Help with Wind uplift load combination etc 2

[SYLK]

I live in Southern California. I am trying to design an open steel frames structure with 3" pipe to place Solar Panels on top. It just have 2 bays and has 6 columns of pipe posts. It is elevated and has a 24^ mono-sloped roof. The wind speed is 110 mph and Exposure B.
I computed based on ASCE 7-16 that the velocity pressure qz = 15 psf and the wind pressure due to the direction of wind @ 0^: Cnw & Cnl for Load Cases A & B were -ve numbers and Direction 180^ were +ve numbers. Does that mean I would only get uplift when the wind is in Direction 0^?

See attached sketch and work sheets.

Is the design load combination for downward force 0.9 DL + Wind? Is the uplift load combination purely the -ve pressure I computed? Someone did suggest I add factored DL to the uplift which I think that doesn't make sense since DL is always acting downward.

Since wind forces act perpendicular to the slope surface, when resolving to vertical force, the resolved vertical force should be greater than the wind normal force, is that correct? Someone also said the vertical load is smaller.

 

[SYLK]

Look like no one knows the answers! A disappointment.

 

[Aesur]

Give it some time, sometimes the forum gets very busy and other posts show up before these. It's good that you responded which brought this back to the top of the list, for others to see.

A few comments on your base wind calculation:
1. When checking C&C and MWFRS, there are two separate values of Kz for exposure B and 15' height (default min height typically). I get 0.57 MWFRS and 0.7 C&C which gives 15.13 psf and 18.43 psf.
2. Double check that you truly are exposure B, many engineers misunderstand B versus C and in my experience most projects designed as B really should be designed as C. ASCE7-16 commentary does a great job explaining open patches etc. now days (older codes was not as easy to understand this). The reason I say this is because typically a canopy or solar structure would be placed in a more open area or parking lot, because I don't know more specifics of your project I cannot comment further on this.
3. If you can, I suggest looking into ground elevation (Ke) factor, it helps a good bit as the air density decreases with elevation.

Wind on open structure:
1. You have C&C and MWFRS to deal with, which are different portions of ASCE. Typically the C&C would apply to the panels, panel connections and purlins, below that I would start looking at MWFRS, however one could argue that C&C applies to even the girders in this instance as there is a note elsewhere in the code that says when a member is greater than 700 sq-ft then you can use MWFRS (30.2.3), these members are less than 700. Use your engineering judgement here.
2. Based on the slope you have given, 24 degrees, yes using the 0 degree direction I get uplift only for both case A, B, Clear and Obstructed. For 180 degrees you get a mix of pressure and uplift depending on Case and Obstruction.
3. Please note that you should be checking the members for both C&C and MWFRS (worst of) where C&C applies.
4. Depending on if you are using ASD or LRFD the load combinations vary, let's assume you are using LRFD (since your combination above is that). For downward you have 1.4D, 1.2D+1.6Lr+0.5W, 1.2D+1.0W+0.5Lr; for uplift you have 0.9D+1.0W. I would use the panel dead load as you know the weight and it's there, however it's usually less than 3 psf, so doesn't help all that much.
5. I'm not sure of your seismic requirements, but you should consider these being in CA.
6. You won't want to hear this as I have seen many engineers ignore this, but there are roof live load requirements for solar, depending on conditions, take a read of IBC section 1607.13.5. I believe CA has updated codes, so this may have moved, but if you search for "photovoltaic" in chapter 16 of IBC you should find it. From what I recall, if there isn't a deck below the panels, you can design for 12 psf roof live.

I have included a PDF showing the psf results I get for the system you describe for your comparison. I took a guess on C&C trib areas using 1/3L*L for the purlins and 1/4 of a 3'x5' panel for panel connections.

 

[goodbusiness]

I am not super familiar with American codes, but Canadian is similar. You have 4 wind load combinations to check (technically 8, but 4 can be omitted due to judgement):

0.9 DL + 1.0 WL (0°, Case A)
0.9 DL + 1.0 WL (0°, Case B)
1.2 DL + 1.0 WL (180°, Case A)
1.2 DL + 1.0 WL (180°, Case B)

You do not mix & match your wind cases; A or B, not both. You do all of them because you need to know what your worst effects are for both uplift and downward wind load combinations.

Pressures are normal to the surface, so -ve coefficients are indeed uplift in your case. Vertical and horizontal components are dependent on the angle of the slope of course. Vertical component is always smaller then the resultant, this is simple trig.

