## General Mechanical

#### Brittle Material Bending Fracture in Static Structural

• jackhero
Subscriber

I am trying to simulate the brittle fracture of concrete under three-point bending. The concrete beam is 40mm x 40mm x 160mm with span length of 120 mm. The beam is not reinforced with any other material and is supposed to have a brittle fracture after some loading (experimental value around 5Mpa).

I would like to ask that how can I model the brittle fracture in static structural? I have defined the ultimate tensile and ultimate compressive load values in engineering data for the material. The simulation load is passing after the defined load values. I request if someone please guide me where I am doing wrong? I have used the PLC crack command to see the crack but I am not getting any thing there. I am only interested in getting the max load on which failure occurs and the crack on the material. The safety factor Mohr-Coulomb material model is optional.

I made the model in both 3D and 2D. Instead of drawing the physical rollers for the supports and loading, I have split the faces for the loading support at the top and bottom face for the two supports.For the 2D model it is quicker to solve but would be helpful if some expert may comment on whether the designed 2d model is correct or not.

• peteroznewman
Subscriber

Jack,

The 2D model is incorrect for a 3 point bending model because your displacement has selected the entire top edge and not a vertex at the center.  You can create a vertex at the center in DesignModeler by splitting the edge.

The 3D model has created a narrow face to push on at the center. It would be better to put a remote point on that face and push down on the remote point, which will distribute the force to the face.

The 3D model also is holding two faces on the bottom with 0,0,0 displacement. That does not represent a 3-point bending test that is typically on rollers. Please review this discussion and come back with any questions.

The material you call concrete has no failure mechanism in it to generate cracks, but you can evaluate the Maximum Principal Stress and compare it with the Ultimate Tensile Strength to decide if a failure has occurred.  The Compressive Ultimate Strength value in the Material is only used by the Stress Tool to plot Factor of Safety.

Regards,

Peter

• Sandeep Medikonda
Ansys Employee

Jack see, if these 2 videos of DrDalyo help:

Regards,
Sandeep
Best Practices to post on the Student Community

• jackhero
Subscriber

1. I would like to confirm that the failure criteria maybe defined as the UTS value only (ultimate tensile strength) in Engineering Data? In static structural, for linear elastic modeling, once this UTS value is reached the simulation would stop or give results that the applied load exceeded material ultimate strength. Is there any option to do something like this? Although the applied boundary conditions, as you have suggested, were not correct I have seen the simulation going pass the defined ultimate values. Maybe this is due to the incorrect boundary conditions but I would like to confirm about the failure criteria could be or could not be taken by ANSYS with the defined material values? I searched for this question in the community and came across to one of your post here, and it seems like we have to manually guess that the material failure has occurred.

2. I have seen some videos in which they have used Solid65 material to model the non-linear behavior along with the cracks (using PLC crack command). However, as you have mentioned that the default Ansys concrete material does not have the failure mechanism to generate cracks. Currently I am modeling the brittle failure only, in future if I were to model the non-linear behavior with cracks, is it like I can only select the Solid65 material instead of the default concrete material for crack observation? Or I have missed something when defining the default concrete material properties to include the crack as well?

3. For the 2D model, I split the top face while designing the beam in the Design Modeler.

I also split the bottom face for two supports.

But I did not get the split part when I had the final sketch. If you have time could you please take a look at where did I do wrong? It would be helpful if you may provide steps for creating the vertex at the center in DM by edge splitting? I have done face split for 3D models but for 2D I can not exactly figure out.

4. In 2D model, if I were to add displacement for the two bottom supports at the two extreme ends isn't it like no need to split the bottom face of the beam as I have done in the image shown above?

5. Are there any tutorials or guide available of 2D model designing for the ansys design modeler? And for creating symmetry, axisymmetry models.

6. For the two bottom supports, you mentioned in the discussion to use Z=0 for the roller bars, with all other displacements and rotations are free.

In one of your 2D designed model for fracture toughness by three point bending, the bottom supports were X=0, Y=0, for one end of the displacement while as Y=0 for the other end of the displacement.

Although the first model is 3D but the displacements applied or suggested to apply are same on both of the bottom supports. However in the 2D model the displacement applied on the bottom are different on both ends. Could you please suggest which one would be better to use?

