ASCE 7-10 Wind Load - Cases 6

[nataliab]

Hi,
I'm beginner user of ASCE Minimum Design Loads.
After analysis of Wind Load Directional Procedure I've got some doubts concerning wind load cases.
Firstly, I thought, that Figure 27.4-1 shows basic Wind Load case, with windward and leeward side wind load for walls and roofs.
But then, I realized that 4 basic wind load cases that are shown on Fig. 27.4-8 should be analyzed and that they refer to Figure 27.4-1.
Could someone explain to me, which load cases should I use to analyze MWFRS? Are these five mentioned above separate load cases?
And what about wind load on roof structure in those basic 4 cases? It's not shown on schemes - does it mean, that those 4 wind load cases affect only walls?

I would appreciate any advice

 

[southard2]

Well I usually use the other procedure (The envelope procedure) but I'll take a stab at helping you. Figure 27.4-1 is what you use to determine the wind pressures PWX, PLX,PWY and PLY shown in figure 27.4-8. Figure 27.4-8 are the four load combinations you must consider when designing your MFWRS elements such as your diaphragm and your connections that have two loaded surfaces such as a joist to wall connection where you are transferring the wind shear through the seat of the joist and haven't elected to use blocking. For the vast majority of flexible diaphragms case I with both directions considered separately is going to control. Remember the torsion gets distributed to the front a real loaded walls so case II will rarely control. Case III won't control if you have a flexible diaphragm because it is essentially the same thing as Case I just with less loads. If you have a rigid diaphragm then the loads transfer to the shear walls based on stiffness and having the two directions acting simultaneously do matter because torsion and funky load distributions will be involved. And again case IV if its flexible then it probably doesn’t control simply because the loads are so much less. So usually it’s no big deal unless you are designing a building with only three shear walls and an open face.

At any rate the tricky part is actually the roof loading shown in figure 27.4-1. Notice that the windward side of the roof's pressure can be positive or negative. Often the negative wind pressure on the windward side of the roof will overwhelm the leeward sides suction. The result is that you have the net pressure on the walls projected surface going one way but the net pressure on the roof's project area is going the opposite way......and that doesn't make any since does it. Hence the purpose of note #9 on figure 27.4-1. If the net roof pressures go in the opposite direction as the wall net wind pressures they are to be neglected. Unless you are talking about metal building frames.......because that industry is cheap cheap cheap. The same problem exists with the other procedure as well but the minimum design load section 28.4-4 helps cover this.

To be honest with you the wind code needs to be simplified. It has become far to complex for simple low rise building design. And for tall structures heck they are just using the wind tunnel procedure anyway to develop their loads. The whole wind code was based primarily on metal buildings so how it applies to tall buildings is ......dunno. The cost of getting 10% more accurate far exceeds any benefit we will get from material savings. In fact I'm willing to bet it actually results in more overdesign. All the tinkering with the codes' accuracy and statistical precision or whatever, is making it so that to be like 10% more accurate we are going to increase the odds of engineering error five fold. Yes I realize we have computers but user error is going way up. And worse I think many have just said to heck with it and they just do whatever. If you are a stickler like I am you'll see things under designed all the time and it will upset you.

I would rather academics just INCREASE the factor of safety a bit and greatly simplify the wind code. I can tell you that workmanship in the field is going to make a much larger difference in whether the building stands up or not than how accurate our wind loading is. And our factors of safety, statistically speaking, are to account for material variances, loading variances, etc... Well they should consider a factor of safety for poor workmanship because believe me it is there. Many don’t' know this but the 0.85 in the typical 0.85 f'c in concrete design you always see is there to account hydraulic lift of the water in the mix after it has been poured. You'll even see the water rise in the testing cylinders.

I'm glad you are trying to learn how to apply the wind loadings correctly. Once you get your spreadsheets worked out and once you have applied it for a few YEARS you will get proficient at it. And you will have figured out by then all the things you missed before. So be diligent. Unfortunately, I find that most structural engineering offices hardly provide any real mentoring or training. So whether you become the guy who is applying the wind loads correct or the guy who just designs everything for 20 psf, or the guy who does nothing other than randomly selecting details will all depend on your character. I'm glad to see you care enough to really dig deep into it.


And sorry for any mis-spellings, typings, etc.......I'm in the running for worst writer ever so there you go.

John Southard, M.S., P.E.

http://www.pdhlibrary.com/

 

[wannabeSE]

STEP ONE: Look in appendix D to see if the building is exempt from the torsional load case.

Figure 27-4.8 shows 4 load cases (figure 6-9 in ASCE 7-05). But the diagram is a simplification because the wind load can be different depending on which side is windward and leeward (the wind loads may be different coming from the north than from the south). There can be 32 cases to consider.

Case 1 = 4 cases: Two cases in x direction and 2 cases in y direction: (Pwx1+Plx1); (Pwx2+Plx2); (Pwy1+Ply1); (Pwy2+Ply2)

Case 2 = 8 load cases: Similar to above except the number doubles because plus and minus eccentricities should be considered.

Case 3 = 4 case: There are 2 x-direction forces and two y-direction forces to consider so it ends up being 0.75(Pwx1+Plx1+Pwy1+Ply1); 0.75(Pwx2+Plx2+Pwy1+Ply1); 0.75(Pwx1+Plx1+Pwy2+Ply2); 0.75(Pwx2+Plx2+Pwy2+Ply2).

Case 4 = 16 cases: similar to case 3 except there are 4 case for each set to account four plus and minus eccentricities (+ex +ey), (+ex -ey), (-ex +ey), (-ex -ey)

Most times you can eliminate some load case by inspection. With regular shaped building the net load is the same when the wind comes from the north as the south. But, building with steps can create all 32 cases.

Note: I do not like figure 27-4.8. However, ASCE 7-10 appendix D provides some relief.

 

[RFreund]

Good replies.

Its interesting to note that for low sloped roofs the wind pressure on the roof is negative for both subcases (in table 27.4-1), which actually "helps" you (to a certain degree) for load cases that apply roof live or snow in combination with wind.

This is where you have to applaud Chicago... 20 psf and I'm not sure the last time a building blew over.

EIT

http://www.HowToEngineer.com

 

[nataliab]

I tried to get proper values of wind load pressure for torsional cases.
However I obtained quite big values and I don't know if its correct approach.
I assumed torsional loading pattern (for semirigid diaphragm).
I used moment & force equilibrium equations, in order to get PWX1 and PWX2.
Input data in this example are:
a=54 ft
b=54 ft
B=108 ft
e=16,2 ft
w=250 lb/ft
After some easy calculations I obtaind:
[highlight #FCE94F]PW2=25 lb/ft[/highlight]
[highlight #CC0000]PW2=475 lb/ft[/highlight]
Is it possible? Should I, in next step apply it to diaphragm?