Assignment 4-RADIOSS Material Laws Challenge
OBJECTIVE-
to carry out simulation of the given starter file and compare the results. there shall be 7 cases ( according to the given parameters ) and they are to be compared primilarly via animation and plots . the file on the worked on are the FAILURE_JOHNSON_0000.rad file and with LAW27_0000.rad file. the cases are as follows:
1) law 2 (file: FAILURE_JOHNSON_0000.rad)- base simulation.
2) law 2( file: FAILURE_JOHNSON_0000.rad) with l fail_sh=1, Dadv=1, and ixfem = 1
3)law 2 (file: starter the file for case 2) with fail/johnson card deleted.
4) law 2 ( file:starter file for case 3) with EPS_max value deleted.
5) law 1 ( file: failure_johnson_0000.rad) with fail/johnson card deleted and values for density , E, nu assigned.
6) law 27 ( file:law 27 _0000.rad) with recommended shell properties.
7) law 36 ( file: law27_0000.rad) with recommended shell properties and given refrence curve.
CASE SET-UP AND EXCUATION-
1)CASE 1-
1) with the hyperwork open , the file is imported via the ' import solver deck ' option through FILE> IMPORT. the FAILURE_JOHNSON_0000.rad file is selected and import is clicked.
2) for this case, we are run to the simulation as is. the file as following material and failure cards.to run the simulation , we go to analysis >radioss from the bottom panel .we then need to save the file. the file is saved as 'law2_epmax_failure_0000.rad' . after saving , we go ahead and check the include connectors box and type is -nt 4 in the option text box , before clicking radioss.
1) Discplcement-
in geometry discplacement a vector whose length is the shortest distance from the initial to the final position of a point undergoing motion.it quantitifies both the distance and direction of the net or total motion along the a straight line from the initial position to the final position of the point trajectory . discplacement may be identified with the translation that maps the initial position to the final postion.
a discplacement may be also discribed as relative position resulting from the motion that is as the final position. the corrospoinding discplacement vector can be defined as the diffrence between the final and initial position.
von moises yeild criterion-
the von mises yeild criterion ( also known as the maximum distortion energy criterion).suggest the yeilding of a ductile material begins when a second daviatoric stress invarient reaches as a critical value. it is part of plasticity theory that applies best to ductile material such as some material response can be assumed to be of a nonlinear elastic,viscoelastic or linear elastic behaviour.
3)velocity- velocity is defined as a vector measurement of the rate and direction of motion . put simply velocity , velocity at which speed of something moves in a perticular direction , such as the speed of car travelling north on measure freeaway . or the speed of rocket travel it launches in space.velocity calculate by using simple formula that uses, rate distance and time.
4) % thickness reduction calculated- maximum %thickness reduction is 7.392E+00 and the minimum is -1.169e+01 at node 3896..
5) plastic strain along the thickness- maximum plastic strain is 1.502E-01 at node 3863 and a minimum 0.000E+00.
6) if plot rigid wall forces , internal energy , hourglass energy, contact energy , the toatl energy of simulation.
CASE 2- Law 2 with fail -johnson and crack visualization.
in this case, a few parameter are added to the defined case to visualize checking or bracking down of elements with the help of 'ixfem' option in fail -johnson failure model card.the necessary changes were made in parametr as follows
maximum energy error in this base simulation case is 4%
energy error is acceptable upto 15%
therefore positive signal for further simulation.
maximum mass error is 0.000
1) displacement-
maximum displacement at node at no.3984 is 1.327E+01 and
minimum displacement at node no. 3697 is 0.000E+00.
2) velocity-
maximum velocity is 2.311E+01 at node 3598 and minimum is 000E+00 at node 3635.
3) von mises yeild criterion-
the von mises stress plot show a peak value is 2.947E+02 at node 4380 and minimum value is 0.000E+00 at node 3697.
4) % thickness reduction calculated-
maximum % thickness reduction is 9.908E+00 at node 56316 and minimum is -1.325E+01 at node 3949.
5) Plastic strain along with thickness-
maximum plastic strain is 1.489E-01 at node 3863 and minimum is 0.000E+00.
CASE 3- Law johnson card without fail.
in this case, the material model in same in case elastoplastic but the same in model card is removed to look for any un results hence, the plas-johnson card has deleted and the simulation was done. as we know the law material model is not capable of simulation compression under forces with realistic results plas-johnson card recommended addition.
maximum energy error in this base simulation case is 0.8%
energy error is acceptable upto 15%
therefor positive signal for further simulation
maximum mass error is 0.000.
1) displacement-
maximum displacement at node no 3844 is 1.858E+01 and.
minimum displacement at node no. 3697 is 0.000E+00.
2) von mises stress-
the von mises stress plot show a peak value is 2.705E+02 at node 3812 and manimum value is 0.000E+ 00.
3) velocity-
maximum velocity is 2.793E+01 at node 3868 and minimum is 000E+00 at node 36.35.
