where to learn to solve this snow load structure question in PDD?

Comments

23 comments

  • Avatar
    Matthew Lawton

    IBC 2012:
    1607.12 Roof loads.
    The structural supports of roofs and marquees shall be designed to resist wind and, where applicable, snow and earthquake loads, in addition to the dead load of construction and the appropriate live loads as prescribed in this section, or as set forth in Table 1607.1. The live loads acting on a sloping surface shall be assumed to act vertically on the horizontal projection of that surface.

    1
    Comment actions Permalink
  • Avatar
    Richard Wilson

    This is a tricky example exam question.

    I'm only used to flat roofs. If the roof slope was increased from 4"/12" to 12"/12" for example, then the tributary length of one side of the slope would increase by about 3.5ft, technically adding an additional 200lbs to each side. So what's the difference between a 4"/12" and 12"/12" slope. Is there a factor for steeper slopes somewhere...

    0
    Comment actions Permalink
  • Avatar
    Li Xing

    Is there anywhere to download 2012/2015 IBC free? Thanks.

    0
    Comment actions Permalink
  • Avatar
    Matthew Taylor-Rennert

    Li, 

    Here is online free versions of 2015 IBC:

    https://codes.iccsafe.org/public/document/toc/542/

    Downloads cost, but check your local library, you can probably check one out there. 

     

    0
    Comment actions Permalink
  • Avatar
    Li Xing

    Matthew:

    Thanks.

    0
    Comment actions Permalink
  • Avatar
    Michelle NCARB

    Hey Matthew and Li -

    Don't forget ARE 5.0 is on IBC 2012, not 2015.  You'll want to use this link instead: https://codes.iccsafe.org/public/document/toc/353/

    And I would also read up on this in the structures-related books in the reference matrix.

    0
    Comment actions Permalink
  • Avatar
    Benjamin Norkin

    Richard,

    Even though there is actually more surface area on the 12:12 roof like you said, not as much of the weight is going straight down. The steeper it gets the more that snow weight wants to slide off and the vertical component gets smaller. I'm rounding, but loosely based on the example above, the 12:12 roof would be a lot heavier than the flat roof, but not actually putting as much weight straight down.

     



    0
    Comment actions Permalink
  • Avatar
    Kurt Fanderclai (Edited )

    This is actually a case of keeping things simple.

    I think that somehow the very first response by Matthew L. has been overlooked, where he cited a critical piece of info for these types of calcs.:

    "The structural supports of roofs and marquees shall be designed to resist wind and, where applicable, snow and earthquake loads, in addition to the dead load of construction and the appropriate live loads as prescribed in this section, or as set forth in Table 1607.1. The live loads acting on a sloping surface shall be assumed to act vertically on the horizontal projection of that surface."
     
    So, no trig required.  5' tributary each side of beam X 2 sides = 10';  then 10 feet times 60 = 600.    
     
     
     
    Some of the reality of an actual snowfall seems tricky as we look at load calcs -- it seems counter-intuitive.  But if you picture 12" of snow falling out of the sky like a horizontal fluffy blanket descending to earth -- flat surfaces obviously get the 12", and the corresponding snow load.  It's harder to picture that a sloped roof gets the same snow load from the 12" blanket of snow.  You can prove this to yourself by drawing a  flat roof and a pitched roof in autocad or whatever.  Draw 12" verticals on both the flat and sloped "roofs".  You'll quickly notice that -- per lineal unit -- the load is the same -- because the thickness of the snow is thinner when measured perpendicular to the slope -- so the individual 12" verticals of snow land on the slope just like they land of a flat roof, but the 12" snow equates to about 8.5" when measured perpendicular to the 12/12 roof surface.  So, same load, due to a different resulting thickness.  So IBC 1607.12 actually makes sense.
     
     
    0
    Comment actions Permalink
  • Avatar
    Sean Ragudo

    I think you are getting a bit wrapped around the axle on this question.  ARE5.0 will ALWAYS provide you with the pertinent information necessary to answer a questions.  The original questions requires a snow load 60 PSF, and a tributary area to be answered.  This is basic statics.

    60PSF x ((5FT + 5FT)x1LF) = 600PLF

    snow load x ((1/2 left side joist + 1/2 right side joist) x (implied distance of load) = total load per linear foot of beam.

    There are some tricky portions notably,

    1. The roof pitch.  This doesn't matter because you're given a snow load (which in actual practice given in the IBC)

    2.The rafter length.  This tricks your mind into thinking something other than "span" matters to solve the equation.

    3.Overhang.  This makes you think that there might be some opposing force that would lighten the beam load... it doesn't

    4.  Lack of length of ridge beam.  Makes you think you don't have enough information to solve, but it's embedded in the question as length per lineal foot on the beam.

    It's meant to be hard, it's meant to make you think.  Don't think it's too hard, don't overthink!  Good luck on the test!

    0
    Comment actions Permalink
  • Avatar
    Benjamin Norkin

    Sean, (and Kurt's deleted comment)

    I agree with what you're saying. Matthew L basically gave the code explanation above. I think Richard's question, and my attempted response, was trying to get at "why" the code only cares about the horizontal distance rather than a way to solve this problem. It's a good question...why do we only care about the horizontal span when our observation would lead us to believe there is more real world weight on a structure. In the case of a 12:12 pitch vs 4:12 pitch, you have more surface area so more actual snow.

