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Non-Uniform Pressure Load

Non-Uniform Pressure Load

Non-Uniform Pressure Load

So I have a simple rectangular tank.
I placed my Pressure load on the inside faces.
The non-uniform distribution would be something like this:

P = ro * g * h

h is the Y+
ro * g is the value I put in to the box.

My question is:  Units
What units does Simulation use when solving this?
Based on the Pressure value I specify? (N/m^2, psi)
If I choose N/m^2, then my Y+ should be solving in meters, correct?
If I choose psi, then my Y+ should be solving in inches, no?

But when I use psi, it doesn't produce the results I expect.
ie. ro*g
9.81 in metric (kg, meters)
13.91 in imperial (lbs, inches)

Devon Murray, EIT [Mechanical]
Solidworks 2011 SP 2.0

RE: Non-Uniform Pressure Load

(1)    P = ρgh

so for water in the IPS system:
       ρ = 62.4 lbm/ft3 / (1,728 in3/ft3) / 386.16 lbm/slinch = 0.000094 slinch/in3
(2)    F=ma    so    F/m=32.18 ft/s2*12 in/ft = 386.16 lb/slinch

which gives, for a one inch water column on in an inertial frame on the earth's surface:

       0.001122 slinch/in3 * 386.16 in/s2 = 0.0361 lb/in2

which is about what you would expect.

†where "slinch" is a slang word for the mass unit in the IPS system since there is no formal word for mass in this system.

Note:    People brought up in the English system confuse force with mass, while many people brought up in the metric system confuse mass with force, eg, torque wrenches calibrated in kg-m.
Nobody said the Imperial (IPS) units were easy.


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