Heavier Residential Footing Design

Outline

1. Introduction
2. Big Thick (Plain) Footing
3. Thinner (Transversely) Reinforced Footing
4. Summary and Discussion
5. Conclusion
6. References

1. Introduction

Now let us consider a more heavily loaded residential footing. The line load to the top of the footing is 3650 plf TL (2350 plf DL + 1300 plf LL). The allowable soil bearing pressure is 1500 psf. The width (thickness) of the foundation wall is 8 in. Our approach thus far in footing design has been basically the following.

1. Make the footing wide enough so the supporting soil doesn't fail,
2. Make the footing thick enough so that it (the footing) doesn't fail in shear, and
3. Add reinforcement (if necessary) so that the footing (projection) doesn't break (from the upward soil pressure).

We will explore two approaches: one will be using a footing that is thick enough so that it can't shear, and in so doing, making sure it doesn't need reinforcement; the other approach being to use a shallower footing, that we know will need reinforcement, and making sure it still doesn't shear. In this second approach the footing thickness will not necessarily be greater than the projection of the footing past the face of the foundation wall, so we will need to calculate shear stresses in the projection. The two approaches are illustrated in the accompanying sketch.

2. Big Thick (Plain) Footing

Our first step in both approaches is the same; we need the footing width. Let's `guess' a footing weight of 300 plf, add it to the line load to the top of the footing, and calculate the minimum footing width. So, TL = 3650 + 300 = 3950 plf.

Using ... f _p = ω / width ... and letting f _p = F _p = 1500 psf,

we get ...

width (min.) = 3950 plf / 1500 psf = 2.633 ft = 31.6 in.

Let's round this up to 32 in.

Now let's see how thick it needs to be. The projection of the footing past the face of the foundation wall is ...

... x = (32 - 8 )/ 2 = 12 in.

So, if the footing is at least as thick (deep) as the projection, it must be 12 in. thick.

Good.

But before we go further, let's make sure this thing doesn't weigh more than our `guess', or at least not way more.

ω _self-weight = γ A = 145 pcf (12/12 ft x 32/12 ft) = 387 plf.

Hmmmm ... heavy thing ...

We better make sure that our width is adequate.

TL = 3650 + 387 = 4037 plf.

... f _p = ω / width = 4037 plf / (32/12 ft) = 1514 plf.

Ughhhhh ... slightly over. Let's go with a 34 in. wide footing.

.. x = (34 - 8) / 2 = 13 ...

So, 34 x 13 footing.

Check again using greater self weight ...

ω _self-weight = γ A = 145 pcf (13/12 ft x 34/12 ft) = 445 plf.

TL = 3650 + 445 = 4095 plf

ω _self-weight = γ A = 145 pcf (13/12 ft x 34/12 ft) = 445 plf.

(I know what you are thinking. Why not be clever and come up with an equation that skips all this iteration and gives us a minimum footing width directly? And I say, go for it!)

Okay, it is wide enough to not fail the soil. It is thick enough to not shear. Now we need to make sure it (the projection) doesn't break off, in bending, like an upside down cantilever.

The length of this so-called upside down cantilever is x = 13 in.

The width of the beam is 12 in. ... (since we like to do these calcs on a per foot of footing basis).

The tributary width of the beam is also 12 in.

The line load on the cantilever is ...

ω = σ x tributary width ... where in this case σ = f _p = 1445 psf.

So, ω = 1445 psf x 1 ft = 1445 plf.

For a cantilever beam,

M = ω ʆ ² / 2 ... where ... ʆ = x, so,

M = 1445 plf (13/12 ft)² / 2 = 848 lb-ft (per foot of footing)

Getting into lb-in. ... 10,175 lb-in.

Using our `flexure' formula, I'll get the extreme fiber tension stress in the concrete. First recall that for `plain' concrete we discount the bottom two inches of the footing (where cast against earth). So, our beam cross section is ... 12 in. wide (from the per foot of footing) and 13 - 2 = 11 in. deep.

... f_b = M/S, ... where S = bh² / 6 = 12 in.(11 in.)² / 6 = 242 in.³

So, f_b = 10,175 lb-in. / 242 in.³ = 42 psi.

I'm going to check it with the `old' ASD allowable stresses for concrete, for which, F_b = 1.6 √f 'c, and let's assume that we have specified 2500 psi concrete.

