huangyhg

weld design
does the aisc spec address how to design a weld that is subjected to in-plane shear as well as out-of-plane shear simultaneously? i can't seem to find any reference on how to handle this.
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i don't believe it does. this is how i do it. my first statement will be that by in-plane shear and out-of-plane shear you mean direct shear and tension. i've never heard of out-of-plane shear (unless you mean an in-plane shear applied at some eccentricity to the centroid of the weld group.
before i spend the time to write it out, can you please clarify exactly what you mean?
two ideas:
1. use the elastic (vector) approach.
2. look in part 8 and see if one of those instantaneous center of rotation tables (table 8-5 for example) matches your case.
sure thing. good thing you asked because i do not mean direct tension (but the theory may still apply).
i have a beam that is at a 10 degree angle. it has normal vertical shear (in-plane). the beam is also part of a diaphragm system and as a result of the angle there is a small horizontal shear (out-of-plane) component which arises from the axial load.
i am trying to design the weld at the end of the beam for the 2 shears (assuming simple supports).
now with more info:
1. elastic method
2. table 8-4, page 8-67, case with k=0. actually, whether it's the 8-67 or 8-68, etc. table depends on the angle of the resultant.
i prefer the elastic method.
assuming there are no eccentricities to consider just find the resultant shear.
[(vx')^2 + (vx)^2]^0.5
divide that by your total length of weld. this will give you a result of some number of k/in. design the fillet weld based on your result divided 0.928k/in/16th in. the result will be the number of 16ths of an inch of fillet weld that you need.
you will need to use the polar moment of inertia, j, of the weld group of your shear from the diaphragm is applying the shear at an eccentricity. you can get that value from blodgett for various shapes.
strleit: why not s and not j?
s would be if the shear were applied out of the plane of the weld. use j if the shear is applied in the plane of the weld but at an eccentricity of the centroid of the weld group.
since he didn't mention any type of tab, i assumed the beam was transferring the load directly into the weld, hence so out-of-plane eccentricity and no need for s. depending on the diaphragm, it could be delivering the load in the plane of the weld, but at an eccentricity to the centroid.
i'm not missing anything there, am i?
i assumed that you were referring to the eccentricity being out of the plane of the weld.
maybe i am missing something. i assumed that the load would be halfway between two vertical welds->therefore no torsion and no need for j.
the way i understand the op, the load goes through the weld group centroid, but is simply inclined. in that case, no need for j or s. if i'm understanding him right, the case looks just like the 13th ed. page 8-67 lower diagram (c with k=0), but with ex=0.
i agree with that for the beam shear. i just thought the diaphragm shear may not be applied through the centroid of the groupl
thanks for the replies... it makes much more sense now.
i agree the elastic method does seem more straight forward and intuitive. to me, it seems as though it's a mohr's circle problem. is that how they come up with the elastic method?
i'm also curious to see how aisc came up with the load-deformation curve for the ultimate strength method.
i'm not too great with weld design yet but if my resultant is 32 degrees- does the effective throat change?
eit- 0.928k/in/16th: where are you getting this number? i use 1.39 k/in/16th (0.6*0.75.*70*0.707/16)

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