Loading on Studs 1

[slickdeals]

Folks,
Please see attached sketch. It is a situation where a steel beam is hung from underneath a slab to pick up a cantilever load.

Will the first row of studs see all the loads or will there be a different distribution? I would have analyzed this as a propped cantilever and got the resulting reactions and designed the studs for T/C loads. But would there be a different behavior in the stud group?

Your thoughts are welcome.

 

[Lion06]

Depending on the member slope at the stud group in tension, I think you're pretty safe designing it with all 6 anchors taking equal load. If you want to be a little conservative, apply a nominal eccentricity to it (say 2"-3") and see what that does to the capacity of the stud group.

What kind of loads are you dealing with? How thick is the slab? If you can't get studs to work with App. D calcs, I often weld rebar to embed plates and get away from the App. D calcs. I typically only do this if studs won't work or for really high loads.

One other note, this probably qualifies as a hanger connection that needs to have the capacity increased by (something like) 1/3 per IBC.

 

[slickdeals]

The reason for this question is this:
As the load travels from the tip of the cantilever to the support points, the group of studs it hits (2 studs) will see the load first. What kind of redistribution (stud elongation/micro concrete cracks or some other) will have to occur before the loads get picked up by the next row of studs and so on.

 

[SlideRuleEra]

IMHO, the (approximate) answer is based on the proportions of the beam / stud spacing:

1. If the stud spacing is "small" compared to the length of the beam, then the studs would act as a group.
As a "first order approximation" I would say that if beam length is at least 20 times the stud spacing, the studs act as a group.

2. Also, if the stud spacing is "relative small" compared to the DEPTH of the beam, the beam will be essentially rigid and the stud will act as a group.
For this case, if beam depth is at least twice the stud spacing, the studs act as a group.

For proportion of this order of magnitude, I'm sure that the answer can be calculated more PRECISELY, but I doubt if the results will be any more ACCURATE.

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[ash060]

Have you considered using hanger rods that are embedded into the concrete above and run thru to the bottom flange and are connected there?

That would pretty much guarantee that the tension connection is tension and does not try to form a couple.

Some horizontal bracing would be necessary to prevent swaying, but I dont have a good feeling about the way it is currently shown.

If that first connection wants to take the whole load, it went from a cantilever with a backspan to a straight cantilever. And I think that first plate will have trouble taking all the tension and the moment as well.

 

[slickdeals]

This is not on a real project. It was a in-house discussion we were having regarding load paths when you have a connection such as this one. Some were saying all studs will see the load, some were saying the first row of studs will try to pick up the load.

The horizontal sway would be picked up by the couple developed between the two welded plates on the flange. Obviously flange bending (and use of stiffeners) would need to be checked.

I wanted to get a feel for how other engineers would approach the problem. Don't ever use such a connection is also probably a valid answer.

 

[ash060]

That is good to hear.

I have never been comfortable with an embedded concrete connection "living" in tension. I am not saying it wouldn't work, but it does give me goosebumps.

 

[csd72]

Personally I would just analyse it as all six studs taking the load evenly but make sure that there is at least 10 to 20% extra capacity to allow for the effects of partial fixity, beam rotation e.t.c.

 

[FFENGINEER]

I thought appendix D took this into account with group factors. I like the idea of DBA's welded on to take the load.

 

[BAretired]

If the beam is very stiff relative to the tributary width of slab, stud #3 takes more load than either #1 or #2, in fact you could make a case for #1 being in compression.

BA

 

[slickdeals]

You mean stud 3 being in compression and stud 1 taking more load than 2 and 3 ?

 

[slickdeals]

Correct me if I am mistaken, but stiffer the more equally the studs will take the load.

The more flexible the beam, the more local effects in a stud group, right? Meaning one of the studs in the group could go into compression.

