GBA Logo horizontal Facebook LinkedIn Email Pinterest Twitter X Instagram YouTube Icon Navigation Search Icon Main Search Icon Video Play Icon Plus Icon Minus Icon Picture icon Hamburger Icon Close Icon Sorted

Community and Q&A

Typical, actual live / dead loads per room?

BrunoF | Posted in General Questions on

Just out of curiosity, I would like to understand what the typical, actual live and dead loads are for residential rooms and construction.  My house was engineered using 40/10 PSF for the living spaces and 30/10 PSF for the limited storage attic area.  I have come across some documents online that indicate a design load of 30/10 is (was?) allowed for sleeping areas which raises questions about what the load variation is between rooms and what assumptions were made to determine that a 40/10 uniform design load is sufficient and with what (if any) buffer was included to ensure safety.

are there any documents that exist to give me an idea of what my actual loads will be so I can estimate how much “head room” I have between actuals and the design load?

thx!

GBA Prime

Join the leading community of building science experts

Become a GBA Prime member and get instant access to the latest developments in green building, research, and reports from the field.

Replies

  1. jollygreenshortguy | | #1

    First to answer your question about materials weights, here is a website with all you should need. There are many others like it.
    https://www.accessengineeringlibrary.com/content/book/9781260128673/back-matter/appendix2

    I've found that most floor assemblies (excluding tile) calculate out to about 8 to 9 psf. Furniture is too variable to predict. Bookcases weigh a lot but are often at walls, with supports almost directly below. But a bookcase on a floor with no wall below may well cause deflection and vibration issues.

    Second, in case you haven't considered, which you may have - Designing a floor system to the code minimum is safe but will result in a floor that bounces and vibrates a lot. Though it may strictly meet code it, too an extent, "fails" because it fails to provide the kind of comfort most people reasonably expect. So engineers typically "over-design" floor systems with respect to code. I'm very much in favor of doing that. The incremental construction cost is minor and results in much more functionality.

    To design the stronger floor system, you can either over-estimate the design loads, or design for less deflection. Either approach works.

    Keep in mind that spaces with tiled floors will be bearing more dead load AND need to be stiffer to avoid tile and joint cracks. Also rooms with lots of built-ins, for example a kitchen with an island, may have much more than the standard dead load concentrated in certain areas.

  2. Expert Member
    DCcontrarian | | #2

    Note that the deflection limit for joists is not about safety -- that's not the point where the joist will snap. It's about comfort, it's disconcerting to walk on a floor that moves noticeably under your weight. It's also about keeping the drywall on the ceiling and tile on the floor from cracking, although for tile you usually want an even stiffer floor.

  3. BrunoF | | #3

    Thx! I’m not too worried about deflection as we went for L/480 minimum for joists in most areas, L/600 min for tile areas and used mainly LVLs for girders.

    I’m just curious how much actual load (vs design load) I may have on some of my point loads in the crawlspace.

  4. Expert Member
    DCcontrarian | | #4

    The stereotypical maximum live load for residential construction is a big house party. Imagine people packed in wall to wall. That could reach 50 pounds/sf live load.

    I've never heard of a house collapsing under those conditions. It's not uncommon for decks to collapse under that kind of load, often with horrific results.

    1. Expert Member
      MALCOLM TAYLOR | | #6

      DC,

      I can't remember ever hearing about a code compliant house that failed due to loading without there being either decay, or some other compromise to the structure.

      1. Expert Member
        DCcontrarian | | #8

        The only structure collapses due to loading I can think of were due to snow. Decks seem to collapse semi-regularly. Usually what you see in houses is sagging long before collapse.

        Note again that joists are sized for deflection, we don't even talk about what their ultimate failure strength is. Beams too. Rafters, posts and walls are sized against failure.

  5. Expert Member
    BILL WICHERS | | #5

    Decks are often built as DIY projects, many times by people with little or no construction expierience. The clasic failure is to nail a joist or header to the side of a 4x4 column, then have that connection fail in shear under heavy load. There is a good reason why nails are usually just there to hold the wood in place, while the load gets carried by wood sitting on other pieces of wood.

    The span tables in the code books are pretty conservative too. Remember that framing lumber is *visually* graded. The typical "number 2" grade used means someone LOOKED AT the piece of wood, and thought it looked grade "number 2". Since wood is a natural product, there are all kinds of factors that come into play regarding the strength of the final boards. The code tables have to ALWAYS be safe, so they are VERY conserative. I wouldn't be surprised if you could load a residential floor up to double the rating and still be OK, although I'd certainly not recommend trying that!

