design of weld -- moment of inertia

[mfstructural]

question about weld design...

when I'm calculating the moment of inertia of a weld, specifically the welds parallel to the axis you are calculating I about, you use parallel axis theorem....I_total = I +A*d^2.

Does the I = 0 since the "height" is cubed? Then you are left with just using Ad^2.

 

[JLNJ]

I is not zero, but its becomes very small compared to the total as "d" increases.

Ad^2 is probably close enough.

 

[dhengr]

Mfstructural:
I think you are over complicating and over thinking this problem, although you do come up with a good answer with your method. I assume you are thinking of something like a line of weld on the t&b flanges on the end of a WF beam, to accommodate a cantilever moment. Try this on for size.... The canti. moment is ‘M’, and the WF depth is ‘d’; replace ‘M’ with a couple ‘T’ and ‘C’ with a lever arm of ‘d’. Then, what length of a given weld size does it take provide ‘T’ and ‘C’? Alternatively, given a flg. width, less .5" near each flg. tip, you have a weld length (bf less 1"), which leads to a weld cap’y. needed per inch of weld length; pick a weld size which provides this shear flow.

 

[Lion06]

I always treat the welds as line elements with zero thickness and then the I is in units of in^3, S is in units of in^2, and A is in units of in. Then when you do your M/S or V/A calculations, you get units of k/in, which compare nicely to the typical fillet weld strengths of k/in.

 

[Hokie93]

I follow the same procedure as Lion06. If you aren't already using one, there are published tables of section properties (S and I, for example) of weld groups of various shapes that will save you a few steps in the calculation process. Tables are available in "Design of Welded Structures" by Blodgett, the PCI Handbook, "Steel Structures: Design and Behavior" by Salmon and Johnson, and the AISI Cold-formed Steel Specification.