Inventor FEA of sliding square tubes with a pin in center

[Anonymous]

So, its quite some time since I have fiddled with Inventor FEA, but I have this frame that is to be loaded in the center with enough weight that the pin has a safety factor of 2 (s235 steel) 

I have taken out the 50x50x3 square tube with a 10mm hole in center, and the 40x40x3 square tube that is to be sliding inside with the equalent hole. 

50x50x3 square tube is fixed in bottom, the 40x40x3 is sliding. The pin OD10mm is the part that keeps the inside square tube in place and dont let it slide. Question is what force is required to keep a safety factor with S235 steel.

Ive used force of 2000N on top of the 40x40x3 tube, only bonded constrains. but displacement do not seems correct, and the stress seems a bit to little.

Does anyone have any input? How do I  upload the file, with the FEA part?

[Anonymous]

..... Or should I use seperation-no sliding for every connection, and use bearing force (1000N on each). The diseplacement looks much better and realistic than using force from top of square tube. 

Von mises shows 95MPa with these specifications.

[Anonymous]

added displacement picture also

[JDMather]

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@Anonymous wrote:

How do I  upload the file, with the FEA part?

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You create a New Folder (if not already existant)

Place the three *.ipt files and one *.iam file into the folder.

Right click on the folder and select Send to Compressed (zipped) Folder.

Attach the resulting *.zip file here.

[Anonymous]

Hi, thanks. 

attached is the files. 

Ive used force 2000N on HUP and seperation-no sliding on all constrains. 28MPa in 1.principal stress. Quick calculation, 2000/((pi/4)*10^2) gives around 26MPa. 

Thanks for the help.

[JDMather]

Instructions were explicitly to Attach *.zip file.

I do not have a rar extractor installed on my clean Inventor machine.

[Anonymous]

Hi.

Yeah, sorry about that. In my world, Zip, winrar, etc. is the same, but I guess thats only in my world.

Sorry for that.

Attached is in zip format. Hope you can open it.

[JDMather]

I have to get some other work done - I don't have time to validate results just now, but if you are testing the pin - I don't think  you need the other components. (see Attached.)

Edit:  I just thought of a similar problem that I will have to run when I get a chance.

[Anonymous]

This will be a silly question, but regarding von mises, for a pure shear stress view, what plane is to be looking at in the IPT that you helped me with? 

2000N in total, average shear stress would be (p/2)/A -> (P/2)/(pi*r^2) -> (2000N/2)/(pi*5^2) = 12,73MPa.

The stress plane XY has a maximum stress of 13.09MPa, thus the closest to the caluclated avg. stress.

- Is the XY plane for shear stress for this bolt?

[lena.talkhina]

Hello @Anonymous  !

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