Validation and design examples for concrete spread footings per ACI 318-14

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Example 1 - Square Concentrically Loaded Footing

Reference: James Wight, Reinforced Concrete Mechanics and Design, 7th Edition, 2016, Pearson, Example 15-2, p. 825

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Inputs

ParameterValue
Column dimensionsbb = 18” and β„“\ell =18”
Footing dimensionsBB = 11’-2”, LL = 11’-2”, HH = 32”
Concrete strengthfcβ€²f'_c = 3,000 psi
Allowable Bearing Pressureqaq_a = 6,000 psf
Depth of soil over footinghsoilh_{soil} = 1 ft *
Soil unit weightΞ³s\gamma_s = 135 pcf *
Reinforcement11-#8 each way, fyf_y =60 ksi, 3” cover
Dowels4-#6
LoadsD = 400 kips and L = 270 kips
*In the example, there is 6” of soil and a 6” concrete slab - we represent this with a 12” layer of soil with the average density of the two materials

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Outputs

ResultExampleClearCalcs
Bearing stress qsq_s *123Β ft2124.7Β ft2=98.6%\frac{123 \text{ ft}^2}{124.7 \text{ ft}^2} =98.6\%5910Β psf6000Β psf=98.5%\frac{5910 \text{ psf}}{6000 \text{ psf}} =98.5\%
X and Y axis moment demandMuM_{u} = 954 kip-ftMuxM_{ux} = 954 kip-ft,
MuyM_{uy} = 954 kip-ft
X-axis moment resistance**Ο•Mn\phi M_n = 1070 kip-ftΟ•Mnx\phi M_{nx} = 1080 kip-ft
Y-axis moment resistance**Ο•Mn\phi M_n = 1070 kip-ftΟ•Mny\phi M_{ny} = 1050 kip-ft
X-axis shear demand**VuV_{u} = 204 kipVuxV_{ux} = 201 kip
X-axis shear resistance**Ο•Vc\phi V_{c} = 308 kipΟ•Vnx\phi V_{nx} = 314 kip
Y-axis shear demand**VuV_{u} = 204 kipVuyV_{uy} = 208 kip
Y-axis shear resistance**Ο•Vc\phi V_{c} = 308 kipΟ•Vny\phi V_{ny} = 303 kip
Two-way shear demandvuv_u = 156 psivuv_u = 156 psi
Two-way shear resistanceΟ•vc\phi v_{c} = 164 psiΟ•vn\phi v_{n} = 164 psi
Development length ***β„“d\ell_d = 54.8 inβ„“d\ell_d = 27.5 in
Concrete Bearing StrengthΟ•Bc\phi B_c = 1070 kipΟ•Bc\phi B_c = 1070 kip
*The example works with net bearing stress and only calculates bearing area, while we calculate gross bearing stress and work directly with stresses. The utilization ratios are thus compared and give the same result (the 0.1% difference comes from rounding in the textbook example). **The example takes the average depth in the X and Y directions, while we consider both individually, hence the slight differences *** The development length in the example is calculated using the simplified equation (ACI 318-14, Cl 25.4.2.2) while we use the full equation (ACI 318-14, Cl 25.4.2.3). This significantly reduces development length as we take advantage of the large confinement provided by the 3 inches of cover. Additionally, we take the allowable reduction based on excess reinforcement (ACI 318-14, Cl 25.4.10) which decreases development length further.

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Example 2 - Plain Concrete Square Concentrically Loaded Footing

Reference: Mahmoud Kamara and Lawrence Novak, Simplified Design of Reinforced Concrete Buildings, 2011, Portland Cement Association, Example 7.8.1, p. 7-22 This is a very basic example and only checks bending and the concrete bearing resistance. No soil or shear checks are performed. This example was written for ACI 318-11, with the only difference being that Ο•\phi was increased from 0.55 to 0.60 in the ACI 318-14 standard.

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Inputs

ParameterValue
Column dimensionsbb = 16” and β„“\ell =16”
Footing dimensionsBB = 9’-10”, LL = 9’-10”, HH = 42.6”
Concrete strengthfcβ€²f'_c = 4,000 psi
Allowable bearing pressureqaq_a = 6,000 psf
ReinforcementPlain concrete
Dowels4-#8 bars
Axial loadsPuP_u = 512 kip

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Outputs

ResultExampleClearCalcs
Required effective thickness*hh = 42.6 inhh = 39.1 in
Concrete Bearing Resistance**Ο•Bc\phi B_c = 479 kipΟ•Bc\phi B_c = 1030 kip
*The total footing thickness minus two inches, per the code. The decrease in required thickness is entirely attributable to the increase in Ο•\phi : 0.6/0.55 = 42.6/39.1. ** The reason for the huge difference in capacity is because the design example does not consider the A2/A1\sqrt{A_2/A_1} factor which doubles the capacity in this case to 958 kip. Then we consider the difference in Ο•\phi: 0.6/0.55 = 1.091, which brings the capacity to 1045 kip. The 15 kip difference can then be attributed to the fact that we remove the area of dowels in the concrete bearing area.

