lateral force restraint through out-of-plane bending

[Prestressed Guy]

I am designing a building that has a stiff flat roofed CMU box with multiple interior CMU shear walls. The adjoining bay is a higher sloped roof with almost zero lateral strength due to steel stud walls at each end with large openings. The sloped roof is long and slender with an aspect ratio of 4.5:1. The low roof is at 15’-6” and the high roof is at 20’-0”.

I am looking at dumping the lateral force into the common 8” CMU wall through out-of-plane bending. Is there any reason this would not be acceptable?

This project is an SDS Cat. D.

 

[COEngineeer]

If Grid 4' can take the bending and you put members to transfer the shear to the flat roof I dont see why not.

Never, but never question engineer's judgement

 

[Prestressed Guy]

I am surprised that I have only gotten one response to this post. Was it due to being asked over the weekend? Anyone care to comment on whether this is a good idea or not?

 

[haynewp]

Is there an R value for that?

 

[Prestressed Guy]

It would be most similar to an inverted pendulum system. That would indicate an R of between 1 and 2.5. These system's typically have very low redundancy whereas this wall will provide almost continuous support. I am planning to use 2.5

 

[haynewp]

I would use R=1.

 

[Structures33]

Can you give us some more information?
1. To clarify, are you proposing that the out of plane lateral force on the 30' high wall will travel through the sloped roof into the top of the 8" cmu common wall?

2. Will the walls of the adjoining bay (framed w/ metal studs) perpendicular to the 8" cmu common wall be able to take/transfer the in-plane shear?

3. What about the lateral force in line w/ the 8" common wall? How will that be resisted? With a flexible diaphragm you'll have a substantial amount of in-plane shear in the 30' high end wall.

 

[Prestressed Guy]

Question #1 Yes. I will be using the LH joists as collectors to transfer the seismic and wind loads to the top of the CMU. Did I say Omega o? ;>)

Question #2 No. The walls perpendicular to the CMU bearing wall at each end of the sloped roof are all overhead door. There are 12” piers with a 16’ x 16’ door. They will just go for the ride. As an added problem the sloped roof has an aspect ratio of 4.5:1 so it is too long and skinny to work as a diaphragm even if the end walls had capacity. The simple truth is that the load is going to the CUM wall on grid 4` regardless of how I try designing it because it is much stiffer. The wall to the adjoining bay is a clerestory wall with a high roof at 30’ and a low roof at 20’. You can’t get much better hinge.

Question #3 Grid 4` wall is designed as a Special Reinforced Masonry Shear Wall. It has loads of in-plane capacity. The end wall of the 3’-4’ bay will be balloon framed for out-of-plane loads but nothing for in-plane.

 

[Prestressed Guy]

haynew, how do you justify an R of 1. Ordinary plain concrete and URM are designed with 1.5.

 

[haynewp]

Plain concretee and URM are not even allowed in SDC C or worse. On second thought, I probably wouldn't even use out of plane bending on that cmu for lateral resistance in SDC D at all.

 

[Prestressed Guy]

I find it strange that you would not consider using this continuous wall taking a relatively light seismic load when no one would think twice about designing a cantilevered RC slab with a wall load on the end of the cantilever. It is strange how we can be so comfortable with continuous large load and be scared spitless about a load with a 2%-50yr.load.

My Cs is .268 W The roof dead load is 15 psf so 25’ x 14’ x 15psf x .268 = 1.41 kip per joist. With Over-Strength Factor of 2.5 the collector load is 3.517 out-of-plane at 14 foot spaces along the wall. I ran a trial design and it looks like #5 @ 24” each face will carry it. Am I missing something?

 

[structuralhawaii]

Even me I will use R=2.5..#5 @ 24" o.c. sounds good in approach of 2006 IBC ASD..

 

[Structures33]

I think the lateral transfer system needs to be evaluated/determined before you get into the design of the 8" cmu wall.

It sounds like you need to brace the roof in such a way that it will serve as a diaphragm - otherwise, how will the loads even get to the 8" cmu wall?

The balloon framed 30' high wall needs to have in-plane capacity because it will have shear load -- the stiffness of the cmu means nothing with a flexible diaphragm.

If the walls perpendicular to the 8" wall are all doors, it sounds like you need some portal frames in there or something.

 

[Prestressed Guy]

Here are my comments on Structures33's response. It would be nice if we could always live inside the box shown in the textbooks but sometimes the architect won't go along.

1. The type B roof deck will serve as a series of short span sub-diaphragms between the LH joists. These joists will be spec’ed for 3.5 kips axial load. The joists seats will be designed to transfer 3.5 kip collector axial load to the CMU. The wall will be designed for the full collector force which I consider very conservative. The omega o factor is to protect against brittle failure and a CMU wall subjected to out-of-plane bending is very ductile.
2. The 30’ tall wall on grid 3` is designed as a Building Frame – Ordinary Concentrically Braced Frame as allowed by footnote f (“Steel ordinary concentrically braced frames are permitted in single-story buildings up to a height of 60 ft (18.3 m) where the dead load of the roof does not exceed 20 psf (0.96 kN/m2) and in penthouse structures”).
3. Portal frames will not work because they are very flexible and the adjoining CMU building is very stiff. The roof has an aspect ratio of 4.5:1 between the end walls. The load always goes to the stiffest element so it is going to be dumped into the CMU walls and there is nothing that I can do to change that. I could design the stiffest portal frame I can imagine but the diaphragm is so flexible that very little load would get to it.