20' HIGH STUD WALL TOWER 3

[youssef91]

I am working on 10'x15'x20' high stud wall tower with flat roof (min. slope for rain).
At one of the 10' sides there is an opng. 5'wide by 5'at 12'above slab on grade.
What I can use for lateral bracing of 20' high stud wall (5' long) . I suggested using a cantilever steel column but the architect does not like it. Any thought are appreciated

 

[FalsePrecision]

72199- Explain this some more. Right now, I see a relativley slender tower, with a 5 foot square opening in one side. Sheathe the whole thing in plywood/OSB, calculate the holdowns, check the shear capacity of the wall shear panel, etc.

 

[jjeng2]

If you are using steel studs you can use flat strap X-bracing. If you are using wood I agree with false precision. See the IBC for calculating shear walls of sheathing.

 

[youssef91]

Thanks Falseprecision and jjeng2you for the reply and let me explain more about the tower, two sides 15' wide by 20' high each has 3'x7' door and other two sides 10'wide x 20' high with 5 foot sq. opng. at only one side at elev. 12' above slab on grade.
My concern is the wall 5'wide x 20' high can't be used as a shear wall(aspect ratio 4:1), seismin zone 4.

Regards,

 

[FalsePrecision]

Sorry, still don't understand the layout of this. Where are you measuring 5' wide?

 

[UcfSE]

Remind the architect he isn't an engineer. ;-) If you are limited in your options due to your seismic zone there is only so much you can do. The 4:1 ratio sounds like a prescriptive requirement rather than an engineered one, but I am not familiar with seismic zone 4 stuff. If you need the steel columns and cannot get studs to work then so be it. You're an engineer not a magician. Aside from that, I would go with the flat straps if you have steel studs or some good sheathing and big mojo hold downs. A steel frame with bracing above the openings would probably work out instead of cantilevered columns.

 

[youssef91]

Thanks FalsePrecision and Ucfse for the reply.
The 10' x 20'high wall has 5' sq. window located at the edge of the wall, window sill is 12' above F.F.,so the window header is 3' below the roof. The remaining solid portion of the wall is 5'wide by 20'high which it does not meet the requirements for shear walls design.
Do you think lateral force at this wall can be resisted by any thing other than cantilever steel column or moment frame.?

 

[FalsePrecision]

72199-
What you have is a shearwall that is 20' high, but 10' wide for the lower 12' of its height, and 5' wide for 8' of its remaining height.
I am not convinced that the intepretation of its diaphragm ratio is 4:1. It gets wider at its lower portion.
I am assuming that you are in CA, and not familiar with the IBC concept of perforated shearwalls, which is a "dumbed-down" version of the APA TR157 report. Try Google, keywords "APA perforated shearwall design".
For added reassurance, put double the shear in the opposite parallel wall, and treat the whole thing like the horizontal diaphragm in rotation. (shear walls on 3 sides only). There is a requirement to check the ensuing vertical diaphragm (shear wall) deflection, but I haven't done that calculation in ages. I am sure it is in the APA website.
Any one who has all this automated on Excell would be welcome if could "trade" some Excell programming.

 

[youssef91]

Thanks Falseprecision for your recommendation to design the diaphragm on 3 shear walls.
I will deisgn the tower for the rotation of flexible diaphragm.

 

[nrguades]

hi 72199

I guess you have to check your diaphragm deflection with respect to the shearwall deflection in order to use the open front principle of rigid diphragm.Sometimes the diaphragm needs to be blocked in order to pass the criteria.
Calculate the center of rigidity and center of weight to properly distribute the direct shear and torsional moments and ensure proper connection for this values.

noel

 

[FalsePrecision]

I don't think this has anything to do with rigid diaphragm theory. Read the APA website info for confirmation.

 

[nrguades]

gentlemen...

With due respect to your discussion, just correct me if I am wrong with my interpretation. I believed what you are talking about shearwall on 3 sides belongs to rigid diaphragm principle.

From section 2305.2.5 (IBC 2003)

Design of structures with rigid diaphragms shall conform to the structure configuration requirement of section 9.5.2.3 of ASCE7.....

Open front structures with rigid diaphrams resulting in torsional force distribution are permitted provided the length L of the diaphragm normal to the open side does not exceed 25 feet......and the l/w ratio (as shown in figure-ibc 2003)... is less than 1.0 for one-storey structures or 0.67 for structures over one story in height.

From section 9.5.2.3.1 ASCE7-02

Diaphragm Flexibility.. A diaphragm shall be considered flexible for the purpose of distribution of story shear and torsional moment when the computed maximum in plane deflection of the diaphragm itself under lateral load is more than two times the average drift of adjoining vertical elements ......

72199...
If your not comportable with open front principle, try considering this...
Your Openning Height is only 12 ft.ryt? That means the shearwall aspect ratio is only 12/5 or 2.4. It is lesser than 3.5 limitation set by IBC 2003 table 2305.3.3. But it exceed the value of 2, so you have to multiply the resistance. by a value of 2w/h. (see ibc 2003 section 2305.3.3)
Therefore you can use it as shearwall provided that, you have to design and detail that shearwall considering force transfer around the opening.(simpson strap will do)
Just see section 2305.3.7.1 and some textbooks for more information about this principle...

I hope this will help

regards

nrguades

 

[nrguades]

72199...

Correction to my thread above, the opening height is only 5 ft... so the ratio is only 5/5 or 1..lesser than 2, so there's no need to multiply by 2w/h.
sorry for that..

nrguades

 

[FalsePrecision]

nrguades,
I don't care what anybody says about you, code-readers are blessed. The UBC codes didn't say anything about requirement for rigid diaphragm. The idea was to compute a torsional moment based on the moment arm as measured form the geometric centroid of the horizontal diaphragm to the shearwall taking the effect of direct shear. So I was correct per UBC, and thank you for bringing to everyone's atention the refinement presented in the IBC.

 

[SacreBleu]

Since this is a small wood structure, why is the horizontal diaphragm required to be rigid? If you compute defelections as shown in the IBC, the horizontal diaphragm can be flexible.

 

[nrguades]

Theoritically, I believed that, the check is necessary in order to ensure that the diaphragm can possibly handle rotation distribution. Try to imagine having a diaphragm having no stiffness as all, would it work in this case?
Bur your ryt, experience will tell us that, for smaller wood structure, deflection problem will not be critical in some cases, so it myt not be necessary to do some tedious calculations just to check the diaphragm actions.

nrguades

 

[SacreBleu]

Just because a diaphragm is wood, doesn't mean it has zero stiffness.

 

[nrguades]

Of corz it has, but I believed, that the code are just giving us a range where it is safe to use it for rotation.
Im sorry about the post above, Im actually not talking of wood having a zero stiffness, but im reffering it to some other materials which would have a very mimimal inherent amount of stiffness.

nrguades