So,

F_b = 1.6 √2500 psi = 80 psi ...

Since f_b = 42 psi ≤ 80 psi = F_b ... GOOD!

It's thick, and weighs a lot, but it checks out ... and without the need for reinforcement.

Now, it could be argued that we shouldn't use the old allowable stress equations. Okay, I'll check it with Strength Design equations in another lesson.

And note, even though I can now argue that we won't need transverse reinforcement, we'll still spec longitudinal reinforcement.

The min. area of longitudinal reinforcement should be 0.0018 (34 x 13) = 0.796 in.² ...

Using # 4 bars ... n = 0.796 / 0.196 = 4.06 ... use 5 # 4 bars ... or ...

Using # 5 bars ... n = 0.796 / 0.31 = 2.6 ... use 3 # 5 bars ...

3. Thinner (Transversely) Reinforced Footing

FOOTING WIDTH ...

Let's try a thinner (but) reinforced footing (transversely reinforced).

I will use a footing that is 9 in. thick, allowing for 6 in. embedment of the V.S and 3 in. clear from bottom. And let's see if we can get back into the 32 in. width.

It will weigh ... ω _self-weight = γ A = 150 pcf (9/12 ft x 32/12 ft) = 300 plf. Oh, good, we know this will work (for width), since that was our original guess on self weight and that is what got us the 32 in.

FOOTING THICKNESS (AND SHEAR)

For reinforced concrete we will switch over to Strength Design to check shear and bending. As such, we need to have factored loads.

Total Live load ... 1300 plf (given) ...

Total Dead load ... 2350 plf + 300 plf = 2650 plf

Our load factors for reinforced concrete design are ... 1.6 for L and 1.2 for D ...

So,

Our factored load at the bottom of the foundation is ...

... 1.2 (2650) + 1.6 (1300) = 5650 plf.

Our factored soil pressure is ...

σ _u = 5650 plf / width = 5650 plf / (32/12 ft) = 1973 psf ...

It is customary to not count the soil pressure acting of the footing `inside of' the potential diagonal tension shearing crack (see sketch). I will detail the steel at 3.5 in. from the bottom of the footing to reinforcement centerline, and this will give us an effective depth of footing, `d', equal to 9 in. - 3.5 in. = 5.5 in. The soil pressure that will be effective at shearing the footing will be taken to be that acting outside of where the diagonal (45º) tension crack crosses this effective depth ... giving us, first in equation form, and then a `number' ...

... ʆ _e = x - d ...

... 12 in. - 5.5 in. = 6.5 in.

We could call this outer 6.5 in. the effective length of the cantilever beam for shear.

The factored shear load on the footing projection is ...

V_u = ω _u ʆ _e ... where, ...

... ω _u = σ _u x 1 ft = 1973 plf x 1 ft = 1973 plf.

So,

V_u = 1973 plf x 6.5/12 ft = 1069 lb.

Our shear strength is ...

V_c = 2 b d √f 'c = 2 (12 in.)(5.5 in.)√2500 psi = 6600 lb.

Our factored shear strength is ... (using our strength reduction factor of 0.85 for shear in reinforced concrete) ...

... φ V_c = 0.85 (6600 lb) = 5160 lb.

So, since factored load = V_u = 1069 lb. ≤ φ V_c = 5160 lb = factored strength ... GOOD!

That is really good news ... at least if we really wanted to use the 9 in. footing. We'll see that adding reinforcement for flexure is relatively easy, but adding shear reinforcement if a footing would be really nasty.

FOOTING (TRANSVERSE) REINFORCEMENT

As a starting place for reinforcement I use the t/s minimum amount, based on the gross area of concrete. Let's use # 4 bar (it develops bond better).

Our `spacing' equation is ...

s = a _s / ( ρ h ) = 0.196 in.² / (0.0018 x 9 in.) = 12.1 in.

Let's use 12 in. o.c. ...

The reinforcement ratio for bending, based on effective depth, is ...

... ρ = a _s / (s d) = 0.196 in.² / (12 in. 5.5 in.) = 0.00297.

NOTE: this is less than the 0.0033 minimum for flexure. The `code' recognizes that, especially in a lot of foundation designs, we are working with reinforcement ratios less than the `flexure' minimum, and will allow us to do so, as long as we are an extra 33% strong, when it's all over with.

