Residential Concrete Foundation Details by Calculation

Outline

1. Introduction
2. Shear
3. Bending
4. Footing Longitudinal Reinforcement
5. Stem wall Horizontal Reinforcement
6. Stem wall Vertical Reinforcement
7. Cover and Spacing Limits
8. Other Features

1. Introduction

Most of `Part 1' (here) we actually talked about loads; it wasn't until the very end that we actually calculated a feature of our foundation. However, obviously, our foundation must be able to carry all the `stuff' above it. Note in the calculations where I rounded `up' in the building material dead loads. In practice, I may even round up `a little bit more' to allow for some changes in materials during construction (and the possibility that the new materials are heavier). Also, I mentioned that there are some areas in which we can `split hairs'. Some of those hairs I split in the design of the foundation, and some I don't, in case there are changes later on (and then I need to split them).

In general, the footing design goes like this:

1. We make the footing wide enough to not fail the soil. (That's what we did last time.)
2. We make the footing thick enough so as to not shear the footing.
3. And we add reinforcement, if necessary, so that the footing doesn't break in bending.

These last two items we will cover now.

2. Shear

The footing for the foundation wall acts like two upside down cantilever beams resisting the pressure of the soil on the footing bottom acting upward. Generally the weight of the soil above the footing is neglected in counteracting this upward force.

Concrete subject to shearing forces (stresses) tends to fail in what is called `diagonal tension'. In other words, if the concrete fails in shear it will crack at a 45 degree angle starting from the edge of the foundation wall. And, only the soil pushing up on the footing beyond this (potential) crack is effective at causing the footing to shear.

So, if the footing is thicker than how far it projects past the wall (h > x), there is no effective shearing force on the footing and it won't shear.

In our example the footing width is 18 in. Using a foundation wall of 8 inches leaves us with a projection of the footing on either side of 18 - 8 divided by 2 = 5 in. So, x = 5 in.

Since the footing thickness h = 8 in. > x = 5 in. ... the footing won't shear; it is thick enough.

In cases where h is not greater than x, we will need to `calculate' whether or not the footing will shear, by actually looking at shear stresses. I will reserve this case for a later example.

3. Bending (of the Footing)

Now let's make sure the footing doesn't break in bending.

If it breaks, it will crack vertically at the face of the wall.

I will check this possibility by calculating the flexural tension stress in the concrete and checking it against an allowable flexural tension stress. This will be an ASD check. Please note that this is for estimation purposes only as the current building codes use Strength Design. But, it will indicate as to whether or not we should use (transverse) reinforcement to resist this breaking effect.

Note: in this design I am assuming that we will be using longitudinal reinforcement, but (hopefully) not transverse reinforcement. (I say `hopefully' because if transverse reinforcement is required, there will be a lot of short pieces of rebar required placed across this 18 in. wide footing along its length.)

To do this check I will introduce the idea of `per foot of footing' ... as that is how we will do a lot of our calculations.

Per Foot of Footing

In a lot of our structural calculations we find it convenient to do our calculations in terms of `per foot' of our structural member, or system. In fact, our line load from above leads us right into this. So, what we will do is look at what is happening along a `foot long' length of our foundation wall. In our present example, then, we have two upside down cantilever beams (upside down diving boards) that are 1 ft wide and 5 in. long.

Let's look at just one of them (since they are identical inside and outside).

The `area load' on the upside down cantilever beam is simply the soil pressure.

From before, f_p = 1492 psf ... which we could now call σ to match our earlier nomenclature.

The line load on one foot of footing acting as a cantilever is ...

ω = σ x trib width = 1492 psf x 1 ft = 1492 plf.

The total force W on the 1 ft wide by 5 in. long cantilever is ...

W = ω L = 1492 plf x (5/12 ft) = 622 lb.

The bending moment of a cantilever beam (see p. 94) is W L / 2 ...

M = W L / 2 = 622 lb (5/12 ft) / 2 = 130 lb-ft.

Let's get it into ... M = 130 lb-ft x 12 in./ft = 1554 lb-in.

Bending Moment Load is ... 1554 lb-in.

For the bending stress ... we need to first discuss some things.

Assuming that we don't use any transverse reinforcement, the footing will be a `plain concrete footing' in the transverse direction. Indeed, this is one of few places in modern construction in which we are allowed to use plain concrete. As such, we need to `not count' the bottom 2 in. of the footing, structurally, where the footing is cast against the earth. (And anyone who has been on a construction site would probably agree why.)

So, the structural thickness of our footing is 8 in. - 2 in. = 6 in.

Our structural section properties, then, of a cantilever beam that is 12 in. wide and 6 in. deep are ...

