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Assignment 4- Comparison of 7 different cases of material card

AAman Kamble5 Apr 2020

Assignment 4 - Comparison of 7 different cases of material card.

Objective-

Compare below mentioned 7 cases

Run the model as it is. Name it Law_epsmax_failure.
In /FAIL/JOHNSON, change Ifail_sh=1, Dadv=1 & Ixfem=1. Run the model as Law2_epsmax_crack.
Delete / Fail/JOHNSON card. Run the model as Law2_epsmax_nofail.
Delete the value of EPS_p_max & run the model with name Law2.
Convert this material model to Law1 elastic with the same density, E, nu & run it. Then Check the results.
Convert the card to a Law36 elasto-plastic based on the details given.
Open LAW27_0000.rad

Change the shell properties to recommended shell properties
Notice the material card Johnson cook parameters but with different values for EPS based failure.
Run the model with name Law27.

On basis of below mentioned criteria’s & prepare a ppt/docx and list down case by case result and your conclusion as to why the failure happened.

Total number of cycles, Energy error, mass error and simulation time.
Notice the animation of all 5 and describe the animations in brief on the
basis of whether the elements are being deleted or cracked.
Plot energies and notice any difference.
Based on all the results, which case would represent the on-field
scenario.

Material properties used for this assignment are-

Material- Aluminium
Density (ρ)- 0.0028 Kg/mm^2
Young’s modulus (E)- 71000 N/mm^2
Poisson ratio (μ)- 0.33
Yields Stress (SigmaY)- 290 N/mm^2

Procedure-

CASE 1- Law_epsmax_failure

First import the file FAILURE_JOHNSON_0000.rad into solver.
Now go to analysis> Radioss> click on radioss. This will give us out files from radioss.
Now go to file FAILURE_JOHNSON_0001.out. From here we will get energy & mass error.

FIg- Above fig shows us mass & energy error.

From above fig,

Energy error is below 5%, so it is an acceptable value.
Now to run simulation, go to Hyperview>import .h3d file> apply contour> select von misses stress, select averaging method as simple> apply.
Now run the simulation.

Fig- Above fig shows von misses stress simulation for case 1

From above fig, our max stress comes as 275.1 Mpa at node 3812.
As we have set EPS_p_max=0.151, it means that elements will get deleted when strain energy in rupture plate reaches 15% of total strain energy.
It can be seen in above image that elements got deleted.
Now plot the energy graphs.

From above graphs it can be seen that internal & total energy goes on increasing with time because plate is resisting the punch force. Max value of both is 26500 J.
It can be seen that internal energy increases till 4ms it is because plate is able to absorb the force applied by punch. After 4ms energy becomes almost constant it is because plate started rupturing.
Kinetic energy is zero, as there is no movement of rupture plate is involve.
Contact energy & Hourglass energy is zero because we have used Ishell=24 (QEPH).

CASE 2- Law2_epsmax_crack.

First import the file FAILURE_JOHNSON_0000.rad into solver.
Now go to Failures>Failure_Johnson_1>change Ifail_sh=1, Dadv=1 & Ixfem=1.

Fig- Above fig shows change in properties of Failure Johnson.

Now go to analysis> Radioss> click on radioss. This will give us out files from radioss.
Now go to file FAILURE_JOHNSON_0001.out. From here we will get energy & mass error.

Fig- Above fig shows us mass & energy error

From above fig,

Energy error is below 5%, so it is an acceptable value.
Now to run simulation, go to Hyperview>import .h3d file> apply contour> select von misses stress, select averaging method as simple> apply.
Now run the simulation.

Fig- Above fig shows von misses stress simulation for case 2

From above fig, our max stress comes as 309.06 Mpa at node 4380. Due to the cracking phenomenon we have observed little more stress that case 1.
As in this case we have set Ixfem=1, this failure criteria will lead to element cracking instead of element or layer deletion.
D_advfunction is, element cracks,

If the element has no failed neighbours and D > 1, then in this case, new crack initialization in element.
If the element has failed neighbours and D > D_adv, then in this case, crack advanced and D_adv used for crack advancement. D_adv will be only used if existing crack arrives to a boundary of an element.
Element deleted, if a second crack arrives to the same element.

Case 2 is like first cracking and then deleting of elements.
Now plot the energy graphs

From above graphs it can be seen that internal & total energy goes on increasing with time because plate is resisting the punch force. Max value of both is 31000 J. In this case max value for energy is more than case 1 as less energy got dissipated. This is because of less no. of elements got deleted as we have used Ixfem & D_advvalues.
It can be seen that internal energy increases till 4.5ms it is because plate is able to absorb the force applied by punch. After 4.5ms energy becomes almost constant it is because plate started rupturing.
Kinetic energy is zero, as there is no movement of rupture plate is involve.
Contact energy & Hourglass energy is zero because we have used Ishell=24 (QEPH).

CASE 3- Law2_epsmax_nofail.