7. In addition, with in that 2D fracture toughness model, the force applied on the top seems to be on the entire edge instead of the vertex as you have suggested in your reply above. If this was done deliberately, it would be helpful if you may explain the reason for it.

• peteroznewman
Subscriber

1. Engineering data holds the Ultimate Tensile Strength, but it has no effect on the solver. The solution will not stop when this value is reached. The only time this value is used is when you put the Stress Tool into the Results and plot the Factor of Safety. Then the UTS is divided by the Stress to compute the Factor of Safety.  If the material model is a linear elastic material, the solution can take the stress result way past the UTS. It is only in Post Processing, that you can make a plot like the Factor of Safety to show if the material is over or under the failure value of UTS.  I don't usually plot Factor of Safety. I prefer to just plot the stress (Max Principal for brittle materials, von Mises for ductile materials), and I set the value on the legend that separates orange from red at the value of UTS then any material colored red has failed. For ductile materials, instead of using a linear elastic material, I add a plasticity material model. That allows the material to stretch at values of stress above yield. Then instead of comparing stress to UTS, I look at Total Strain and compare that to Elongation. In this case, I set the threshold on the legend between orange and red to the value of Elongation and any material colored red has failed.

2. Solid65 is an element type not a material.  Solid65 elements are designed to work with the Concrete material.  When you use concrete and solid65 elements together, the material has failure mechanisms built in and the ability to simulate cracking and crushing.  The problem with the concrete/solid65 combination is the results are extremely sensitive to the mesh used in the solution. You have to use very small elements and very slow loadings and a slightly different mesh can give a very different result.

Some people are more interested in the gradual failure of a concrete beam and a plasticity model is more useful than the cracking model used with Concrete/Solid65 approach.  A plasticity model has been developed called Microplane that is very robust and gives accurate results across a wide range of meshes. You can read the Microplane example in the Technology Guide in 19.1.  Here is a discussion on that topic.

3. You can put a plane at the center and use Slice, then put the two pieces in a Multibody part using Form New Part in DM.

4. You don't need any overhang, you can just apply the supports on the corner vertices.

5. Look on YouTube for tutorials on 2D design modeler.

6. In a 2D model without using symmetry about the center, you need one support to have X=0, Y=0 and the other support to have just Y=0.   If you have symmetry at the center, then the one support only needs Y = 0 because the symmetry prevents motion in X.

7. You can apply a force to the entire top surface of a beam. That is called a distributed load. That is not how a 3-point bending test is done, but it is a valid load.  Applying a displacement to the entire top surface is not a valid load.  A typical 3-point bending test is to apply a displacement load to a point or small area at the center.

Regards,
Peter

• jackhero
Subscriber

3. You can put a plane at the center and use Slice, then put the two pieces in a Multibody part using Form New Part in DM.

I can not exactly follow this it would be helpful if you may show the steps graphically (screenshots).

I created a new plane at  the center.

Right click on new plane > insert > slice

Then I get these options and for some of these I can not select the target body or the face for slice.

• peteroznewman
Subscriber

The XY axis of a plane does the slicing.

If you copy the YZ plane to the center, then that will be correctly oriented to do the slicing.

You can change the Base Plane to YZ or you can rotate the plane you have at the center.

• jackhero
Subscriber

I followed the instructions and ended up with this.

However, while applying load on the top surface I can't find the vertex using the vertex selector. I turned on to show all the vertex and the point I created is not shown in there. Have I missed something in the DM or I am applying the loads in incorrect way in Mechanical?

• peteroznewman
Subscriber

In one of your earlier screen shots I saw a Solid that was suppressed and a Surface Body that was not.

In your recent screen shots, I see the Solid has been sliced into two, but the Surface Body has not.  Delete the Solid Body and try the Slice on the Surface Body.

• jackhero
Subscriber

I re-modified the design to slice the Surface Body. After Slice I selected both parts and formed a new single part. This is the final design I get in the DM.

I am not sure if I have followed your instructions correctly but I am ending up with this in the Mechanical. I had to separately assign Material Properties to both of the Surface Bodies under Part 2. Is it normal to have like this or I again have missed something in the DM?

• peteroznewman
Subscriber

Yes, each body needs a material assignment. That is one of the benefits of Slice.  Instead of slice, you could have used Concept, Split Edge, and created a vertex at the 50% point on the edge.

