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ASCE 7-16 12.2.3.2 Two-Stage Analysis Question

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MJC6125

Structural
Apr 9, 2017
120
I have a question about how exactly to analyze the lower portion when using the Two-stage seismic analysis procedure per ASCE 7-16 (section 12.2.3.2).

The way I see it, there are two possible ways to determine the seismic force distribution among the lower portion's levels and I'm not 100% sure which is correct. For an example case let's say you have a project that is 3 elevated levels of concrete slab (top level being a podium) w/ concrete shear walls and then 3 levels of wood framing (wood shearwalls) on top of that. 6 levels total.

For designing the lateral system on the lower levels do you use option 1 or 2 below:

1. Weight of upper portion is applied as a weight at the podium level. Then determine vertical distribution factors (Cvx) for all of the lower levels based on that weight being applied at podium level?

2. Weight of the upper portion is never explicitly included when determining the distribution of seismic forces in the lower levels. You determine lower level vertical distribution factors (Cvx) simply based on the weights of the concrete levels only. Then add in the seismic base shear of the upper levels at the podium as an amplified shear force (R/ρ of upper portion over R/ρ of lower portion).


I think option 2 is what the code says to do. But part of me thinks option 1 makes more sense for some reason. In the case I'm checking I get higher shear forces at the podium level when I do option 1 (more overturning & therefore more conservative for most likely everything besides diaphragm design at the lowest levels of the lower portion) The results weren't crazy far off from each other either way.
 
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Without studying the code more closely, I'd say I agree that #2 is the intended approach according to the language they used, strange as it may seem.
 
It's #2. Seismic forces are not external, they're internal inertial forces generated by ground movement. So the wood upper levels are very flexible and absorb a lot of the energy as they move - not as much is left to be felt as an internal force at the base (transition from wood to concrete). The base shear of the upper level should be what the analysis says it is, and that's the force that needs to be resisted at the top of the podium. Then the podium has its own mass and has its own inertial forces to which you add the inertial forces you already calculated for the upper levels.

Think of it this way: when you design a 4 story wood building on a spread footing, do you design the wood for R=6.5 and then switch to R=2 just because it's sitting on a CMU foundation wall?


Still #2, and my logic was sound, but my code interpretation was a bit off - my apologies.
 
I agree that it's number two but it's not only the upper structure base shear that gets added to the low structure. It's also the upper structure base moments. If you're doing that, option two should usually produce greater overturning in the low structure.

It makes sense that option one would produce the higher podium shear since you're taking the upper structure mass and pretending that it vibrates at the higher frequency of the lower structure. This would be conservative if, and only if, you're still applying the upper structure base moments.
 

Option 2 but just want to remind some points with excerpts from the standard..


12.2.3.2 Two-Stage Analysis Procedure
....e. The upper portion is analyzed with the equivalent lateral force or modal response spectrum procedure, and the lower
portion is analyzed with the equivalent lateral force procedure.



C12.2.3.2 Two-Stage Analysis Procedure.
....An example would be a concrete podium under a wood or steel-framed upper portion of a structure. The upper portion
may be analyzed for seismic forces and drifts using the values of R, Ω0, and Cd for the upper portion as a separate structure. The
seismic forces (e.g., shear and overturning) at the base of the upper portion are applied to the top of the lower portion and
scaled up by the ratio of ðR∕ρÞ upper to ðR∕ρÞ lower. The lower portion, which now includes the seismic forces from the upper
portion, may then be analyzed using the values of R, Ω0, and Cd for the lower portion of the structure.


IMO, the standard is clear enough..
 
Thank you for the responses. It does seem clear enough what to do. And I think I'm understanding the why behind it. Also, I neglected to mention it, but I know you have to for sure include the overturning forces too.

One final question/example:

If you have a building that is just the lower portion (3 elevated levels lets say (levels 2 - 4)), and there is no wood framed portion above. This will give you some story shears at levels 2 & 3. Now if you have a different building with the same exact lower portion plus a wood framed upper portion, using the 2-stage lateral analysis you will get the same exact story shears at levels 2 & 3 as the first building. Why is that the case? Is it that the lower levels of those two buildings will accelerate back and forth the exact same amount and seismic force = mass x acceleration? Or the mode shape (displacement pattern) of both buildings will be the same in that area, so the forces therefore will be the same?
 

No.. not really.. For the second case you are expected to analyze the rigid lower portion (with ELF ) with adding the effects of upper flexible portion.

That is , for the second case, V = Cs X W + V upper portion .

A two-stage equivalent lateral force procedure may be used for structures that have a flexible upper portion above a rigid lower
portion. Literally , the governing modes of two portions too far .

I will suggest you to perform RSA for the whole structure to see the effective modes for each portions. In your case, 3 storey rigid basement+ 3 storey flexible wood framing,I guess almost 80-90 % of timber storey masses will participate to 1st mode while the basement storeys will participate at higher modes ( say 10 th mode and above for 3D analysis ) having short period..
 
MJC said:
Is it that the lower levels of those two buildings will accelerate back and forth the exact same amount and seismic force = mass x acceleration? Or the mode shape (displacement pattern) of both buildings will be the same in that area, so the forces therefore will be the same?

That's right. We're overestimating the acceleration of each floor by not accounting for the period elongation that we"d get if the flexibility of the the structure as a whole were taken into account. The disparate stiffnesses of the upper and lower floors is what makes this approximation reasonable. The seismic motion of each portion of the structure is quasi-agnostic to the presence of the other.
 
MJC6125 said:
Now if you have a different building with the same exact lower portion plus a wood framed upper portion, using the 2-stage lateral analysis you will get the same exact story shears at levels 2 & 3 as the first building.

No you wont. You'll get the same shears, plus the entire "base shear" from the upper portion multiplied by the ratio of redundancy and response modification factors. You analyze the upper portion as though it's its own building. You get a whole bunch of reactions - usually various horizontal and vertical reactions from a wood framed building. You multiply all those by the R & rho ratios if it's greater than 1. You analyze the lower building with all those extra loads on top.

 
OP said:
This will give you some story shears at levels 2 & 3.

I suspect that there's a slight terminology problem there that's causing confusion.

Story shear = aggregate shear in the lateral system for most folks. This would not be identical for the two hypothetical cases.

I'm guessing that OP really meant the seismic load to be applied at each diaphragm which would be identical for the two hypothetical cases.
 
My bad on the terminology. I did mean seismic load per story. This one from ASCE 7-16

Fp_pw4kas.jpg


Not cumulative seismic load.
 
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