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Shed construction /w a plant roof

  1. May 4, 2009 #1
    Hey,

    I am building a storage shed in the backyard and would like to build a flat roof which houses plants instead of a shingle roof. The slope of the roof will be around 1 foot rise per 6 feet run.

    The dimensions of the building are, 12 feet by 16 feet. floor is built with 2x10's and the floor joists are 24 inch o.c. The walls are 2x4, studs 16 inch o.c.

    I am a bit confused with the roof because i am not sure how much wet soil weighs. The roof will be covered equally with 5 inches of soil. I was thinking of using 2x6's to make the rafters but how close should they be? 12 inch o.c. or is that an overkill?

    The location is Florida, so there is a potential for a lot of rain and wind during hurricane season.
     
  2. jcsd
  3. May 4, 2009 #2

    minger

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    Let's see, quick calculations....you will have 80 ft³ of wet soil. A quick Google gives 'dry' soil as a density of between 75 lbf/ft³ (1200 kg/m³) and 100 lbf/ft³. The question about wet soil is difficult I would think. Perhaps your best bet is to actually measure it.

    So sake of argument, let's assume twice as much, giving a total force of 12,000 lbf. If the load is evenly distributed, then the load that each beam takes is 12,000 lbf/n, where n is the number of beams.

    Now, for your case, I would 'think' that the best case would be a uniformly distributed beam with an added axial loading. Essentially rotate your roof beams such that its horizontal. You now have two components of your force, one transverse, and the other axial.

    The transverse loading (per unit length) will be
    [tex]W = \frac{5}{6}\frac{F}{l} = 625 lbf / ft-n[/tex]
    While the axial loading will be
    [tex] F_{ax} = \frac{1}{6}F = 2000 lbf[/tex]
    For end constraints, I would assume fixed-fixed. We will also need the moment of inertia for your beam cross section.
    [tex]I = \frac{bh^3}{12} = 1.736e^{-3} ft^3[/tex]
    We have the maximum moment at x = l/2
    [tex] M_{max} = \frac{w}{k^2}\left[\frac{kl/2}{\sin(kl/2)}-1\right][/tex]
    Where
    [tex] k = \sqrt{\frac{P}{EI}}[/tex]
    For the stresses, you'll need a decent value for the yield strength of the wood you'll be using. This may be more difficult than you think. From there, I think that you can just use a combined loading for stress
    [tex] \sigma = \frac{My}{I} + \frac{P}{A} [/tex]
    Where P is the axial loading, and A is the cross sectional area. This gives you a stress value.

    From the looks of the equation, it's not purely linear. The axial component of the load in the moment equation is not linear. This means you cannot simply take this stress, divided by the stress you want, to get the number of beams. As a sure bet, do that, but then plug the load numbers back in to make sure you get something that makes sense.

    Now this is pretty idealized, so if you are going to do a methodology like this, include a good factor of safety (e.g. 10 or so), because you don't know how age, humidity, or weather is going to affect things like this. So be careful, and use an analysis like this only as a starting point.

    Good luck,
     
  4. May 4, 2009 #3
    Is there a translate to english button somewhere? lol

    so if 1 cubic foot of wet soil weighs x and i need 12x12x4 per sq ft. What i need would be x/3? So i need to calculate based on that measurement to figure out what size ceiling joists and the spacing i need, right?

    Also, I was thinking of just measuring out 1 cf of soil and weighing it and then adding a lot of water and weighing it again.

    What about live and dead loads? The dead load is the weight normally on and the live load is extra weight from snow or water. Right? Most charts only go up to like 20 for the dead weight. So would i figure the weight and go by the live load chart?
     
  5. May 4, 2009 #4

    minger

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    I'm a little confused by what you said, let me try and put it into more layman's terms.

    You have a loading x, that you will figure out. Weight a cubic foot of wet soil, then multiply that number by the cubic feet of volume you plan to have soil on. You have a loading.

    You have a beam in mind. It will be a 16' long 2"x6" wood beam. You know the orientation of the beam (i.e. 1' of rise for 6' or run). You also know how the beam is attached as the ends (e.g. pinned, simply supported, fixed). Based on these parameters, an engineer says
    If I assume that I only have one single beam, I can calculate a maximum stress. Almost certainly this stress will be too high. Let's say for arguments sake that the strength of our wood beam is 1000 psi.

    We calculate our stress to be 10,000 psi. That means that we will need roughly 10 beams in order to not fail. However, as I mentioned, those equations are nonlinear, which means that you cannot simply divide like this. You will then need to recalculate the loading (it will now be the total loading / 10) and rechug a stress value. Hopefully it will be close. Iterate a few times until you reach a steady value.

    However, having said that, there are MANY uncertainties when dealing with an unsteady loading, particularly on wood, which already has inherit uncertainties. You will need to include a large factor of safety in order to account for these (i.e. overkill is good).

