I am currently a graduate student doing structural studies with mixed meshing (in SolidWorks Simulation 2009-2010)

Solid meshes are used on relatively thick isotropic parts, and shell meshing is used on thin plate composite parts, specifically carbon fiber from vendor "X." I used data from that vendor to make my own custom linear elastic orthotropic material. I defined composite shells using that custom orthotropic material in a stacked sequence of ply orientations.

The issue I am having is understanding how SolidWorks calculates the FOS (factor of safety). I fully defined the orthotropic material's yield properties as follows...

1. tensile strength in x (I will call it XT)

2. compressive strength in x (I will call it XC)

3. tensile strength in y (I will call it YT)

4. compressive strength in y (I will call it YC)

5. shear strength in xy (S12)

The first time around, I got good FOS results on the isotropic solid parts, but a FOS of 0.0 across the entire composite shell parts!

I first investigated how SolidWorks calculates composite failure criteria (Tsai-Wu/Tsai Hill for instance, check out this link for example)...

It is clear that first and foremost, SolidWorks for thin composite shells assumes a plane stress problem (all stresses out of plane are zero). As explained at that link, what SolidWorks does is uses the above five yield stresses (XT, XC, YT, YC, S12) in conjunction with your nodal principal stresses to calculate a so-called "Factor Intensity," and the FOS = 1/Factor Intensity. The Factor Intensity is a function of those five yield stresses, which I had defined. So why was I getting an FOS of 0.0 indiscrimintaly across all of the composite shells?

After looking back at material properties for the custom orthotropic part, I saw an entry for **yield strength**, which I had left blank because I didn't think it mattered for Tsai Wu/Tsai Hill. I decided to give it a shot and put a non-zero number equal to XT (as a guess to see what happens). Voila! I finally get an FOS distribution for the composite shells (rather than 0.0 everywhere). After playing around with that **"yield strength"** number, the FOS for the composite shells is highly dependent on that number! Yet, the SolidWork explanation does not even mention how this **"yield strength"** in the orthotropic material definition plays a role in Tsai Wu/Tsai Hill.

What is this **"yield strength" **number representing, and how does SolidWorks use it when performing Tsai Wu/Tsai Hill failure criteria?

Found your question in a search and my dilemma is similar. I am analyzing a laminated wood component and having difficulty trying to figure out the best failure criteria so I can determine FOS and whether or not this component will fail. As is fairly common with SolidWorks, trying to get this indepth with the software results in a dead end in the documentation. Does ANYONE have any links or hints as to where to look for some useful information? I have the USDA Wood Handbook (excellent resource for wood mechanical properties, but the way), and it does not address this issue.

Any help would be much appreciated.