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BFstr (Structural) (OP)
12 Aug 09 9:30
Hi everyone,

I am working on wind load for a CMU bldg which the wall height is 24ft including 2ft parapet. I started using Fig. 6-3 page 42 ASCE7-05 Components and Cladding - Method 1.

The building footprint is 42ft by 48ft. Wind speed 90MPH.
Can any one please tell me what is "Effective wind area" that I have to go on this table?  10 or 20 or 50 or 100?

What if the effective wind area of building doesn't match with the above given values in the table 6-3 page 42 ASCE 7-05?

Also a *?@# question, why they call this  Figure 6-3 isn't it a Table?

I appreciate your response.


Thanks,
BFstr.  
enginerding (Structural)
12 Aug 09 9:35
From ASCE 7-05: "The area used to determine GCp.  For component and cladding elements, the effective wind area in Figs. 6-11 through 6-17 and Fig. 6-19 is the span length multiplied by an effective width that need not be less than one-third the span length.  For cladding fasteners, the effective wind area shall not be greater than the area that is tributary to an individual fastener."
JAE (Structural)
12 Aug 09 10:15
I agree with enginerding.

 
BFstr (Structural) (OP)
12 Aug 09 10:20

 Dear enginerding (Structural),

Thanks for your response, I have seen this definition in code but I don't know how to apply the definition in my
case and also I am talking about Figure 6-3 which is forcing you to pick one of the given Effective wind areas. What if the number that I come up doesn't match with the given ones in Figure 6-3?

May I ask anyone to help me out determining the correct value for Effective wind area in my building?

Is it possible effective wind area becomes a different value for roof compare with the wall?

I appreciate any feed back from anyone.

Regards,
BFstr
Helpful Member!  ChipB (Structural)
12 Aug 09 11:12
Personally, I don't design CMU buildings as "Components and Cladding" as I think of the CMU as the MWFRS.  Your horizontal reinforcing is going to aid in the distribution, even if it is laddur @ 16".

Perhaps I've been doing it wrong?

If I were to design under components and cladding, I would use 100 for the walls (100<(22)^2/3<500 => use 100)  

On smaller jobs, which is what I would consider a 42x48 CMU building, I've always gone quick and dirty way and haven't plunged into ASCE 7 (unless I really needed to):
Ht of bldg: 24'
Ht of Parapet: 2'
Wind Speed: 90mph
Shape factor of Parapet: 2.0
Basic wind pressure:
  0.00256 * (v)^2 * (H/33)^(2/7)=
  0.00256 * (90)^2 * (24/33)^(2/7)= 18.9psf => use 20psf for design

Then I find the moment in the wall and design the vertical reinforcing steel around it:
M= 1200 ft-lbs (1170.32)
Reaction at bottom of wall (Rb): 216.36 lbs
Reaction at top of wall(Rt): 303.64 lbs
(See attached: Output results are in pounds not kips)
Don't forget to check the interaction of the axial load with the bending of the wind load.

Then I figure out the maximum tension in the bond beam at the top of the wall:

W=1.6*Rt = 1.6(303.64) = 485.8 plf =>use 490plf

M=(WL^2)/8
where L=Maximum length of wall (48')

Tension (F)=M/D
where D=distance between top & bottom chords(12" block assumed)
D=44'-(1'/2)-(1'/2)=43'

Lately, I have found that seismic tends to control in my area.  (Since the code has changed)
Helpful Member!(2)  DaveAtkins (Structural)
12 Aug 09 12:39
When the CMU is acting as a shear wall, it is MWFRS.  For wind perpendicular to the wall, it is C & C.

I get an effective area of 192 sf--you could interpolate between the values for 100 and 500 sf.

ChipB is doing MWFRS calcs and C & C calcs using the same wind load.  This is not correct.  The tension in the bond beam should be calculated using MWFRS loads, but the wall itself should be designed for C & C loads.

DaveAtkins

ChipB (Structural)
12 Aug 09 13:18
Thanks Dave! Now I know!

At least I go the conservative way versus liberal. (Politics not involved)

As I said, quick and dirty and out the door.

Just so you know, I went through the numbers back in 2003 (ASCE 7-02), and figured out that based on an Exposure Classification of C (which is what the 0.00256V^2(H/33)^2/7 is based on), once you run through all of the windward/leeward calculations vs the quick and dirty, for the design of the diaphragm chord, components and cladding, it was pretty much a wash.  

More indications that the codes are making it more and more difficult.  That formula is just too simple for us engineers to use.  We need something more difficult to understand just so we can justify our high (cough, cough, hack, hack) fees.

I don't know about the rest of you, but I became an engineer because I liked crunching numbers, not reading words.
BFstr (Structural) (OP)
12 Aug 09 13:26

Thank you so much ChipB (Structural) for explaining and attaching the nice drawing and calcs. I appreciate you nice reply.

And thank you very much DaveAtkins (Structural) for giving your valuable comment. I just have a quick question that might be so basic and that is the value you came up with 192 is it good for roof effective wind area and is a same value for the roof effective area that is for wall?
I appreciate to hear your comment.

Once again I think you both and all people that help a new person like me.

Regards,
BFstr
JAE (Structural)
12 Aug 09 18:53
BFstr,

Component and cladding wind loads are calculated separately for each individual component based on that component's exposed wind area.

So if you are designing a wall, you would use the tributary area of that wall to determine a wind pressure and then design the wall with that.

If you then need to design a roof purlin, you would calculate the tribuary area of that purlin and determine a separate wind pressure for it - then design the purlin.

 

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