element formulation includes a definition of the shape function, which defines what the strain is everywhere in the element, including the corner nodes. The gauss points are not very relevant during post processing.nGauss points are defined in the element formulation and can have an effect on the solution process and the resulting nodal displacements. You may have seen keyopts that can change how many gauss points are used in an element. That might be done to avoid problems that can occur in the solution such as hourglassing or volumetric locking.n

March 26, 2021 at 6:54 amRameez_ul_HaqSubscriber,I read here: https://www.quora.com/WhatisGausspointinFEMnthat the guass points are called integration points because these are the points within the element which are used to compute the element's stiffness matrix and then they all are summed up to compute the overall global stiffness matrix. I think you also wrote similar words in one of your previous comments. But you said the integration at these guass points is carried out by making use of the displacements at these points, and the displacements at these points are calculated by interploating the displacements at the nodes. But to calculate the displacements at the nodes, we already need to have the global stiffness matrix and local elemental stiffness matrix assembled and simultaneous equations solved. Why the paradox?n

March 26, 2021 at 7:41 amRameez_ul_HaqSubscriber,and also, is there a relation in the usage of guass points within an element (like 1 or 4) and the type of element used (like linear or quadratic)?n

March 26, 2021 at 12:00 pmpeteroznewmanSubscribernThere are more Gauss points in a quadratic element that in a Linear element.nWhen assembling the stiffness matrix, the displacements are unknown. They appear as a variable in the equation, along with all the constants for terms such as Young's modulus. The equations use the location of the Gauss point within each element as a place to do the integration required to have equilibrium in the solution.nDuring solving, only the unknown displacements are solved for. Stress and strain are not solved for, but the equations that have stress and strain terms in them were included during the assembly phase of the stiffness matrix, before the solving started.nAfter the system of equations is solved, the displacements become known and now values can be put into equations to extract out values of stress and strain.nIt is a three step process. 1. Assemble the stiffness matrix. 2. Solve for the unknown displacements. 3. Postprocess solved displacements to get stress and strain. n

March 26, 2021 at 2:36 pmRameez_ul_HaqSubscriberArray, so basically, the guass points itself are a function of the displacements of the nodes, therefore while carrying out the integration to assemble the element stiffness matrices and to have the required equilibirum in the solution, it all drops down to having only nodal displacements of the whole model, as the unkowns. Is that correct, sir?nStress and strain are not solved for, but the equations that have stress and strain terms in them were included during the assembly phase of the stiffness matrix, before the solving started. These strain and stress will also simplify to only having constants and unkown nodal displacements, right?n

March 26, 2021 at 5:05 pmpeteroznewmanSubscribernthe Guass points itself are a function of the displacements of the nodesnNo, that is not correct. The Gauss points have a location within the element, which is defined by the location of the nodes. Go to ANSYS Help and under Mechanical APDL, search for Gauss points.nThese strain and stress will also simplify to only having constants and unknown nodal displacements, right?nA system of equations for stress, strain and equilibrium gets written out in terms of unknown displacements during the assembly of the stiffness matrix.n

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