Lateral Force Distribution Collector Question

[damascus]

Quick question on horizontal forces in collector.

Given: 3 bay frame, middle bay is steel brace frame, each bay is 20' long, lateral force is 10k at this frame (wind or seismic)

Vs= 10k/60' = 0.166 k/ft (force through diaphragm)

Vr=10k/20' = 0.5 k/ft (brace frame resisting)

Difference = 0.333 k/ft

My question is do we do a axial diagram with distributed forces I have given above or do we do it as point forces.

As distributed forces the axial force goes up from 0 to +3.33k to -3.33k to 0, and therefore the collector only feels a max axial force of 3.33k

If I do it as a point force it goes from 10k to 0 when we reach the brace frame, and therefore the collector on one side feels 10k of axial force.

Does this analysis of force in collector differentiate between wind and seismic? 

I attached a picture for a more graphic representation of what I am trying to describe.

Thanks for the help.

 

[Mithrandir918]

You have it right using the unit shear forces.  You can't use the collector/struct reaction (10k) to determine the drag force.  The drag forces is derived from the lag that the shear force from the diaphragm transferring the load to the vertical force resisting system thus you need to have them in terms of unit shear forces.  

Note you also have to consider overstrength factors and the maximum considered collector design loads per ASCE 12.10.2.1, this is a good place where the test makers can trick you, especially in buildings afternoon.

 

[damascus]

Thanks for the response, follow up question:

If you have a 4 bay frame with 2 brace frames (both 20') and considered to have the same rigidity are the lateral loads split evenly between them. 

Therefore each would take 5k of the total 10k of the last problem. And if the frames are different lengths but same rigidity do they take on the respective force in respect to there length. 10' brace frame would take 3.33k and the 20' frame would take 6.67k?

Sorry if these questions are simple or obvious, I just never have been corrected if I have been doing this the wrong way. 

 

[Mithrandir918]

Thanks for the response, follow up question:

If you have a 4 bay frame with 2 brace frames (both 20') and considered to have the same rigidity are the lateral loads split evenly between them. 

Therefore each would take 5k of the total 10k of the last problem. And if the frames are different lengths but same rigidity do they take on the respective force in respect to there length. 10' brace frame would take 3.33k and the 20' frame would take 6.67k?

Sorry if these questions are simple or obvious, I just never have been corrected if I have been doing this the wrong way. 

Yes if they are both the concentric frames you have shown and the other two bays are not rigid moment frames then yes the load would get split evenly.

For rigid wall/frame systems the load is spit bases on the relative stiffness or relative rigidity or the systems resisting the load (R factor), for light framed wood walls you split the load based on the relative lengths of the walls.

 

[McEngr]

Is the brace a tension-only member?  If so, the drag load diagram is totally different.  It collects all the way to the far end of the braced bay, drops immediately, then tapers back up.