SA52: Frame Analysis under Wind Load (Airplane Hangar)

Please, Where the coefficient: 0.45; -0.69; -0.42; -0.35, for Load case A (East-West)????. I found on the code at page 313. Figure 28.3-1, the coefficient should be: 0.4;-0.69; -0.37; -0.29, for roof angle 0-5 degree

How to make these type of videos. and whcih software you use. please guide. Thanks

Thank you Dr. structure for the comprehensive video explaining the application of the wind pressures calculated by ASCE 7 procedure which I could not find anywhere on the web or elsewhere. I have to say finding those required pressures is a piece of cake but how you applied those to the building is none existence. A lot of videos out there are elementary knowledge and have no value at all. I was stuck for months on that and now I can proceed!. Thanks.

thanks a lot for your kind information... can you give link asce7-10, 16 standard code,,,,,

For the life of me, I find it so unbelievable that the leeward side can ever experience anything close to even half the windward side.
I work on projects both here in the US and in Canada, and I'm constantly calculating Wind Loads (Windward WW) of Leeward W=.625WW, Parallel W=.71WW, Roof W=.585WW.
That just seems insane to me. Because I've been on the Leeward and parallel sides of buildings during high winds. I've never seen anything close to those kinds of loads.

Released what couldn't Learnt at college learnt here. Thank you very much

An explanation is more than wonderful, but if I want to use an artificial intelligence network to predict the response of this structure to wind loads at different speeds, is there a Matlab code or an analysis program that can find the response of the structure by simply changing the parameters affecting the wind load and its speed

So detailed and very clear, thank you so much!

WHY USE METRIC UNITS ? AMERICA DOESN'T USE METRIC UNITS. QUIT TALKING METRIC WHILE THINKING IMPERIAL. NOBODY HAS THE FEEL OF KN/METER SQUARE.....EVERYBODY KNOWS LBS/SQUARE FEET. I DON'T HAVE TIME TO CONVERT AND THEN GET THE FEEL. KILO-NEWTON TO ME = 1000 ISSAC NEWTONS.

Excellent lecture.

I am so impressed with this video, the quality of work, and how concisely it it conveyed.

Thanks you sir , if we have a canopy or extension member how to design winload ?

0.534 * 58^2 is not the result of 1196.4 N/m^2.
It is 1796.4.

Excellent presentation

I love poeple doing quality works, it shows how much they respect themselves and the others, thanks!

Excellent. Thank you.

You are the best online professor Dr.Structure

The internal coefficient ci = 0.18 seems to be too low for hanger type structure. Should it be 0.55 instead (partially enclosed building, or, open front structure)? The difference is not insignificant. Please confirm. Thanks.