• jackhero
Subscriber

I ran the simulation and request guidance on the following queries;

(1). In the experimental result, the deflection at (brittle) failure was 0.16mm with a maximum stress value of around 3Mpa. The test was carried out at 0.5mm/min of the displacement rate. I would like to ask when running a simulation, and the aim is to verify the experimental results, how to decide or calculate how much of the displacement or load in boundary condition should be applied to get the almost same result as that of the experiment?

(2). I applied the displacement of 0.16 mm and ran the simulation, I get the maximum principal value as of around 114Mpa. The resultant maximum load is too high then the expected value.

I then ran number of times with different displacement boundary conditions until I had the maximum load value of around 3Mpa in the simulation. I had this value on the application of 0.004mm of displacement boundary condition.

I would like to ask,

(2-i). Is it the correct approach to verify the results?

(2-ii). Whether the approach is correct or not, in both of the displacement loaded boundary condition the maximum stress has occurred on one of the side support. Shouldn't I be getting the maximum stress somewhere at the middle of the beam or near to the area where the displacement load has been applied, in actual work this is where the beam had generated a crack and then split apart.

(3). In one of your reply on another thread, you suggested to use Solid185 or 187 element instead of solid 65. Also you have suggested to use microplane damage plasticity model for gradual damage. I would like to ask that,

(3-i). I I designed the beam on 2D and the element ANSYS is using to solve the simulation is Plane183. Can you suggest which element should better to use if I would like to see the crack on the beam in case of 2D analysis as Solid 65 or 3D elements are not valid for my case.

(3-ii). The microplane damage plasticity model is I think added in Ansys 19.1 version. I am using Ansys 19.0 version. Could you please suggest a failure mechanism that I shall to enter in Engineering Data for my simulation? So far, I have only added Linear Elastic properties of the material and I could not find any failure mechanism. For this simulation work I only intend to verify the maximum load and the cracking pattern. There is no non-linear behavior present as the material has brittle fracture.

Thank you

• peteroznewman
Subscriber

Hello Jack,

(1) In the experiment, you can measure deflection, but you can't measure stress. Where are you getting the 3 MPa stress value from.  The experiment applies a force, what was the force just before fracture? It would be best if you can get the force-displacement curve from the experiment. In the simulation, you can plot a force-displacement curve, export that to Excel and overlay the two curves.

(2) The maximum stress in your simulation is coming from the support, so is irrelevant, because your support is on a vertex, which is an idealization that causes high stresses that should be ignored. One technique to be able to easily ignore the stress at the support is to slice the body at the 25% and 75% of the length, and put those three bodies in a multibody part and use Shared Topology to mesh them back into a single beam. However, you still have 3 bodies, so when you request a stress output, just pick the center body and the maximum stress should be in the center.

(2-i,ii) Do the multibody part method.  You might need to split the top edge to make a small length to apply a remote displacement to avoid a peak stress at a single vertex on the top.

(3) Solid 65 can show cracks. Microplane does not explicitly show cracks but does accurately show the curvature in the force-deflection curve leading up to the ultimate failure.

(3-i) You will have to go to 3D to use Solid65.

(3-ii) If the experiment shows a sudden, brittle fracture, is it sufficient to predict the load when that happens?  Do you need to show cracks? The simplest failure prediction model is linear elastic. You plot Max. Principal Stress vs Load.  When the Max. Principal Stress = Ultimate Tensile Strength of the material, that is the load when fracture is predicted. That may over-predict the load when fracture occurs. Fracture Mechanics is an approach that predicts the fracture load when a small initial flaw or crack exists in the sample being loaded. ANSYS can calculate the critical load for a given edge crack length.

• jackhero
Subscriber

Thank you for your reply.

(1). From the experiments I had the load-displacement values which I later converted to stress in excel to compare with the output of maximum principal in ansys workbench. If I could verify the load-displacement curve it is also ok. Attached is the load-displacement curve of the experimental test. The maximum load is around 1120 N. With this result how shall I calculate the value of remote displacement which I shall to apply to verify the simulation and experimental results?

(2). If the current geometry can verify the experimental result I think it is enough for me. If not then I will modify the geometry to multi-body parts (by slicing) as per your suggestion. I drew the vertex but I applied displacement instead of the remote displacement. I will re-run with the remote displacement at the top to see if there is any difference.

(3-ii). If I could get to show the crack occurrence on my 2D geometry it would be preferable along with the maximum load that the beam can sustain before failure.

