Announcements

The Autodesk Community Forums has a new look. Read more about what's changed on the Community Announcements board.

Convert Surface Mesh into Solid Mesh

Anonymous

Convert Surface Mesh into Solid Mesh

Anonymous
Not applicable

Hi Everyone,

 

I have simple plate which I have fixed on when end and apply a load to the other as seen in the image. (Okay I haven't included the load here.)

Surface Mesh 2.jpg

 

 

My issue arises when I mesh the plate, I seem to always generate a mesh on the surface of the object, however the mesh does not penetrate through the object see image below. My question is how on earth do you create a 3D mesh that penetrates right through the object not just a surface mesh that wraps around the plate?Surface Mesh .jpgw.

Reply
Reply
0 Likes
Reply
Accepted solutions (1)
1,199 Views
10 Replies
Replies (10)

Stefan2653
Advocate
Advocate
Accepted solution

Hi Nikhilmakan02,

I assume you have run at least a model check or performed a full analysis, which would have invoked the solid mesher. Once you have done that you can view the internal mesh in both the editor or results windows as follows:

 

Editor window: View/Object Visibility/Internal Mesh

 

Results window: Results Options/Show Internal Mesh

 

Stefan

Reply
Reply
0 Likes

Anonymous
Not applicable

Thanks Stefan!

 

That was quite simple, 

 

I'm fairly new to simulation mechanical and I'm concerend about the results I'm getting from the simulation.

My material is steel galvenised and its 16 mm thick I have applied a load of 2,2 tons but the results I'm obtaining are quite extreme.

 

Do you see anything wrong with the way I have setup teh simulation?

Loads.JPGResults.JPG

Reply
Reply
0 Likes

Anonymous
Not applicable

I should have mentioned I'm trying to simulate the force exerted by a shackle that hangs from the the hole in the plate and carriers a load below it.

Reply
Reply
0 Likes

Stefan2653
Advocate
Advocate

Hi Nikhilmakan02,

I did a rough hand calc with some assumptions on width of load applied for a net area of load application of 16mm x 16mm area, and I get a stress of about 85 N/mm^2. So, it appears something is not quite right here.

 

I see that you have applied perhaps 100 nodal forces. Make sure you have divided your total applied load by the number of nodes (e.g. 2,2 tons/100) so that the sum of the nodal forces equals your intended load. If you set the nodal force to 2,2 tons, then each node will carry 2,2 tons.

 

Stefan

Reply
Reply
0 Likes

Anonymous
Not applicable

Well you are correct, I was making the assumption that the software was already dividing the forces.

 

I have a total of 47 Nodes, which works out to about 459 N on each node. ((2.2*1000*9.81)/47).

 

However this yields 449 MPaResults 2.JPG

Reply
Reply
0 Likes

Anonymous
Not applicable

And you are quite correct, it is roughly a 16mm x 16 mm area.

Reply
Reply
0 Likes

Stefan2653
Advocate
Advocate

Nikhilmakan02,

You may be seeing a localized stress concentration at the edge nodes along the top and bottom edges of the hole. If you apply equal forces to all nodes, the edge nodes will cause higher stresses as there are no elements adjacent to those nodes to carry load. Two ways to handle that:

 

You can halve the nodal forces on the edge nodes. Just make sure the sum of all nodal forces is still the total applied load.

 

You can define a unique surface where your load bears on the hole (16mm x 16mm) and apply a surface pressure that will provide the equivalent total load.

 

Other than using FEM methods, you'll also want to perform some hand calculations for bearing and shear tearout failure. With small hole-to-edge distances those effects should be considered as well.

 

Stefan

Reply
Reply

Anonymous
Not applicable

Hi Stefan

 

Thanks for the tip, I went with option 1 and came to much better result of 79 N/mm2.

Now I just need to understand exactly why we halved the forces at the edges is this because we missing essentially half the nodes?

 

And am I correct in saying that the stress concentration is created by the program itself due to the lacking adjacent nodes and not due to there being an actual stress concentration in the model.

Corrected.JPG

Reply
Reply
0 Likes

Stefan2653
Advocate
Advocate

Hi Nikhilmakan02,

Think of it as loading the elements rather than the nodes. Internal nodes can be shared by several elements (say, 4 in your case). Edge nodes will be shared by fewer elements, 2 in your case. If you divide a given nodal load by the number of elements that share the given node, the resulting load/element values should all be the same.

 

Stefan

Reply
Reply
0 Likes

Anonymous
Not applicable

Would it not be better to apply a bearing load. This will simulate more wat you are trying to achieve.

Reply
Reply
0 Likes