Assignment 4-RADIOSS Material Laws Challenge

Objective:

Compare 7 cases as discussed in last 3 slides, in a tabulated format on the basis of:

1. Total number of cycles, Energy error, mass error and simulation time.
2. Notice the animation of all 5 and describe the animations in brief on the
basis of whether the elements are being deleted or cracked.
3. Plot energies and notice any difference.
4. Based on all the results, which case would represent the on-field
scenario.
5. Prepare a ppt/docx and list down case by case result and your conclusion
as to why the failure happened.

CASE-1 LAW2_EPSMAX_FAILURE:

• In this case Change cardimage AS M2_plas_johns_zeril.

                          

 

• shell element properties and property card values are entered
• then go to Analysis

• Maximum energy error is 0.8%.upto 15% error is acceptable.so it is possitive signal for further simulation.maximum mass error 0.000.
• Total number of cycles are 49379.
• simulation time is starter runtime is 4.1s, ENGINE RUNTIME = 263.36s (00:04:23)
  STARTER+ENGINE RUNTIME = 267.87s (00:04:27)

1.Displacemennedt:

Displacement is  the Relative position resulting from the motion that is final position of apoint relative to its intial position

• In this case Maximum displacement 1.586E+01 at Node number 3984 
• minimum Displacement 0.000 at node number 3635 

2.velocity:

velocity is defined as a vector measurment of the rate and direction of motion.scalr magnitude of velocity is the speed of the motion.In calculus velocity is the first derivative of position with respect to time.velocity formula is displacement/time.

• Mximum velocity is 2.389E+01 at Node number 3994
• Minimum velocity is 0 at Node number 3635

3.%THickness Reduction calculated :

• %Thickness reduction is maximum at node number 3863 is 7.227E+00
• minimum value is -1.126E+01 at node number 3883

4.Plastic strain along thickness:

Plastic strain is the strain that occurs after yield point which do not reverse after removing load.

• maximum plastic strain 1.492E-01 at node number 3863
• minimum platic strain 0 at node number 3697

5.Strain:

 

strain(Elastic) is the ratio of amount of deformation experienced by the body in the direction of force applied to the initial dimensions.Elastic strain which is developed in elastic region with in yield point.this can be reverse after removing the load.

• Maximum strain 1.404E-01 at node number 4016
• minimum strain 0 at node number 3697

6.Vonmises stress:

vonmises stress ia a value used to determine if a given material will yield or fracture.the vonmises yield criteria states that if the vonmises stress of material under load is equal or greater than the yield limit of the same material under simple tension then the material will yield.vonmises stress is the stress developed in all directions.

• Maximum value of stress is2.752E+02 at node number 3812
• minimum value of stress is 0 at node number 3697.

6.Plot rigid wall forces, internal energy, hourglass energy, contact energy, total energy of the simulations.

•  open Hypermesh
• change to hyper graph 2D

• Internal energy is 0 intial stage of simulation but as the simulation go on internal energy will increase due to deformation increases.in this case there is no hour glass energy.contact energy is constant i.e 0 through out simulation because there is no sliding action.kinetic energy is 0 through out the simulation.total energy=internal energy+kinetic energy.so total energy is as same as internal energy.
• To plot Rigidwall forces select types as Rigid wall forces< requests 4rigid wall1

Rigid wall forces:

• Tangent resultant force is 0 through out the simulation.Resultant normal force is intially zero but after that it is increased and fluctuating between 6000kN to 8000kN upto 4ms then suddenly there is a drop of Resultant normal force to below 1000kN then it fluctuates 4to 5 ms then comes to zero.

CASE2 LAW2_EPSMAX_CRACK:

In this case change values in failure_johnson_1 as Ixfem=1,Ifail_sh=1,Dadv=1.

• Run the simulation note down the values of energy error,mass error,total number of cycles,simulation time.