    I would typically be more concerned with the connections at the ends of the joists than I would with the span itself in terms of "what might fail" stuff.

    Bill

  6. BrunoF | | #7

    Thx folks. It more of a question relating to the calculated point loads and pier footing size / soil bearing capacity. If the point loads on the crawlspace piers are all based on a hypothetical house parties occurring on the first and second floor plus the attic…then there is probably a healthy margin of error built in.

  7. maine_tyler | | #9

    "If the point loads on the crawlspace piers are all based on a hypothetical house parties occurring on the first and second floor plus the attic…"

    Is that how foundation point loads are designed though? By 'that' I mean by adding the design load of the floors (and roof). I would be somewhat surprised if that is true since, as described above, floor design load has more to do with deflection. If what you say is true, the foundation would actually 'fail' before the floor joists. I would think foundation specs should exceed the additive floor design specs, but I don't know.

    I would also be surprised to see structural failure occur right AT design load (in the case of both the foundation and the floors).

    Or said differently: Is there no safety factor built into design load itself? Is it truly the point of failure and the only 'wiggle room' comes from design load being higher than is actually expected to EVER occur?

    1. Expert Member
      BILL WICHERS | | #11

      The pound per square foot load is supposed to be carried down to the foundation, through the supporting columns/headers/etc. The floor is supposed to be able to support the design load. In the commercial world, I like to explain this as "if the tenant wants to fill the square footage of their suite with full file cabinets, the floor should be able to support that without any trouble.".

      Where you run into problems are with unusually high loads (i.e. a rooftop hot tub), or fancy structural stuff (usually very long unsupported spans requiring large beams).

      Point loads are limited by the subfloor's resistance to puncture, which is typically pretty high (I don't have a number handy though). The final load gets distributed over a larger aera though. It still all gets carried down to the foundation. Think of all the loads as arrows pointing downwards. If a column is centered on 50 arrows of 50 pounds each (each arrow represents 1 square foot), then that column supports the total of those (2,500 pounds), or the load is shared with other nearby columns with the total shared proportion dependent on the distance from the column. You end up with a vector problem. This type of work is known as a "statics" problem. In statics, all the arrows pointed down are essentially cancelled (supported by) arrows point up. Everything balances out, and the structure stays up.

      If the arrows do NOT balance/cancel out, then we have a dynamics problem, instead of a statics problem. The unbalanced forces mean THINGS ARE MOVING. That's bad in the construction world, as it means our building is falling down. We much prefer statics problems in construction :-D

      Obviously I'm simplifying things a bit, but this is basically what's going on here.

      Bill

      1. BrunoF | | #12

        Bill, thank you. Just to clarify though…is it a fair assumption that the loads used to calculate all of this would be the 40/10 design loads for each floor plus the 30/10 design load for the the attic, plus the roof? I’m not worried about the roof in my case as it is stick build and the vast majority of the roof load is carried by the exterior walls and perimeter foundation. My focus area is on some of the piers / footings in the crawlspace.

        1. Expert Member
          BILL WICHERS | | #13

          Yes, total load on the support columns are additive. Basically you start from the bottom, and it's (roof dead+live load) on the uppermost floor support columns/walls, then the next floor down is ((roof dead+live load) + (upper floor dead+live load)), then on the next floor down is (((roof dead+live load) + (upper floor dead+live load) + (current floor dead+live load))), etc. You keep doing that until you get to the lowest floor, the one right on top of the foundation, at which point its ((all the upper levels dead+live loads) + (lowest floor dead+live load)), and that gives you the load on the footing/piers in the foundation itself.

          Everything transfers downwards from the top, adding up as you go, until you get to the final bearing surfaces of the foundation elements. Right under that, you are working with the pound per square foot bearing capacity of the soil you're building on, which is a geotechnical issue. The job of the foundation is to spread the load of the upper level framing over a sufficiently large area of soil that the structure doesn't settle over time. This is why really big skyscrapers like to be built on areas with relatively shallow bedrock like Manhattan island.

          Bill

  8. maine_tyler | | #10

    As far as 'what are actual expected loads' no one here knows what you will do in your house.

    A 200 lb person spreading their weight over 5 sq. ft. is 40 psf. That's a square of about 2.25 feet. So if the room was filled with 200 lb people on 2.25' grided centers, standing statically (not sure how or if dynamic loading is taken into consideration) then you've hit your 40 psf.

Log in or create an account to post an answer.

Community

Recent Questions and Replies

  • |
  • |
  • |
  • |