Column unbraced length question when the concerned beam falls on beam just 6" far from column face 4

[NewbieInSE]

Hello,
I have a practical question. The following is the scenario. I have attached two figures below, one is beam layout, the other is the construction photo.
I would like to know whether the beam connected 6" or so apart from the face of the column would provide bracing effect to the column so that the unbraced length in a story for that column could be considered 1 or so??

Thanks........

 

[BridgeSmith]

Assuming the stub from the beam to the column has adequate strength to brace the column (which doesn't take alot of strength), it should be able to be considered a brace point for the column.

 

[NewbieInSE]

How to convince someone that this stub has enough strength or stiffness to laterally brace the column? Is it possible to develop some sort of calculation? Is it to show that the stub is capable to resist 2% of the column compressive capacity as a shear force acting at its end?

 

[JAE]

I'd say about 2% would be adequate but I would guess that the beam stub could take much more than 2% of the column axial load.
In both directions I would add, assuming there's proper reinforcement developed into the column.

 

[HTURKAK]

Effective length factor (k) from ACI 318-14

k is Effective length factor and

Ψ = Σ(EI /lc ) / Σ(EI /lb )...... ( EI/L is bending stiffness) ..

in this case , you should add the torsional stiffness ( JG/lb) of short beam (6in or so ) to the nominator

Ψ =( Σ(EI /lc) + JG/lb )+ / Σ(EI /lb )






EDIT= TORSIONAL STIFNESS OF SHORT BEAM SHALL HAVE NEGATIVE IMPACT SO ADDED TO NOMINATOR.

I cannot give you the formula for success, but I can give you the formula for failure..It is: Try to please everybody.

 

[BAretired]

Bracing is provided if the anchorage to the column is capable of resisting a lateral force of 2% of the column load in any direction. If the black line shown in the photo is a separation, the column is not braced at that point.

 

[NewbieInSE]

Thanks Hturkak. I'll calculate it and share.

BAretired, haha, no that's not a crack. It is a black rope. You can see it hanging below.

 

[BAretired]

I'll have to take your word for it because I can't see it hanging below.

 

[NewbieInSE]

in this case , you should add the torsional stiffness ( JG/lb) of short beam (6in or so ) to the nominator

Ψ =( Σ(EI /lc) + JG/lb )+ / Σ(EI /lb )

Ok. But why not Ψ = Σ(EI /lc)/ (JG/lb)
by EI/lb, do you mean the bending stiffness of the beam which rests on the stub?

 

[Tomfh]

The 2% rule is a fairly standard approach. If this little stub can happily carry the column loads plus 2% of the column load (applied laterally), then I’d consider it braced.

 

[NewbieInSE]

If this little stub can happily carry the column loads plus 2% of the column load (applied laterally), then I’d consider it braced.

SO in this case, I need to find 2% of column Pu probably, and then apply it laterally (shear in the minor axis) which will produce shear force and some bending moment considering the stud length 6" supported one end continuous. I need to show the stub ok for these load effects. Is that correct?

 

[Tomfh]

Yes I’d be checking it for something like that.

Those stubs do look like the weak point (hence BARetired’s concerns about a crack). As a slightly separate matter, with the short backspan you can end up with more (or less) vertical shearing force (and column load) than what you anticipated.

 

[BulbTheBuilder]

Looking at this image, can the beam's soffit be considered as the unbraced length? From my understanding I see the slab to be providing a lateral support to the column than the beam due to its in-plane stiffness. (Not related OP's question but I was wondering for this scenario)

Also from the OP's picture, the slab is away from the column (no connection). How do you justify the unbraced length (lu) for in-plane direction? (not effective length factor; k)


Aside the chart you can use these equations for k,

k is Effective length factor and

REF ACI 318-05
k=0.7+0.05(ΨA+ΨB)=<1.0
k=0.85+0.05Ψmin=<1.0