Strength Design of a Plain Concrete Footing

Outline

1. Problem Statement
2. Approach
3. Solution (Calculations)
4. Conclusion
5. References

1. Problem Statement

We saw in the last example (here) that even with the minimum amount of reinforcement we were way over strength. Over-strength isn’t necessarily bad, but it does raise the question: what about less reinforcement? Well, according to the `code’, if we have `less than minimum’ reinforcement, we have `plain’ concrete (no reinforcement at all). So, let’s see if the (relatively thin) footing from the previous example is adequate with no reinforcement. We found that a relatively thick footing was adequate without reinforcement; but could we go with less?

2. Approach

From the earlier example we found that for a thickness of 9 in. we could use a width of 32 in., based on the allowable soil bearing pressure and the weight of the footing and the loads from above. I will stick with 9 in. for a minimum thickness as it allows 6 in. of embedment of the vertical reinforcement from the wall and satisfies 3 in. clear cover from the ends of these embedded bars and the footing bottom. Since the footing will be cast against the earth / bearing material, we will discount the bottom 2 inches, structurally, giving us an effective depth (thickness) of 7 inches. We will solve this problem using Strength Design, so we must take the loads and factor them up with load factors. And we will use the `Plain Concrete’ portion(s) of the Code to calculate the plain concrete footing strength.

Loads

Our load factors are … 1.6 for Live … and 1.2 for Dead (and self weight). We’ll get our loads from the other example.

Strength

Our strength and strength reduction factors we’ll get from the ACI Code, Chapter 22 (and 9.3.5) …

Shear strength (nominal), V_n = (4/3) âÂ?Â?f ‘c b h … and …

flexural (tension) strength (nominal) is M_n = 5 âÂ?Â?f ‘_c S.

The strength reduction factor (�) both shear and flexural tension is 0.55.

3. Calculations

Factored Soil Pressure

From the example our loads to the top of the footing were 1350 plf Live and 2350 plf Dead. Our 9 x 32 plain concrete footing will weigh,

… 9/12 ft x 32/12 ft x 145 pcf = 290 plf …

(This is just slightly less than the weight calculated for the same size but with the `reinforced concrete’ specific weight of 150 pcf.)

Our total Dead load is, thus … 2350 + 290 = 2640 plf.

Our factored load is, thus, … 1.2 (2640) + 1.6 (1300) = 5248 (plf).

Our factored soil pressure is � _u = 5248 plf / (32/12 ft) = 1968 psf.

Shear

We will again work with a `per foot of footing’ basis, so each foot of footing has a projection that has a tributary width of 1 ft, so,

… ÏÂ? _u = ÏÂ? _u x 1 ft = 1968 psf x 1 ft = 1968 plf.

For our shear load we will only count the upward soil pressure that acts `outside’ of a potential diagonal tension crack from the face of the wall and intersecting the footing 2 in. up from the bottom. (Remember, we don’t count the bottom 2 inches.) From our sketch we get an effective length for shear of 5 in.

Our (factored) shear load, V_u,is … ÏÂ? _u Ã?Â?_e = 1968 plf (5/12 ft) = 820 lb.

Our (factored) shear strength is … ÏÂ?V_n = 0.55 (4/3) âÂ?Â?2500 (12 in.)(7 in.) = 3080 lb.

Since V_u,= 820 lb âÂ?¤ ÏÂ?V_n = 3080 lb … Good!

And we would hope this to be the case, because we didn’t have any shear reinforcement in the 9 in. thick reinforced footing anyway … even though a plain concrete footing is not exactly the same as a footing without transverse reinforcement.)

Bending

For flexure our `effective cantilever length’ is (in this example) … 12 in.

So,

for our load … M_u = ÏÂ? _u Ã?Â?_e ² / 2 = 1968 plf (1 ft)² / 2 = 984 lb-ft = 11,800 lb-in.

And for our strength, …

S = bh² / 6 = 12 in. (7 in.)² / 6 = 98 in.³, and …

… ÏÂ? M_n = ÏÂ? 5 âÂ?Â?f ‘_c S = 0.55 (5) âÂ?Â?2500 psi (98 in.³) = 13,500 lb-in.

M_u = 11,800 lb-in. âÂ?¤ ÏÂ? M_n = 13,500 lb-in. … GOOD.

So, … wow, we have just shown that our 9 in. deep footing does NOT need transverse reinforcement. I think we will find that this is a `symptom’ of the relatively low soil bearing capacity and thus low factored soil pressure on the footing. This does NOT suggest that any9 x 32 footing does not need transverse reinforcement. We have run this example on relatively weak soil. Wide footings resulting from higher loads on stronger soils will result in higher factored loads in excess of the factored strength.

For example, let’s say we have total loads of 1750 plf Live and 3500 plf Dead and also an increased allowable soil pressure of 2000 psf.

Our necessary footing width is …

… width = ÏÂ? / F_p = (1750 + 3500) plf / 2000 psf = 2.625 ft = 31.5 in. … use 32 in. (Yes, I rigged it that way).

The factored load on the soil is …

… 1.6 x 1750 + 1.2 x 3500 = 7000 plf.

The factored soil pressure is …

… 7000 plf / (32/12 ft) = 2625 psf …

The factored line load on 1 foot of footing projection, acting as a cantilever, is, 2625 psf x 1 ft = 2625 plf.

Using the same footing …

M_u = ÏÂ? _u Ã?Â?_e ² / 2 = 2625 plf (1 ft)² / 2 = 1313 lb-ft = 15,750 lb-in.

Since I just said we’re using the same footing (same strength) …

M_u = 15,750 lb-in. is NOT âÂ?¤ ÏÂ? M_n = 13,500 lb-in. … NOT GOOD.

We would need transverse reinforcement in this (even) more heavily loaded scenario.

4. Conclusion

In this example we showed that a 9 in. x 32 in. footing could be used without transverse reinforcement for our previous `more heavily loaded residential footing’ example. But we also showed that being able to use such a wide footing without transverse reinforcement is a function of the relatively `weak’ soil (low allowable bearing pressure and resulting low soil pressure on the footing). For a great number of foundations higher allowable soil bearing pressures are be used, and where the loads are such as to require wider footings (in this case 32 in.) we will need the transverse reinforcement.

Note: you’re probably sitting there thinking, “… yeah, but 9 in. is still pretty thick for a 32 in. wide footing.” If we go with an 8-in. thick footing, we will need transverse reinforcement (even for the lighter loads, allowable soil pressures, and lighter soils pressures on the footing). I like 9 in. because it allows for 6 in. of embedment of the vertical wall steel and 3 in. of clear, and the footing can be formed using nominal 10 in. dimension lumber. If you wanted to use 8 in. in this example, fine! You will need the transverse reinforcement and you will need to somehow talk the building official into overlooking embedment and / or cover issues for the vertical reinforcement embedded in the footing.

5. References

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