The Fracture Mechanics as you have suggested works on an initial flaw or existing crack in the sample. Should I to include the crack or flaw in the geometry? This approach is mostly followed for measuring the crack or stress intensity factor (SIF) (fracture toughness). One supplementary question I may ask is can I, without experimental result, predict the stress-intensity factor value (fracture toughness) by using Fracture Mechanics if I include the crack of certain size in the current geometry? I saw one of a video in which there have description regarding the Fracture Mechanics modeling for SIF (

). The issue is the loading boundary condition input value (of remote displacement) if I run a simulation to predict SIF without any experimental result.

• peteroznewman
Subscriber

Hello Jack,

(1) What is the published Ultimate Tensile Strength for the material?  Let's say it is 3 MPa, which is what you got in your experiment. Request stress output from the vertex at the bottom center of the beam, which will exclude the supports. This is a small deflection, linear analysis, so all you need is one point on the stress-load curve to draw the straight line from the origin. Let's say you got a simulation result of 6 MPa stress at a load of 2400 N. Just take the ratio to calculate the load at failure = 2400*3/6 = 1200 N.

Do a ratio to calculate the deformation at failure. Take the FEA result of load and deflection at failure and plot that point on the experimental graph. That is a comparison between FEA and Experiment.

Calculate the percent error between the failure load of the experiment and the FEA result.

Percent Error = 100*(F.exp - F.fea)/F.exp = 100*(1120 - 1200)/1120 = -7%

If the Percent Error is < 10% that is a pretty good agreement. If you get a large error, then check all the inputs to the FEA model. Confirm that the FEA dimensions match the measured dimension of the experimental sample, including the distance between the supports as well as the thickness and depth. Confirm that the material properties match.

(2) Applying load at the vertex on top is okay because you will get the stress from the vertex at the bottom.

(3-ii) Showing cracks is going to be a lot more work.

If the FEA result predicts a load that is say 2400 N, an error that is 114%, you can insert a crack at the bottom vertex and use Fracture Mechanics to calculate the crack length needed for the beam to fail at 1120 N instead of the flawless 2400 N result.

Regards,
Peter

• jackhero
Subscriber

Thank you for the detail explanation.

One confusion I would like to ask about the (remote) displacement boundary condition value on the top. How shall I estimate/calculate how much (remote) displacement I shall apply to verify/get the FEA result?

I assumed that this (remote) displacement value may be equal to the experimental deflection value at break. Is it correct?

Mostly in FEA if I want to verify the experimental results for different type of structural mechanic simulations (mostly related to concrete) I struggle to estimate/determine the loading (whether be displacement or force) boundary condition value selection. Is there any specific rule or theory to determine the loading boundary condition value for FEA from the experimental results?

• peteroznewman
Subscriber

I suggested remote displacement on an edge of some small length to spread out the stress concentration of the point load on the top vertex. Then I thought to ask for stress not on the whole body, but just on the bottom vertex, so now I don't care about the stress concentration on the top vertex, it will not pollute the stress output at the bottom vertex.  So for this model, forget about remote displacement.

Since your model is linear, it doesn't matter what force or displacement you apply to the model because you are going to take the ratio I described above. You can push more or less than the experiment. It doesn't matter to the ratio, you will get the same result no matter what value you use in the model.  If you want to make a nice picture of the stress at break, you can rerun the model at the value calculated by the ratio and you will have it.

The rule for boundary conditions in FEA is to make them look as close to the experiment as possible. Use symmetry if you can accept only symmetrical results. Remote displacement is a good way to push a line or area down by a displacement, while allowing that line or area to tilt a little. Remote displacement enables that because a single point is being displaced, and the rotations can be left free. You can't get that effect if you apply a displacement directly to the line or area.

Regards,
Peter

• jackhero
Subscriber

To get the Force Reaction at the bottom vertex I tried selecting the bottom vertex but I got a question mark next to the Force Reaction.

I selected the Force Reaction of Top Displacement and following results are all according to this Force Reaction results.

The first simulation I ran at top Displacement of 0.16mm and I had the following Force Reaction (152.54 N) and Max Principal Stress (114.45 Mpa) values.

For the load ratio I followed as,

154.54 / 114.45 * 3 = 3.99N

where 3 Mpa is the experimental max stress value.