    When I say factor of safety, it means that I want my calculated stresses to be much lower than the strength of the material. For a factor of safety of 10, with a strength of 1000 psi, I do not want any stresses to be higher than 100 psi.

    Now, I am personally not really familiar with structures such as this; I do not know how to take into account things like wind load, which while may be small by itself, may cause loadings and reactions to suddenly become non-uniform which may cause problems.

    I would imagine that there is a standard written for this type of appliation, and you may want to call a friend/contractor to see if you can get that reference.
     
  6. May 4, 2009 #5

    nvn

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    Meseria: Saturated soil density can go up to about 1900 kg/m^3. Plus I would allow for an additional water depth (rain) on top of the soil of 40 mm. Therefore, for a 127 mm soil depth, your soil, water, and rain load, per unit of rafter length, works out to be w2 = (13.28 N/mm)/n, where n = number of 3658-mm-long rafters minus 1. I assume the roof self weight, per unit of rafter length, is w1 = (3.464 N/mm)/n. For wind gusts up to 220 km/h, I'm currently getting a roof wind load, per unit of rafter length, of w3 = (1.030 N/mm)/n. Therefore, the current rafter load is w = w1 + w2 + w3 = (17.77 N/mm)/n.

    Therefore, 38 x 140 x 3658 mm rafters, southern pine (SYP), grade number 2, at a spacing of 305 mm, currently have a stress level of R = 167.4 %, which exceeds 100 %, meaning the applied stress exceeds the allowable stress. (We want R less than or equal to 100 %. R > 100 % indicates the component is overstressed.) Likewise, 38 x 184 x 3658 mm rafters at a spacing of 305 mm currently have a stress level of R = 101.2 % > 100 %. If the 38 x 184 mm rafter spacing is reduced to 287 mm, the current rafter stress level becomes R = 95.2 %. This assumes the rafters will not be wet. (If the rafters will be wet, let us know, because it would increase the R values.) Don't forget to put hurricane straps on your rafters.

    Also, ensure your walls can withstand this high roof load, which is more than double a typical roof load. Therefore, double the compressive strength of your walls, and double the diagonal bracing.
     
    Last edited: May 5, 2009
  7. May 4, 2009 #6
    my walls are going to be 2x4 spruce studs at 16"o.c. Will this be strong enough to bear the weight of the roof? Do i need to use 2x6's instead or use closer spacing or is this adequate?

    I did a little test with some soil outside and found a cubic foot weighs around 120 lbs when fully saturated. So at 4 inches of depth the load would be 40 psf.

    If i do a flat roof with a 12 foot span, what size joists do i need to use and at what spacing? I would like to use 2x6 pine as it is what i have.

    Also, my floor is going to be made with 2x10's at 24" o.c. It will be sitting on 8 cinder blocks equally spaced along the outside.
     
  8. May 5, 2009 #7

    nvn

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    Meseria: So your saturated soil density is 1922 kg/m^3; and you decreased your soil depth to 101.6 mm. Therefore, the current rafter load is w = w1 + w2 + w3 = (3.464 + 11.26 + 1.030)/n = (15.75 N/mm)/n. Therefore, 38 x 140 mm rafters at a spacing of 203.2 mm currently have a stress level of R = 99.0 % < 100 %. (As explained in post 5, we want an R value that does not exceed 100 %.) Alternately, 38 x 184 mm rafters at a spacing of 325 mm currently have a stress level of R = 95.7 % < 100 %. I don't have comments on your walls at this moment.
     
  9. May 5, 2009 #8

    RonL

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    Don't forget to include the weight of wood and decking in addition to the soil.
    I would think 2"X10" or even 12" on 16" centers for the roof, 2X6 wall studs on 16" centers, sandwiched with 1/2 or 3/4 plywood.
    One other thing to consider, is all this weight is supported by the outside pier and beam locations, they will need to be increased by ground area or quantity.

    Just my thoughts, nothing professional. You might look at sites or forums where others have built their own structures.
     
  10. May 6, 2009 #9

    nvn

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    Meseria: Here is a summary of rafter cross-sectional dimensions, maximum allowable rafter spacing, and stress level, for rafter load w = (15.75 N/mm)/n and rafter length 3658 mm (southern pine, grade number 2). This particular w value, which was explained in posts 5 and 7, corresponds to an applied roof pressure loading of 3230 Pa.

    Size (mm)..Spacing (mm)..R (%)
    38 x 140....203.2........99.0
    38 x 184....325..........95.7
    38 x 235....443.4........91.2
    38 x 286....609.6........91.5


    Regarding your walls, what is the wall height? Will your walls have large structural sheathing panels attached? If so, will the structural sheathing panels be on the inside and outside of the walls, or only on the outside?
     
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