Stress Analysis - XX, YY, ZZ....which to use?

[samstreet101]

Dear all,

Please excuse the large amounts of images in this post, as well as the amount of text but I feel it necessary to fully describe the problem. It's one I've often had with Inventor and it is that of reconsiling results from FEA stress analysis and those of traditional hand calculations.

I have a part (see attachment) which will be fixed at one end (in the file I have constrained it such that it can float in the x axis as I believe it would be able to in reality).

A load is then applied to the other end (51.5mm from the point at which is fixed). This is admitedly not perpendicular but at an odd angle..this may have something to do with the problem.

As you can see from the above image a load of 2kN creates a moment of 103 Nm (M=Fd ...M=2000 x 0.0515m).

I then refine the mesh and change the convergence settings until convergence (0.5%) is acheived (screenshot available if proof is required).

The results I get are not necessarily unweildy as 2kN may be an unfairly large load for a part such as this. My problem is reading the results and knowing which results relate to which kinds of stresses.

Here, stress in the YZ plane which I would assume to be bending stress, is 327 MPa. However, I have worked this out by hand and it's a little off. I make it 715 MPa by the general bending equation...(stress = My/I where M is moment, y is distance to the neutral axis, and I is second moment of area). I believe the weakest section, where maximum bending stress will be, to be across the section above where the cross sectional area is 12mm high x 6mm across. Therefore I = (b x (d)^3/12).... = (0.006 x (0.012)^3/12) = 0.864x10^-9 m^4. I make y (distance to neutral axis) to be 6mm or 0.006m. Therefore the bending stress using the equation I mentioned in the earlier sentence to be 715 MPa.

Now the stress in the linear axis ZZ is much closer to my calculated result, coming in at 730 MPa and the stress in YY is 223 MPa (see screens below). Given the angle at which this part is constrained in, this could be the case but my question again is how to know which stress results relate to which real world stresses (axial stress, bending stress, shear stress etc...)?

[JDMather]

It is unclear to me where you are taking your information for hand calculations.

Create a workplane at the location and new sketch - Project Cut Edges. That will help me understand the cross section location you refer to.

Also, scan and include the hand-calcs.

And - is that one image superimposed over Autodesk ForceEffect image (I am trying to figure out how you did that with only Inventor if ForceEffect was not used).

[blair]

Unless you did a Split-Surface where you applied the force load, Inventor will apply it over the face of the surface so your calculation to a fixed distance won't be the same.