Big Column Footing Part 2

Draft - Comments Welcome

1. Introduction

Continued from Part 1 (here) where we already determined the footing width and with an assumed footing thickness demonstrated that it won't shear (beam shear). The footing thickness is 12 in. and we assumed that the flexural reinforcement would be located at an effective depth of 8.5 in. The compressive strength of the concrete is 3000 psi and we already determined a factored soil pressure (acting upward) of 2338 psf.

2. Approach

Here's the approach ...

1. Determine the effective length for flexure,
2. Determine the factored line load on either the whole `side' breaking off, or a 1 ft wide strip,
3. Determine the factored bending moment, M_u,
4. For reinforcement start with the temp / shrinkage minimum amount,
5. Determine the reinforcement ratio for flexure (if the t/s min amount of reinforcement ends up less that the min. amount for flexure, which it often does, then the footing needs to be an extra 33% over-strong, which it often ends up being),
6. Determine the factored strength of the footing,
7. Do the design check ... is M_u ≤ φ M_n ... or ... 1.33 M_u ≤ φ M_n? ... (Is factored load less than or equal to factored strength?)
8. If the answer is `no', then we need to do something; if the answer is `yes' ... we're good. (Unless we are so `way' good that it might be worth investigating making something more modest, such as thickness).

Let's do it ...

If the footing breaks in flexure it will do so at the face of the support. Thus the length of the footing acting like an upside down cantilever beam is 19 in. (half the footing width minus half the column bearing distance). Let's do the calc on a `per foot of footing' basis, like we did beam shear in Part 1 of this example.

So, ω _u = σ_u x 1 ft = 2338 plf.

... l _{eff, flex} = 19 in. = 1.58 ft.

M_u = ω _u (l _{eff, flex})² / 2 = 2338 plf (1.58 ft)² / 2 = 2931 lb-ft ... = 35,200 lb-in.

Now for strength ...

We will put at least the min. amount of reinforcement for temp/shrinkage effects (at least that's what I do) ...

Temp/shrinkage reinforcement is based on the full thickness of the footing ...

Using Gr. 60 rebar ... ACI 318 specifies a min. ratio of 0.0018.

Looking for a minute at a section of the whole footing, ...

A_{s min} = 44 in. x 12 in. x 0.0018 = 0.95 sq. in.

We can get this with 0.95 in. sq. / 0.196 sq in per # 4 = ... 4.84 # 4 bars (use 5 - # 4) ... or ...

... 0.95 / 0.31 = 3.06 ... use 4 - # 5 bars.

Four bars (each way) will be about 11 in. o.c. ... let's do it ... (use the 4 - # 5s).

Now let's calc the flexural reinforcement ratio, based on rounding up to a whole number of bars, and putting the rebar at d = 8.5 in.

... ρ _flex = ( 4 x 0.31) / (44 x 8.5) = 0.0033.

Whoa, that is sweet ... 0.0033 is the min. amount of flexural reinforcement required for Gr. 60 rebar.

So, we don't need to require that the footing be an extra 33 percent strong.

Our equation for factored flexural strength is ...

φ M_n = φ b d² R

... where

φ = the strength reduction factor for flexure ... 0.90,

M_n = the nominal or perfect world concrete strength in flexure ... either for the whole side of the beam, or per foot, whatever we decide,

b = the whole width, or 12 in., whatever we decide,

d = effective depth, 8.5 in. in this case,

R = ρ f_y (1 - 0.59 ρ f_y / f '_c)

where R is the so-called `Resistance Factor',

f_y is the yield strength of the rebar, and

f '_c is our concrete 28-day compressive strength.

We have already started the bending calc on the per foot of footing basis, so let's continue.

R = 190 psi. The above equation for R is cumbersome enough that you will want to either use a table of already-cranked-out values or program a spreadsheet or other program to crank it out for you.

So,

φ M_n = φ b d² R = 0.90 (12 in.)(8.5 in.)² 190 psi = 148,300 lb-in.

Is M_u = 35,200 lb- in. ≤ φ M_n = 148,300 lb-in.? ... Yes! ... WAY YES.

We are so much stronger in both shear and flexure that we might want to investigate a shallower footing, or, if it would serve some other purpose, raising the rebar up a bit.

But, either before, or after, we also need to investigate punching shear.

Conclusion

Nice; our 44 in. x 44 in. x 12 in. footing with 4 - # 5 Gr. 60 rebar each way placed 3 in. clear from the bottom is GOOD.

Note that had we picked the 5 - # 4 bars we would probably have fell just a bit under the 0.0033 min. for flexure. Then we would have had to check the strength against the 1.33 times the load, which still would probably have been good. Another possibility would be to have raised the rebar up a bit, reducing the `d' and thus increasing the reinforcement ratio. Sometimes we end up tempted to do stuff like that ... making something a tiny bit (or big bit) weaker, just to make the numbers come out more pleasant. In real life I shy away of purposefully making something weaker.

So good, in fact, that we might want to investigate 10 in. thickness. (Homework problem ... :).

References

Big Column Footing Part 1, Jeff Filler, Associated Content.

Punching Shear Example 2, Jeff Filler, Associated Content.

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