I = bh³/12 = 12 in. (6 in.)³ / 12 = 216 in.⁴

and S = bh²/ 6 = 12 in. (6 in.)² / 6 = 72 in.³

The flexural stress is,

f_b = M/S = 1554 lb-in. / 72 in.³ = 22 psi.

The allowable flexural tension stress is ... given as ... F_b = 1.6 √f '_c ...

Assuming a specified footing concrete strength of 2500 psi ...

The allowable flexural tension stress, F_b = 1.6 √2500 = 80 psi.

Since f_b = 22 psi is way ≤ F_b = 80 psi ... we don't expect the footing to break, either.

Good, we don't want all those little pieces of transverse reinforcement, unless we really need them.

4. Footing Longitudinal Reinforcement

If the stem wall is relatively short, and if the footing is thicker than the projection of the footing past the face of the stem wall, then the foundation acts mostly like a (stout) compression member. But we will still provide reinforcement to control cracking.

In so doing, the foundation will also be able to act as a lightly reinforced (very long) beam ... to spread loads arriving from the superstructure along the foundation to the bearing soil below, and accommodate potential unevenness in the bearing ability of the soil.

For the footing, we will provide `shrinkage and temperature reinforcement' per the American Concrete Institute (ACI) Code.

Temperature and Shrinkage Reinforcement

Ref: ACI 318 Ch. 7.12.

For slabs and footings, unless a greater amount is required by `flexure' calculations, minimum amounts of reinforcement are required as follows, to control cracking caused by temperature effects and shrinkage of the concrete after being cast. The amounts are shown in terms of ratios of area of steel to gross (total) area of concrete. The areas considered are the with regard to the section across the concrete element being considered.

... for Grade 40 deformed bars ... 0.0020

... for Grade 60 deformed bars ... 0.0018

I generally use the nomenclature ρ _t/s ... where

ρ = A_s / A_g

where

A_s is the total cross section area of the steel at the section

A_g is the total area of the concrete element (gross area, including the area of the steel)

and

t/s, obviously, stands for temperature and shrinkage.

Hence, in the example of a footing that is 8 x 18 ... reinforced with Gr. 60 reinforcement,

0.0018 = A_s / (8 in. x 18 in.) ...

A_s = 0.0018 (8 x 18) in.² = 0.26 in.²

So, we need a total area of steel equal to or exceeding 0.26 square in. of steel reinforcement in the footing.

Common rebar sizes and areas will be discussed in greater detail later.

For now, however, we will limit our size `selection' to #3, #4, and #5.

These # refers to `eights of an inch in diameter' in the English (lb-in. / lb-ft) system.

Hence, a #4 bar is 4/8^th in. = 0.50 in. in diameter.

The cross section area of a #4 bar is ...

a_s = π d² / 4 = π (0.5 in.)² / 4 = 0.196 in.² ( ... often rounded to 0.20 in.²), where,

in general,

a_s is the area of a single bar,

π is our familiar 3.14159,

and,

d is the bar diameter.

The minimum number of bars (n) to meet the t/s requirement in this example, if we want to introduce and use yet another equation, is

A_s = n a_s,

or,

n = A_s / a_s = 0.259 in.² total steel / 0.196 in.² per bar = 1.32 bars (of #4).

We are not going to deal with fractions of bars, so we'll round up to ... 2 bars of # 4.

Since the reinforcement is to control cracking, I generally detail or specify it as follows:

"2 - # 4 bars evenly spaced at mid-depth"

Alternately, we could have `picked' 2 bars of # 4 from the attached table/figure.

Note: we could have also picked a single # 5 bar to meet the requirement. In fact, some prescriptive footing designs do use a single # 5. However, in my opinion, a single bar down the middle(?) of a footing that is 18 in. wide leaves me a bit concerned that it is (not) protecting the whole footing from cracking or breaking up. So, I generally tend to use more but smaller bars.

5. Stem wall Horizontal Reinforcement

I consider a stem wall a `wall' and design in accordance with ACI 318 Ch. 14.

Minimum Reinforcement: (H.S.)

a) ρ _min = 0.0020 for bars not larger than # 5 and spec. yield strength not less than 60,000 psi (Gr. 60)

b) ρ _min = 0.0025 for other deformed bars

and,

bars shall not be spaced farther apart than 18 in. or 3 times the wall thickness.

For the H.S. we may determine the number of bars / spacing in the same manner as the footing, or as I will illustrate with V.S. below.

Following the footing calc example, and limiting ourselves to # 4 or # 5 Gr. 60 bars,

A_s = 0.0020 (30 in. x 8 in.) = 0.48 in.²

Choosing # 4 ...

n = A_s / a_s = 0.48 / 0.20 = 2.4 bars ... round up to 3 - # 4 H.S.