I re-ran the simulation with 1mm of Top displacement and had the Max principal (715.3 Mpa) and Force Reaction (953.39 N).

For the load ratio I followed as,

953.39 / 715.3 * 3 = 3.99N

The load ratio, if I have calculated correctly, is the same for both simulations which you have also mentioned that the ratio will be same. Does it mean that according to FEA the load at failure is 3.99N?

You suggested to do a ratio to calculate the deformation at failure. The max deformation values (which I think are corresponding to the load at failure or the maximum load) in both simulations are same as the value of the top-displacement which I input as boundary condition (0.16mm and 1mm respectively for both simulation run). How shall I get the ratio of deformation at failure?

• jackhero
Subscriber

The rule for boundary conditions in FEA is to make them look as close to the experiment as possible.

If we talk about the loading boundary condition only, not the fixed support(s). As a general rule (not only for the brittle material simulation which we are discussing in this thread, but generally for structural simulations), if in experimental test I applied displacement at a certain rate (in mm/min) and in result I had the load-displacement data values. Then in Static Structural simulation, to verify the experimental results by FEA, I apply displacement (in mm) then this displacement should correspond to which experimental value?

(i) Will it be the final value of the displacement from the experimental load-displacement data values? Which i think shall be the case for non-linear analysis. OR
(ii) corresponding displacement value (from experimental load-displacement data set) on which the failure happens? which i think shall be the case of linear analysis.

• jackhero
Subscriber

@Peter

Hope to have your suggestions/recommendations soon.

Thank you

• peteroznewman
Subscriber

Hello Jack,

I am not on this site as frequently as I was in the past due to a big ANSYS project at work, so sorry to keep you waiting.

Reaction Force is only available on Displacement or Fixed Boundary Conditions, which is why you can get it at the top.

When requesting Max Principal Stress result, you must change the geometry filter to Vertex and pick the vertex at the bottom of the center of the beam. The screen snapshot above shows the whole body, which includes the stress concentration at the supports that we want to exclude.  Looking at your contour plot, once you do the vertex, the answer will be between 12 and 25 MPa.  Let's say it is 20 MPa. So at a force of 152 N, you got 20 MPa at the bottom of the beam.

What is the published Ultimate Tensile Strength for the material?
Let's say it is 3 MPa.  That means you pushed too hard in the model.
Force at Failure would be  152.54 * 3/20 = 22.9 N.
The displacement that would give that force is 0.16 * 3/20 = 0.024 mm.
The above calculations are assuming the Ultimate Tensile Strength = 3 MPa.

I will answer the other question in a new post.

Regards,
Peter

• peteroznewman
Subscriber

For either linear or nonlinear, it is best to configure the simulation end displacement to go past the experimental displacement at which failure occurred.  Though for linear, it is possible to go less and use the ratio, but for nonlinear, you must go past. The solver may stop due to convergence issues before the end simulation displacement is reached, but if that was because of material failure, that is okay.

Regards,
Peter

• jackhero
Subscriber

I am unable to select the center vertex at the bottom of the beam for Maximum Principal, and even for the Deformation have the same issue. Can you please provide any suggestion for this?

For the time being I selected the bottom center node for the results. The max principal value is 3Mpa with the suggested application of 0.0219mm displacement. Since I only selected the node for the results of Max Principal and Deformation therefore in the output I could only see that node. In one of the result I selected All bodies for the total deformation and I had the image like shown below. The maximum deformation, in case of All Bodies, has been shown to occur at the center and this is where, i think, the maximum was expected. But could I get same in the Maximum Stress image? So that I can show the stress concentration or max stress at or right before the point of failure. Maybe since I have currently selected bottom node so the results are like this, but if I select bottom center vertex maybe I could get image to illustrate the failure.

Currently, I have the output values of Max Principal (at node) which I can plot with Total Deformation (at node) but the deflection value of experimental outcome (0.16mm) is different than that of the FEA Total Deformation (0.0219mm). Is it the correct way to plot the x-axis of FEA Total Deformation values and y-axis of FEA Max Principal along with the Experimental Stress values? OR plot the experimental x-axis of deflection values and on y-axis plot the FEA Max Principal and Experimental Stress values?

Although the verification value is the Max Stress results, which for both FEA and Experimental output are same. But theoretically wouldn't be the two different x-axis deflection values be of any concern/question? And with the given 2d Model and FEA outcome, the only possible result I could mention/show to compare the FEA and experiment results, with FEA value reaching 3Mpa and we assume that the failure had occurred in FEA as well?