First import the file FAILURE_JOHNSON_0000.rad into solver.
Now go to model>failure> & delete the Failure_Johnson_1.
Now go to analysis> Radioss> click on radioss. This will give us out files from radioss.
Now go to file FAILURE_JOHNSON_0001.out. From here we will get energy & mass error.

Fig- Above fig shows energy & mass error for case 3

From, above fig

Energy error is below 5%, so it is an acceptable value.
Now to run simulation, go to Hyperview>import .h3d file> apply contour> select von misses stress, select averaging method as simple> apply.
Now run the simulation

Fig- Above fig shows von misses stress simulation for case 3

From above fig, our max stress comes as 270.5 Mpa at node 3812.
As in this we haven\'t specified any failure card, so EPS_p_max is set to 0.151 & Ixfem=0 by default.
It can be seen in above image that elements got deleted. Max stress induced in this case is little less that case 1.
Now plot the energy graphs.

From above graphs it can be seen that internal & total energy goes on increasing with time because plate is resisting the punch force. Max value of both is 27000 J.
It can be seen that internal energy increases till 4ms it is because plate is able to absorb the force applied by punch. After 4ms energy becomes almost constant it is because plate started rupturing.
Kinetic energy is zero, as there is no movement of rupture plate is involve.
Contact energy & Hourglass energy is zero because we have used Ishell=24 (QEPH).
Total energy is also little bit more than case 1.

CASE 4- Law2

First import the file FAILURE_JOHNSON_0000.rad into solver.
Now go to materials>Aluminium>set EPS_P_max to zero.
Now go to analysis> Radioss> click on radioss. This will give us out files from radioss.
Now go to file FAILURE_JOHNSON_0001.out. From here we will get energy & mass error.

Fig- Above fig shows us mass & energy error

From above fig,

Energy error is below 5%, so it is an acceptable value.
Now to run simulation, go to Hyperview>import .h3d file> apply contour> select von misses stress, select averaging method as simple> apply.
Now run the simulation

Fig- Above fig shows von misses stress simulation for case 4

From above fig, our max stress comes as 249.3 Mpa at node 3574. Which is less than all above cases.
In this case EPS_P_max is kept zero but the failure card is still present having Ifail_sh value =2 default which means for each integration point, the stress tensor is set to zero when D=∑Δεpεf≥1\">D=∑Δεpεf≥1D=∑Δεpεf≥1. The shell is deleted or cracked when D=∑Δεpεf≥1\">D=∑Δεpεf≥1D=∑Δεpεf≥1 for all integration points or layers.
EPS_P_max=0 alone will not let the elements to be get deleted.
Now plot the energy graphs.

From above graphs it can be seen that internal & total energy goes on increasing with time because plate is resisting the punch force. Max value of both is 38000 J.
Energy is increased till the end of the cycle.
Kinetic energy is zero, as there is no movement of rupture plate is involve.
Contact energy & Hourglass energy is zero because we have used Ishell=24 (QEPH).

CASE 5- Law1 elastic

First import the file FAILURE_JOHNSON_0000.rad into solver.
Now go to model>failure> & delete the Failure_Johnson_1.
Now go to materials>Aluminium>1) Change the image card to M1_ELAST. 2) Entre the values of density, Poisson’s ratio & young’s modulus.

Fig- Above fig shows change in material properties

Now go to analysis> Radioss> click on radioss. This will give us out files from radioss.
Now go to file FAILURE_JOHNSON_0001.out. From here we will get energy & mass error.

Fig- Above fig shows us mass & energy error

From above fig,

Energy error is below 5%, so it is an acceptable value.
Now to run simulation, go to Hyperview>import .h3d file> apply contour> select von misses stress, select averaging method as simple> apply.
Now run the simulation.

Fig- Above fig shows von misses stress simulation for case 5

From above fig, our max stress comes as 10860 Mpa at node 3812.
In this case stress induced is high because there is no failure criteria & and we have defined the material as elastic in nature. Thus the rupture plate act as elastic material & whatever deformation is there will take place in elastic phase only. Thus more strain will cause more stress.
It can be seen in above image that elements neither deleted nor ruptured.
Now plot the energy graphs.

From above graphs it can be seenthat internal & total energy goes on increasing exponentially with time because plate is resisting the punch force till last travel of the punch. Max value of both is 800000 J.
Kinetic energy is zero, as there is no movement of rupture plate is involve.
Contact energy & Hourglass energy is zero because we have used Ishell=24 (QEPH).

CASE 6- Law36 elasto-plastic

It is a different type of Law, it consist isotropic elasto-plastic material using user-defined functions for the work-hardening portion of the stress-strain curve for different strain rates.
It is widely used in industries because of its special facility to feed lab stress strain data into material card.
Thus results from this material card is near to actual conditions.
Now, first import the file FAILURE_JOHNSON_0000.rad into solver.

