huangyhg

cover plates
i need to reinforce some wide flange girders and columns with cover plates welded to the flanges. i am welding the cover plates to the flanges with a 1/4" fillet, 3" long, spaced at 12" o.c.. i did this because i've seen this "skip" weld called out elsewhere. but how do you check if that weld is sufficient? i guess i could be really conservative and call out two 40' long 1/4" fillet welds per cover plate, but then the welder would probably hate my guts.
check the design code, for one thing.
you can also calculate the shearing stress at that point, and find average longitudinal shear in the welds.
if it's a compression flange, you'd have to have adequate welding to prevent the plate itself from buckling.
be sure to thoroughly read the last sections of chapter b in the aisc specification. this section governs the design of cover plates, specifically how you terminate the ends with solid welds.
the design of a stitch weld is simply taking the capacity of the stitch and converting it to a capacity per foot for comparison with the horizontal shear that's developed between beam and plate.
for lrfd for two stitch fillet welds using e70 welds:
φrn = 2(0.707)(weld size)(stitch length per foot)(1/12)(0.75)(0.6)(70)
so for a 3" fillet each side at 12" o.c. you would have:
2(0.707)(.25)(3/12)(0.75)(0.6)70 = 2.78 kips/ft.
you're off by a factor of 12. it would be 33.4 kips per foot.
(2 welds)(0.707 factorless)(0.25 in)(3 in/12 in)(0.75 factorless)(0.6 factorless)(70 ksi) = 2.78 kips/in (follow the units).
(2.78 kips/in)(12 in/ft) = 33.4 kips/ft
nutte - thanks for that correction. sorry for the confusion. it should have been simply kips/in at the end of my calc.
for the girder, you would need to design the intermittent welds to transfer the change in shear along the length by vq/i. the end welds are designed for mq/i to allow the bending stresses to flow from the beam section only to the full cover plated beam cross section. the aisc commentary discusses this in detail.
if a column has mainly axial load, intermittent welds are needed to prevent the cover plates from buckling and end welds are need to allow the axial stresses to flow from the column section only to the entire cover plated column section.
another consideration for the using skip welding wherever possible is the propensity of the cover plate to warp fairly significantly if you try to weld all the way down it.
it bothers me a bit that you used skip welds because you have "seen this 'skip' weld called out elsewhere". while there is certainly much to be learned from examining similiar conditions and others' details, you should always base the actual engineering on a sound understanding of fundamental behaviour.
jike, would you mind posting where you found the mq/i? i've not run across that equation yet. you mention aisc. can you give a chapter or page number?
ucfse - i believe this was in older aisc manuals when cover plating was much more common. for instance page 2-89 of the 6th edition manual says to use the equation 12mq/i to determine the "cut off" weld criteria (jike's equation but with m in ft-kips and q and i in inches so the moment units conversion is built in) for cover plates.
i have typically just used enough weld on the ends to fully develop the yield strength of the plate. however with a huge plate or for instance a wt it is not practical. in that case the 12mq/i works fine. this equation can be derived as follows:
1. at the terminating ends of the cover plate, assuming strain compatibility between the cover plate and the beam flange, stress (fs) at the centroid of the cover plate attached to the beam is equal to (mc/i)*12"/ft with m in ft-kips where c is the distance from the neutral axis of the combined section to the neutral axis of the cover plate and i is the combined moment of inertia of beam and cover plate.
2. generally a cover plate is thin enough that the stress calculated at the centroid can be assumed to be constant through the thickness of the plate.
3. force in plate is therefore equal to fpl = fs*a where a is the area of the cover plate.
4. so substituting back in for fs yields fpl = (mc/i)*12*a
5. noting that q of the plate is equal to a*c and subbing back into the previous equation yields: fpl = (mq/i)*12.
the welds are then designed to transfer this much force into the plate.
thanks willis. i learned something today!

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