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- Re: PRESSURE LOADS IN AUTOCAD INVENTOR

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1125 Views, 7 Replies

11-02-2012 04:31 AM

When adding pressure loads to a perpindicular face inventor states that the pressure is applied uniformly to the selected face.

As an example a disc of 300 mm dia(area= 70685 sq mm) with a uniform pressure load of 10MPa.

The Question is this !

Is the 10MPa load input construed by inventor as 10MPa per square mm or 10MPa over the entire disc area.

which is the correct interpretation?

Solved! Go to Solution.

Solved! by blair. See the answer in context.

There is a difference between "Pressure" and a "Force" load in Inventor. A pressure load of 10 psi on a 12" x 12" plate will produce a total "load" of 1440 lb. A "Force" load of 10 lb on a 12" x 12" plate will produce a total "load" of 10 lb.

The pressure load will always be acting perpendicular to the surface whereas a force load can be in any direction (vector) to the surface.

The area that Inventor will use is dependent of the units used. If you are using Imperial and apply a 10psi, then the load force will be the 10 lb on every square inch.

So a Pa is force of one Newton per square Meter and a MPa is 1,000,000 Pa

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11-02-2012 05:23 AM in reply to:
methoni

10MPa total over the area.

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11-02-2012 05:40 AM in reply to:
JDMather

Thanks for the reply,if that is the case then i will have to rethink the plate thickness,the deflection seems awfully

high.

just for the record the input is as follows.

Plate dia = 488mm

thickness say 12 mm

pressure on plate area 29 psi(0.199 Mpa)

plate constraint on perimeter.

material .mild steel

analysis shows max displacement =0.48

If you wish have a look.

Regards

methoni

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11-02-2012 05:57 AM in reply to:
methoni

oops, let me wake up a bit and look at this again.

...back in a while.

In any case - it should be really easy to do a couple of virtual experiments to verify.

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11-02-2012 06:54 AM in reply to:
methoni

The answer is in your question. What's a MPa?

A MPa is 1,000,000 N/m² or 1 N/mm²

which is an expression of pressure.

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11-02-2012 07:28 AM in reply to:
rhinterhoeller

Thanks for your input.

What started out as a simple problem now seems to be confusing.

So are you saying that if you enter a figure in inventor of say 10MPa,inventor knowing the surface area of plate(from original sketch input calculates the total load over the plate area.

The deflection/stress is then derived from this extrapolation.

Does this make sense ?

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11-02-2012 11:30 AM in reply to:
methoni

That makes sense.

As JD suggested, experiment. Try some simple exercises for which you can determine the results using classical methods until you're comfortable with the FEA interface.

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11-03-2012 09:38 AM in reply to:
methoni

There is a difference between "Pressure" and a "Force" load in Inventor. A pressure load of 10 psi on a 12" x 12" plate will produce a total "load" of 1440 lb. A "Force" load of 10 lb on a 12" x 12" plate will produce a total "load" of 10 lb.

The pressure load will always be acting perpendicular to the surface whereas a force load can be in any direction (vector) to the surface.

The area that Inventor will use is dependent of the units used. If you are using Imperial and apply a 10psi, then the load force will be the 10 lb on every square inch.

So a Pa is force of one Newton per square Meter and a MPa is 1,000,000 Pa

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