4) % thickness reduction calculation-
the maximum thickness reduction 7.410E+00 at node 3985 and minimum thickness reduction 1.166E+01 at node 3697
5) plastic strain along the thickness-
plastic strain along the thickness value maximum 1.494E-01 at node 3989 and minimum 0.000E+00 node 3697.
CASE 4- Law 2 with infinite eps_p_max value.
in this case 4 the eps_p_max which defines the maximum permitted plastic strain is changed to 0 which input value of 10^3 as default. this make the material to behave like deal elastoplastic such as that starin in computed but not failure du to plastic deformation beyond the limit is exprienced.
maximum energy error in this base simulation case 1.1%
energy error acceptable up to 15%
therefore positive signal for further simulation
maximum mass error is 0.000
1)displacement-
maximum displacement at node no 3999 is 5.629E+00 and
minimum displacement at node no. 3697 is 0.000E+00
2) von mises stress-
the von mises stress plot show a peak value 2.5316+02 at node 3574 at minimum value is 0.000E+00.
3) velocity-
maximum velocity is 3.031E+01 at node 3688 and minimum is 000E+00 at node 3635.
4) % thickness reduction calculation-
maximum% thickness reduction is 50429E+00 and the minimum is -1.169+01 at node 3896.
CASE 5- Law 1 elastic material-
here the material is elastic as an ideal elastic which liniearly follows hook law upto the elastic limit and when the goes the plastic stage througout any stress no matter the magnitude. the basic E,nu rho parameter are input and the analysis is done. here no plasticity is expected from the comonenet material.
maximum energy error in this basic simulation case 0.3%
energy error is acceptable upto 15%
therefor positive signal for further simulation.
maximum error is 0.000
1) displacement-
maximum discplacement at node no 3688 is 4.999E+00 and
minimum displacemet at node no 3697 is 0.000E+00
2) von mises stress-
the von mises stress plot show a peak value is 1.086E+02 at node 3688 and minimum value is 0.000E+00
3) % thickness reduction calculated-
maximum %thickness reductiion is 4.584E+00 and the minimum is -1.169E+01 at node 3896.
CASE 6-
in this case, the material is plas_brit which is an elastic-plastic material with additional parameter to define in a nature of material such as eps_t1, t2,eps_f1,f2 eps_m1,m2 and dmax1,2 which are tensile failure in strain. maximum tensile failure in strain at element detection and maximum tensile failure, tensile strain at element detection and maximum tensile damage to limit in two direction.to start with law27 is nothing but law 2 with addfitional paramertr hence, much detail analysis can be done.
maximum energy error is this base simulation case is 1.1%
energy error is acceptable upto 15%
therefor positive signal for further simulation.
maximum mass error is 0.000.
1) discplacement-
maximum displacemet at node no.3990 is 6.013E+00 and
minimum dispalcement at node no 3697 is 0.000E+00
2) von mises stress-
the von mises stress plot show a peak value is 2.76E+02 at node 3820 and minimum value is 0.000E+00
3) velocity-
maximum velocity is 6.906E+01 at node 3990 and minimum is 000E+00 at node 3635.
4) % thickness reduction calculated-
maximum % thickness reduction is 6.669E+00 and the minimum is -1.169e+01 at node 3896.
5) plastic strain along the thickness-
maximum plastic strain is 1.689E-01 at node 3697 and minimum is 0.000E+00
CASE 7- Law 36 plas_tab material-
in this material model , the input parameter previously discussed eps_p_max,eps_m and eps_f in a single direction are defined. the important feature of the model is a user can input a stress v/s strain plot as curve and callit in the card from which the strain rate parameter is capturing. also, scaling factor, hardening coeficients etc can also be added to give further detail for better analysis . the necessary input are gathered from vidio lecture 6 and 7 are added mode.
maximum energy error in this base simulation case is 4.2%
energy error is acceptable upto 15%
therefor positive signal for further simulation
maximum mass error is 0.000
1) displacement-
maximum displacement at node no. 36880 is 5.000E+00E+00 and
minimum displacement at node no 3697 is 0.000E+00
2) von mises stress.
the von mises stress plot show peak value is 3.03E+02 at node 3688 minimum value is 0.000E+00
3) velocity-
maximum value is 1.404E+01 at node 3594 and minimum is 000E+00 at node 3635.
4) % thickness reduction calculated-
maximum %thickness reduction is 4.589E+00 and the minimum is -9.229+01 at node 3626.
5) plastic strain along with the thickness-
maximum plastic strain is 5.94E-01 at node 3688 and minimum is 0.000E+00.
CONCLUSION-
as stated, in a summary foir the current problem where the material in quetion ia alluminimum alloy case1, case26 with the proper input 7 is our realistic results .law 27-plasbrit material model is the best option with given input with more diteailed result than law-2 elastoplastic with fail_johnson card . at least plas_tab material model being sesitive to its input (stress vs strain curve) , plastic starin parameter coud be another option but need accurate data preferbly from physical testing.
