Assignment 5-RADIOSS Interfaces & Study of Effect of Notches Challenge

OBJECTIVE:

PART 1:

Mesh the bumper assembly model with target element size as 6mm.

PART 2:

Run the crash tube model with 6 different cases as given below and compare.

1. Run the crash tube model as it is.
2. Change the Inacti=6 and run.
3. Create the type 11 contact and run.
4. Remove both notches and remove boundary condition on rigid body node then run.
5. Create a new notch in the middle, select the whole section and run.
6. Create a new notch with nodes only from opposing 2 faces and run.

PROCEDURE:

PART 1: Meshing bumper assembly:

• Import the model to the hypermesh file. FILE>OPEN>MODEL>CHOOSE THE FILE FROM LIST>OPEN.
• Change to by topo view to check for the errors and surface deformations.
• We can find some errors on the connection of surface edges. These errors can be togggled away by using toggle edge option from quick edit tool.

• The surface can be attached by using trim with surface/planes option from surface edit tool.
• After the surface cleanup is done, our model is ready for meshing.
• Meshing can be done using automesh option from 2d tool.
• For meshing, the target element size is given as 6mm since it is asked in the question.
• Click on align and size options for getting better mesh flow and connectivity.
• Select the surface required to mesh one by one and conduct the meshing.

FINAL MODEL:

PART 2:

Recommended interface parameters:

• Igap: Determine how the size of the gap is calculated.
• Gapmin: Minimum gap for activation of interface.
• Inacti: Action to take if initial penitration exist.
• Istif: Affects how the stiffness of interface is calculated.
• Iform: Fricton formulation.
• Stmin: Minimum stiffness to use in the interface.
• Idel: What to do with the slave nodes and master segments if the elements fail(deleted) that they are attached to.

CASE 1: Run the crash tube model as it is.

In this case, we are going to run the model as it is. The parameters applied are listed below.

• Igap=0, The gap will be contant and the gap is defined by gapmin.
• Gapmin=1, The distance between master and slave are 2.1mm and the Gapmin=1. So the master have 1mm gap and the slave also have got 1mm gap. If we add them, they sum up to 2mm only. Hence there is no initial penetration.
• Inacti=0, No action takes place at all.
• Istif=4, Interface stiffness is computed from both master and slave characteristics.
• Iform=2, Stiffness formulation is done, sliding force are computed using stiffness parameters of the interface.
• Stmin=1, Minimum stiffness to use in the interface.
• Idel=0, No update, the master and free nodes are kept and may lead to instability.

SIMULATION:

• From the simulation we can understand that, after the initial crash, the elements approach to the bottom and the elements are crushed again.
• In the begining of the crush, a very little amount of penetration occurs and a small amount of contact force is applied on the slave.
• As the time increases, crush is also increasing and the slave tries to penetrate into the master, which leads to the increase in the contact force to the end of the simulation.
• Initial crush occurs in the region where the notches are present.
• After the crush, the tube completely relaxes and moves away from the rigid wall.

INTERNAL ENERGY GRAPH:

• From the energy graph we can understand that, the kinetic energy is maximum in the initial stage and as the crush occurs and the elements are deforming, the kinetic energy decreases gradually. At the end of the simulation, we can find that the kinetic energy increases a bit after completely reducing to zero. This is because of the slight spring like movement of the tube away from the rigid wall soon after the crash.
• The kinetic energy is absorbed by the deformation and is converted into internal energy. Hence, as the kinetic energy decreases, the internal energy increases.(Energy can neither be created nor destroyed. It can only be transffered from one form to the another).
• The contact energy occurs only when there is a penetration. In the begining there is no penetration. But at the end of the simulation some penetration occurs and the contact energy slightly increases.
• The hourglass energy is zero since we have assigned the ishell value as 24 QEPH.
• The total energy is the sum of all the energies.

RIGID WALL FORCES:

• As the force is applied on the rigid body, for the initial crush the normal force is within the range of 200KN. When it crushes for the second time the normal force suddenly increases to a maximum of 1337.9KN. Increase in force is experienced on the rigid body at the 2nd crush.
• After reaching 1337.9 KN the crushed tube move backwards and results in the sudden fall in rigid wall force.

CASE 2: Change the Inacti=6 and run:

• In this case we changed the value of Inacti, ie; Inacti=6
• Since inacti=6, The gap defined will be changed. The new gap will be 5% of the difference between the total gap and initial penetration(Po).
• In inacti=6 the gap is reduced and scaled, hence this value is the recommended value and is more accurate.

SIMULATION:

• When we checked the simulation for energy graph and force graph, we can notice that there is no difference in any of the values. This is because the change was made for Inacti value. Inacti value only affects the initial penetration. Since in this case there is no initial penetration, the 'inacti=6' is not applicable here. thus no change in values.
• Igap=0 and Gapmin=1. The slave and master is having a distance of 2.1, which is above the required gap.
• so no initial penetration occurs.

INTERNAL ENERGY GRAPH:

• From the energy graph we can understand that, the kinetic energy is maximum in the initial stage and as the crush occurs and the elements are deforming, the kinetic energy decreases gradually. At the end of the simulation, we can find that the kinetic energy increases a bit after completely reducing to zero. This is because of the slight spring like movement of the tube away from the rigid wall soon after the crash.
• The kinetic energy is absorbed by the deformation and is converted into internal energy. Hence, as the kinetic energy decreases, the internal energy increases.(Energy can neither be created nor destroyed. It can only be transffered from one form to the another).
• The contact energy occurs only when there is a penetration. In the begining there is no penetration. But at the end of the simulation some penetration occurs and the contact energy slightly increases.
• The hourglass energy is zero since we have assigned the ishell value as 24 QEPH.
• The total energy is the sum of all the energies.

