Scan-and-Solve for Rhino

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Hi,

What is the best way to take into account the different yield strength for steel due to different thickness?Of course the conservative way is to choose the lowest, but is there another way. I have solids with varying thickness.

BR Patrik

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Hello Patrik,

For clarity, is the yield stress changing in the material as an effect of the manufacturing process? It may be helpful to include a picture of your model.

If you are interested in modeling the solids as separate pieces with their own material properties, you can use the Scan&Solve Pro Beta version which supports multiple components and materials in a model.


- Will

Will,

The difference in yield strength is guided by the thickness as below:

"Here is the table showing reduction in yield strength based on thickness
Up to 16 mm - 355 Mpa
16 mm < t  ≤ 40 mm -  345 Mpa
40 mm < t  ≤ 63 mm -  335 Mpa
63 mm < t  ≤ 80 mm -  325 Mpa
80 mm < t  ≤ 100 mm -  315 Mpa
100 mm < t  ≤ 150 mm -  295 Mp"

"However the yield strength reduces when you go up in thickness above 16 mm for flat products & hollow sections"

The above applies to welded steel structures made of flat steel, and it is relevant in generic steel construction as well as< house construction. Not sure if it comes from regulations or material suppliers process, or both.

I am designing strucutres of flat steel and hollow sections welded together.

BR Patrik

In this case, since Scan&Solve only uses the yield stress to calculate danger level, you can instead have Scan&Solve show the Von Mises stress in the material (under advanced options in View).

For each section of different thicknesses, you can find the maximum Von Mises stress in the section using section view and point query tool in the View tab. Then, with this data and the table for yield strength, you can manually calculate the peak danger level as the maximum Von Mises stress divided by the yield strength in each section.

Let me know if this helps,

Will

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