Questions on PPD

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    Daniel Brooks

    Hey Ronaldo, See the image on the bottom right

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    Ronaldo Guimaraes

    Thank you very much Daniel.

    Very helpful.

    Regards.

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    Ronaldo Guimaraes

    Here is another question that I ma looking for additional help from anyone on call...

    "A structural engineer has told you that the total pressure on the retaining wall you designed cannot exceed 1500 plf. How high can you make your wall? Assume the pressure is 30lbs/ft^3.

     
     
    Here is the answer: " the answer is 10
    P = ½ (30 lbf / ft^3) h^2
    1500 plf = ½ (30 lbf / ft^3) (x ft)^2
    x = 10 ft.
     
    I just couldn't figured out this math.
    I am little confused with this symbol "^" before #3 and #2. 
     
    Thanks.
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    David Kaplan

    Ronaldo,
    I have zero clue how to answer this question, but I AM certain that h^2 means "h squared" and "ft^3" means "feet cubed."

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    Jessica Deaver

    David is right.

    Here is how I did the math. Hope this helps and someone please correct me if I'm wrong.

     

    1500 = 1/2 (30) (h^2)

    1500 = 15 (h^2)

    Divide both sides by 15

    100 = h^2  and since 10 x 10 = 100

    h = 10

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    Kristina Blazevski-Charpentier (Edited )

    Reading this thread as I prepare for PPD, why is 1/2 first in determining solution?

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    Jessica Deaver

    Kristina,

    In the problem we are given the Plf or P for the equation.

    The unsolved value is h, so to find h we have to get it alone on that side of the equation. 1/2 x 30 gives us a single value that we can use to divide and bring to the other side of the equation. Then we can solve for h.

     

    Hope that helps.

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    Ronaldo Guimaraes

    Jessica and David.

    Thank you for the clarifications.

    I was wondering why we should use 1/2 first in determining solution.

    Based on Jessica's explanation it seemed logical, however I was just wondering if there would be another way to solve this problem.

    Thanks for all. 

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    Ronaldo Guimaraes

    Any hints & tips regarding heating and cooling systems?

    For all the study material I have been using it seems to be very broad and complex.

    Besides the technical approach of all systems available, they also relate to life-cycle and cost criteria.

    Any clue how to summarize all types in a very efficient and memorizing way?

    Another question is regarding delta & wye connections. There are so many variances and applications that is easy to be caught by surprise.

    I have used MEEB, PPI2PASS, Amber book, BS Arch Exam Prep and all other kinds of resources.

    Still struggling but, I am gonna get there, eventually.

    Any help is welcome.   

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    David Kaplan

    Ronaldo,

    For me, MEP systems was sufficiently and greatly summarized and explained well in the Architect's Studio Companion.  It was more or less the sole resource I used to study MEP systems for both PPD and PDD and it was great.  It includes an explanation of each type of system, the benefits/disadvantages of using them in a building, and has graphic depictions of the general setup and components of each system.  In the front of the MEP chapters, there is a really good summary page which I think gets into the "memorization" aspect you're looking for.  It has a list of different criteria - one example being "owner wants to minimize the first cost of the system" - and tells you which MEP systems best meet that.  It has other criteria such as "no ceiling plenum available" or "least amount of long-term maintenance."  I read those summary pages probably 10 times in studying for my exam.

    Same thing for structural systems too.  If you don't have it, I strongly recommend it.  It isn't listed in the NCARB Handbook resources, but it should be. 

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    Ronaldo Guimaraes

    Thank you David.

    Very helpful your directions.

    Regards.

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    Ronaldo Guimaraes

    Please can anyone explain the "privacy wall" being included on the typical net floor area occupant load calculation?

    Where did the 40sf come from?

    Any help, please

     

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    David Kaplan

    Ronaldo,

    That is a head scratcher, but I have two points to offer you here:

    1) I "think" perhaps that in this scenario, the privacy wall is being treated like a freestanding element in the room, thus, its square footage is being deducted from the NET number because it's not usable space.  No different than the fireplace - it too takes up a footprint that I can't occupy, so it's square footage is deducted.