Fig- Above fig shows material data & Stress strain data for Law 36 material

Now go to materials>Aluminium>1)Change the image card to M36_PLAS_TAB. 2)Entre the values of density, Poisson’s ratio, young’s modulus, EPS_P_max, EPS_t & EPS_m
Here,

EPS_max/EPS_F- It is a value at which after whichever element crosses it gets deleted from the system.
EPS_t- It is strain value at which strain damage of material starts & stress starts reducing.
EPS_m- It is a value at which stress becomes completely zero.

Now we have to define a function which will represent the lab stress strain curve
To do so, go to model> Function>Create new> entre the values of stress strain curve> click update.

Fig- Above fig shows curve creation (function) for M36 PLAS TAB

Now Select the created function as a function ID & set Fscale=1 & Eps_Dot=1

Fig- Above fig shows Law36 material card creation

Now go to analysis> Radioss> click on radioss. This will give us out files from radioss.
Now go to file FAILURE_JOHNSON_0001.out. From here we will get energy & mass error.

Fig- Above fig shows us mass & energy error

From above fig,

Energy error is below 5%, so it is an acceptable value.
Now to run simulation, go to Hyperview>import .h3d file> apply contour> select von misses stress, select averaging method as simple> apply.
Now run the simulation.

Fig- Above fig shows von misses stress simulation for case 6

From above fig, our max stress comes as 975 Mpa at node 3564.
Also our stress value is more than Ultimate stress of 425Mpa
In this case as we have specified the value of EPS_max=0.16 so, after this value % of strain elements have been deleted.
Now plot the energy graphs.

From above graphs it can be seen that internal & total energy goes on increasing exponentially with time till 3rd ms, at this point strain reaches EPS_t & value of stress starts reducing thus energy graph becomes constant after 3rd ms. Max value of energy for both is 45000 J.
Kinetic energy is zero, as there is no movement of rupture plate is involve.
Contact energy & Hourglass energy is zero because we have used Ishell=24 (QEPH).

CASE 7- LAW27

This law combines an isotropic elasto-plastic Johnson-Cook material model with an orthotropic brittle failure model. Material damage is accounted for prior to failure. Failure and damage occur only in tension. This law is applicable only for shells.

Fig- Above fig shows properties of PLAS_BRIT material

Here,

EPS_t1- It is a tensile failure strain at which stress starts to reduce.
EPS_m1- Maximum tensile failure strain in principal strain direction 1 at which the stress in the element is set to a value dependent on d_max1
D_max1- Maximum damage factor in principal strain direction 1 it controls the damage between EPS_f1 and EPS_t1.
EPS_f1-Maximum tensile strain for element deletion in principal strain direction 1
EPS_t2-ensile failure strain at which stress starts to reduce in the principal strain direction 2
EPS_m2- Maximum tensile strain in principal strain direction 2 at which the stress in the element is set to a value dependent on d_max2
D_max2- Maximum damage factor in principal strain direction 2 it controls the damage between EPS_f2 and EPS_t2.
EPS_f2- Maximum tensile strain for element deletion in principal strain direction 2.

First, set all recommended shell properties.
Now go to analysis> Radioss> click on radioss. This will give us out files from radioss.
Now go to file LAW27_0001.out. From here we will get energy & mass error.

Fig- Above fig shows energy & mass error for case 3

From, above fig

Energy error is below 5%, so it is an acceptable value.
Now to run simulation, go to Hyperview>import .h3d file> apply contour> select von misses stress, select averaging method as simple> apply.
Now run the simulation.

Fig- Above fig shows von misses stress simulation for case 7

From above fig, our max stress comes as 278.5 Mpa at node 3820.
In this case it shows a brittle failure of elements.
As we have specified the value of EPS_f1=0.151 so, after this value % of strain elements have been deleted.
Now plot the energy graphs.

From above graphs it can be seen that internal & total energy goes on increasing. It is almost linear up to 4.2ms but after that its slope of line is reduced it may be due to the starting of element failure. Max value of energy for both is 40000 J.
Kinetic energy is zero, as there is no movement of rupture plate is involve.
Contact energy & Hourglass energy is zero because we have used Ishell=24 (QEPH).

Learning-

Learned about different types of material cards & there properties.
Learned about failure cards.
Learned reasons behind element failure.

Brief comparison-

For brief comparison please visit below link.

https://drive.google.com/open?id=1gGFCWsS2UVLS2RjCLRzSUhXtG7nHkZJg

Conclusion-

Elements gets deleted based on value of EPS max.
If element gets cracked or deleted then the stress & internal energy induced is less.
Case 2, case 6 & case 7 shows more on field scenario out of which case 6 is more realistic as we have freedom to give inputs from lab data. Case 2 & case 7 will give realistic result if the inputs given are more accurate.
Card selection is biased on its application.

Gdrive link-https://drive.google.com/open?id=1ptJNimY8uyGhLia_3YTy8WVUhWTDhEcl

Assignment 4- Comparison of 7 different cases of material card

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in your own course.