Force at the center of the table

[Anonymous]

I just need an advice . Please see the following pic 1 :

This the table of length 160*80 cm and the load is applied at the center of the table of 30 kg.

Similar to the above pic 1 i have designed the table in pic 2 in Inventor which is as follows :

The Force in the above pic , i assumed that it is equal to the weight in the center of the table as in pic 1  . So just want to clarify whther my assumption is correct or not ? and if not then do i have to use any split tool for that ?

Your advice in this regards will be highly appreciated .

Best Regards

[JDMather · Accepted solution]

To apply the force as shown in the picture you must Split the face of the table in Inventor.

Otherwise the force is spread out across the entire top face rather than concentrated where the weighted object is placed on the table in your image.

[Anonymous]

So , for that specific concentrated load do i have to make it for the same size as of image ? or just a small area can be split ?

[JDMather]

P=F/A

If you make the diameter of the split 1mm, what is the pressure?

If you make the diameter of the split 1000mm, what is the pressure?

[blair · Accepted solution]

Yes, you would need to create a sketch on the table top face with a Circle that matches the contact face diameter of your lower weight. Use the Split-Face to set this as a separate Face and then apply your 30Kg load to the circular face.

I would use the Fixed Constraint for the base of one of the table legs and use the Constrain (Vector) to fix the other 3 Leg Face bases. This closer represents the legs on a floor. Allows for a better analysis of the Table, Frame and Legs. I would normally apply a coefficient of friction on the 3 leg face bases because they are not truly "fixed". 

[Anonymous]

Yes , there will be a difference . Thank you for your advice 🙂