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lastrally
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| Joined: | Mon Sep 1st, 2008 |
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Posted: Tue Sep 2nd, 2008 03:14 am |
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Hi all,
I'm looking for help in understanding diaphragm calcs. I understand the basics but Kaplan has me thoroughly confused. For WIND design, Kaplan goes through the solving of Base Reaction and Roof Reaction at a parapet and the calcs just don't look right. (see attached)
Can anyone shed some light? I'm testing for LF on Friday....!
Thanks!
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londoner
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| Joined: | Sun Jul 6th, 2008 |
| Location: | London, United Kingdom |
| Posts: | 44 |
| Exams Taken: | GS, LF, CD | | Exams Passed: | | | Describes Me: | determined |
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Posted: Wed Sep 10th, 2008 01:29 pm |
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I am totally confused as well by the Wind Design chapter. My eyes go crossed and I just don't understand how we're supposed to understand that. I'd be interested to hear if anyone can shed some light on this.
Marc's course went over wind design and he uses a simple formula for calculating wind pressure P=Qs x Cq x Ce x I and I thought that was it. Then I read Kaplan. Totally lost!
Qs= .00256V^2
Cq=sum of the structural parts
Ce= height, exposure and gust factor
I= importance factor
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Coach
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| Joined: | Tue Mar 4th, 2008 |
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Posted: Sat Sep 13th, 2008 11:43 pm |
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| I can't read the diagram and it's been ages since I've done calcs, but see if turning it clockwise and thinking of it as a beam with a cantilever helps.
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FLA 3000
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| Joined: | Fri May 2nd, 2008 |
| Location: | Fort Lauderdale, Florida USA |
| Posts: | 9 |
| Exams Taken: | PD, GS, LF, ME, BD/MM, CD, SP, BP, BT | | Exams Passed: | PD, GS, LF, ME, BD/MM, CD, SP, BP, BT | | Describes Me: | ARCHITECT |
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Posted: Sat Sep 20th, 2008 12:29 am |
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| The cantilever beam comparison is quite helpful. The equation that is circled is the moment equation taken about the base to solve for the reaction at the roof. Then, by the looks of it, the building is 150' wide at the windward side. So the reaction of 165 was multiplied by 150' and divided by 2 to give you the reaction to each of 2 shear walls. The circled equation at the bottom is just the equilibrium equation the give the reaction at grade. Although you have probably already figured this out, This has been a good exercise for me. I have my LF exam tomorrow morning, and for some reason have not studied a parapet situation like this. The problems I studied were solvable by using the tributary area. So, thank you.
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mango61
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Posted: Thu Sep 25th, 2008 04:40 am |
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Yeah ,I agree kaplan doesnt even refer to that wind calc formula. However, my archiflash does and so does a 10 yr old Ballast book as well. is this formula outdated,
or are the boys at Kaplan going way to deep once again. The best is, in both their GS and LF books they must quote about a dozen times," These calculations are far to in depth that one should expect to see on the exam". HELLO!!! Stick to he meat and potatoes people. Dont they realize we hang onto to every strand of info that will possible put us over the edge.
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