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Message 1 of 5
isaacc
223 Views, 4 Replies

Weights?

If I put a force on a bolt pattern by clicking all 6 bolts holes and making the force 500 pounds, does it spread it out across the bolts or is it 500 pounds per bolt?

4 REPLIES 4
Message 2 of 5
dean.rose
in reply to: isaacc

When applying a force load, the load is evenly distributed across the selected features. The features allowed to have a force load applied to them are surfaces, vertices and edges.

 

For assistance on applying different kinds of loads please view this video.

 

If my post answers your question, please click the "Accept as Solution" button. This helps everyone find answers more quickly!



Dean Rose
Message 3 of 5
isaacc
in reply to: dean.rose

Well, I thought I had this figured out yesterday. This post says the load is broken apart into the number of surfaces and the forces are listed out (in my case 1-9 for the number of bolts). The model I did yesterday balanced out when I divided the load by the number of bolts. Also, when I typed the load as 500lbs, the deflection was off the charts (because it was 9x the 500lb load). So, which is right?

Message 4 of 5
dean.rose
in reply to: isaacc

This may be slightly confusing. The surface load of 500 will be distributed on each surface resulting in a combined load. nsurfaces*500=total load. The 500 magnitude load you define will be spread evenly across each surface that has been selected. Keep in mind this is only applies to the surface load.

 

 

Does this make it a little more clear?

 

If my post answers your question, please click the "Accept as Solution" button. This helps everyone find answers more quickly!



Dean Rose
Message 5 of 5
isaacc
in reply to: dean.rose

I gotcha! The load is physically spread to all of the bolt holes, but the actual force is the value you type multiplied by the number of holes the load is spread between. So, you have to divide the load you want by the number of holes and type that in at the force value. Thank you!

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