design of weld -- moment of inertia
design of weld -- moment of inertia
(OP)
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.
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.






RE: design of weld -- moment of inertia
Ad^2 is probably close enough.
RE: design of weld -- moment of inertia
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.
RE: design of weld -- moment of inertia
RE: design of weld -- moment of inertia