• peteroznewman
Subscriber

I don't know why you can't pick a vertex. The purpose of selecting the bottom node for a stress output was to get a single precise number for the ratio calculation. Now that you have that, you can plot stress on the body.  On the Legend, you can type a new number for the boundary between orange and red.  Type 3 MPa to represent the Ultimate Tensile Strength.  Instead of Total Deformation, plot Directional Deformation and pick the Y axis.  You can use Reverse Rainbow for the colors on the Legend.

Now you have to search for the error between the experiment and the model. The experiment broke at 0.160 mm while the model predicts fracture at 0.022 mm. Check all the inputs to the model, what is the value of Young's modulus, the thickness of the part.

Please reply with a description (photos?) of how the experimental displacement was measured. If the displacement was measured from the concrete floor, the deflection of the beam supports was also being measured. Please reply with a description of the force measurement transducer. When was it last calibrated?  What is it's full scale range?

I suggested earlier that an initial crack in the physical beam sample can significantly weaken the beam and cause the fracture to occur at a stress much lower than the theoretical strength of a flawless specimen. You can use Fracture Mechanics to introduce an initial crack length of 3 mm and calculate the failure load. Use different crack lengths to find the one that matches the experimental result.

How many specimens were cracked in the experiment?  There is variation in any experiment and it is important to have many or at least several samples tested to compute the mean and standard deviation.

Regards,
Peter

• jackhero
Subscriber

I changed the value on the Legend but on the top I am still getting the default automatic value of 15Mpa. Is there any way to remove that one? I tried but I couldnt edit the automatic legend value.

The deflection (displacement) was measured by the movement of the actuator (shown by the red arrow in the image below). Please mind the top grip that was later replaced with another one (not shown in image) for bending with one support in the middle.

The force was measured by the load-cell with a capacity of 100KN, however the calibration date is two years old.

For the Fracture Mechanics I will run the simulation soon and will update on it. In the experimental work, I did not have any initial crack on the beam, so if I have to match the Fracture Mechanics results (of different crack lengths) with the experimental result, as you have suggested, then how shall I match without experimental result values?  Or is there a way to estimate the result theoretically by having the experimental result value of flawless specimen?

A total of 5 specimens were tested for the brittle failure of the beam under three point bending.

Thank you

• peteroznewman
Subscriber

You can create a Named Selection of elements from the center 50% of your beam and plot the stress using the Named Selection instead of the body.  The other way I mentioned is to slice the body into 3 bodies, then you can plot the stress of the center body.

Thank you for the description and photo. The experimental setup is missing the displacement transducer under the center of the beam. This is a standard method of recording data in a three-point bending test and typically uses an LVDT displacement transducer.  Here is one paper on the errors in using cross-head displacement.

The Deflectometer or Displacement LVDT transducer should have been mounted to the base where the supports are. A direct measurement of the deflection relative to the supports is the most accurate measurement. Measurements made using the displacement sensor on the moving crosshead measure the combined displacement of the machine frame and the sample. What if half the measured displacement came from the flexing of the machine frame, and half was the flexing of the sample?  Maybe it's not that much, but you don't know.  One way to get an idea of the flexibility of the machine frame is to have no sample, and run the center pusher into the Machine Base. Run the machine up to the same load that was used to fracture the sample. What is the measured displacement of the machine frame?

Another source of displacement error is the settling of the sample into the supports and nosepiece pushing on the top. Did the displacement output values have the offset subtracted from them?  Below is the output from a tensile test that had a small offset.

On the topic of cracks, it's possible to have subsurface cracks that you can't see. It's also possible to have cracks that are present, but difficult to see.

Regards,
Peter

• jackhero
Subscriber

Thank you for your reply.

The details provided in the link are informative. Thank you for sharing.

I am attaching the data file with raw data values (here) and the calculated deflection and load values. I think maybe I did not do the fitting /offset correctly (the data doesn't include the fitted values). I had to subtract the initial actuator position value from the recorded displacement values to get the actual displacement. Apart from the displacement and load values, time is also given in seconds. In your knowledge could you please mention any link/guide on correctly offsetting / removing the idle part from load-displacement graph?

