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    Member
    Posts: 3
    Registered: ‎11-02-2012
    Accepted Solution

    PRESSURE LOADS IN AUTOCAD INVENTOR

    329 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?

     

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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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    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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    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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    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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    blair
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    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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