C&C Loads on CMU Walls

[CBlaker114]

Hello All,

I'm running my checks for C&C loads on exterior CMU walls. When I go to find my applicable wind pressure, I'm wondering what tributary area applies when designing CMU... The majority of my previous experience is in wood and steel design, so please forgive me if this is rather elementary... but do you typically use the area of the entire wall, which in this case would exceed 700 sf and I could therefore use MWFRS wind pressures, or would you use the spacing of grouted cells x the story height.. or (story height)^2/3.. I'm essentially viewing the grouted cells as I would studs in a wood framed wall in this approach. Seems like CMU would distribute the load between grouted cells so I'd be able to use the full area of the wall... but I've got a little shoulder angel with lots of experience with wood framing giving me the side eye...so yeah..

Thanks all..

 

[mike20793]

I use the height of the wall squared divided by three. I believe the commentary has some good information about walls.

 

[jayrod12]

is it the end of the world to use the 10 sq ft C&C number?

I think the spacing of the grouted cells would be a good place to start.

 

[CBlaker114]

I'd say the 10 sq ft C&C number would be overly conservative. Floor height is 10.5' so the minimum I'd have to use is 10.5^2/3= 36 sq ft. Either way, seems like the both responses so far agree with my shoulder angel. Thanks fellas.

 

[sandman21]

It is not over 700sf and would not fall under MWFRS. Grout spacing by the height.

 

[wannabeSE]

For ASCE 7-10, see the definition of effective wind area in chapter 26. I use the same as Mike.

 

[DaveAtkins]

I also use height squared divided by three.

Unless designing a window or door jamb--then I use the actual tributary area.

Oh...and I typically ignore that comment in ASCE 7 about designing all fasteners for the tributary area to the fastener. If you follow that rule, all fasteners, connectors, etc. would need to be designed for the 10 SF load.

DaveAtkins

 

[UcfSE]

I use the minimum effective area, span*span/3.