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Example 3 - Rectangular Eccentrically Loaded Footing (Uniaxial)

Reference: James Wight, Reinforced Concrete Mechanics and Design, 7th Edition, 2016, Pearson, Example 15-4, p. 833

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Inputs

ParameterValue
Column dimensionsbb = 16” and β„“\ell =16”
Footing dimensionsBB = 10’, LL = 12’, HH = 26”
Concrete strengthfcβ€²f'_c = 3,500 psi
Allowable bearing pressureqaq_a = 4,000 psf
Depth of soil over footinghsoilh_{soil} = 1 ft *
Soil unit weightΞ³s\gamma_s = 235 pcf *
Reinforcement10-#7 for X-axis bending, fyf_y =60 ksi, 3.625” cover**
Axial loadsPDP_D = 180 kip and PLP_L = 120 kip
Moment loadsMDM_D = 80 kip-ft and MLM_L = 60 kip-ft
*In the example, there is 6” of soil and a 6” concrete slab, and a 100 psf surcharge - we represent this with a 12” layer of soil with the average density of the two materials and add 100 pcf to represent the surcharge. **The example uses a value of dd of 22 inches, which corresponds to a cover of 3.625 inches with #7 bars.

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Outputs

ResultExampleClearCalcs
Bearing stress qsq_s *11Β ft12Β ft=91.7%\frac{11 \text{ ft}}{12 \text{ ft}} =91.7\%3640Β psf4000Β psf=91.0%\frac{3640 \text{ psf}}{4000 \text{ psf}} =91.0\%
X axis moment demandMuM_{u} = 563 kip-ftMuxM_{ux} = 564 kip-ft
X-axis moment resistanceΟ•Mn\phi M_n = 580 kip-ftΟ•Mnx\phi M_{nx} = 579 kip-ft
X-axis shear demand**VuV_{u} = 147 kipVuxV_{ux} = 139 kip
X-axis shear resistanceΟ•Vc\phi V_{c} = 234 kipΟ•Vnx\phi V_{nx} = 234 kip
Two-way shear demand***vuv_u = 132 psivuv_u = 138 psi
Two-way shear resistanceΟ•vc\phi v_{c} = 177 psiΟ•vn\phi v_{n} = 177 psi
*The example works with net bearing stress and only calculates the length of footing required, while we calculate gross bearing stress and work directly with stresses. The utilization ratios are thus compared and give essentially the same result - the 0.7% difference comes from rounding errors, and the example ignoring the weight of soil replaced by the column, while we consider it. ** The discrepancy comes from the example conservatively using a rectangular stress distribution to simplify calculations, whereas we use the more accurate trapezoidal distribution. ***This discrepancy comes from the example using a depth of 22 inches, which we set as the X-axis reinforcement depth, however, when considering the Y-axis reinforcement which will be above the X-axis bars, we get an average shear depth of 21.5 inches. Using an average depth of 22 inches we also find vuv_u = 132 psi.

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Example 4 - Soil Pressure for Footings Under Biaxial Loading

Reference: Sanket Rawat, Ravi Kant Mittal and G. Muthukumar, Isolated Rectangular Footings under Biaxial Bending: A Critical Appraisal and Simplified Analysis Methodology, 2020, ASCE. This paper compiled 13 different design examples from various sources for footings under biaxial bending. We picked 5 examples, outlined in the table below, which cover a variety of biaxiality conditions. An emphasis added on Zone 2 examples as they are by far the most complex. Only the maximum bearing pressure is provided, but it remains a good validation source. Since only the footing plan dimensions, total axial load and moments are provided, the following parameters are set for all examples, and only LL, BB and loads are changed as indicated. Note that the units in the paper are all SI-based, but we convert them here to imperial units.

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Inputs

For all examples:
ParameterValue
Concrete densitywcw_c = 0 pcf (set using β€œcustom” weight classification)
Soil densityΞ³s\gamma_s = 0 pcf
Per example:
ParameterGurfinkel (1970)Bowles (1982)**Kâseoğlu (1975)Wilson (1997)Chokshi (2009) Ex 2Chokshi (2009) Ex 3
BB (ft)*2068.22216.419.7
LL (ft)*1064.9268.216.4
PDP_D (kip)4006089.9111292281
MDxM_{Dx} (kipΒ·ft)100012088.569.11202070
MDyM_{Dy} (kipΒ·ft)4001201114071330553
*BB and LL are inversed in the paper’s terminology. We are using ClearCalc’s definition, where B is the dimension parallel to the X-axis. **In the Rawat et al. paper, there appears to have been some round-off error when converting the values from the original values in imperial units to SI units. Using the original imperial values however yields a perfect match.

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Outputs

Outputs are given in the format: (Example result / ClearCalcs result). Since the examples are all results from different approximations, there are some discrepancies. These are most pronounced where the approximation was made with graphical methods.
ResultGurfinkel (1970)Bowles (1982)Kâseoğlu (1975)Wilson (1997)Chokshi (2009) Ex 2Chokshi (2009) Ex 3
Max Bearing stress qsq_s (psf)17491750\frac{1749}{1750} 2250022500\frac{22500}{22500}77977796\frac{7797}{7796}22642281\frac{2264}{2281}75237528\frac{7523}{7528}1566215661\frac{15662}{15661}
Loading Zone*2 / 25/52/22/23/34/4
*The paper uses different numbers to identify eccentricity zones. To match from ClearCalcs zones to the paper: 1β†’21 \rightarrow 2, 2β†’52 \rightarrow 5, 5β†’15 \rightarrow 1 (ClearCalcs zone numbers are on the left).