RIGID WALL FORCES:

• As the force is applied on the rigid body, for the initial crush the normal force is within the range of 200KN. When it crushes for the second time the normal force suddenly increases to a maximum of 1337.9KN. Increase in force is experienced on the rigid body at the 2nd crush.
• After reaching 1337.9 KN the crushed tube move backwards and results in the sudden fall in rigid wall force.

CASE 3: Create type 11 contact and run.

For this case, we need to create a type 11 contact card.

1. Click on groups in model browser.

2. Right click and choose contact.

3. Change the card image to Type11.

4. Choose the components for slave and master lines.

5. Give recommended parameter values.

6. Save the model.

RECOMMENDED VALUES FOR TYPE 11:

• Igap=3
• Gapmin=0.5
• Inacti=6
• Istif=4
• Stmin=1
• Idel=2
• Iform=2

SIMULATION:

• For this simulation type 11 is used, which implies edge to edge contact. Type 11 interface can be added to capture the physics correctly. This type is defined by identifying master and slave lines.
• In type 7 we could notice some penetrations to occur even when the gap conditions are satisfied. Hence we use type 11 for correctly getting the penetrations. As this type is based on edges, no penetrations can pass without knowing.
• Also we can notice a small increase in the contact force in this case compared to the others because some penetrations which are not caught by the type 7 is noted in the type 11.

INTERNAL ENERGY GRAPH:

• In energy graph we cant see much of a difference.
• Internal energy is increasing as the K.E is decreasing.
• Contact energy increases as the penetration happens.
• zero Hourglass energy due to use of QEPH.

RIGID WALL FORCES:

• Resulatant Force increases as the crush occurs and proceeds with time.
• The resultant force for the second crush increases compared to the force for the first initial crush.

CASE 4: Remove both notches and remove boundary conditions on rigid body node then run.

• First we have to remove all the notches from the model.
• This can be done by using align tool(F7).
• Click on first end and second end nodes to check for how the nodes should align like followed by the nodes that need to be aligned.
• After removing the notches i have removed all the boundary conditions.
• By going to the solver browser and click on BCS.
• Then I deleted the BCS option completely.

SIMULATIONS:

• Here we can understand that the removal of notch has resulted in reducing the contact force.
• Here the maximum recorded contact force is 6.743E+01N, which is almost half of the contact force in other cases.
• Usually initial deformations occur in the region where notches are present.
• But here the deformations start from the bottom part.

INTERNAL ENERGY GRAPH:

• In energy graph we cant see much of a difference
• Internal energy is increasing as the Kinetic energy is decreasing.
• Contact energy increases as the penetration happens. But contact energy is lesser than in other cases.
• zero Hourglass energy due to use of QEPH.

RIGID WALL FORCES:

• Normal forces increases as the crushing proceeds. At the 1st crushing normal force is less but it suddenly increases as the 2nd crush take place.
• Here we can see that the Max. normal load is 1202.85kN which is the lowest Max normal in the 6 cases.

CASE 5: Create a new notch in the middle, select the whole section and run.

• In order to create a notch in the middle, we can use translate option.
• Select the nodes to be translated and give the direction in which we want the notch to be ceated.

SIMULATION:

• Since we have created the notch in the middle, the initial deformation will occur near the notch.
• The contact energy increases due to the presence of notch.
• Since there is only one notch, the contact energy will be less compared to the first and second case.

INTERNAL ENERGY GRAPH:

• The kinetic eergy decreases as the internal energy increases.
• Sine the notch is present, we can notice an increase in contact energy.
• Hourglass is still zero due to the Ishell 24.

RIGID WALL FORCES:

• We have noticed that when the notch was removed, the normal force was reduced.
• Since we have created a notch in the middle, the normal force has increased.
• The maximum normal force is 1203.61 kN.

CASE 6: Create a new notch with nodes from opposing 2 faces and run.

• In this case, we have to crate the notches only on the opposite faces.
• This can be done by following the same procedure for the previous case.
• Only difference is that the opposite faces should be translated.

SIMULATION:

• In this case, we can experience the highest contact force.
• In this case, the initial deformation occurs in the place where the notch is located.

INTERNAL ENERGY GRAPH:

• Compare to all other cases, the maximum contact energy is experienced in this case.
• Internal energy increases as the kinetic energy decreaes.

RIGID WALL FORCES:

• The normal force in this case is almost same as that of case 5.

How does the notch affect the results?

Cases with Notches:

When the notches are present we can identify that the model started initially deforming from the region where the notches were peresent. Also we can notice an increase in the contact forces in the model with notches. When notches are present the stiffness is also reduced.

Cases without Notches:

When the notch is removed the contact force and contact enegry is reduced and the stiffness is more in these parts.

Conclusion:-

• Case 4 is more acceptable since, the notch is not present the contact forces and energy is reduced which means the penetration happening in that case is lesser compared to the other cases, also the stiffness of the body is increased.
• Also i found that the notches in the tube accepts the deformation first so that the rest of the portion behind will get less impact .
• Notches affect the results because without notch the structure behaves more stiffer due to more stiffness so the vonmesis stress also increases.