Rupture plate

Rupture Plate:

Aim

1. Become familiar with format of RADIOSS Engine and Starter files
2. Know how to use HyperWorks Solver Run Manager to execute a RADIOSS simulation
3. Analyze starter and engine output files
4. Be able to post-process key results from an explicit analysis with HyperView
5. To learn about materials law 2,law 27 & law 36 with its various failure modes.

Case 1:

Open Failure Johnson model.Run the model as it is. Rename it as Law2Eps_maxfailure.

About the model

The material undergoes isotropic elasto-plastic behavior which can be reproduced by a Johnson-Cook model with or without damage (/MAT/LAW27 and /MAT/LAW2, respectively).The tabulated material law (/MAT/LAW36) is also studied.

The model is an aluminum plate with 3mm thick.

This law represents an isotropic elasto-plastic material using the Johnson-Cook material model. This model expresses material stress as a function of strain, strain rate and temperature. A built-in failure criterion based on the maximum plastic strain is available.

The values a,b & n are calculated.

The EPSpmax are given as 0.151.So the rupture will happen at the failure strain of 0.151.

/FAIL/XFEM_JOHNS (Obsolete) is an XFEM (eXtended Finite Element Method) failure model.A failure model is created.Ifail_sh(Shell failure flag) = 2. (For each integration point, the stress tensor is set to zero, if D=∑Δεpεf≥1\">D=∑Δεpεf≥1D=∑Δεpεf≥1 ; i.e the platic strain is greater than the failure strain so the shell is cracked (not deleted), for all integration points or layers.)The constant D1,D2 & D3 which are real constants are also provided in order to calculate the εf (failure strain).

The material also fails when plastic strain becomes equal to rupture strain

So when the material is loaded, the von mises increases upto the EPS limit, then rupture takes place.

The plate initially has a rupture and the damages the element, the plate at the centre breaks into small fragments

From the graph, we get to know about the effect of energy involved.

Initially, The rigid wall is moving towards the aluminium plate,the plate deforms,a punch force which is created due to the initial velocity,so the internal energy gradually increases, upto to the rupture and then it becomes constant.Here the internal energy is the same as the external work,no hourglass energy is present.

Once the plate is ruptured,the punch force decreases become it has no resistance.The kinetic energy decreases and there are some ripples because,the it has to pierce the plate and the elements are cracked.The internal energy constant becomes constant, because of no resistance in the plate.

Case 2:

In Fail/Johnson, change Ifail_sh = 1,Dadv = 1, Ixfem = 1.Run the model as Law2epsmaxcrack

Where Dadv is the criterion for crack propagation

Shell failure flag = 1, Shell is cracked (not deleted), if D=∑Δεpεf≥1\">D=∑Δεpεf≥1D=∑Δεpεf≥1 for one integration point or layer.

XFEM formulation (Ixfem=1) is only compatible with BT Q4 (I_shell=1, 2, 3 or 4), and QEPH (I_shell=24) shell elements. If XFEM is activated (Ixfem=1), the failure criteria will lead to element cracking instead of element or layer deletion.

In this case, the whole element thickness is considered as a single layer. The failure criterion is calculated in each integration point, but only one single crack can appear in this element. This approach is compatible with all values of the shell flag (I_{fail_sh}=1 or 2). The crack direction is determined by the principal constraints in the last failed integration point.

So when the material is loaded, the von mises stress increases upto the EPS limit, then rupture takes place.

The plate initially has a rupture and the damages the element, the plate at the centre makes a cut into 4 parts,here the plate is much more stiffer than case 1,the elements hold together and just makes way for the spherical ball to pass through

From the graph, we get to know about the effect of energy involved.

Here the energy graphs when compared to case 1 has a smooth failure, which is in the sense that the plate at the centre splits into 4 equal halves,instead to small fragments, so the the crack propagation is uniform,less resistance when compared to case 1.The rupture gets delayed when compared with case 1 & no hourglass energy

Case 3:

For case 1 & 2 we used Fail/Johnson model

Using the Johnson-Cook failure model /FAIL/JOHNSON along with the Johnson-Cook material model provides more user control of the failure
Johnson-Cook failure strain definition:

Failure of the element happens when 1 integration point reaches εpmax

The previous models uses cumulative damage to compute failure.

The energy dissipation is similar to the previous cases.

Even though the failure card is deleted but still the failure occurs due to EPS_p_max

1. EPS_p_max - Plastic Strain beyond this will cause element deletion

2. EPS_t1 - Strain beyond which the line connecting t1 and t2 will be used

3. EPS_t2 - Strain beyond which the element will be separated from the connected elements

Since the model is modelled with plastic strain Epsmax values not zero, it ruptures

Case 4:

Delete the EPS_p_max and run the model as Law2

It means that failure strain is not defined,the plate once it is loaded , is subjected to elastic and plastic strain,the plate has more elastic strain than plastic strain,so the internal energy gradually increases and goes on and kinetic energy increases and becomes stable, hourglass energy is zero & the punch gradually increases and never drops because of no failure.

This is is mainly because the EPS_p_max is zero, where the failure takes place, the plate behaves much more ductile, internal energy is stored in the plate