    2) This being said, this question does not give you any way to calculate the square footage of that privacy wall.  You don't know its length.  I have no way to tell you how they came up with 40 SF.

    3) YOU WILL NOT HAVE THIS LEVEL OF CONFUSION ON THE ARE.  Period.  You will not be given a question where you find yourself not able to calculate the square footage of something.  Also, the question would give you information about the privacy wall, including its dimensions, possibly its height as well.  You would be able to discern the fact that it is an element in the space the footprint of which needs to be calculated and deducted, because you can't occupy its footprint.

    Chock this one up to a poor example on wherever you got this diagram from.  It's not you - it's them.

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    Derek Mason (Edited )

    The wall is 80'-0" long and 6" wide. 

    80 x .5 = 40'-0"

    Why? Read what David said. 

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    Ronaldo Guimaraes

    Thank you David and Derek.

    This figure was taken from the IBC 2018 Code and Commentary, which caused me a big surprise to see such an example with a confusing explanation. Regarding the wall square footage I got it but why it wasn't also considered the cloak room walls? Any way, I am just overthinking which is not appropriated for the purpose I am in now, pass the ARE exams., which David brought to very wise point.

    Again thank you both for sharing your helpful thoughts.

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    David Kaplan

    Ronaldo,

    I think that 12' x 5' Cloak Room info is intended to INCLUDE the walls but again, it's not clear.  You have no way of knowing that.  It needs to be clearer and it will be clearer on the ARE. 

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    Ronaldo Guimaraes

    I came across this question and got it wrong.

    I was just wondering if there is a mathematical or graphic explanation to clarify the answer.

    As I am not satisfied by only knowing the answer, I would like to understand the answer, and I couldn't find arguments for the correct answer.

    Does it mean that the structural design and all elements in a building (structural and non-structural???) should be calculated to overcome the seismic force so the building would be safe or at least bear some stress before failing?

    Any help is welcome.

    Thanks.

     

     

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    Derek Mason

    Ronaldo,

    Are you familiar with the FEMA 454 document? You can find a free copy off of a link that someone has posted in the forms, or do a search like I did and download a copy. It does a great job at explaining the concepts.

    I wouldn't worry about any mathematical equations other than f=P/A (FEMA 454 discusses this) and the general structural formulas we all learned in school. 

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    Ronaldo Guimaraes

    Hi Derek.

    Yes I have been studying on FEMA docs.

    Thank you for your directions.

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    Ronaldo Guimaraes

    Any help clarifying the question below?

    Not sure why heat should be collected is correct and decreasing interior permitted temperature is not a correct answer. 

    What does "interior permitted temperature" mean in this context? 

     

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    Jessica Deaver

    Hi Ronaldo,

    The key word in that question is "permitted temperature." In hot climates (like TX where I live) you want to allow a higher permitted temperature inside because according to the psychometric chart, in hot weather our bodies tolerate higher temperatures. This question didn't mention humidity so you're only dealing with temp.

    On that note, because humidity isn't mentioned, collecting heat to be dispersed when the temp drops as you would do in a hot arid climate like Arizona would be associated with thermal mass for passive heating/cooling. Read more here:  https://sustainability.williams.edu/green-building-basics/passive-solar-design

     

    Hope that helps, Jes

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    Allison Hora

    Ronaldo,

    Collecting the heat in say a trombe wall, would store the heat and then radiate it out once the exterior temperature is lower than the interior.This improves enegry efficiency by decreasing the amount of "bought" energy you will need (ie: gas or electric heat you run at night).

    To be energy efficient and send less money to your electric company for running your AC in a hot climate you would want to increase the interior temperature of a building. The closer your interior temperature is to your exterior temperature means the less work your AHU will be having to do to maintain whatever the allowable interior temperature is. For instance, if it is 80 degrees outside and you want to keep your home more comfortable at 74 but realize that may be making your electric bill too high, you may increase the interior temperature a bit to 76 or so. Now your AHU wont be working as hard to maintain your interior temp. Make sense?

    Allowable interior temperature is basically whatever temperature you set your thermostat to.

    I hope that helps you out a bit!

     

     

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    Ronaldo Guimaraes

    Thanks Jessica and Allison for the clarifications.

    Very helpful.

    Regards.

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