You do indeed add the factored DL to your uplift cases, but its factor is 0.9 which is a reduction; DL still acts to resist uplift.

EDIT: There are actually way more load combinations with wind, but for simplicity and discussion, I only talked about wind as the principal load and ignored all the other companion loads.

 

[SYLK]

Thanks Aesur & goodbusiness for the responses. Appreciate your comments.
I graduated 25 years ago with good grades in Structural courses but never work on any Structural project at work. Past couple of months having been reviewing my books and from youtube.
I refer to a page from my Timber Design book by our Calpoly Professor, Donald Breyer (see attachment earlier with my questions). The resolution of normal forces on a sloping member to vertical components. The vertical component force was bigger. In the resolution triangle, shouldn't the vertical component be the hypotenuse, and therefore larger?

 

[Tomfh]

shouldn't the vertical component be the hypotenuse, and therefore larger?

No.

 

[goodbusiness]

I refer to a page from my Timber Design book by our Calpoly Professor, Donald Breyer (see attachment earlier with my questions). The resolution of normal forces on a sloping member to vertical components.

I believe your confusion is coming from what you are trying to find. In your example, you are trying to determine the forces acting on a sloped member. Dead load and snow load always act vertically, but you need to resolve your forces to act normal and parallel to your member being analyzed to determine moment, shear, and axial reactions. So your triangle is oriented to the member (and that's why your resultant/hypotenuse is vertical). However, in your wind load calcs, wind load already acts normal to your member so finding your horizontal and vertical components mean your triangle is oriented according your basic x and y axis.

Basically, you need to pay attention to your frame of reference.

 

[SYLK]

Thanks goodbusiness for the explanation on my confusion on why resolving the normal force to vertical was larger. That really helps. Thanks again.

 

[SYLK]

Good morning Aesur,
Thanks for taking the trouble to run the data I provided. I looked at the computer generated outputs you provided, look like the interpolation from 22.5^ to 24^ were not correct. They all fall outside the range.
See attached and my hand computation
Indeed my site Exposure is B given by the AHJ's Building Official and the location is in an urban area all sides surrounded by buildings. Three sides are shielded by building taller than my proposed structure. The only unshielded area is Direction 180^ because the grade on that direction was lower. Frankly speaking, there is no wind from Direction 0^ as the grade are higher and farther North is a mountain.

 

[Aesur]

@SYLK - you are getting different values because I include my gust factor G (0.85) to the the final interpolation. I forget why I built it this way 10+years ago, but there was probably some reason I applied it here. If you divide my value by 0.85 it matches your interpolation.

 

[SYLK]

@ Aesur – In spite of seeing the number different in interpolations. I don’t have to multiply 0.85 again (as I already did with P = qhG Cn already) and the PSF load you obtained already matched my hand calculated numbers.SEE attached..
As a novice in this area, thanks for educating me. I only have access to ASCE Ch 26 & 27 and I could not find what is a, a^2, 4a^2 you used in C&C but you did provide the solution to them. Thank you.

I believe total length & width of solar panels should be used and not the length of the girdle (21’) supporting the solar panels by the rafters (purlins). I should have make it clear to you that the solar panels covered area is 22.7’ x 14 ‘ and not 21’x 9’. And “h”, the roof mean height is 10 feet not 15 feet.
Can I ask you a favor by re-running your program with these new numbers so that I can use your number to do the rest of the computation. Thanks

 

[SYLK]

@ Aesur - I researched and found what is "a, a^2 and 4a^2" I asked you earlier.
However, the sloping rafter is 14 feet long but in plan view the Horizontal projected length is 12.79 feet. So in plan view, the solar panel area is 22.5' x 12.79'. Thanks.

 

[SYLK]

@goodbusiness,
I have questions on using the wind pressures calculated from MWFRS and C&C.

Am I using the wind pressure from MWFRS to calculate the uplift and downward force? Assuming I am doing that, do I use the total horizontal roof area to calculate the total uplift or downward force and then because I have 6 columns, I divide the number by six to design one of the foundation?

 

[goodbusiness]

Sorry but I'm not very familiar with MWFRS and C&C because we don't use that precise a methodology in Canadian Code. I would assume since you are resolving loads to your foundation, you would use the MWFRS wind calcs because your columns would be part of your MWFRS. Personally, I would just use whatever gives your larger reactions.

You should also base your column loads on the tributary area around each column, so you would not just divide by six. Your interior columns should see more load than the outer columns.