Wind load Uplift concept 1

[SYLK]

I need help with the wind load concept. I am a civil engineer but not working in building structures. I am designing an elevated open steel structure to install solar panels for my home. The structure is 13 feet tall and is mono slope roof at 29^. I used ASCE-16 and got a horizontal wind load of 21.68 psf and that translates into 24.79 psf normal to the solar panels. Is the uplift forces also 24.79 psf?
The structure is supported by 6 posts (2 rows 3 bays). Is the uplift resistance analysis based one column or all together? Is the uplift resistance force the tributary Dead load plus the weight of the foundation?
Thanks.
Also how would I consider whether seismic base shear or wind load is the critical load? Thanks

 

[ThomasH]

Regarding the wind load concept.

The wind load is due to a pressure difference between surfaces. The "surfaces" can be the front wall of a building and the back wall of the same building. The air flow causes "pressure" on the front and "suction" on the back. The "surfaces" can also be the two sides of your solar panels . And the pressure always acts perpendicular to the surface.

Perhaps some reasoning based on this can help you with the remaining questions.

 

[SYLK]

Thanks ThomasH for the education, I do understand the basic concept, but how to approach the uplift threw me off course.
I am designing and open mono-slope roof open steel frame structure for placing solar panels. Using ASCE7-16, I have computed wind pressure for both directions -0^ & 180^ for Load Cases A & B for Cnw & Cnl.
Is the largest -ve number indicate the uplift pressure from that specific direction and largest +ve number as the wind load on the roof?
And in analysis the uplift, do I just take one post/column into consideration or I have to take all the columns together? I guess , it is then comparing the total Dead load including weight of footings to the total wind load. Is that correct?

 

[sponton]

SYLK

you analyze the whole structure, wind is not uniformly distributed in the roof so your total maximum uplift load will change, so it depends on the direction of the framing, the wind direction, bdg dimensions.. etc

 

[SYLK]

Can you elaborate a little bit more?

I have 6 pipe posts in 2 bays configuration. There is no wall, just an open building. Wind would blow directly towards the sloppy roof or at 180^. The other 3 sides are blocked by buildings.
The wind pressures form Load Case A and B yield +ve and -ve numbers. Is the -ve number multiply by the tributary area be the uplift forces for the column ?

 

[Tomfh]

Provide a sketch.

As for your questions, the wind loads specified are the largest pressures (either positive, or negative) that you can expect on each surface. The structure must have sufficient capacity to resist these loads, with an acceptable factor of safety.

You are generally assessing uplift on a per column basis. So you’ll check your worst case column, and design for that.

You also need to check lateral loads, ie check it won’t blow over. This may well be critical.

As noted by ThomasH, wind loads are generally perpendicular to surfaces. For assessing uplift you are checking the vertical component of these forces. For lateral loading, the horizontal component.

 

[SYLK]

Thanks Tomft.
I believe I am on the right path but not too sure as I am assessing the middle column which carries the most tributary load.
I went through all the steps I learnt from Youtube in getting the wind pressure.

For 110 mph wind speeds I got qz of 15 psf. And finally for for 24^ roof slope, Load case B, I obtained the highest -30.86 psf and +29.07 psf.
Does that mean, the uplift force for that column would be resolving 30.86 psf into vertical component and multiply by the tributary area of the column?
See attached sketch and calculation I did. Thanks for your comments.

 

[Tomfh]

There’s a few mistakes. (Could be more)

In your 180 degrees diagram you tag the windward side as “Cpl”.

You are also resolving your vertical component incorrectly. You have drawn the triangle wrong. The vertical component of the normal force cannot exceed the normal force.

Also, you aren’t combining loads correctly. You need an uplift load case, in which you use the reduced gravity load, and a downwards case, in which you increase your gravity loads. You seem to have applied the uplift combination (0.9DL + Wind) to the downwards load case.

 

[SYLK]

Thanks Tom for reviewing my work.
I tried to digest what you said and here are my followup questions.
you said: In your 180 degrees diagram you tag the windward side as “Cpl”.
Are you saying I should applied separately the Cnl and Cnw as CPL (concentrated point loads) not their udls? I took the largest number, I should used both separately.
You said:
You need an uplift load case, in which you use the reduced gravity load, and a downwards case, in which you increase your gravity loads
What are the factored load combinations for uplift and downward loads? I only find 0.9 DL x Wind Load from books and nothing else said from anywhere?
Why would we add any DL to uplift load since the DL is always acting downward?
Last question:
I used our Professor Donald Breyer's Wood Structure Design Book to resolve the wind normal forces to vertical.

It vertical load is largest than that of normal forces.
That makes sense to me because the slope length is longer and we tried to covert the applied load to vertical on shorter horizontal span. Don't you think the vertical force should be greater so that the total load on the element added up to be the same. I maybe wrong. I attached the page from the book for your reference.

 

[Tomfh]

Are you saying I should applied separately the Cnl and Cnw as CPL (concentrated point loads) not their udls? I took the largest number, I should used both separately.

No, I am saying that you have the “cpw” and “cpl” labels back to front in the 180 degrees case. Cpw is the side the wind is coming from, ie the windward side. If you’re using the worst case it won’t make a difference.

What are the factored load combinations for uplift and downward loads? I only find 0.9 DL x Wind Load from books and nothing else said from anywhere?
Why would we add any DL to uplift load since the DL is always acting downward?

Because in some load cases the wind is pushing down on the roof, in which case you generally assume a larger dead load. Hence some of the pressures having positive numbers.

Typical combinations in Australia are 90% of dead load plus full wind load for wind uplift (ie you reduce dead load when it is working in your favour), and 120% of dead load plus full wind load for wind downwards action (ie you assume a larger dead load, when dead load is working against you).

My point here is that you seem to be mixing up your signs, and don’t seem to be clear on which way direction gravity and wind are pointing.

It vertical load is largest than that of normal forces.

That example is snow loads, which apply vertical loads to start with.

But you are starting with a normal force in the wind load case, in which case the vertical component of the force is less than the normal force.