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11-27-2016 06:02 AM - edited 11-27-2016 06:13 AM

Hi All,

Currently I'm trying to check my hand calculations using FeMap. At one end a tube is fixed, at a distance of 22.6mm from the fixation there is a hinge. At the other end there is a force of 1500N.

When I try to analyse the beam it gives the warning message 9142. I have no clue on what I did wrong.

Can anyone please help me?

With kind regards,

Emiel van Setten

Solved! Go to Solution.

6 REPLIES

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11-27-2016 01:07 PM

Dear Emiel,

Take a look to my post https://community.plm.automation.siemens.com/t5/3D-Simulation-Femap-Forum/FEMAP-1D-Tube-with-interna...

The **CTUBE** element you use in your FEMAP model is the same as the **CROD** element except that its section properties are expressed as the outer diameter and the thickness of a circular tube. The CROD element of NX NASTRAN is a straight prismatic element that has only axial and torsional stiffness, **do not support bending efects, OK?**. **The CROD element -and then the CTUBE- is ideal when you need an element with only tension-compression and torsion.**

In summary, to solve your problem you need to use a genuine 1-D beam element, for instance CBAR or CBEAM, ok?. The CBEAM element includes extension, torsion, bending in two perpendicular planes, and the associated shear. The CBEAM uses two grid points and can provide stiffness to all six DOFs of each grid point.

For the **HINGE**, please caution: a cantilever BEAM as yours with a hinge in the middle is a MECHANISM, you will have a rigid body motion and then your stiffness matrix will be singular: **NX NASTRAN will give you error for sure**!!.

In summary, please setup correctly **your problem concept**, with FEA we solve structural problems, not mechanism, the structure should be properly constrained.

In any case, to learn how to define a HINGE in a CBAR/CBEAM element please visit my website in the following address: **http://www.iberisa.com/soporte/femap/cbar.htm**

Best regards,

Blas.

Blas Molero Hidalgo, Ingeniero Industrial, Director

IBERISA • 48004 BILBAO (SPAIN)

WEB: http://www.iberisa.com

Blog Femap-NX Nastran: http://iberisa.wordpress.com/

IBERISA • 48004 BILBAO (SPAIN)

WEB: http://www.iberisa.com

Blog Femap-NX Nastran: http://iberisa.wordpress.com/

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11-28-2016 05:47 AM

Thank you for you're reaction. I understand the first part on the CTUBE vs CROD. But your explanation on the hinge I don't understand. At one end it's fixed so it can't move right? So it isn't a mechanism? Or is this a wrong assumption?

I included the 3D model situation and sketch maybe this help in finding the solution?

With kind regards,

Emiel

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11-28-2016 11:00 AM

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11-28-2016 02:00 PM

Dear Emiel,

Seeing the picture I understand the problem: this is not an internal hinge, but a pinned support, there you need to make sure to have a node and apply an external constraint with TX=TY=TZ=0: this approach is only valid when you isolate the beam, but if you try to study the full assembly (beam + support) then you will need to use **PIN FLAG** to remove rotations.

To explain better the PIN FLAG concept in the next image you have two braces connected by grid points 7 and 16 to a horizontal span. At the end of the braces you have a HINGE, ie,** the rotational degree of freedom, θz is released**.

In the following image you have the detail of the job done in FEMAP to define Element Releases (ie, PIN FLAG): Beam element 30 is the brace element that is connected to the horizontal span at grid node#7. Note that in the case with releases, **there’s no moment transfer to the brace** (grid point 7) at these location. The moments, however, are transferred across the horizontal span (beam elements 6 and 7).

The following picture shows the displacement results on the deformed shape of bridge with release at brace: not moments are transferred to the horizontal beam from the end of the brace!!. **The rotation at the end of the brace θz is fully released!!.**

Best regards,

Blas.

Blas Molero Hidalgo, Ingeniero Industrial, Director

IBERISA • 48004 BILBAO (SPAIN)

WEB: http://www.iberisa.com

Blog Femap-NX Nastran: http://iberisa.wordpress.com/

IBERISA • 48004 BILBAO (SPAIN)

WEB: http://www.iberisa.com

Blog Femap-NX Nastran: http://iberisa.wordpress.com/

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11-30-2016 11:13 AM - edited 11-30-2016 11:14 AM

Thank you for your response it kinda works! I did my calculations by hand two ways and checked them with my teachers the result is a reaction force of 18000N. The Femap software give's 9000N so I'm assuming I did something wrong in FeMap? But can't find it?

Hoping you can also help me with this little problem?

With kind regards,

Emiel van Setten

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11-30-2016 03:08 PM

Dear Emiel,

Your **material properties are in wrong units**, you are using Steel in Psi units for EX and inches for length, please configure FEMAP properly, use the following link to setup FEMAP correctly, if not your results will be useless, make sure to select material library **mat_eng_mm-N-tonne-degC-Watts.esp** that is in MPa, N and mm, OK?: http://www.iberisa.com/soporte/femap/femap_tips_tricks_preferencias.htm

Also you keep using CTUBE element, wrong!!, I told you to use CBEAM element instead. Also, be generous with mesh density, in your model I see only ONE element per segment!!.

In summary, here you are the FEMAP model in NEUTRAL format version 10.3 (you must upgrade your FEMAP version, now we run V11.3.2), I have applied FY=1500N at the free end and the resultant reaction force is in fact RFZ=-1500N:

Best regards,

Blas.

Blas Molero Hidalgo, Ingeniero Industrial, Director

IBERISA • 48004 BILBAO (SPAIN)

WEB: http://www.iberisa.com

Blog Femap-NX Nastran: http://iberisa.wordpress.com/

IBERISA • 48004 BILBAO (SPAIN)

WEB: http://www.iberisa.com

Blog Femap-NX Nastran: http://iberisa.wordpress.com/

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