Another composite question

[Lion06]

Does anyone know why (in the green book) that in order to get the number of studs for 100% composite action you find the smaller of 0.85f'cAc/2 and FyAs/2?? Why are they both divided by 2? This makes a pretty big difference in the % of composite action you get for a given number of studs (you get a much higher % composite for a given number of studs compared to the 13th ed spec) or the the number of studs needed for a given % of composite action (you can use fewer studs to gain a specific % of composite action compared to the 13th ed spec). This obviously makes a difference in deflection calcs as well.

 

[haynewp]

I don't have the green book with me but I think that is the old safety factor.

 

[LobstaEata]

Taking the lesser of the steel capacity and 0.85% of the concrete capacity ensures ductile failure.

 

[LobstaEata]

85% I mean

 

[Lion06]

haynewp-

I thought that might be it, but that approach makes it so you need FEWER studs to get 100% composite action (which is UNconservative). If the stud values were factored down by 2 as well, then it would make more sense, but the stud values are only factored down by 1.5.

LobstaEata-
Ductile/nonductile failure doesn't have anything to do with what I'm asking. I'm asking why the values are divided by 2. BTW, the 0.85 isn't to ensure a ductile failure it's the value always applied to concrete in compression (no different than a concrete beam).

 

[LobstaEata]

EIT

I thought the following was your first question

Does anyone know why (in the green book) that in order to get the number of studs for 100% composite action you find the smaller of 0.85f'cAc/2 and FyAs/2??

It seemed to me that the fact that this addresses ductile failure is what your question was asking. Sorry if I understood your question incorrectly.

Oh and BTW in any concrete (or composite design), it is critically important to check for ductile failure of the member. This is why we compare the concrete crushing limit with the steel yield limit.

 

[haynewp]

That's the factor to get the values down to estimated working load level.

 

[Lion06]

lobsta-

I was specifically asking why they are divided by 2, not why those two specific criteria are considered. I understand why those two criteria are considered, I just don't understand why they are both being divided by 2. haynewp has stated an opinion, but I address that below. Also, these two criteria have nothing to do with ductility as there are no ductility requirements (that I'm aware of) for a composite steel beam. Additionally, while it is good practice to check for ductility in a concrete beam design it isn't required. That's all taken into account with the sliding phi factor.


haynewp-
I originally thought that it was to get everything down to working loads, but then why are the nominal stud strengths reduced by 1.5 and not 2? It seems to me that this would underestimate the strength at an actual failure condition, no?

 

[haynewp]

My old steel book says the studs strengths are reduced by a factor of 2.5 to get working load values.

 

[Lion06]

That would make sense then, but the green book uses AsFu/1.5.

 

[haynewp]

What page?

 

[Lion06]

pg 5-59 gives the allowable horizontal shear loads per connector. All of those values listed are the nominal shear values (per the 13th ed) - AsFu - divided by 1.5.

 

[haynewp]

Those values change with increasing concrete strength so bearing is controlling the table values and not AsFu/1.5, right?

 

[haynewp]

Are you comparing NW concrete values in the green book to NW concrete stud values in 13th?

 

[SlenderBeam]

StructuralEIT,

Not familiar with the Green Book (I'm in Australia), but it would be odd if there is already a 'phi' factor (material/strength reduction factor) as well as the halving which you describe.

The only thing that would make sense is that the conc/steel force component is being 'prepared' to be applied in the strength/lever arm calc's where the force is applied at 'half' the depth of the compression or tension zone.

?

 

[miecz]

StructuralEIT

I'm comparing AISC to the formulation shown in AASHTO'w Standard Specification. AASHTO doesn't divide the concrete and steel strength by 2, and uses the ultimate shear strength of the shear connectors. AASHTO's formula for the ultimate strength of the shear connectors results in roughly twice the allowable shear load listed in AISC's Table 1.11.4.

 

[hokie66]

StructuralEIT,

What you are resisting is the horizontal shear at the point of maximum moment, and half of it goes each way. Sorry to be coming in so late with such a simple answer.

 

[haynewp]

That is correct but I disagree based on his question. The tabulated value per stud is different if you are comparing ASD vs LRFD tables.

 

[hokie66]

The formulas he stated have been that way since at least the 1980 Spec. Maybe there is an error in the new book, but the 1/2 is not a safety factor.

 

[strucguy]

StructurlEIT -

If I am not wrong the values you mentioned (ie. 0.85f'cAc/2 and FyAs/2) for number of studs, correspond to 25% composite action. The smaller of the values mentioned here is divided by capacity of stud to determine the number of studs.

- strucguy