Plan check and redundency factor question

[shacked]

I just received plan check corrections for a two story light wood framed residential remodel that I did the engineering for located in Southern California.

Scope included removal of an existing shear wall at one location in the house, and elements of the vertical resisting system changed as well.

We have the as built Structural plans that were approved in 2001 and they clearly labeled all of the existing shear walls and their capacities. This falls under the 97 UBC

I addressed the removal of the existing shear wall by using the plans to determine the rated capacity of that SW and multiplying that by the SW's length to get the total lateral force along that line that needs to be resisted by 2 new shear walls. Unfortunately this creates a vertical type 4 irregularity. Therefore I designed the new floor beams for the uplift and compression forces to include the Omega factor.

One of the plan check corrections states that the Seismic design shall be based on the equivalent lateral force procedure using a redundancy factor rho = 1.3.

3 questions:

1) What are your thoughts on the design approach I used to account for the removal of the existing shear wall?

2) I know that in the 97 UBC they utilized a redundency factor but in order to determine that now I would have to perform a complete lateral analysis and vertical distribution of seismic forces which is what I am trying avoid by using the existing structural plans.

3) There is an exception in ASCE 7-10 (12.4.3.1) for the use of the Omega factor. For anyone who is familiar with this exception is this saying that I don't have to design the beams supporting the shear walls for the uplift/compression forces with omega = 2.5 as long as I design the beam for the maximum forces that can be developed in that shear wall sitting on top of the beam?

Thanks for your help.

 

[wannabeSE]

1. What are your thoughts on the design approach . . .
It seems reasonable but requires engineering judgment after reviewing the plans and having an understanding of the loads. However, this method may not meet building code requirements.

2. I know that in the 97 UBC they utilized a redundancy factor . . .
Plan check is asking for two things: determine the loads using Equivalent Lateral Force and use ρ = 1.3. Since they are asking for the building to be analyzed, I don't know what will be gained by using the 97 UBC ρ rather than using the current code.

3. There is an exception in ASCE 7-10 (12.4.3.1) for the use of the Omega factor . . .
The trick is determining the maximum force that can be developed. In some cases like foundation uplift, it might be straight forward. But, in other cases, the maximum force can be hard to quantify and justify. Also, it can be much greater than using Ω₀.

 

[jdgengineer]

1) The approach makes sense but I don't think it technically meets the code requirements. Any new lateral elements need to be designed to the current building code. Therefore designing the shearwalls to have am equal capacity as the walls they are replacing makes sure that the system is not weaker than before it does not guarantee that the new walls are designed for the full requirements of the new code. You are also trusting that the original design was executed correctly. Something that may or may not be true.

2) calculating the redundancy factor is not that challenging. For simplicity I usually default to 1.3. However, there is a prescriptive approach (need x number of walls at building exterior) and a calculated approach. Typically with the calculated approach you can tell by inspection whether you will meet the requirement or not. I believe there are two triggers 1) not reducing the story strength by a certain amount and 2) not creating a torsional irregularity. The story strength is relatively easy. If there are a lot of walls in the same direction this won't be an issue. The torsional irregularity takes some judgment with a flexible diaphragm if the walls is on a diaphragm boundary and is the only wall. Typically in these cases I consider the diaphragm "unstable" and use p = 1.3

3) I don't have the code in front of me but what you are saying sounds correct. However it is based on the "expected" strength not the allowable strength. A holdown may be rated for 3100# but in reality has a safety factor of 3 and is good for 10,000#. You'd have to use this value to ensure that the beam is not the fuse in the system.

 

[shacked]

Thanks for your inputs. After some thought I ended up just providing a complete lateral analysis and bring the loads down to that floor. Funny though, is that this approach produces an almost exact lateral load as the approach that I previously used.