This Thing Called Tributary Width

Of all the structures related material that I have taught engineering and architecture students one thing that routinely hangs them up is this thing called `tributary width'. It's nothing too complicated really; it's simply the `width' of some space that contributes load (or something else) to a structural member. Consider, for example, a house (or portion thereof) that is 32 ft wide, with a floor spanning wall to wall, and also resting on a beam that goes down the center of the house (see sketch). Imagine plywood floor, carpet or whatever over that, and the plywood supported by sets of joists that each span from the wall to the beam. In this scenario the joists `hang' off the sides of the beam.

Now consider how the `loads' of the building and potential occupants are transferred to the beam (or the walls). The weight of the floor itself and occupants is supported by the joists, and the joists carry the loads to the walls and the beam. Plus, of course, the joists must carry themselves, plus any ceiling materials nailed or screwed to their bottoms, insulation, etc. How much of what the floor (or a joist) is carrying goes to (contributes to the load on) the wall, and how much to the beam? Well, silly, half to each, of course. Half the load on each joist goes to a wall, and half to the beam.

The `tributary width' then, for the beam is ½ of the total span on one side (say, the left side), and ½ the total span on the other side (say, the right side). In terms of numbers,

... tributary width for the beam is ½ of 16 ft + ½ of 16 ft = 2 x ½ x 16 ft = 16 ft.

Strictly speaking, the spans of the joists are going to be a bit less than 16 ft, due to the particulars of how they frame onto the walls, the width of the beam, etc. However, with regard to the beam, it will half to carry half the spans of the joists plus the load directly above it, so whether we go from end-to-end of the joists, or center-of-beam to center-or-edge of wall, or whatever, the differences are either small, or nil.

So, a common `structures' problem would be to determine the applied loads to the beam, from which we would determine internal loads, and then, depending on our design approach, internal stresses. In this example let's consider the loads from the weight of the floor itself, plus the loads due to use or occupancy (say a `party' of people). And we will also have to include the weight of the beam itself. Building codes dictate the occupancy loads that are used, and they are typically given in terms of `weight or load per square foot'. I call them `area loads', and denote them by σ. For Residential occupancy a common load is 40 psf (40 pounds of weight per sq ft). The load on the beam is equal to the area load multiplied by the tributary width (the width of floor contributing to the beam). In equation form,

... ω = σ x tributary width,

where ω is what I call a `line load', and its dimensions (units) are in pounds per foot of length of beam (plf).

In this example, then, ω = 40 psf x 16 ft = 640 plf.

We could say that each foot of beam carries 320 lb from one side, and 320 lb from the other.

You might notice that the equation above `pictures' the load from each side continuously being applied to the beam, whereas in real life (we have already pointed out) that it arrives via joists that are some `spacing' apart. (And you are correct to notice that.) By example we will show that in most cases taking the load to act continuously along the beam will give us essentially the same answer (for the beam) as if we applied the loads discretely at the points of joists (unless the beams are short and the joist spacing large).

We also need to consider the weight of the floor system. One way to handle building weights is to look up `per square foot' equivalents (or totals) found in literature tabulated for various construction types. Our text, for examples, in Table 4.1 on pages 153-154 has values for common construction materials.

So, let's say for our example the floor is made up of the following materials with the `area' weights provided by Table 4.1.

Carpet and pad ... 3 psf

3/4 in. plywood sheathing @ 3 psf / inch ... 2.25 psf ... (see under `Roofs')

2 x 12 joists @ 16 in. o.c. ... 3.2 psf

½ in. gypsum (sheetrock) ceiling ... 2.5 psf

And let's say some insulation ... 0.5 psf

Total, 9.210.95 psf, say 11 psf.

Note that these are average values. Certainly with joists 16 in. apart there are 1 ft x 1 ft `areas' that don't have any joists at all!

So, 9.211 psf ... often with wood frame floors we use a `presumptive' value of 10 psf (often with 2 x 10 joists). If, however, we know that we have components that are heavier (tile, lightweight concrete overlay, whatever), we would add them in for a heavier floor.

This, for our example, the load from the floor materials (we call Dead load) is ...

... ... ω _DL = σ _DL x tributary width = 1011 psf x 16 ft = 160176 plf.

Note that I added the `DL' subscript. No big deal. We could have subscripted the 40 psf occupancy load calc with `LL' (for `Live' load).

Then, finally, the beam must also carry itself (its own weight); we call this `self-weight'. We'll illustrate that in numerous other examples.

Before I close let me say a couple things. First, while the building codes dictate a minimum area design load of 40 psf (for Residential occupancy Live load), they also generally recognize that the greater the area involved (`tributary area') in all likelihood it will not (all) be loaded to 40 ... and thus allow a reduction. We will cover this in detail later also. Second, we have considered joists framing to the sides of the beam (ending and terminating at the beam). If the joists span continuously over the beam (as with modern manufactured wood I-joists) the load on the beam will be greater than that computed using tributary widths that go out half way along the joists. We will deal with that later also.

Often we design beams using `design software'. Often the software will ask for the (whole) tributary width (in this case 16 ft). In some cases is will ask for the tributary width on each side (8 ft and 8 ft in our example).

So, question: what is the tributary width for each joist (joists @ 16 in. o.c.)? Answer: 16 inches, of course (1/2 of 16 from one side, and 1/2 of 16 from the other).

References

Minimum Design Loads for Buildings and Other Structures, ASCE Standard ASCE/SEI 7, American Society of Civil Engineers, www.asce.org.

Simplified Engineering for Architects and Builders, Ambrose, J. and P. Tripeny, 10^th edition, John Wiley & Sons, Hoboken, New Jersey.