"Stiffeners" at torsional connection

[a7x1984]

Have a torsional connection at the top of a concrete footing (supporting a relatively light item with a lot of surface area). Therefore, wind is my evil loading...See attached sketch.

Torsional moment is 9.5 k-ft - not large, but trying to get an estimated weld length has been difficult. The only way around it, that I could see, was to flare "stiffeners" from the HSS section to get the required weld length.

Basically (3) things I am unsure of:
1. Now that I have introduced these elements, is it even worth checking their capacities? If yes, is there a rational procedure floating around? My intuition says this small loading can be handled by inspection with not much additional analysis.
2. I am concerned that as the HSS begins to torque there will be an uneven distribution of stress in the welds, maybe causing the 'front-loaded' welds to crack. Any ideas in configuring this connection in a different manner?
3. I can't find a prequalified weld between the HSS and the base plate. The closest I could find was a flared weld, but it didn't show the HSS bearing on the plate to be welded to.



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[renderu]

I would assume all torsion is taken up at the first stiffeners. Now it's simple.

 

[BAretired]

You said it was a torsional moment, but your double headed arrow shows otherwise. If it is a torsional moment, the arrow should be pointing along the axis of the HSS. Using the right hand rule, the thumb points in the direction of the arrow and the fingers point in the direction of the moment. Please clarify.

BA

 

[a7x1984]

BA: Yes, I oriented the arrow incorrectly.

renderu: That is a thought. I am trying to be economical as well, as the work is for a nonprofit group.

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[renderu]

I think my suggestion is economical. You could do a calc on my assumption within, at least an hour. Trying to figure out the stresses going on in your arrangement is complicated. Material for the extra gussets is small. Labor would be an extra hour or so. You probably spent that much just posting and discussing the problem on this forum. If you're not charging then that's all good. How many of these are you constructing? Just one? That's important also.

 

[a7x1984]

renderu: I agree it is complicated. If I could find a way to avoid the use of those stiffeners, it would be good - there are four of them at the moment (two displays, two per display). My time spent is immaterial as it is a 'not-to-exceed' LS - I expect to spend some personal time on more interesting projects.

At the moment, I am wait to hear if we can get rid of the beam in torsion and extend the vertical members (not shown) into the footings. That option is highly dependent on the displays, as they have already been constructed by the artist.

Do you see an option that avoids the stiffeners?

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[renderu]

I don't have any specific advice on the arrangement at this moment. I don't have that reference info at home. Torsion in welds and in non-continuous (open) sections is tricky. I would need to go back to my textbook. But it is ultimately simple.

 

[BAretired]

You could weld the HSS directly to the plate. M= 9.5'k = 114"k. It requires 114/5 = 22.8k each side. That is probably not an approved welding procedure, so use a connector plate each side of the HSS about 6" long . Fillet weld to the HSS and the base plate as required.

If you want to stiffen the HSS, use a cap plate on the end.

BA

 

[renderu]

I agree with BAretired, partly. Except that I would consider the peak stress at the start of the weld. I would need to think about it a bit more to be sure, but I think that in this case I've previously used a weld stress along the length of the HSS that has a linear profile from start to finish.

 

[a7x1984]

BA: I would love to do that - 22.8k/(1.392*1/2") = 32.7" for a transversely loaded 1/2" fillet weld. If I lose the constraint on the footing dimension that would work well.

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[a7x1984]

renderu: I wonder if this attachment would also apply. They are discussing longitudinally loaded fillet welds - I wonder what logic would change for the maximum length of a transversely loaded weld.

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[a7x1984]

Never mind that: AWS indicates effective lengths for only end-loaded weld lines.

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[BAretired]

a7x1984, are you sure you are reading that weld capacity correctly? In Canada, a 1/4" weld 6" long would have a factored shear value of 1.28*150mm = 192 kN or 43.2k. That is equivalent to an allowable shear of 28.8k. I doubt that your weld values are much different.

BA

 

[a7x1984]

BA: See attached pdf from AISC Manual. First page has longitudinal capacity, second page has 50% increase for transverse loading.

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[BAretired]

Another possibility is to use a 4x4 angle welded to each side of the 5x5HSS with a couple of anchor bolts each. In that way, the base plate may be omitted entirely. If the bolts are spaced at [2.5+5+2.5] = 10" o/c, the tension per bolt is 9.5*12/10/2 = 5.7k, a light load for a 3/4" dia. A307 bolt.

BA

 

[BAretired]

BA: See attached pdf from AISC Manual. First page has longitudinal capacity, second page has 50% increase for transverse loading.

Look again. 1.392DL for 1/2" weld would be 1.392*8L or 11.1kips per inch length of weld. D is given as the number of sixteenths in the diameter, so 1/2" is 8 sixteenths.

BA

 

[a7x1984]

I like that one too, BA.

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[wannabeSE]

BAretired,
With strength design in the US, a 70 ksi fillet 70ksi weld is good for 1.392 kips per 1/16". I am guessing a7x mixed up 1/2" in the denominator with 8 (for 8/16=1/2").
For a 1/16" fillet weld [Φ]0.6 F_exxA_w = 0.75(0.6)(70)(0.0625)/sqrt(2) = 1.392 kips.
If the applicable code allows, a flare bevel groove weld (with reinforcing fillet if need) may work.

 

[a7x1984]

Good lord. I need to go back to wood design.

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[BAretired]

D is given as the number of sixteenths in the diameter, so 1/2" is 8 sixteenths.

should read:
D is given as the number of sixteenths in the weld size, so 1/2" is 8 sixteenths.

BA