*Draft – Comments Welcome*

**Background**

We have been designing a concrete house per the 2006 *International Residential Code*. The design has been `prescriptive’. By that I/we mean that if we simply follow the rules (`prescription’) of the *Code* … we won’t need to hire an engineer. That is good. In fact, that is the *intent* of the `R’-Code … to provide a way to design and construct residential stuff without necessarily requiring engineering. (We engineers are busy – save hiring us for the fancy stuff, the tall stuff, the stuff that really does require specific engineering.) We have run into a few snags, but we have found our way around them. But here is another snag: the Owner wants to go with a concrete Main floor system.

Our prescriptive path so far has limited us to `light frame Main floor and roof systems’. Growing in popularity, however, at least around here, are concrete floors with radiant heat in them. In the past (around here) the concrete floors have been non-structural concrete slabs poured over wood framing (and subfloor). Recently, however, an ICF *floor* system has come on the scene … totally cool. The system utilizes insulating forms that produce a monolithic concrete joist-slab system. And when the concrete cures, the insulation (below) remains and becomes substrate for ceiling attachment. It may be that the building official will require engineering certification on the concrete floor system design, but for now let’s set that aside, or to the future, and (for now) at least be able to calculate appropriate footing sizes. Obviously the concrete floor will be heavier. Also, our interior bearing wall will also need to be concrete.

With some research we will find that a joist-slab system with 8 in. deep ribs (joists) and 4 in. slab will easily carry our residential design loads out to and past 16 ft, and the system, with assumed finishes, is given to weigh 85 psf. This includes 2 in. of form below the ribs. So, the system is 4 in. + 8 in. + 2 in. = 14 in. deep. We could actually calculate this number (the 85 psf), but for now I just `extracted’ it from the manufacturer estimating data. The estimated rebar required for the system is also provided by the manufacturer, but they note that *that* will need to be certified by an engineer.

Let’s make our interior bearing wall of concrete, 6 in. core thickness. And now calculate our footings.

Recall that our allowable soil pressure for this example is 1500 psf ( … as for native Silty-Clay).

**Exterior Wall Footing**

From before (here) our `line loads’ are …

Recall our structure is 32 ft wide, with 18 in. eaves, 10 ft floor-to-ceiling heights, and so on. We are going to put the foundation on Silty-Clay, which has low bearing capacity, and thus requires big footings. For the wood framed floor and roof we came up with 30 in. wide footings, both prescriptively (here) and by calculation (here).

Roof Snow load … 640 plf

Eave Snow load … 120 plf

Roof Dead load … 263 plf

Our wall weights will change, since our floor system is deeper and our building actually a bit taller to accommodate the added floor thickness.

So, let’s use our Main level wall being 10 ft floor-to-ceiling plus the 14/12 ft floor depth … gives 11.2 ft. And let’s add 10 psf for finishes.

Main Story Wall DL … 11.2 ft x ( 6/12 of 150 pcf + 10 psf ) = 11.2 ft x 85 psf = 952 plf.

Now let’s do the concrete Main floor …

The concrete floor will be a two-span continuous system. As such, the ends will carry a little bit less than `half the span width’ … and the center support (interior bearing wall) will carry a bit more. For the condition of both spans (both sides of the interior support) uniformly loaded, the end reactions are 3/8 ω L. This loading / support condition is regularly found in structures texts and manuals and is Case 10 on p. 94 of our text. And both spans loaded is appropriate for Dead load …

So … 3/8 (85 psf ) 16 ft = 510 plf …

Main floor Live …

Now, in this case, the `worst’ case will be the near span loaded and the far span not loaded … in this case, our reaction at the near end is 7/16 ω L (see the *Timber Construction Manual*, 5th ed., page 443, Case 29b) … (the near end carries a bit less than half the span’s Live load). Note: if you just used ½ ω L you would be a tiny bit non-conservative, which would be fine!

So … 7/16 ω L = 7/16 (40 psf) 16 ft = 280 plf.

Basement wall DL …

… say 8 in. concrete core plus 6 psf of finishes, and the wall height includes the 4 in. basement slab on grade thickness … so …

Basement wall DL … ( 8/12 x 150 pcf + 6 psf ) x 10.33 ft = 1095 plf.

Let’s assume a footing size … 12 x 36 …

Footing weight … 12/12 x 36/12 x 150 pcf = 450 plf.

Now let’s add them up …

SL … 640 + 120 = 760 plf SL …

DL … 263 + 952 + 510 + 1095 + 450 = 3270 plf DL …

LL … 280 plf …

Grand Total … 760 + 3270 + 280 = 4310 plf.

Let’s see if the 12 x 36 is wide enough …

… is fp = 4310 plf / (36/12 ft) = 1437 psf ≤ 1500 psf = Fp? … Yes! … Good!

Is it deep enough? … in other words, is P ≤ h? … Let’s see …

P = (36 – 8 )/ 2 = 14 in. Is P = 14 in. ≤ h = 12 in.? … NO! Yikes! … What should we do?

We could make the footing 14 in. thick … then P = 14 in. = h = 14 in. … and we would add a bit to the weight of the footing … but we’d still probably be good.

Longitudinal reinforcement … using the temp./shrink. Min. of 0.0018 …

As = 0.0018 (14 x 36) = 0.91 … we can get this with 3 – # 5 bar.

OR …

We could add transverse reinforcement …

If we look at our Ambrose text, p. 480, Table 16.1 … we see that for 1500 psf, a footing of dimensions 10 x 36 will carry 4125 plf. Ughhhh … But wait! … it says that it excludes the weight of the footing. So, if we take 4310 and back out the weight of the footing, 450, we get … 4310 – 450 = 3860 plf … GOOD! (3860 ≤ 4125). The footing is only 10 in. thick, … *but it has transverse reinforcement.*

**Answer …** 10 in. x 36 in. footing with 3 – # 4 long steel and # 3 @ 11 in. o.c. transverse steel.

YEAH! … BABY!!!

Notes:

1. Earlier we saw (in our basement retaining walls) that the Code tables gave us one bar size and spacing, and we were allowed to use a different bar size if we made sure we still had the same cross section area of steel. Generally the bar size given was `big’ and we were allowed to use smaller bars but more of them (smaller spacing). In this example we are given # 3 bar, which is pretty small. And, the footnotes further read … f’c = 2000 psi and Gr. 40 ksi steel. We could explore bigger bars, farther apart, and Gr. 60 vs Gr 40. But there are `issues’ going to larger bar size, which we have not tackled yet. So, for now … let’s just say …

… **10 x 36 footing w/ # 3 or # 4 @ 11 in. o.c. (Gr. 40 or Gr. 60), placed 3 in. clear from bottom, and 3 – # 4 longitudinal steel, evenly spaced, tied above the transverse steel.**

**Or …**

**14 x 36 w/ 3 # 5 long steel and # 3 or # 4 @ 16 in. o.c. placed 3 in. clear from bottom.**

I’ll let you do the interior bearing wall footing.

**References**

*International Residential Code*, 2006, International Code Council, 4051 West Flossmoor Road, Country Club Hills, IL 60478.

Concrete House Footings by Prescriptive Design, Jeff Filler, Associated Content.

Calculated Footing Width for a Concrete House, Jeff Filler, Associated Content.

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

*Timber Construction Manual*, 5th Edition, 2005, American Institute of Timber Construction, Published by John Wiley & Sons, Hoboken, New Jersey.