ASCE 7-16 Sect. 29.4 Trussed Towers

[RattlinBog]

I've read the other threads on eng-tips about this but could use some clarity on a couple things. Attached is an example I put together quickly--it's not a real structure.

1. When using Fig. 29.4-3 in ASCE 7-16, why does a larger ε=(solid area/gross area) ratio lead to a smaller Cf (and therefore smaller design force, F)? For example: ε=0.15 --> Cf=3.2 but ε=0.5 --> Cf=2.05. A larger ε means there's more solid area facing the wind. For some reason, I was thinking more solid area would lead to a larger wind force...

2. After you calculate the design wind force, F, how do you typically distribute that to the structure? Do you just take F and divide it and apply it equally at each panel point?

I'm probably missing something simple. Thanks

 

[straub46]

The coefficient is smaller but you'll still get a higher wind load since you're multiplying Cf * A. Note 1 in Table 29.5-3 says to use the solid area of the tower face projected on the plane, not the gross projected area. That should significantly reduce your wind load. The 113 kip wind load you have is equivalent to 71 psf projected over the entire area of the tower. That's higher than a solid building would have.

I usually add the wind load to each member locally. Taking the total lengths of members and dividing the total load by that, then putting in a distributed plf load on each member. If I were designing by hand, I'd put it at the panel points.

Go Bucks!

 

[RattlinBog]

Thank you--that makes a lot more sense. I noticed the 71 psf too, which seemed insane. I was expecting <30 psf.

I don't have ASCE 7-16 in front of me right now, but I think the definition of Af under the F = qz*G*Cf*Af equation tripped me up. I believe it's defined as the "projected area normal to the wind except where Cf is specified for the actual surface area." That "except" part of the definition didn't make sense to me at the time.

Now that we're straightened out, I think my example would be F = (26 psf)(0.85)(3.2)(235 ft^2) = 16.6 k. F/Agross = 16,600 lbs / 1600 ft^2 = 10.4 psf. Seems right?

 

[straub46]

The 16.6 kips sounds way more reasonable for the tower you have. And going back to your earlier example, if you had a 50% solid area ratio, the wind load is 36.2 kip, even with the lower Cf.

Go Bucks!