 

• Maximum aenergy error is 4.1%.upto 15% error is acceptable.so it is possitive signal for further simulation.maximum mass error 0.000.
• Total number of cycles are 49217.
• simulation time is starter runtime is 4.06s, ENGINE RUNTIME = 252.72s (00:04:12)
  STARTER+ENGINE RUNTIME = 256.78s (00:04:16)

1.Displacement:

 

• Maximum Displacement 1.127E+01 at node number 3866
• minimum displacement 0 at node number 3697

 2.velocity:

• Maximum velocity 2.268E+01 at node number 3829
• minimum velocity 0 at node number 3835

3.% Thickness reduction :

• %Thickness reduction is maximum at node number 5316 is 9.886E+00
• minimum value is -1.312E+01 at node number 3949

4.Plastic strain along thickness:

• maximum plastic strain 1.508E-01 at node number 4500
• minimum platic strain 0 at node number 3697

5.Strain:

• Maximum strain 1.426E-01 at node number 3816
• minimum strain 0 at node number 3697

6.vonmisess stress:

• Maximum value of stress is2.951E+02 at node number 4380
• minimum value of stress is 2.158E-01 at node number 5388 .

6.Plot rigid wall forces, internal energy, hourglass energy, contact energy, total energy of the simulations.

•  open Hypermesh
• change to hyper graph 2D

 

 

• Internal energy is 0 intial stage of simulation but as the simulation go on internal energy will increase due to deformation increases.in this case there is no hour glass energy.contact energy is constant i.e 0 through out simulation because there is no sliding action.kinetic energy is 0 through out the simulation.total energy=internal energy+kinetic energy.so total energy is as same as internal energy.
• To plot Rigidwall forces select types as Rigid wall forces< requests 4rigid wall1

Rigid wall forces:

• Tangent resultant force is 0 through out the simulation.Resultant normal force is intially zero but after that it is increased and fluctuating between 6000kN to 8500kN upto 4ms then gradually there is a drop of Resultant normal force to 0 from 4to 5 ms.

CASE3:LAW2_EPSMAX_NO FAIL

IN this case delete failure_johnson card then run the simulation

• Maximum aenergy error is 0.8%.upto 15% error is acceptable.so it is possitive signal for further simulation.maximum mass error 0.000.
• Total number of cycles are 49407.
• simulation time is starter runtime is 4.72s (00:00:04), ENGINE RUNTIME = 270.60s (00:04:30)
  STARTER+ENGINE RUNTIME = 275.32s (00:04:35)

1.Displacement:

 

• Maximum Displacement 1.932E+01 at node number 3989
• minimum displacement 0 at node number 3635

 2.velocity:

• Maximum velocity 2.153E+01 at node number 3985
• minimum velocity 0 at node number 3697

3.% Thickness reduction :

• %Thickness reduction is maximum at node number 3985 is 7.456E+00
• minimum value is -1.218E+01 at node number 3950

4.Plastic strain along thickness:

• maximum plastic strain 1.509E-01 at node number 4000
• minimum platic strain 0 at node number 3697

5.Strain:

• Maximum strain 1.358E-01 at node number 3824
• minimum strain 0 at node number 3697

6.vonmisess stress:

• Maximum value of stress is2.707E+02 at node number 3812
• minimum value of stress is 0 at node number 3697 .

6.Plot rigid wall forces, internal energy, hourglass energy, contact energy, total energy of the simulations.

•  open Hypermesh
• change to hyper graph 2D

 

• Internal energy is 0 intial stage of simulation but as the simulation go on internal energy will increase due to deformation increases.in this case there is no hour glass energy.contact energy is constant i.e 0 through out simulation because there is no sliding action.kinetic energy is 0 through out the simulation.total energy=internal energy+kinetic energy.so total energy is as same as internal energy.
• To plot Rigidwall forces select types as Rigid wall forces< requests 4rigid wall1

Rigid wall forces:

• Tangent resultant force is 0 through out the simulation.Resultant normal force is intially zero but after that it is increased and fluctuating between 6000kN to 8000kN upto 4ms then suddenly there is a drop of Resultant normal force to below 0 then it fluctuates 4to 5 ms from 0 to 2500KN then comes to zero.