For a detail, consider placing one bar not farther than 6 in. from the bottom, another bar not farther than 6 in. from the top, and the third bar at mid-height.

This will place the bars about 30 - 6 - 6 / 2 = 9 in. apart. Sweet.

If we are on sloping ground where the stem wall may change height, we might want to add language that will still provide the minimum reinforcement if the wall height increases, for example,

H.S.: # 4 with the top bar 6 in. from the top, the bottom bar 6 in. from the bottom, and 9 in. o.c. (3 bars min.).

or

H.S.: # 4 @ 9 in. o.c. with the top bar not less than 6 in. from the top, and the bottom bar not less than 6 in. from the bottom.

or

H.S.: 3 - # 4 with the top bar within the top 6 in., the bottom bar within the bottom 6 in., and in no case spaced more than 9 in. o.c.

While determining the exact number of bars is quite precise, developing a set of details and specifications that is precise, yet also flexible and manageable in the field, is somewhat an art. (It involves people.)

6. Stem wall Vertical Reinforcement

Ref: ACI 318 Ch. 14 (Walls)

Minimum Reinforcement: (V.S.)

a) ρ _min = 0.0012 for bars not larger than # 5 and spec. yield strength not less than 60,000 psi (Gr. 60)

b) ρ _min = 0.0015 for other deformed bars

and

not spaced farther apart than 18 in. or 3 times the wall thickness.

Now, in the case of vertical reinforcement, the `section' involved is a long one, namely, the length of the wall and it's thickness. For calculations, we generally `switch' from using the whole wall as a (gross) section (with some huge number of bars) to the idea of spacing.

Here's an equation.

... s _max = a_s / (ρ _min h)

where

s _max is the maximum spacing of individual vertical bars

a_s is the area of a single bar (the bar size of our choice)

and

h is the wall thickness.

So, in our example, and choosing Gr. 60 steel, and, let's say, this time, # 5 ...

s _max = 0.31 in.² / (0.0012 * 8 in.) = 32.2 in.

(where the 0.31 is calculated or obtained from the figure)

Indeed some designers will use 32 in. (and `disregard' the 18 in. maximum) ... I like to stick to the 18 in. max. spacing.

In that regard # 4 would also work at 18 in. o.c. max ... let's check:

s _max = 0.196 in.² / (0.0012 * 8 in.) = 20.4 in.

Hence, consider,

V.S.: # 4 @ 18 in. o.c., embedded 6 in. into the footing with standard hooks alternating directions.

7. Cover and Spacing

The ACI Code provides limits on bar spacing and minimum cover of concrete.

For cast-in-place concrete refer to ACI 318 7.7 (Concrete protection ... COVER),

and

7.6 for SPACING.

From those sections of the Code, pertinent to the present design example,

Cover requirements are ...

1-1/2 in. for # 5 bar and smaller, exposed to earth or weather ...

3 in. for concrete cast against ... earth (we saw this in our footing design)

Spacing ...

for parallel bars ... min. clear spacing is equal to the bar diameter but not less than 1 in.

In our example spacing limits are easily met ...

And we can handle the cover requirements in a number of ways:

a) dimensions on our detail drawing

b) language on our detail

c) or on our Specifications sheet

8. Other Features

We have now dealt with a lot of `stuff' related to our Residential foundation. And there is a lot more stuff to consider, but for now we need a rest. But, before we take this rest, a few things are worth mentioning. First, we want the horizontal wall and footing longitudinal reinforcement to be continuous around the structure (continuous around corners), for integrity. Foundations are generally a lot longer than pieces of rebar, so the rebar will need to be `spliced'. We'll deal with splicing later. However, it should be noted on the Foundation plan or on General Specifications sheets. Also, by now you have seen mention of reinforcement `grade'. Reinforcement is generally available in Gr. 40 and Gr. 60, referring to 40,000 psi and 60,000 psi yield strengths, respectively (lb-in. system of units). Gr. 60 is becoming commonplace (and yet is 50% stronger), and I generally specify it in almost all my designs. The last time I checked, Gr. 60 was only a tiny bit more expensive than Gr. 40 ... and by now it might even be less expensive. Generally I do not specify more than one GRADE of rebar on a job. (Doing so is only asking for trouble.) Along those lines I generally specify only a few sizes of rebar. Further, while we can calculate the minimum bar spacings, min. footing widths, etc. to inches and decimal inches, for our designs we will round up, or down (in the case of spacing) to the nearest inch, or even inch.

A whole `list' of foundation details is covered in a later lesson.

References

Calculated Footing Width for Residential and Light Commerical Structures, Jeff Filler, Associated Content.

Building Code Requirements for Structural Concrete, ACI 318, American Concrete Institute, P.O. Box 9094, Farmington hills, Michigan, 48333.