Almost all of the raw data had the offset/idle part on the graph before the actual load began and I had to fit the data to remove the idle part. Your suggestion on error in offsetting deflection value is correct. It would be much helpful if you may provide any guidance on correctly off-setting the raw data. The raw data curve had this kind of plot.

Thank you

• jackhero
Subscriber

I suggested earlier that an initial crack in the physical beam sample can significantly weaken the beam and cause the fracture to occur at a stress much lower than the theoretical strength of a flawless specimen. You can use Fracture Mechanics to introduce an initial crack length of 3 mm and calculate the failure load. Use different crack lengths to find the one that matches the experimental result.

May be I didn't explain properly. Regarding the cracks, in my last post, I meant with reference to your above quoted reply. In the experimental work, I did not have any initial crack (or initial flaw) on the beam, so if I have to match the Fracture Mechanics results with the experimental result, as you have suggested, then how shall I match without experimental result values?  As currently the only experimental result I have is of the flawless specimen or beam, not of the beam having initial flaw in the form of certain crack length.

In Ansys, I can introduce an initial crack of certain length, say 3mm, but how shall I estimate the failure load?

• peteroznewman
Subscriber

When I plot the raw data in the spreadsheet you provided in the link, I get very different values of displacement compared with the image of the plot in your post.

Below is the plot of the spreadsheet data. Why does this data go out to 0.7 mm while the data you show above goes to 0.4 mm? The post from 2 weeks ago has an even shorter displacement of 0.16 mm.

I fitted two lines to show how to extrapolate from a linear portion of the curve back to the Deflection axis to find a zero point. There was a softening of the beam between the green and red lines. That could have been when the first crack appeared.

Is it possible to trace the load-deflection of the testing machine when the nose is pushing on the support base without any sample in the machine?

I will post another reply to comment on the other post.

• jackhero
Subscriber

If I get access again I may run the machine without any sample. But I think without any sample, there would not be any load (in Newtons) to be sensed by the machine. Would it be ok if I run the machine, at the same displacement rate, for the same amount of time (in seconds) as that of the sample which we are discussing here. And get the data of load-displacement for the respective period of time.

For the speciemen data I uploaded I will check again and then update this post.

• peteroznewman
Subscriber

Jack, you run the machine in the closing direction until the pusher starts pushing on the base of the machine, then it will sense a load. If the machine won't close to the point when that happens, you move the supports so close together that they are touching and put a thick steel plate across them to push on.

You are trying to measure the stiffness of the testing machine frame without a sample to bend.

• jackhero
Subscriber

The last data I uploaded is also from the same test but not the same specimen whose graph I upload previously. Please see the file here for the raw data. The first three columns is the raw output from the machine. The initial actuator position value was subtracted from the recorded displacement values to get the actual deflection given in column E. The resulted graph is shown as black in the image (having 0.4 mm deflection). I fitted the graph (data given in the file with red header) and resulted in red graph shown in the image (0.16mm deflection). I think the fitting is not correct and this is where I get the wrong deflection value.

According to Ansys simulation I may get the maximum load around 3MPa with 0.022 mm, with the given Engineering data values. Issue is if the experimental deflection value, for corresponding maximum load, is not correctly interpreted the Engineering Data values also become  doubtful.

The maximum load value is almost same for all of the specimens tested. However, the raw data deflection value vary, two of the samples we have seen here had raw deflection of 0.4mm and 0.7mm.

• jackhero
Subscriber

(2) The maximum stress in your simulation is coming from the support, so is irrelevant, because your support is on a vertex, which is an idealization that causes high stresses that should be ignored. One technique to be able to easily ignore the stress at the support is to slice the body at the 25% and 75% of the length, and put those three bodies in a multibody part and use Shared Topology to mesh them back into a single beam. However, you still have 3 bodies, so when you request a stress output, just pick the center body and the maximum stress should be in the center.

You can create a Named Selection of elements from the center 50% of your beam and plot the stress using the Named Selection instead of the body.  The other way I mentioned is to slice the body into 3 bodies, then you can plot the stress of the center body.

I followed the instructions of slicing the body into 3 to plot the stress of center body, and had the following output. I hope you meant to have the same.

• peteroznewman
Subscriber

Yes, that is what I meant. If you turn on the Max flag, where is it?

The stress along the bottom seems to show values that are around 3 MPa, but the Tabular Data shows 12 MPa. Is the Max value at the contact point above?