 

[slickdeals]

"stiffer the beam"

 

[BAretired]

If the beam is a rigid body, it rotates an angle theta when loaded. The slab has to bend in an 'S' shape between support points because it is continuous beyond the left support. It requires a counterclockwise moment at the right support to align it with the beam. This reduces the tension in Stud #1 and increases the tension in Stud #3.

BA

 

[hokie66]

I don't see that, BA, not as a practical matter. If the steel beam is not loaded, it just goes along for the ride.

 

[BAretired]

hokie,

From a practical perspective, I would assume studs #1, 2 and 3 carry the same tension, but somebody suggested that #1 would carry the load until micro-cracking, etc. took place to pass the load on to the other two.

From a strictly theoretical point of view, #3 carries more load than either of the other two if the steel beam is a rigid body and the slab is flexible.

For all three studs to carry the same load, consider the slab subjected to two loads, an upward reaction at the left stud group and a downward reaction at the right stud group. Determine deflection and slope at each reaction point, then select a steel beam having the exact stiffness to match the slope of the slab at the right reaction.

If the steel beam is rigid, it will have a smaller slope at the right support than the slab and will exert a counterclockwise moment on the slab in order to make the slopes compatible. This is similar to prying action on a baseplate.

If the beam stiffness is such that the slope of the beam and slab are equal at the right support, then Studs #1, 2 and 3 will carry precisely the same load. If the beam is too flexible, #1 will carry more than the average. If the beam is too stiff, #3 will carry more.

BA

 

[Splitrings]

slickdeals,
I think this is the calculation you are looking for. I am having trouble getting my hands around the sign convention. I think you can convince yourself the method is correct by trying it with only two bolts and summing the moments and forces.

 

[BAretired]

Splitrings,

If both the concrete slab and the beam are rigid bodies, your method is correct but your sign convention is screwed up. For all six bolts, the P term should be -10/6 = -1.666.
For #1 the compression should be -10/6 + 85*3.5/55 = 3.74k.
For #2 it would be -10/6 + 85*3/55 = 2.97k.
For #4 it would be -10/6 - 85*2.5/55 = -5.53k.
For #6 it would be -10/6 - 85*3.5/55 = -7.08k.

If the slab and beam are permitted to flex, your approach is wrong as it does not consider potential prying action on the slab.

BA

 

[paddingtongreen]

There is too much information that I don't have. I would like, ideally, to have a hinge at that point. Since that is not practical, and if the slab is thick enough, I would use long headed studs and a stiff plate, allowing the strain in the studs to help distribute the loads.

Michael.
Timing has a lot to do with the outcome of a rain dance.

 

[dhengr]

Splitrings & BA:

RE: your 11MAR 12:12 & 12:44 posts, I think your both nuts . I know structural engineering has changed a lot since I started, and I’m gettin confuseder, every day, with old age, but has simple statics been disproved too, now? Doesn’t R4-6 = 10(11.5/6) = 19.167k tension & 19.167/3 = 6.389k for each stud 4, 5 & 6. And, R1-3 = 10(5.5/6) = 9.167k compression on the three studs 1, 2 & 3. The only thing that’s got a moment of inertia in this problem, in my world, is the stl. bm.

I don’t really like the studs in tension either, as someone above mentioned, and I would make darn sure they had plenty of embedment and confining reinfg. around them and up into the slab. I believe Slickdeals’ OP questioned the flexure of the loaded canti. tending to pull more on stud 6 than it would on stud 5, etc. and might stud 6 start to fail in distributing load to stud 5. I think this is certainly a valid question. And, for that very reason I would put some load factor on R4-6 to intentionally oversize it. He could also turn his weld plates to be across the length of the canti. beam, instead of on each side of the web, out on the flange tips, thus softening that detail’s action on the studs. I would make it one cross plate centered on stud 5, with a couple web stiff. pls. under the top flange and half way down the web.

We can analyze the crap out of this with FEA or today’s bolt group in a base pl. approach and the structure has no idea what deep thinking we are doing, it just keeps acting the same old way it did forty years ago. Even if we told it to comply with today’s rigorous methods.