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Simplified Static Procedure (UBC 1629.8.2 and 1630.2.3)

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**This is a new thread based on a string of responses
(only slightly off-topic) to Dennis Wish's post of July
10 entitled "Plwd: Rigid Diaphragm Analysis - Opinions
Wanted".  I feel these responses deserve their own thread,
and I have tried to summarize the main points.

My comments follow the snips, and are bracketed
by ** at each end.  These are my own opinions and
not necesarily those of my employer.

Mark Swingle, SE
Oakland, CA
Bruce Resnick wrote:

> It is for all of the complications outlined in your posting that we are
> currently sticking with the Simplified Static method as noted in my posting a
> week or so ago.  The forces are a bit high, but the analysis is still
> "flexible diaphragm" and relatively easy to figure.

Michael Cochran wrote:

> You still would have to perform the rigid diaphragm analysis would you not?
> Or if we are not checking deflection, then we don't know if the diaphragm is
> rigid, therefore we don't have to do the rigid diaphragm analysis?

Dennis Wish wrote:

> What confuses me is your reference to a simplified static approach. Where in
> the code does it allow the engineer to ignore the diaphragm deflection
> calculation, the need to design for torsion or the need to design shear
> elements based on rigidity?

**(MS) I believe you must still comply with 1630.6 and
determine relative rigidity of diaphragm with respect
to shear walls, as Dennis and others point out.  Dennis'
original post regarding diaphragm analysis of irregular
buildings with plywood diaps is still relevant to the
Simplified Static Procedure.
Dennis Wish wrote:

> It seems that the only simplified approach suggested in the code is for the
> calculation of Rho and Omega such as in a residence where it is not required
> unless you are dealing with mixed structural systems or transfering shear
> from a wall to a frame or columns below.

Gerard Madden wrote:

> The redundacy is not effected by the short period clause ... It still
> must be check and is a pain.

Bill Allen wrote:

> First of all, whether or not "redundancy factor" rarely becomes an issue,
> you still have to calculate it. This is one of my anxieties over the so
> called "Simplified" Static approach. If the Code authors really intended to
> provide a "Simplified" alternative, albeit with higher design forces, the
> method should have accounted for this redunancy factor so we would not have
> to determine if it was an issue or not.
> Secondly, the only point of being "well behaved" that I'm familiar with is
> related to the cap of Na. Big deal. Based on 1629.4.2, we still have to
> establish that rho is not greater than 1.0 (it doesn't say you don't have to
> calculate it) and we also have to be satisfied that the soil is not type SE
> or SF and that there are no vertical irregularities of type 1,4 or 5 (table
> 16-L) or plan irregularities of type 1 or 4 (table 16-M). I would be
> surprised if there are any multiple story wood framed residential "real
> world" structures that would qualify.
> So far, the two wood framed structures I have prepared calculations for did
> in fact have a calculated rho less than one, so maybe it will be a mute
> point once we get more familiar with the methodology.

**(MS) I believe rho still needs to be calculated
with Simplified Static Procedure.  If rho=1,
then you can limit Na to 1.1 if other criteria
apply per 1629.4.2 (as Bill points out).

Personally, I don't think rho is such a big deal.
I belive most wood-frame buildings with shear
walls will have rho =1.  But I will leave that to
a separate thread....
Bill Allen wrote:

> From what I've read, the only things that the simplified static method saves
> is looking at the fault maps to determine Na and the vertical redistribution
> calculation.

**(MS) I don't think there is a sectionthat voids 1630.1.1, so Na must still be
determined to find Ca.
Gerard Madden wrote:

> The simplified method is exactly the same as the normal method except
> that the following requirements not not be checked or preformed:
> SECTION 1630.1.2:       Modeling Requirements
> SECTION 1630.1.3:       P-Delta Effects
> SECTION 1630.2.1:       Design Base Shear (non-simplified) (still
> applies w/ simplified force)
> SECTION 1630.2.2:       Structure Period
> SECTION 1630.5: Vertical Distribution of Force (non-simplified) (still
> applies w/ simplified force)
> SECTION 1630.9: Drift
> SECTION 1630.10:        Story Drift Limitations
> SECTION 1631:           Dynamic Analysis
> and the Maximum Inelastic Drift (Delta M) shall be equal to .01*Story
> Height
> That's it. All other stuff still applies like redundancy, overstrength
> for collector design, and you design for 20% higher forces. I feel that
> if you design your buildings for a 20% higher force and your competitors
> spend an extra day or two doing calcs, your clients will gradually go
> elsewhere. I feel that engineering fees are nothing compared to
> construction costs. I recommend (not that I'm some authority) to do the
> extra calcs, save your client some money, and you'll also know the code
> a little better. If your buildings cost 20% more than the competition,
> you'll lose.

Alfonso S. Quilala Jr. wrote:

> Providing extra 20% higher force doesn't translate to 20% more costs.
> Sometimes doubling the force will not even increase the costs by 10%.
> Assuming that after calculating using the long method you end up with shear
> equals 275 lbs/ft. providing shear panel with 360 lbs/ft capacity.  Then
> using the simplified method you will come up with 330 lbs/ft, you will then
> use the same 360 lbs/ft capacity.

Gerard Madden wrote:

> Your comments are true when designing a small building with small forces
> on good soil, but the higher forces apply to all elements in the LFRS.
> The 20% cost increase was just a correlation (not having much info on
> costs) but it would seem that the extra time spent in the design
> outweighs the construction cost increase.  I work mostly on tilt-up
> structures that are priced in cost per square foot of coverage in a
> general sense. If I'm designing diaphragms, wall panels, shearwalls,
> footings, collectors all to a 20% higher force level I think the cost
> can be significant. Don't forget that our forces are already swelling
> compared to the previous code and an additional 20% force level won't
> translate well to an established developer who bases the costs on
> present construction. If you are designing a house, I grant you the fact
> that the costs may not be that significant.

**(MS) Wonderful exchange.  I believe it depends
on the building.  Every building is different,
some buildings will be affected tremendously
by a 20% increase, and some will not.

After we get through the technical stuff, this
exchange is the real point: is it worth it to do
a "Simplified" procedure that is difficult to
interpret when there are so many consequences
that seem to be so poorly understood?**

Mark Swingle, SE
Oakland, CA