turn on suggestions

Auto-suggest helps you quickly narrow down your search results by suggesting possible matches as you type.

Showing results for

Close

Inventor General Discussion

- Autodesk Community
- >
- Inventor
- >
- Inventor General Discussion
- >
- Re: PRESSURE LOADS IN AUTOCAD INVENTOR

Topic Options

- Subscribe to RSS Feed
- Mark Topic as New
- Mark Topic as Read
- Float this Topic to the Top
- Bookmark
- Subscribe
- Printer Friendly Page

- Mark as New
- Bookmark
- Subscribe
- Subscribe to RSS Feed
- Highlight
- Email to a Friend
- Report Inappropriate Content

1017 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

- Mark as New
- Bookmark
- Subscribe
- Subscribe to RSS Feed
- Highlight
- Email to a Friend
- Report Inappropriate Content

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

10MPa total over the area.

Please mark this response as "Accept as Solution" if it answers your question.

-----------------------------------------------------------------------------------------

Autodesk Inventor 2014 Certified Professional

Autodesk AutoCAD 2013 Certified Professional

Certified SolidWorks Professional

Inventor Professional 2015-SP1 64-bit

http://www.autodesk.com/edcommunity

http://home.pct.edu/~jmather/content/DSG322/inventor_surface_tutorials.htm

-----------------------------------------------------------------------------------------

Autodesk Inventor 2014 Certified Professional

Autodesk AutoCAD 2013 Certified Professional

Certified SolidWorks Professional

Inventor Professional 2015-SP1 64-bit

http://www.autodesk.com/edcommunity

http://home.pct.edu/~jmather/content/DSG322/inventor_surface_tutorials.htm

- Mark as New
- Bookmark
- Subscribe
- Subscribe to RSS Feed
- Highlight
- Email to a Friend
- Report Inappropriate Content

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

- Mark as New
- Bookmark
- Subscribe
- Subscribe to RSS Feed
- Highlight
- Email to a Friend
- Report Inappropriate Content

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.

Please mark this response as "Accept as Solution" if it answers your question.

-----------------------------------------------------------------------------------------

Autodesk Inventor 2014 Certified Professional

Autodesk AutoCAD 2013 Certified Professional

Certified SolidWorks Professional

Inventor Professional 2015-SP1 64-bit

http://www.autodesk.com/edcommunity

http://home.pct.edu/~jmather/content/DSG322/inventor_surface_tutorials.htm

-----------------------------------------------------------------------------------------

Autodesk Inventor 2014 Certified Professional

Autodesk AutoCAD 2013 Certified Professional

Certified SolidWorks Professional

Inventor Professional 2015-SP1 64-bit

http://www.autodesk.com/edcommunity

http://home.pct.edu/~jmather/content/DSG322/inventor_surface_tutorials.htm

- Mark as New
- Bookmark
- Subscribe
- Subscribe to RSS Feed
- Highlight
- Email to a Friend
- Report Inappropriate Content

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.

IV 2013 Product Design Suite 64 Bit

Win 7 64 bit

Win 7 64 bit

- Mark as New
- Bookmark
- Subscribe
- Subscribe to RSS Feed
- Highlight
- Email to a Friend
- Report Inappropriate Content

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 ?

- Mark as New
- Bookmark
- Subscribe
- Subscribe to RSS Feed
- Highlight
- Email to a Friend
- Report Inappropriate Content

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.

IV 2013 Product Design Suite 64 Bit

Win 7 64 bit

Win 7 64 bit

- Mark as New
- Bookmark
- Subscribe
- Subscribe to RSS Feed
- Highlight
- Email to a Friend
- Report Inappropriate Content

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

Did you find this reply helpful ? If so please use the Accept as Solution or Kudos button below.

IV2015 SP1 PDSU / Sim Mech 2015 r1 /

Win7-64

EVGA X79 - Classified, iCore7 3930k 32Gb Quad-Channel

950Gb (2 x 500Gb Sata III SSD RAID0 Adaptec 6805E Controller)

Nvidia GTX-690 Classified - 344.60

SpacePilot Pro 3.17.7, 6.17., 4.11

Delta Tau Chi ΔΤΧ

IV2015 SP1 PDSU / Sim Mech 2015 r1 /

Win7-64

EVGA X79 - Classified, iCore7 3930k 32Gb Quad-Channel

950Gb (2 x 500Gb Sata III SSD RAID0 Adaptec 6805E Controller)

Nvidia GTX-690 Classified - 344.60

SpacePilot Pro 3.17.7, 6.17., 4.11

Delta Tau Chi ΔΤΧ

Search This Board

Auto-suggest helps you quickly narrow down your search results by suggesting possible matches as you type.

Showing results for

Post to the Community

Have questions about Autodesk products? Ask the community.

The Knowledge Network

Access a broad range of knowledge to help get the most out of your products and services.

Download & Installation

Getting your Software

Downloads & Upgrades

Serial Numbers & Product Keys

Installation & Licensing

Activation & Registration

Network License Administration

Subscription Management

Sign In / Create Account

Maintenance Subscription Help

Desktop Subscription Help

Cloud Service Subscription FAQ

Announcements

Feel free to drop by our AU topic post and share your plans, plug a class that you're teaching, or simply check out who else from the community might be in attendance. Ohh and don't forgot to stop by the *Autodesk Help | Learn | Collaborate* booths in the Exhibit Hall and meet our community team if you get a chance!

- Privacy | Legal Notices & Trademarks | Report Noncompliance | Site map | © Copyright 2014 Autodesk Inc. All rights reserved

Except where otherwise noted, this work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Please see the Autodesk Creative Commons FAQ for more information.