So, let's hope this will be the case, and continue.

For bending we will look at the footing potentially cracking at the face of the wall, so the effective length in bending (the length of our cantilever beam) is ... x, our projection distance.

So, ... for the load ...

M_u = ω _u ʆ_e ² / 2 = 1973 plf (1 ft)² / 2 = 987 lb-ft = 11,838 lb-in.

For strength ...

... ρ = 0.00297

R = ρ f_y (1 - 0.59 ρ f_y / f '_c) = 171 psi ... (using Gr. 60 reinforcement and 2500 psi concrete)

M_n = 61,960 lb-in.

... φ = 0.90 for flexure ...

... φ M_n = 55,760 lb-in. = 4650 lb-ft.

So, we're not quite done, but let's make sure we're headed in the right direction.

Is our load increased 33% (due to our low reinforcement ratio) ... less than or equal our strength? ...

Well, let's see ... is 987 lb-ft x 1.33 = 1313 lb-ft ≤ 4650 lb-ft? ... Yes, Good.

We need to make sure we have enough length of bar to develop the needed bond. The available length to develop the bars is the distance from the face of the wall (where bond most needed) to the end of the bars. We must terminate the bars 3 in. from the sides of the concrete (required concrete cover), so, the available length is 12 in. - 3 in. = 9 in.

From Chapter 12 of the ACI Code (ACI 318) the required development length is ...

ʆ _d = { (3/40) (f_y /√f '_c) (α β γ λ ) / [ (c + k_{t r}) / d_b ] } d_b = 14.4 in. (where [ (c + k_{t r}) / d_b ] is taken to be the maximum value of 2.5) ...

But, the development length may be reduced by multiplying by the amount of steel needed divided by the amount of steel provided provided.

So, ... 14.4 in. (M_u /φ M_n) = 14.4 in. (987 / 4650) = 3.1 in.

So, we need 3.1 in. and we have 9 in. ... GOOD! ... we can develop the needed reinforcement.

The 9 x 32 footing with # 4 @ 12 in. o.c. transverse reinforcement will work!

Now, for longitudinal steel ...

... A_s = 0.0018 (9 x 32) = 0.52 in.² ...

We could get this with 3 # 4 bars or 2 # 5 ...

In my opinion more # 4 bars would better protect the footing ... plus, for convenience it is the same size as the transverse reinforcement.

4. Summary and Discussion

We have now come up with two solutions.

The first solution, that of a thick footing without transverse reinforcement, is ...

13 x 34 w/ ... 5 # 4 or 3 # 5 longitudinal steel.

Our second solution is ...

9 x 32 w/ ... # 4 @ 12 in. o.c. centered 3.5 in. from the bottom transverse, and ... 3 # 4 longitudinal steel.

The first solution will require more excavation (4 in. more); also it will require more concrete. The 13 x 34 footing has a gross area of 3.1 sq. ft. ... versus the second with 2 sq. ft. For a footing length of 250 ft this difference would amount to almost 10 cu yd of concrete.

The thinner footing would require tying some 250 pieces of (32 - 3 - 3 = 26 in. long # 4 bars) across the footing (250 ft long). In the days where rebar was cut on site - this would be the unhappy task of some laborer. However, it is common practice now for rebar to be ordered already cut - so this might now be not so formidable. Indeed, in my practice I have found that for the larger (residential and light commercial footings) the shallower and reinforced option to be favored.

5. Conclusion

In this lesson we have looked at a couple options for a footing design. The first option was for a thicker footing without lateral (transverse) reinforcement. We took advantage of the concept of making it so thick that the projection or toe distance (x) was less than the thickness, so it couldn't shear. Then we checked to make sure it wouldn't also break (in upward bending), and our calcs checked (that it wouldn't). Our second option was to consider at thinner reinforced footing. The reinforcement we added was used to resist the breaking of the footing upward in flexure. And we found that by adding the minimum of reinforcement we were way strong enough. Since the thinner footing was not as thick as the projection, we had to calculate that it wouldn't shear, and our calculation checked (that it wouldn't). The thinner, reinforced footing will save us excavation and concrete, but it will require acquiring and tying a bunch of short pieces of rebar across the footing. For these wider footings (say wider than 24 in.) it seems to be the preference of contractors to use the shallower, reinforced option.

6. References

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