OBSERVATIONS-
CASE 1-
In this case, it has the fail/johnson card as well as EPS_P_max value assigned, with ixfem value as 0. this means that the elements are deleted as soon as they cross the EPS_P_max threshold, which is what we can see above in the simulation.
internal energy as well as rigid wall forces at rection to the force genrated by the sphere on the sheet .energy increase linearly untill a large number of elements are deleted at around the 4ms threshold. this probably create some vibration and that result in surge of kinetic energy some of the same point.
the rigid wall forces fluctuate du to elementb deletion when a large chunk of element are deleted a 4ms the graph drops,meaning there aren't many element opposing the forces of the sphere.most of the elements in it's path have been deleted at that point.
CASE 2-
make use of ifail_sh,Dadv, and ixfem as part of the fail/johnson card.ixfem is especially important here beacause it decides the way the element react with its value equal to 1here , it means the elements will crack and fail, along with Dadv, crack propogation can also be seen here, with elements being deleted only when crack reaches them. as a result , this is more realistic depeciation. it can be noticed that in both case 1 and 2, the internal energy graph are also similar. they differ come into timetamp at which is they plateu and this crack propogation being a big factor case 2 creates the difference.
CASE 3-
with the fail/johnson card deletd in this case , the EPS_P_max to the factor for elements deletion . as we can see in the following screenshot, is's value is 0.151, meaning will be elements deleted if the strain my integration point with said elements reaches 15.1% of the plastic strain . the elements, in this case , fail rapidly when compared to the other elements in case 2 the XFEM crack formulation there.
there is linear increase internal and total energy untill a bulk of the elements are are deleted from the model. this expected since the energy absorbed by these elements go to waste du to deletion their. this also explain the sudden drop in rigid wall forces at the same timetamp du to most of elements that were contact in with the sphere during collision being deleted. the rigid wall forces is measured being opposing force genrated by the metal sheet against the sphere. fluctuation occure in elements are deleted there are no elements to oppose the incoming sphere , there won't be any rigid wall forces.
CASE 4-
law 2 (plas_johns) expresses material stress as a function of strain , strain rate and tempreture. the metal sheet is able to withstand a large amount of stress and avoid elements deletion primarly du to the fact that the value for EPS-p-MAX was deleted, since it wass deleted radioss assign the default value of 10^30. in addition to that , fail/johnson card was deleted . the material behave like elasto plastic material since it has very high EPS_p_max threeshold and elements cannot be deleted du to no failure cards.
with the value of c ( strain rate coeficient) =0 there is no strain rate ( neigther does tempreture play a role here) as a result , the stress in this case is purely a function of strain genrated in the material.
added the kinetic energy genrated in this case is almost negligibble (2.5-4mj) compared to the internal ( and total) energy here . the total energy increase almost linear over time.
CASE 5-
in this case , the plate are defined with a law-1 material which defines the basic properties of the elastic material. the model consider only the elastic phase of the material , there is no plasticity involved which result in linear deformation.
this case make use of the M1_elast card as we can see , the metal sheet absorb the full impact to the sphere. this material law is used to model purely elastic material. this explain why there no element deletion or ruptures. du to elasticity , it is also is able to absorb and extermly large amount of energy, which expain the peak at end of kinetic energy plot below. absorbing a lot of energy also result in very high stresses - maximum stress is observed metal sheet is 10890 mpa , which is a masive value compared to the same in law 2.
also du to the elastic nature , the internal energy ( and rigid wall forces ) keep increasing exponetially. the internal energy of the metal sheet increases du to absorbtion of the kinetic energy from the impacting sphere and drop down once the plate return to the orignal shape.
CASE 6-
case 6 makes use of M27_PLAS_brit MATERIAL card that is used to isotropic elastoplastic johnson-cook material model with an anthroplastic brittle failure model. the sheet ( compared to case 7) absorbs more force and resorts to more deformation before the elements are deleted ( at round ms)
with the loss energy through deleted elements are deleted elements, we notice the graph at 4ms mark. withsevral elemts are deleted and none to stop the incoming the sphere , the rigid wall forces graph drop at same timetamp of 4ms. the spikes are formed whene an elemet are deleted and is characerized by a through and crest but with so many elements deleted . the through masive at the 4ms mark.
CASE 7-
in this case, from the simulation ,we can say that this material exihits strong brittle behaviour. it does not bend as much in response to the applied force and immediately distingrates . there seems to be not much cracking on the region of element deletion take the shape of the sphere . another thing that we can notice is that the stress cocentartion is localized around the region of impact and this stresses are not dissipated throughout the whole metal plate . taking a look at a graphs,
there is an almost linear relation between internal energy and the time till pass the 3ms mark where a large number of elements get deleted at once. when the element are deleted , the energy on each elemet go to waste , which result in the elasto plastic on the curve.
on the topic of energy wastage and elements deletion. it is more perticurly visible in the rigid wall force graph . each of the spikes is due to an element deletion . the resulatant force gradually incrases the past the same 3ms mark which is when the sheet gives a way and lets the sphere through , with not match on the in the way , the rigid wall forces are drastically reduced and that result drop at the 3ms mark. the first drop occure when the sheet rupture the first time.