Inventor General Discussion

Inventor General Discussion

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Accepted Solution

PRESSURE LOADS IN AUTOCAD INVENTOR

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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?

 

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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Re: PRESSURE LOADS IN AUTOCAD INVENTOR

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

10MPa total over the area.

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Re: PRESSURE LOADS IN AUTOCAD INVENTOR

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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Post 4 of 8

Re: PRESSURE LOADS IN AUTOCAD INVENTOR

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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Post 5 of 8

Re: PRESSURE LOADS IN AUTOCAD INVENTOR

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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Post 6 of 8

Re: PRESSURE LOADS IN AUTOCAD INVENTOR

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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Re: PRESSURE LOADS IN AUTOCAD INVENTOR

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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Post 8 of 8

Re: PRESSURE LOADS IN AUTOCAD INVENTOR

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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