Need a book? Engineering books recommendations...

Return to index: [Subject] [Thread] [Date] [Author]

Re: ASCE 7 Gable Wind Loads

[Subject Prev][Subject Next][Thread Prev][Thread Next]
Brian:

I forgot to mention that all my numbers where based upon Exp B.
Presumably, going to Exp C or A or D would not significantly change the
relative differences but would change the actual Kh/Kz values.

Regards,

Scott
Adrian, MI


On Fri, 17 Mar 2006, Scott Maxwell wrote:

> --------------------------------------------------------------------------
> Your following message has been delivered to the list
>   seaint(--nospam--at)seaint.org at 03:16:49 on 17 Mar 2006.
> --------------------------------------------------------------------------
>
>
> Brian:
>
> One first thing to keep in mind is that the windward pressure determined
> per section 6.5.12.2.1 varies with height, while the windward pressure
> determine per section 6.5.12.2.2 will be based upon the roof's mean
> height.  Thus, the "average" wind pressure for the first case will be
> lower than the pressure used for the second case ASSUMING that you use the
> same velocity pressure exposure coefficients (Kz and Kh).  Note, however,
> that you DO NOT use the same Kz/Kh values for the two methods.  The second
> method requires the use of the "Case 1" Kz/Kh values while the first
> method requires the use of the "Case 2" values.  And the Case 2 values are
> LESS THAN the Case 1 values.  So, the end result is that while the GCp
> values might be lower for the second case, the Kh/Kz values will be larger
> for the second case.
>
> In addition, keep in mind that for the second method (6.5.12.2.2), the
> GCpf value changes (increases) at the ends of the walls, while this does
> not happen for the more "rigorous" method when doing MWFRS.  So, in
> reality your effective GCpf value that you should use for comparison
> purposes would be slightly higher than what you stated.
>
> Based upon your situation, I get this:
>
> For section 6.5.12.2.1 case:
>
> windward:
>
> G = 0.85
> Cp = 0.8
> Kz = 0.57 for 0-15 of height
>    = 0.62 for 20 ft of height
>
> ignore all other factors as they will be the same for both methods
>
> thus, "effective" or average Kz = +/- 0.576
>
> Thus, combined factors are about 0.85*0.8*0.576= 0.39
>
> leeward:
>
> G = 0.85
> Cp = -0.5 (this assume your L/B ratio is less than 1 but you did not give
> the other dimesion, but you implied this by saying you were using the .5
> value)
> Kh = 0.62
>
> Thus, combinded factors are about 0.85*-0.5*0.62= -0.26
>
> For section 6.5.12.2.2:
>
> windward:
>
> GCpf = 0.40 for most of the wall
>      = 0.61 for the last 8 ft at each end of the wall (this assumes that
> 192 ft is the least horizontal dimension)
> Kh = 0.70
>
> Thus, your effective GCpf is about 0.45
>
> Thus, your combined factors are about 0.45*0.7= .32
>
> Leeward:
>
> GCpf = -0.29 for most of the wall
>      = -0.43 for the last 8 ft at each end of the wall
> Kh = 0.70
>
> Thus, your effective GCpf is about -0.3
>
> Thus, your effective combined factors are about -0.3*0.7= -0.21
>
>
>
> So, the results for windward is about 0.39 for section 6.5.12.2.1 vs.
> 0.32 for section 6.5.12.2.2.  Thus, the second method is about 82% of the
> more rigorous method.  For leeward is it about -0.26 vs. -0.21.  This
> results in the less rigorous method being about 81% of the more rigorous
> method.  And both those are within in your stated tolerance of 80% to 90%.
>
> The point is that you need to look at the WHOLE picture.  You focused in
> on the differences between the obvious G/Cp/GCpf differences, but
> neglected the other subtile differences in those values as well as the
> difference between determining the Kh/Kz values in the two methods.
>
> HTH,
>
> Scott
> Adrian, MI
>
> On Thu, 16 Mar 2006, Brian S Bossley wrote:
>
> > I have a question about wind loads on a gable roof.  The building is 20' tall and 192' wide with a " / ft roof slope.  Using the method described in 6.5.12.2.1, I would have a windward Cp of 0.8, a leeward Cp of 0.5.  Using the method described in 6.5.12.2.2, I have a  windward GCpf of 0.4, and a leeward GCpf of 0.29.  This translates to a windward Cp of 0.47, and a leeward Cp of 0.34.
> >
> >
> >
> > Basically, this means that the lateral wind load is 62% of what it would have been had I used the tried and true formulas for every other type of building out there.
> >
> >
> >
> > Does anyone know why this is?  I wouldn't question if the lateral load was 80 to 90% of the other formulas, but when it gets this low, I tend to think I'm missing something.
> >
> >
> >
> > Brian S Bossley, PE
> >
> > Ventura Engineering
> >
> > 7610 Olentangy River Rd
> >
> > Columbus, OH 43235
> >
> >
> >
> >
>
> ******* ****** ******* ******** ******* ******* ******* ***
> *   Read list FAQ at: http://www.seaint.org/list_FAQ.asp
> *
> *   This email was sent to you via Structural Engineers
> *   Association of Southern California (SEAOSC) server. To
> *   subscribe (no fee) or UnSubscribe, please go to:
> *
> *   http://www.seaint.org/sealist1.asp
> *
> *   Questions to seaint-ad(--nospam--at)seaint.org. Remember, any email you
> *   send to the list is public domain and may be re-posted
> *   without your permission. Make sure you visit our web
> *   site at: http://www.seaint.org
> ******* ****** ****** ****** ******* ****** ****** ********
>
>

******* ****** ******* ******** ******* ******* ******* ***
*   Read list FAQ at: http://www.seaint.org/list_FAQ.asp
* 
*   This email was sent to you via Structural Engineers 
*   Association of Southern California (SEAOSC) server. To 
*   subscribe (no fee) or UnSubscribe, please go to:
*
*   http://www.seaint.org/sealist1.asp
*
*   Questions to seaint-ad(--nospam--at)seaint.org. Remember, any email you 
*   send to the list is public domain and may be re-posted 
*   without your permission. Make sure you visit our web 
*   site at: http://www.seaint.org 
******* ****** ****** ****** ******* ****** ****** ********