CASE4:LAW2-

In this case delete the value of EPS_P_MAX and run the simulation

• Maximum energy error is 1.1%.upto 15% error is acceptable.so it is possitive signal for further simulation.maximum mass error 0.000.
• Total number of cycles are 49303.
• simulation time is starter runtime is 3.82s (00:00:03), ENGINE RUNTIME = 266.68s (00:04:26)
  STARTER+ENGINE RUNTIME =270.50s (00:04:30)

1.Displacement:

 

• Maximum Displacement 5.626E+00 at node number 3999
• minimum displacement 0 at node number 3677

 2.velocity:

• Maximum velocity 6.704E+00 at node number 3999
• minimum velocity 0 at node number 3697

3.% Thickness reduction :

• %Thickness reduction is maximum at node number 3688 is 5.402E+01
• minimum value is -2.282E+01 at node number 3635

4.Plastic strain along thickness:

• maximum plastic strain 3.121E-01 at node number 3688
• minimum platic strain 0 at node number 3697

5.Strain:

• Maximum strain 4.012E-01 at node number 3999
• minimum strain 0 at node number 3697

6.vonmisess stress:

• Maximum value of stress is2.451E+02 at node number 3803
• minimum value of stress is 1.239E+00 at node number 3999 .

6.Plot rigid wall forces, internal energy, hourglass energy, contact energy, total energy of the simulations.

•  open Hypermesh
• change to hyper graph 2D

 

 

• Internal energy is 0 intial stage of simulation but as the simulation go on internal energy will increase due to deformation increases.in this case there is no hour glass energy.contact energy is constant i.e 0 through out simulation because there is no sliding action.kinetic energy is 0 through out the simulation.total energy=internal energy+kinetic energy.so total energy is as same as internal energy.
• To plot Rigidwall forces select types as Rigid wall forces< requests 4rigid wall1

Rigid wall forces:

• Tangent resultant force is 0 through out the simulation.Resultant normal force is intially zero but after that it is increased and fluctuating upto maximum 9000kN upto  then it it will decrease to 6500KN.

CASE-5-LAW-1 ELASTIC :

In this case change material card name as LAW1_Elastic.and give E,Nu,Density values as previous case. Then Run the simulation.

• Maximum aenergy error is 1.3%.upto 15% error is acceptable.so it is possitive signal for further simulation.maximum mass error 0.000.
• Total number of cycles are 47968.
• simulation time is starter runtime is 6.82s (00:00:06)), ENGINE RUNTIME = 253.61s (00:04:13)
  STARTER+ENGINE RUNTIME =260.43s (00:04:20)

1.Displacement:

 

• Maximum Displacement 4.999E+00 at node number 3688
• minimum displacement 0 at node number 3677

 2.velocity:

• Maximum velocity 7.815E+01 at node number 3902
• minimum velocity 0 at node number 3697

3.% Thickness reduction :

• %Thickness reduction is maximum at node number 3688 is 4.525E+01
• minimum value is -1.305E+01 at node number 3898

4.Plastic strain along thickness:

• maximum plastic strain 0 at node number 3697
• minimum platic strain 0 at node number 3697

5.Strain:

• Maximum strain 2.010E-01 at node number 3688
• minimum strain 0 at node number 3697

6.vonmisess stress:

• Maximum value of stress is1.089E+04 at node number 3688
• minimum value of stress is 0 at node number 3697 .

6.Plot rigid wall forces, internal energy, hourglass energy, contact energy, total energy of the simulations.

•  open Hypermesh
• change to hyper graph 2D

 

 

• Internal energy is 0 intial stage of simulation but as the simulation go on internal energy will increase due to deformation increases.in this case there is no hour glass energy.contact energy is constant i.e 0 through out simulation because there is no sliding action.kinetic energy is 0 through out the simulation.total energy=internal energy+kinetic energy.so total energy is as same as internal energy.
• To plot Rigidwall forces select types as Rigid wall forces< requests 4rigid wall1

Rigid wall forces:

 

• Tangent resultant force is 0 through out the simulation.Resultant normal force is intially zero but after that gradually it is increased to 5E+05KN.

