Rigid Diaphragm and Bracing Loads 1

[Fatalxception]

I have a 3 story building that uses the same steel brace size and configuration for it's lateral resistance(wind) from top to bottom, and has conc. floor diaphragms.

On the 3rd floor there is for example 3 braces (one on each end of the bldg. and one in the middle of the building).

On the 2nd floor, there is one at each end, one in the middle, and one that is added half way between the middle and one of the sides (say 30 feet from the middle brace).

At the first floor there are 2 at each end, one in the middle and one on each side of the middle (this time 20 feet away from the middle brace).

My question is will the diaphragm be able to distribute the accumulating force between all the braces equally on each floor or will the brace (for example the middle brace)which is lined up one on top of another, accumulate more force than the others as it would in a non rigid diaphragm?

 

[whyun]

It is not accurate to assume that the force that gets delivered to the brace at the upper level will stay in that brace and transfer to the brace immediately below. The traditional approach (hand-calc approach) of analysing multiple story rigid diaphragm structure considering each level as a one story structure, stacking them up and adding up the forces does not reflect the true structural behavior.

Portion of the force at the base of the 3rd floor exterior brace for example will transfer thru the diaphragm and delivered to the intermediate brace line at the lower level. Diaphragm must be checked for the uniform lateral load plus this additional shear.

 

[Fatalxception]

So the brace at the bottom floor (in the middle) which has two floors of braces directly on top of it, with floor beams, can be safely designed as if it is going to equally share all the second and third floor loads with the other braces on the botttom floor? Because of the diaphragm distribution?

 

[JAE]

I think whyun is right. When in doubt - always look at Hooke's Law: Deflection and Force are related to each other directly.

Once you calculate your three forces in each brace between the third and second floor, these forces are applied at the locations of each of those three braces on the second floor diaphragm. The added lateral forces applied due to the mass of the second floor diaphragm creates a new "center of force" applied to the second floor. This totality of forces is resisted by the cummulation of the braces below, in accordance with their relative rigidities.

A floor diaphragm is pretty stiff compared to the braces - in fact, you can imagine the entire "stiff" floor deflecting laterally with minimal horizontal bowing...although there will be some. Thus, all four braces below the second level will deflect the same.....
and Hooke's Law - equal brace and equal deflection = equal force.

Now this doesn't take into account the possibility of center of mass not equalling the center of rigidity and the resulting torque on the diaphragm. That should always be included.

But I wouldn't simply direct all the upper brace forces into the lower braces that are directly aligned below. This ignores the fact that some of the force will be distributed across the diaphragm and the braces, in reality, are all working together, deflecting pretty much the same.

 

[whyun]

Thank you, JAE, for the additional comments. Our methods of analysis (on anything) are based on certain assumptions and we try to attempt to portray reality as accurately as possible in the shortest amount of time possible (I am a consulting engineer and not a researcher).

Often the assumptions include something being infinitely rigid or flexible, which often is not true. So regardless of which method is used to derive at the forces, they are merely our best "prediction" of the forces in the members. As to how close they are to reality depends on the method selected.

Traditional method for load distribution is only accurate enough for a very regular buildings with stacking braces. For multi-level rigid diaphragm buildings with irregular lateral system layout, hand calc may be done using cummulative center of mass and center of rigidity starting at the roof level and going down the structure. The result from this will still be a bit different if compared to computer output using programs like ETABS or SAP.

We can analyse something to death but if your loading is not accurate, all computer generated output is garbage.

 

[JAE]

Consulting Engineer:
Work, think, fret, worry, struggle, race, hurry, frantic, etc.

Researcher:
Show up, continue process, cup-o-java, meetings, 8 hr. days, "what budget?", etc.


Perhaps the above is like whyun's statement: either infinitely rigid or infinitely free.

 

[whyun]

RIGHT ON, JAE!

 

[trainguy]

I second that JAE...