Page 709 - 02. Subyek Computer Aided Design - Beginner’s Guide to SOLIDWORKS 2019- Level 1 by Alejandro Reyes

Beginner's Guide to SOLIDWORKS 2019- Level I

It is important to know that, by the very nature of the Finite Element

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Analysis process, the results are an approximation to the actual

physical phenomenon and solving such a large number of

simultaneous equations will most likely return slightly different values for

stress and displacement results. However, the resulting values should be, in

general, very close to the results shown in this book, assuming that the

analysis is defined the same way {geometry, material, loads, and restraints)

and the mesh size is the same.

The accuracy of any analysis will depend on how well the geometry is

modeled, as well as how closely material properties, loading, mesh size, and

restraining conditions match the actual physical conditions. An analysis

done with SimulationXpress should only be considered as a general guide

and never a substitute for a complete and properly modeled analysis using

accurate loading and restraining conditions.

34.14. - In these results, we can see the Maximum Stress in the model is

approximately 72,000 psi. The Yield Strength of Aluminum 7075-T6 is 73,200 psi.

Based on these results, the lowest Factor of Safety in our model is almost 1.01.

The color-coded stress scale helps to identify stresses in the rest of the model.

Remember that stresses close to fixed faces may not be accurate.

Results

The Factor of Safety is calculated

by dividing the material's yield strength by

the maximum stress in the model. We are

interested in finding the lowest factor of

safety value because if any area of the

model has a factor of safety below 1 , this Show where factor of safety J

means the stress is higher than the ~ (FOS) is below: 1

material's Yield Strength; if this is the case,

our model will have permanent deformation Based on the specified parameters, the lowest

and therefore is considered to have failed. factor of safety(FOS) found in your design is

1.01221 ... __ _

Note that we are talking about permanent deformation and not breaking. To
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yield does not necessarily mean that it will break; it will break if we reach

the Breaking Stress. For most practical purposes, a permanent deformation

will almost always be considered as a failed design because the part will no longer

have the shape it was supposed to have, compromising the integrity, functionality

and safety of the product and/or end user.

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