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02-24-2015 11:52 AM

Ok guys here comes a tricky question.

How to analyze a structure that BC are replaced by "enforced forces" not displacements.

I work in academia and I am trying to teach student how to verify classic textbook problems with FEMAP/NASTRAN. The major problem is that the vast majority of problems replaces the BC by forces.

7 REPLIES

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02-24-2015 12:48 PM

Hello!,

Well, I do not understand very well your problem, the word "*enforced forces*" do not exist, in FEMAP we use the term "**enforced displacements**" to prescribe non-zero displacements in the structure as force loads. Please inform if is this what you need?.

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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02-24-2015 01:26 PM

Blas,

I was aware of the zero "enforced displacement" option in FEMAP but what I will need is an "enforced force" option.

To better explain the problem,

1- it will be like substituting the BC by the value of the reactions at those points. The value of the forces will take are of the reaction moments as well.

2- In that case NASTRAN has to solve for internal forces/stresses so that Newton 1st las exists sumF=0

Does that explains it better.

Thank you for your time and considerations

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02-24-2015 01:46 PM

Hello!,

The what you need to activate with the NX NASTRAN solver is the "**INERTIA RELIEF**" method.

Inertia relief is an advanced option that allows you to simulate unconstrained structures in the linear static solution SOL 101, and in the linear static solution portions of SOLs 105 and 200. Typical applications of inertia relief include modeling an aircraft in flight, an automobile on a test track, or a satellite in space.

Static analysis by the finite element method assumes that the model contains no mechanisms (you forgot to apply constrains) and may not move as a rigid body (strain free). If either of these conditions exists in a conventional finite element analysis, the stiffness matrix for the model becomes singular. When NX Nastran attempts to decompose a singular matrix, a fatal message or unreasonable answers result.

Consequently, you can't perform conventional finite element static analysis on unconstrained structures. However, a method called **inertia relief** is provided in NX Nastran for analyzing these conditions. A simple description of inertia relief is that the inertia (mass) of the structure is used to resist the applied loadings, that is, an assumption is made that the structure is in a state of static equilibrium even though it is not constrained. Two examples are a spacecraft in orbit or an aircraft in flight. In these cases, the structure is in state of static equilibrium, although it is capable of unconstrained motion.

To implement this procedure in FEMAP simply acivate the **PARAM,INREL,-2** in the NASTRAN Bulk Data Options window during the analysis definition. **Then no need to apply constraints, the only condition is that the structure to be in equilibrium between loads & reactions, OK?.**

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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02-26-2015 01:07 PM

Blas,

Thank you for your diligent response but I have tried the PARAM,INREL,-2 (even the -1 with some supports) in the past and it does not work with me. To my best knwoledge the INREL solution is that the inertia (mass) of the structure is used to resist the applied loadings [D'Alambert principle sumF-m.a=0].

Could you please better explain your comment becasue that may be my issue:

**Then no need to apply constraints, the only condition is that the structure to be in equilibrium between loads & reactions, OK?.**

Going to my original comment:

In theory in FE solves the problem {u}=inf[K]*{F} and then obtains stresses from the displacement fields. This is what is known as the displacement based FE formulation.

Theoretically, I do not see where will be the problem in adding an "enforced forces" option in the loading conditions and solve for {u} by finding the forces on each element that will put the system in equilibrium.

I am attaching the a file so that you have a better idea of the problem I am trying to solve.

In addition, I was born and grew up in Spain (Madrid) but I am now working at a US academic institution. I am familiar with your excellent IBERISA web page that I have used to get some ideas for a finite element course.

Buenas tardes y muchas gracias por tus comentarios

JM

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03-03-2015 02:48 PM

Dear JM,

According developers "**the only issue here is the reference frame for the displacement vector**". In this example it seems that using "inrel,-2 (automatic)" the reference frame is not obvious, nor is it easy to understand, then solution arrived using the automatic method (inrel,-2) is not correct.

But we have two ways to invoke the inertia relief with NX NASTRAN:

The first method is to specify the

**SUPORT**entry explicitly by including “**PARAM,INREL,-1**” in the Bulk Data Section.The second method (recommended) is to let NX Nastran select the SUPORT degrees of freedom automatically by including “

**PARAM,INREL,-2**” in the Bulk Data Section.

This is your baseline solution with the loads & constraints defined in your original FEMAP model:

The following picture shows the freebody diagram of your solution. These results will be the input for loadings (additionaly to the service loads) when activating the inertia relief effect:

Using command "**Model > Load > From FreeBody**" it creates loads directly from the selected freebody on the current output set. If I run "**Tools > Check > Sum Forces**" command you will see that everything is *practically* in equilibrium:

Check Sum of Forces Summation of Forces, Moments, Pressures and Body Loads for Set 2 (CSys 0) Nodal Force FX = 0.00048828 FY = 0. FZ = 0. Nodal Moment MX = 0. MY = 0. MZ = 734057. Pressure Force FX = 0. FY = 0. FZ = 0. Body Translational Accel FX = 0. FY = 0. FZ = 0. Body Varying Trans Accel FX = 0. FY = 0. FZ = 0. Body Rotational Accel FX = 0. FY = 0. FZ = 0. Body Rotational Velocity FX = 0. FY = 0. FZ = 0. Totals (CSys 0) About Location X = 0. Y = 0. Z = 0. Forces FX = 0.00048828 FY = 0. FZ = 0. Moments MX = 0. MY = 0. MZ = 0.10938

Using **inrel,-1** solution and picking one of the corners as the support location it creates a reference frame that is easier to understand and give displacement values closer to the original solution.

Here you are the picture to show you where I create the Constraint Set#2 named "Inrel,-1" that I will use to define the SUPORT entry (you can use any name you like):

And here you are the NASTRAN Bulk Data window to define the **PARAM,INREL,-1** entry:

And finally the Boundary Conditions window where you see not any constraint set is defined but a SUPORT1 entry is defined:

And here you are the displacements results of the INREL,-1 study, practically identical to the original solution. According developers "*if you plot the internal forces, then all 3 solutions (baseline; inrel,-2; inrel-1) get exactly the same internal forces and stresses. This means the actual deformed shapes are the same and relative displacements of all the nodes are the same*".

Espero que te sirva de ayuda, y si tienes cualquier problema no dudes en consultarme, encantado de ayudarte, ¿OK?.

Saludos,

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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03-03-2015 04:38 PM

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03-04-2015 10:10 AM - edited 03-05-2015 11:48 AM

Blas and others,

Here we go again guys

I completely agree and understand the solutions you have submitted but there are based on the initial model with fixed boundary conditions at right corners (Baseline)

If I understand properly what you are doing is solving in 2 different ways the baseline problem by:

-replacing BC by reactions forces and moments INREL,-2

-and then using INREL,-1 by calculating reactions forces after applying inertia effects with one end constraints.

However, the actual baseline model should have no BC whatsoever, just the forces. The analytical shear flows (what I want to get) are q=12 N/mm for circular cell,q=18 N/mm for top rectangular cell and q=2 N/mm for the other

The model with no boundary conditions from initial model (see modified initial file) is attached.

Once again, what the solution should be doing is balancing the extrenal forces (12000 N and 8000 N) with the internal shear flows q (12,18 and 2 N/mm) without having to defined any BC.

By hand calulations the problem is solves as follows

1- Solve shear flows q1 by assuming forces go through shear center (No rotations)

2- Solve shear flows q2 due to moment create by forces about shear center.

Final shear flows q=q1+q2

The boundary conditions are never used.

Blas muchisimas gracias.

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