Static Stress - Some components are using a material with invalid properties

[Anonymous]

I have a design with materials of "Plywood, Finish" and "Softwood, Finish". Both of these materials appear to have invalid properties. In fact, as far as I can tell, all the wood materials have invalid properties. No indication of what properties are invalid. Can someone point me in the right direction?

[Mike.Grau]

Hi @Anonymous,

Thank you for posting your question in the Forum.

The answer to your question is that the current solver require isotropic material behavior like steel, aluminum, copper etc.

Wood materials shows orthotropic behavior. Therefore, they can not be used and appear as invalid.

The answer in this forum post here is very helpful in case you want to use wood material properties in your simulation.

http://forums.autodesk.com/t5/design-validate-document/simulation-error-some-components-are-using-a-material-with/m-p/6393950#M61293

Thanks

Mike

[Anonymous]

Thanks. I ignored that answer because the questioner was asking specifically about thermal simulation. Perhaps you could annotate the answer in some way to show that it applies to all simulations.

So there's no way to model stresses in wooden panels?

[Mike.Grau]

Hi @Anonymous,

Thanks for keeping me updated.

Yes, you are right, it would be very helpful to have an answer to the question

which applies to all simulation types. Our Learning content is referring to the Poison ratio, with the link here

http://fusion360.autodesk.com/learning/learning.html?guid=GUID-B13CC32E-E173-420E-B898-7FB512FD006C

You can use wood properties in a static simulation by simply copying steel, modifiy the physical properties to wood properties.

I will create a screen cast for which shows how to do that.

Thanks,

[Anonymous]

Hi @Mike.Grau,

I've had a chance to review this now.

I think all these values are approximations anyway because it will actually depend on the exact type of ply (e.g. what sort of glue was used) and how many layers there are. But I may be able to get some indicative numbers for my prototype. After all, we are going to build this thing and test it out, so I just need to know in one or two places whether I should allow for the possibility of more support.

However, I do need a little help interpreting the values.

For plywood, there are two shear values (one for the X and a lower one for Y&Z). Can you explain (or point me to the appropriate help) how these directions should be interpreted. Thinking about a sheet of ply, there is one direction that is through the flat faces of the sheet, the other two are edges. I'm guessing, that the X value is the face of the sheet, though I find this counterintuitive. But it does make sense that the lower shear value for the other two directions indicates that the shear strength is lower when the force acts to move the laminated sheets across one another. Is this right? Or do the figures refer to the axes in the model, taking no account of how the material is placed? If that is the case, then I will just have to apply both numbers in turn.

The softwood is different since it has values for Strength along or against the grain. Note that these are not affected by changing it from orthotropic to isotropic. I'm hoping that I can create two materials, depending on how I think the more important forces will be acting on particular bodies. Or even trying out both materials to see how it affects the behaviour. However, I need a little help interpreting these parameters, too.

The Yield Strength & Tensile Strength are both set to 0. I am assuming this is because they will be different depending on how the material is placed. So, can I choose from the parallel or perpendicular to grain figures? In other words, for the Yield Strength choosing one of the two Compression figures and for the Tensile Strength choosing one of the two Tension figures. Am I comparing like for like? I.e. is the Compression Parallel to Grain number the same thing as the Yield Strength in that direction?

I'm not going to hold you to this! If it all fails, it all fails. If the prototype bows more than it should, I will just chalk it up to experience, but I'm interested in trying.

Thanks,

Mark.

[Mike.Grau]

Hi @Anonymous,

You could may change the physical properties of wood from orthotropic to isotropic

like in the picture below. That might be very quick for the first simulation.

I hope this information helps. Please, keep us updated if your simulation worked.

Thanks, 

[innovatenate · Accepted solution]

This looks like a fun Friday question! I cobbled together some information from previous discussions that may help with this one.

If you review the Materials dialog in the Simulation environment (from the ribbon in Simulation, not in the model workspace) you will only see a limited number of values on the right hand side when you expand the properties. 

These are the values you should be concerned with for the Static-Stress type simulation. Some a knowns and some are calculated values.

Poisson's ratio,  Nu = User Input/Material known (Given)

Young's Modulus (Modulus of Elasticity), E = User Input/Material known (Given)

The Shear Modulus is calculated from the given inputs above:

Shear Modulus (Modulus of Rigidity), G = E / 2(1+Nu) 

Note that you cannot input the Shear Modulus in the Simulation environment and that only the Young's Modulus is listed. I'm not 100% sure how it is handled in Fusion Simulation, but I imagine the value from the Model Environment is ignored. I'm extrapolating from the Inventor Stress Analysis Environment which shares some components here.

I can find out more about this, confirm and circle back with an update.

The Yield and Ultimate Tensile Strengths will be used to calculate safety factor results. These should probably not be zeros! This values should also be provided from known material properties.

You can make the distinction between the two Safety Factor Calculations in the materials dialog within the Simulation environment:

For Brittle materials (e.g. cast iron), the Ultimate Tensile Strength is the recommended Safety Factor Calculation (Mohr's theory):

SF = ABS[UTS/(P1-P3)] 

Where:

SF = Safety Factor

ABS = Absolute Value function

UTS = Ultimate Tensile Strength

P1 = 1st Principal Stress

P3 = 3rd Principal Stress

Note: If both P1 and P3 are positive, we use P1 in the denominator. If P1 is negative, we use just P3 in the denominator.

For Ductile Materials (e.g. Alloy Steel or Aluminum), the Yield Strength is the recommend Safety Factor Calculation (Von Mises Criterion):

SF = YS/(Max VMS)

Where:

SF= Safety Factor

Max VMS = Maximum Von Mises Stress

I should note that the assumption with Static-Linear Analysis is that the material is loaded within the linear portion of the stress-strain curve and not beyond! Safety Factor calculations based on either criterion (ductile or brittle) are not always valid since any results and corresponding safety beyond the yield point of a material cannot be accurately calculated. Other assumptions for Static Stress Simulations:

• The deflection and stress are linearly proportional to the load. If you double the load, the deflection and stress double.
• Material properties are linear. The stress-strain curve is a straight line with the stress remaining proportional to the strain. There is no yielding of the material.
• The loading is static and it is applied slowly. Dynamic loading effects, such as a sudden load application or impact, are not considered.
• Temperature has no affect on the part geometry or material properties.
• The deformation of the part is small compared to the dimensions of the part. Large deflection requires a nonlinear analysis to account for changing part and load geometry, and it is not considered during linear analysis.
• Buckling is not a concern.

Since the Autodesk Material Libraries can be shared across a range of other products, some of the values in the Material Editor (especially if the material is not isotropic) from the Model workspace don't apply in the Simulation (at least for today - see the roadmap for more info). Also, if you try to use some of these advanced material properties (such as anisotropic material properties), you will get the error message about invalid material properties. 

I hope that information helps! Good luck with your simulation!