    As a caveat, I have zero confidence that my diagram above is the reasoning behind why the code is written...but I think it makes sense? With structures is seems that we're always trying to simplify things to their most basic form. A pin connection isn't really a pin connection, but it performs close enough that we can calculate it that way, that kind of thing. So instead of requiring all these extra calcs, the code might assume that the steeper a roof gets, the less live load would be bearing directly down, so we just assume the uniform weight is only coming from the span distance and that's close enough and give some safety factor.

    Maybe there's a PE lurking on this forum?

    0
    Comment actions Permalink
  • Avatar
    Kurt Fanderclai

    My comment above appeared, disappeared, then re-appeared. Then appeared to be "pending".

    0
    Comment actions Permalink
  • Avatar
    Kurt Fanderclai

    You can conceivably attach more load to the longer length of a pitched roof member, but as my "pending/deleted" post said -- that is not what happens when snow falls. More snow does not fall on a pitched member. Granted, you could go up on the roof and attach an even 1" of steel weights to flat parts and sloped parts, and then the sloped parts would be carrying more load when viewed like the posted example problem. But normal snowfall doesn't attach as such.

    0
    Comment actions Permalink
  • Avatar
    Kurt Fanderclai (Edited )

    You can conceivably attach more load to the longer length of a pitched roof member, but as my "pending/deleted" post said -- that is not what happens when snow falls. More snow does not fall on a pitched member. Granted, you could go up on the roof and attach an even 1" of steel weights to flat parts and sloped parts, and then the sloped parts would be carrying more load when viewed like the posted example problem. But normal snowfall doesn't attach as such.

    When I'm not on a phone, I'll show a diagram of this.  

    0
    Comment actions Permalink
  • Avatar
    Benjamin Norkin

    Ok, I think I see what you're saying. Let me update my diagram to be more accurate:

     

    0
    Comment actions Permalink
  • Avatar
    Kurt Fanderclai

    This thread is a bit of a mess! Mostly due to me.
    By the way, Benjamin -- a huge congratulations on passing PDD!

    0
    Comment actions Permalink
  • Avatar
    Benjamin Norkin

    Yeah we ran this one into the ground. Thanks man. When you testing? Seems like been studying forever. I have PPD on Friday, trying to get my momentum back.

    0
    Comment actions Permalink
  • Avatar
    Kurt Fanderclai

    Benjamin -- I'm scheduled for one week after yours. I did my first 4 exams in three months, but my approach to PPD has been total BS. Very disorganized study methods. I'm going to take a crack at PPD and hope for the best.

    0
    Comment actions Permalink
  • Avatar
    Xiaotian Huang

    Sean -- I don't really agree on overhang doesn't create any opposing force unless the overhang/stud framing joint is a hinge joint which is obviously not because otherwise it won't stand that way.

    Using shear force diagram on the simplified continuous beam to calculate the 3 supporting forces, the outcome is actually 600ish on each stud framing and 300ish on Ridge Beam because of the overhang.

    So please explain why we should ignore that overhang portion with in reality makes this much difference.

    Thanks.

    0
    Comment actions Permalink
  • Avatar
    Sean Ragudo

    Xiaotian,

    1.  LOAD, not shear or bending stress is being asked for

    2.  You get the wrong answer by using your math.

    Cheers,

    Sean P. Ragudo RA, LEED AP

    0
    Comment actions Permalink
  • Avatar
    Xiaotian Huang

    Sean, I know very well it's asking about LOAD.

    Maybe shear method isn't the best way (but it shouldn't be a wrong method, the sum shear force at the joint point is the load to the support).

    Let's try using torque balance this time:

    Taking the stud frame as single point of support, the torque created by load(q) on overhang(x) should equal to the torque created by load on roof side(l) minus torque created by half of ridge beam support force(F), then:

    F=q(l^2-x^2)/2l

    Which again gives me something close to 300lbs.

    I've been trying to convince myself that I overthought and just trust the study guide and you guys. But every other method I tried proved the load on overhang isn't something we can ignore. I'm also asking some of my civil engineer friends about their opinions.

    0
    Comment actions Permalink
  • Avatar
    Sean Ragudo

    Xiaotian,

    It's not about an equation or the reaction.  It's about determining what the first reactor is.  You've been solving what the beam member is doing in resistance to the load.  The question is asking "what is the load?"  Those are two different questions.

    Question asked is: "what is the LOAD?"

    Answer you're giving: "The beam needs to be able to handle this reaction"

    You need to answer the question asked, not what question you think is being asked.  This is a classic way that the test "tricks you"  don't overthink it... answer the question asked.

    Sean P. Ragudo RA, LEED AP

     

    0
    Comment actions Permalink
  • Avatar
    Xiaotian Huang

    Hi Sean,

    I think my problem was understanding "LOAD". The strict terminology of "LOAD" is exactly the force to the structure or component.

    But the given in question is "design snow load" and per IBC it's already considering unbalanced load and other scenarios. Which also means it already simplifies the problem to a "(maybe designed) worst case" and enables us to simply use tributary area to do the calculation.

    So thank you for your patience and help. This also reminds to me read questions more carefully.

    Cheers,

    0
    Comment actions Permalink
  • Avatar
    Jonathan Chertok

    wanted to thank li xing for starting the thread and messr. ragudo for his input on it. i always have a hard time with non-visual solutions and find i have been having to put a lot of pieces together on some of these. thought to post this as i think it explains the "implied distance of load" which i gather is a 1' section. so the tributary area is 10' x 1' as pictured. hope it is of use to someone.

    0
    Comment actions Permalink

Please sign in to leave a comment.

Powered by Zendesk