Riddle-like question
Riddle-like question
(OP)
Hello! I have a question for you. I am not sure myself about the answer, but I have some analysis results that make me totally confused. Therefore, the question will look like a quiz-show question or a riddle. Here it goes:
- Let's consider two buildings with the same fundamental period, say 0.5s. One of them has 3 stories, the other has 6 six stories and they both have the same structural system type (say, moment resisting frames). In which of the buildings would we have larger interstorey drift ratios when the two buildings are subjected to the same ground motion accelerogram? Let's assume that the mass is uniformly distributed along the height.
- Let's consider two buildings with the same fundamental period, say 0.5s. One of them has 3 stories, the other has 6 six stories and they both have the same structural system type (say, moment resisting frames). In which of the buildings would we have larger interstorey drift ratios when the two buildings are subjected to the same ground motion accelerogram? Let's assume that the mass is uniformly distributed along the height.






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Professional Engineer (ME, NH, MA) Structural Engineer (IL)
American Concrete Industries
https://www.facebook.com/AmericanConcrete/
RE: Riddle-like question
Professional Engineer (ME, NH, MA) Structural Engineer (IL)
American Concrete Industries
https://www.facebook.com/AmericanConcrete/
RE: Riddle-like question
Now enlighten us as to how that is!
RE: Riddle-like question
Given natural period/frequency remains the same, the ratio of K/M or M/K must remain constant.
If we double the height, we double mass
To keep the same K/M ratio the 6-story building must be twice as stiff as the 3-storey building.
Since P=Kx and we have same fundamental forcing frequency (and thereby acceleration) our acceleration is the same.
3-Storey Building (Say P=weight of 3 storey building*spectral acceleration, K=lateral stiffness of 3-storey building)
P=K*x
x=P/K
6-Storey Building (Say P=weight of 3 storey building*spectral acceleration, so 2P=weight of 6 storey building*spectral acceleration, )
2*P=2K*x
x=P/K
Storey Drift Will Be the Same.
How'd I do?
Jeff
Pipe Stress Analysis Engineer
www.xceed-eng.com
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I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.
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I don't beleive that the base shear is the same for both buildings, since with a fundemntal period that is identical (equal spectral acceleration), the taller structure will have a greater base shear as it is heavier.
Jeff
Pipe Stress Analysis Engineer
www.xceed-eng.com
RE: Riddle-like question
RE: Riddle-like question
I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.
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But the densities of the buildings do not have to be constant. The shorter building must be heavier and/or less stiff. The shorter building may have mass distributed towards the roof. I would say that there is no reason to assume the same drift. I would not go any further than that.
--
JHG
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Are the drifts in the taller building always smaller?
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3 stories all stories mass 10 kip
6 stories all stories mass 10 kip
both assumed to have a period equal to 1 second
based on the force distribution from the equivalent lateral force procedure, the cumulative drifts are as follow to confirm the 1 second period with the rayleigh method:
3 story building 0.25in, 0.5in, 0.75in
6 story building .125in, 0.25in, 0.375in, 0.5in, 0.625in, 0.75 in
so the interstory drift of the 6 story building is half that of the 3 story building, and they have the same total drift.
This is all based on the very simplified ELF procedure of ASCE, and the quite simplified rayleigh method to estimate fundamental period. To say it could be a general rule from this, I'm not so sure.
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Thank you for your precious time. I ran time history analyses (linear and nonlinear) and they all showed that the drifts in the building with 6 stories are smaller.
Now I invite you all to find a single example/scenario in which the drift ratios would be larger for the taller building, and maybe we can all share a Nobel prize afterwards
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1) Subscript 1 = short building; subscript 2 = tall building
2) h2 = 2 x h1 [building heights]
3) Replace discrete floor buildings with vertically cantilevered beams with distributed mass m.
4) As is appropriate for short buildings, only consider shear deflection. No flexural straining.
5) F2 = 2 x F1 [total seismic load on each respective building]
6) K2 = 2 x K1 [building stiffness relationship required to produce equivalent periods (K/M ratio)]
7) (GA)2 = 4 x (GA)1 [ratio of beam shear stiffness per unit length based on #6 for each building]
8) Recognize that for a first mode only shear building, the peak story drift ratio (DR) occurs at the base of the building and is equal the the unit shear strain of the substitute beams (F/GA).
9) DR2/DR1 = ((F2 / (GA)2) / (F1 / (GA)1) = ((2 x F1 / 4 x (GA)1) / (F1 / (GA)1) = 0.5 = structSU10's result-ish
I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.
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1) Accept Jeff's clever proof that the roof level drifts are the same.
2) Recognize that if the the roof drifts are the same then the whole building drift profile for the taller building is a vertically scaled up copy of that for the shorter building.
3) Recognize that the inverse of drift profile slope = story drift ratio so #2 implies that the taller building has a lower drift ratio on all floors.
I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.
RE: Riddle-like question
The base shear coefficients would be the same but not the actual base shears.
I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.
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Yup.
The spectral acceleration will be the same. But you'll be accelerating twice the mass. Thus twice the base shear.
I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.
RE: Riddle-like question
The ratio of the total building masses does not have to be two, or any particular value, so long as the distribution of mass is uniform for both buildings. You could easily repeat my derivation for any ratio of building heights and masses. Or you could repeat it with a variable representing that ration to keep things uber-general.
I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.
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No worries. I'm a little quick on the draw sometimes.
I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.
RE: Riddle-like question
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Sure. The twos and fours just become other numbers accordingly.
Not possible based on the assumptions that I set forth: first mode, shear deflection only, uniform shear stiffness per unit length (just added that now). To the extent that those assumptions are violated with cantilever action, fixed base plates, etc things may change. That's the thing with rules: they only apply within the limits of when they apply.
I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.
RE: Riddle-like question
Here it goes:
Let us make the following definitions:
1) α = m1 / m2 = k1 / k2 (condition for equal fundamental periods)
where m1 is the uniformly distributed mass of building 1, m2 is the distributed mass of building 2.
2) β = h2 / h1
where h2 and h1 are the building heights.
Now, following KootK's formulation, if S is the spectral acceleration (equal for both buildings):
3) F1 = α m2 S
F2 = m2 S
4) since k1 = αk2 (by definition):
GA1 / h1 = α GA2 / (βh1)
therefore:
GA2 = βGA1 / α
5) DR2/DR1 = (αm2S * GA1 ) / (β GA1 * α m2 S) = 1/β = h1 / h2.
Besides the simplifications and limitations, I think this is sufficiently correct. Do you have any objection?
RE: Riddle-like question
I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.