Dynamic event stimulation

[saiguru-phanidhar_gorli]

Hi,

I am doing dynamic event stimulation in simple a impact test on my enclosure, so as per the norms a steel block of 1.7 kg should drop on it from a height of 290 mm. so I created every thing in the work area but when I gave my initial linear velocity(magnitude) which should be v= 2.426 m/s as per the calculation as a freely drop block aiming onto the enclosure, it is not event coming close to the enclosure. If I give 350 or above 300 m/s it comes to contact with my enclosure. I am not getting where am going wrong or my approch is wrong. Can you please help me with this. how much magnitude should I give or any additional set up i need to implement.

Thankyou and looking forward to your reply. 

[John_Holtz · Accepted solution]

Hi @saiguru-phanidhar_gorli 

There is not enough information here to understand what the results are, so it is hard to know what the problem is in the setup or analysis. For example,

• What is the duration of the event? (You entered that under "Manage > Settings".)
• What is the calculated displacement? The result should be 2.426 m/s * duration.
• What is the gap between the steel block and the enclosure in the model? It should be 0 (or a very, very small value) because event simulation can only solve for very short durations (such as 0.001 seconds), and including a free-fall drop of 290 mm is a waste of analysis time.

Fusion puts a limit on how long the analysis is allowed to run. (It may be 12 hours in the current version.) It is important to do what you can to minimize the runtime so that the limit is not exceeded. This article describes the relationship between the mesh, event duration, and runtime. See Explicit Dynamic analysis run time is long in Nastran. (Note that this article applied to Inventor Nastran and Fusion, so some of the commands may be different between the two programs.)

John

[saiguru-phanidhar_gorli]

Thankyou for the responce.

The event duration is 0.001 s and yeah I gave the initial linear velocity as 2.38 m/s. The distance between the block and the enclosure is 290 mm. So, here is my question the as per the norms the  steel block should drop from 290 mm, and so as it will have same velocity while hitting the enclosure you suggest me to keep the block distance at 0 mm? 

or do I need to add any additional boundary conditions ? please let me know if you need any further information, I will provide you.

Thank you!

[John_Holtz]

Hi @saiguru-phanidhar_gorli 

Sorry for the delay. There is a (temporary?) problem with the forums in that I am not receiving a notification when someone replies. Please use @ followed by my name (which is shown in a pop-up) so that I know there is something I need to look at.

You can calculate what the velocity is when the steel block reaches the height of 0 and use that value. (It's not clear if the velocity is 2.38 m/s at a height of 290 mm, and it continues to accelerate as it falls the last 290 mm, or if the velocity is 0 when it is dropped and reaches 2.38 m/s after it falls 290 mm.)

Regarding the "additional boundary conditions", you need to apply any constraints and loads that exist in reality, or anything that makes the analysis simpler to solve.

• There's probably no constraints on the falling block. 🙂
• Some type of constraints on the enclosure prevent it from moving. (Even if the target was just "sitting" on the ground and technically free to move horizontally, you often put a constraint in the horizontal direction just for stability of the analysis. This is important in a static analysis but not as important in a dynamic analysis.)
• Gravity exists, but you can ignore it if it has a negligible effect on the analysis. (The weight of the steel block may be important as the enclosure is compressed. If so, you need to include gravity. If most of the damage is due to the velocity, then gravity may not be needed.)

John