 Case-6-LAW_36:

 In this case create material plas_tab 36 with given values.then create a curve with values given below and Run the simulation

In this case change card image as M27_PLAS_BRIT.and run the simulation.

 

• Maximum aenergy error is 0.8%.upto 15% error is acceptable.so it is possitive signal for further simulation.maximum mass error 0.000.
• Total number of cycles are 49379.
• simulation time is starter runtime is 3.24s (00:00:03), ENGINE RUNTIME = 258.19s (00:04:18)
  STARTER+ENGINE RUNTIME = 261.43s (00:04:21)

1.Displacement:

 

• Maximum Displacement 1.586E+01 at node number 3984
• minimum displacement 0 at node number 3635

 2.velocity:

• Maximum velocity 2.389E+01 at node number 3994
• minimum velocity 0 at node number 3635

3.% Thickness reduction :

• %Thickness reduction is maximum at node number 3863 is 7.227E+00
• minimum value is -1.126E+01 at node number 3883

4.Plastic strain along thickness:

• maximum plastic strain 1.492E-01 at node number 3863
• minimum platic strain 0 at node number 3697

5.Strain:

• Maximum strain 1.404E-01 at 4016
• minimum strain 0 at node number 3697

6.vonmisess stress:

• Maximum value of stress is2.752E+02 at node number 3812
• minimum value of stress is 0 at node number 3697 .

6.Plot rigid wall forces, internal energy, hourglass energy, contact energy, total energy of the simulations.

•  open Hypermesh
• change to hyper graph 2D

 

 

 

• Internal energy is 0 intial stage of simulation but as the simulation go on internal energy will increase due to deformation increases.in this case there is no hour glass energy.contact energy is constant i.e 0 through out simulation because there is no sliding action.kinetic energy is 0 through out the simulation.total energy=internal energy+kinetic energy.so total energy is as same as internal energy.
• To plot Rigidwall forces select types as Rigid wall forces< requests 4rigid wall1

Rigid wall forces:

 

• Tangent resultant force is 0 through out the simulation.Resultant normal force is intially zero but after that it is increased and fluctuating between 6000kN to 8500kN upto 4ms then suddenly there is a drop of Resultant normal force to  0 then it fluctuates 4to 5 ms from 0 to 2000KN.

CASE-7-LAW-27 :

In this case change card image as M27_PLAS_BRIT.and run the simulation.

 

• Maximum aenergy error is 0.8%.upto 15% error is acceptable.so it is possitive signal for further simulation.maximum mass error 0.000.
• Total number of cycles are 49506.
• simulation time is starter runtime is 4.02s (00:00:04), ENGINE RUNTIME =266.44s (00:04:26)
  STARTER+ENGINE RUNTIME = 270.46s (00:04:30)

1.Displacement:

 

• Maximum Displacement 1.669E+01 at node number 3863
• minimum displacement 0 at node number 3635

 2.velocity:

• Maximum velocity 1.955E+01 at node number 3863
• minimum velocity 0 at node number 3635

3.% Thickness reduction :

• %Thickness reduction is maximum at node number 3688 is 1.789E+01
• minimum value is -1.281E+01 at node number 3896

4.Plastic strain along thickness:

• maximum plastic strain 2.323E-01 at node number 4053
• minimum platic strain 0 at node number 3697

5.Strain:

• Maximum strain 1.607E-01 at node number 3834
• minimum strain 0 at node number 3697

6.vonmisess stress:

• Maximum value of stress is2.875E+02 at node number 3829
• minimum value of stress is 0 at node number 3697 .

6.Plot rigid wall forces, internal energy, hourglass energy, contact energy, total energy of the simulations.