The way to avoid a peak stress on the top is to slice the center block again to make a small length (area) on the top of the beam and use a remote displacement scoped to that edge to spread the force over an area and reduce the stress.

Regards,
Peter

• jackhero
Subscriber

Please see the attached image for the Legend with Max flag. The Max value is same as that of the Tabular Data. The Max value on the Legend and Tabular is 12MPa while as for my model is around 3Mpa. Except the Max value I am able to edit the other values shown in the Legend.

I tried applying the load on the top edges of the center block (as shown in the following image). Although in three-point bending the load is at the center, but for demonstration purpose I ran the simulation.

• jackhero
Subscriber

As per your suggestion, I tried slicing the center block to make a small length or area at the top of the beam (followed instruction given in here). But while meshing I am getting an error, I must have missed some step while slicing. The design is shown below,

Due to the error in the mesh, the solution also could not be run. Hope to have your guidance on it. Thank you

• peteroznewman
Subscriber

The graphics look like there are three bodies but I only see two bodies in the Outline. How is that possible?

You can make two planes and cut top to bottom and get 3 bodies, then put all three in a multibody part.

• jackhero
Subscriber

I followed the following video for slicing, I added the link in my previous post but it didn't work so I am adding it here again.

The original geometry on which I was working before your suggestion on slicing the top for small area, is shown below.

According to the video instructions, the first step was to select Freeze, which in my above shown geometry I couldn't do. Therefore, couldn't follow the rest of the steps. I saved the geometry and then imported it in new Static Structural Schematics. When I imported, the DM considered the four bodies, of the previous geometry, as a single body (shown as the first Surface Body under Parts in the image below). Since the geometry is imported that's why I have the Unfreeze in the tree line, I had to generate after import to get the geometry. I followed the instructions from the above mentioned video to slice and make the small area (shown as the second Surface Body under Parts in the image below). This is how I had total of two bodies, instead of actual five.

I searched for making a small area by slicing but I could not find except this video, which i thought maybe relevant to my case/model. If I am going in wrong direction, please guide me on the mistake

Thank you

• peteroznewman
Subscriber

You don't have to make a small area, you need a short edge on the top. After you slice into three bodies, a left, center and right body, you can get a short edge on the top of the center body by using Split Edge instead of Slice.

• jackhero
Subscriber

I hope I have followed the instructions correctly. Please see the image for the Split Edge on the top of the center block (red arrow pointing towards that edge) and I have three bodies (left, center and right).

I re-run the model and had the Maximum Stress values as following, I tried modifying the legend to around 2.9MPa

The Maximum Stress value on the center block is 6.9MPa, lower than that of the previous result (shown in first image of the stress in my post here) and nearer to the value we are trying to achieve. However, the Max stress value is still higher than that of the second image in the same post.

• peteroznewman
Subscriber

Delete the Displacement on the short edge and replace with Remote Displacement. For the Behavior of the Remote Displacement, there are two choices, Rigid and Flexible. Make sure to choose Flexible.  Enter only a Y component of displacement and leave all others free. If this doesn't work, make the short edge longer.

Since you seem to be able to slice complete planes, you could make the following slices and break the short edge at 50% to apply the load, then you would have a small piece at the top of the center block to exclude from the plot, as well as the end blocks.

• jackhero
Subscriber

Yes, I have used the Remote Displacement on the top and Displacement on bottom vertices as support. I assume by Flexible you meant to choose the Deformable behavior.

I changed the short edge length number of times until I got the Maximum Principal value of around 3MPa.

Since you seem to be able to slice complete planes, you could make the following slices and break the short edge at 50% to apply the load, then you would have a small piece at the top of the center block to exclude from the plot, as well as the end blocks.

I haven't yet followed these suggestions, could you please confirm that these instructions were suggested if I couldn't get the Maximum Principal value in the desired range by varying the top short-edge length?

Jack, you run the machine in the closing direction until the pusher starts pushing on the base of the machine, then it will sense a load. If the machine won't close to the point when that happens, you move the supports so close together that they are touching and put a thick steel plate across them to push on.

You are trying to measure the stiffness of the testing machine frame without a sample to bend.

Thank you for your suggestion, but I regret to say that recently I do not have access to the machine. I may not be able to try it. If there could have any other possible workaround, then it would be much helpful.

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