in the Static Study can anyone tell me what the difference between Applying a force and Applying a pressure? if this question is any where else i apologize i couldn't find any.

in the Static Study can anyone tell me what the difference between Applying a force and Applying a pressure? if this question is any where else i apologize i couldn't find any.

i applied force to a square, and then calculated the PSI to the same square so they would be the "same amount" and im getting different answers

Tom Puttkammer wrote:

i applied force to a square, and

**then calculated the PSI**to the same square so they would be the "same amount" and im getting different answersHow did you do this calculation?

Do you have a *.sldprt file that you can attach?

This is a simple formula P=F/A If you re-arrange the equation F=PA.

I am doing a demonstration of this tomorrow and need a helper for the demo.

I weigh 165lb.

We will calculate the area of the bottom of my shoe.

I will stand on your chest with one foot while you lay on the floor.

Then I will attempt to balance that same 165lb load on the head of a lag screw with the pointed end of the screw is on your chest.

I think you will then understand the equation P=F/A where F is constant but the A changes (approaches zero).

Actually, the rest of the class will understand the demo - you might have to go to the hospital. (it can be a messy demo)

Force applies a load to the part. It can be from a point to a large area but if you select 5lbf it will be 5lbf no matter what. Pressure makes the force depend on the area it is applied to, so 5psi applied to a 1in^2 area would be 5lbf distributed load and a 5psi pressure applied to a 10in^2 area would be 50lbf distributed load

Hey Tom,

You will find a lot of detailed information on simulation loads within the SOLIDWORKS Help page... 2018 SOLIDWORKS Help - Structural Loads

Force applies a load to the part. It can be from a point to a large area but if you select 5lbf it will be 5lbf no matter what. Pressure makes the force depend on the area it is applied to, so 5psi applied to a 1in^2 area would be 5lbf distributed load and a 5psi pressure applied to a 10in^2 area would be 50lbf distributed load