•  open Hypermesh
• change to hyper graph 2D

 

 

 

• Internal energy is 0 intial stage of simulation but as the simulation go on internal energy will increase due to deformation increases.in this case there is no hour glass energy.contact energy is constant i.e 0 through out simulation because there is no sliding action.kinetic energy is 0 through out the simulation.total energy=internal energy+kinetic energy.so total energy is as same as internal energy.
• To plot Rigidwall forces select types as Rigid wall forces< requests 4rigid wall1

Rigid wall forces:

 

• Tangent resultant force is 0 through out the simulation.Resultant normal force is intially zero but after that it is increased and fluctuating between 6500kN to 8500kN upto 4ms then suddenly there is a drop of Resultant normal force to  0 then it fluctuates 4to 5 ms from 0 to 4500KN.

 Comparison of Results:

s.no	parameters	Law2_EPSMAX-fail	Law2_EPSMAX-crack	Law2_EPSMAX-No fail	Law2	LAW_1_Elastic	Law_36_Plastab	Law 27_Brit
1	Energy error	0.8%	4.1%	0.8%	1.1%	0.8%	0.8%	0.8%
2	no of cycles	49379	49217	49407	49303	47968	49379	49506
3	internal energy	26000	31000	27000	37000	8.0E+05	27000	30000
4	kinetic energy	0	0	0	0	0	0	0
5	hourglass energy	0	0	0	0	0	0	0
6	contact energy	0	0	0	0	0	0	0
7	Resultant tangent force	0	0	0	0	0	0	0
8	Resultant normal force	6000-8500 (Sudden drop)	peak 8500 (Gradual drop)	6000-8500 (Sudden drop)	peak 9500 (No drop)	5.0E+05	6000-8500(UPTO 4MS) FLUCTUATES 0-1000	6000-8500(UPTO 4MS) FLUCTUATES 0-4500
9	displacement	1.586E+01	1.127E+01	1.932E+01	5.626E+00	4.999E+00	1.586E+01	1.669E+01
10	velocity	2.389E+01	2.268E+01	2.153E+01	6.704E+00	7.815E+01	2.389E+01	1.955E+01
11	% thickiness reduction	7.227E+00 -1.126E+01	9.886E+00 -1.312E+01	7.456E+00 -1.218E+01	5.402E+01 -2.282E+01	4.525E+01 -1.305E+01	7.22E+00 -1.126E+01	1.789E+01 -1.281E+01
13	plastic strain	1.492E-01	1.508E-01	1.509E-01	3.121E-01	0	1.492E-01	2.323E-01
14	strain	1.404E-01	1.426E-01	1.358E-01	4.012E-01	2.010E-01	1.404E-01	1.607E-01
15	vonmisess stress	2.752E+02	MAX 2.951E+02 MIN 2.158E-01	2.707E+02	MAX 2.451E+02 1.239E+00	1.089E+04	2.752E+02	2.875E+02

conclusions:

CASE-1(LAW2_EPSMAX_FAILURE):

In this case the element fail after reaching EPSmax value.elements will deleted after reaching the EPS MAX value. 

CASE-2(LAW2_EPSMAX_CRACK):

In this case element will get crack due to Ixfem=1 means It will try to break the element like crack under given loading conditions.if IXFEM is 0 then the whole element will delete.

CASE-3(LAW-2_EPSMAX_NOFAIL):

In this case element wont get crack.the element will take as EPSMAX consideration.

CASE-4(LAW-2):

In this case the element will not get delete because of EPSMAX IS 0.so there is no heat decipation.

CASE-5(LAW1-ELASTIC):

In this case the stress strain curve is linear up to elastic limit means if we remove load it will come back to its original position but in crash it is not suitable.In this maximum internal energy is increaing as defomation increases.There is no plastic deformation thats why plastic strain is 0 in this case.due to continuous deformation the rigid wall forces are also high in this case.vonmisess stress developed is high because strain is high.

CASE-6(LAW-36):

In this case we will give additional information of lab tested curve so the stress strain curve will follow the given curve.

 

CASE-7(LAW-27):

There is no nonlinearity in this case the element will get fail directly there is no yielding criteria in this case.