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 ell\\ell =18”
Footing dimensionsBB = 11’-2”, LL = 11’-2”, HH = 32”
Concrete strengthf′_cf'\_c = 3,000 psi
Allowable Bearing Pressureq_aq\_a = 6,000 psf
Depth of soil over footingh_soilh\_{soil} = 1 ft *
Soil unit weightgamma_s\\gamma\_s = 135 pcf *
Reinforcement11-#8 each way, f_yf\_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 q_sq\_s *frac123textft2124.7textft2=98.6\\frac{123 \\text{ ft}^2}{124.7 \\text{ ft}^2} =98.6\\%frac5910textpsf6000textpsf=98.5\\frac{5910 \\text{ psf}}{6000 \\text{ psf}} =98.5\\%
X and Y axis moment demandM_uM\_{u} = 954 kip-ftM_uxM\_{ux} = 954 kip-ft,
M_uyM\_{uy} = 954 kip-ft
X-axis moment resistance**phiM_n\\phi M\_n = 1070 kip-ftphiM_nx\\phi M\_{nx} = 1080 kip-ft
Y-axis moment resistance**phiM_n\\phi M\_n = 1070 kip-ftphiM_ny\\phi M\_{ny} = 1050 kip-ft
X-axis shear demand**V_uV\_{u} = 204 kipV_uxV\_{ux} = 201 kip
X-axis shear resistance**phiV_c\\phi V\_{c} = 308 kipphiV_nx\\phi V\_{nx} = 314 kip
Y-axis shear demand**V_uV\_{u} = 204 kipV_uyV\_{uy} = 208 kip
Y-axis shear resistance**phiV_c\\phi V\_{c} = 308 kipphiV_ny\\phi V\_{ny} = 303 kip
Two-way shear demandv_uv\_u = 156 psiv_uv\_u = 156 psi
Two-way shear resistancephiv_c\\phi v\_{c} = 164 psiphiv_n\\phi v\_{n} = 164 psi
Development length ***ell_d\\ell\_d = 54.8 inell_d\\ell\_d = 27.5 in
Concrete Bearing StrengthphiB_c\\phi B\_c = 1070 kipphiB_c\\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 phi\\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 ell\\ell =16”
Footing dimensionsBB = 9’-10”, LL = 9’-10”, HH = 42.6”
Concrete strengthf′_cf'\_c = 4,000 psi
Allowable bearing pressureq_aq\_a = 6,000 psf
ReinforcementPlain concrete
Dowels4-#8 bars
Axial loadsP_uP\_u = 512 kip

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Outputs

ResultExampleClearCalcs
Required effective thickness*hh = 42.6 inhh = 39.1 in
Concrete Bearing Resistance**phiB_c\\phi B\_c = 479 kipphiB_c\\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 phi\\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 sqrtA_2/A_1\\sqrt{A\_2/A\_1} factor which doubles the capacity in this case to 958 kip. Then we consider the difference in phi\\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 ell\\ell =16”
Footing dimensionsBB = 10’, LL = 12’, HH = 26”
Concrete strengthf′_cf'\_c = 3,500 psi
Allowable bearing pressureq_aq\_a = 4,000 psf
Depth of soil over footingh_soilh\_{soil} = 1 ft *
Soil unit weightgamma_s\\gamma\_s = 235 pcf *
Reinforcement10-#7 for X-axis bending, f_yf\_y =60 ksi, 3.625” cover**
Axial loadsP_DP\_D = 180 kip and P_LP\_L = 120 kip
Moment loadsM_DM\_D = 80 kip-ft and M_LM\_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 q_sq\_s *frac11textft12textft=91.7\\frac{11 \\text{ ft}}{12 \\text{ ft}} =91.7\\%frac3640textpsf4000textpsf=91.0\\frac{3640 \\text{ psf}}{4000 \\text{ psf}} =91.0\\%
X axis moment demandM_uM\_{u} = 563 kip-ftM_uxM\_{ux} = 564 kip-ft
X-axis moment resistancephiM_n\\phi M\_n = 580 kip-ftphiM_nx\\phi M\_{nx} = 579 kip-ft
X-axis shear demand**V_uV\_{u} = 147 kipV_uxV\_{ux} = 139 kip
X-axis shear resistancephiV_c\\phi V\_{c} = 234 kipphiV_nx\\phi V\_{nx} = 234 kip
Two-way shear demand***v_uv\_u = 132 psiv_uv\_u = 138 psi
Two-way shear resistancephiv_c\\phi v\_{c} = 177 psiphiv_n\\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 v_uv\_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 densityw_cw\_c = 0 pcf (set using “custom” weight classification)
Soil densitygamma_s\\gamma\_s = 0 pcf
Per example: |Parameter|Gurfinkel (1970)|Bowles (1982)**|Köseoğlu (1975)|Wilson (1997)|Chokshi (2009) Ex 2|Chokshi (2009) Ex 3| |---|---|---|---|---|---|---|---| |BB (ft)*|20|6|8.2|22|16.4|19.7|| |LL (ft)*|10|6|4.92|6|8.2|16.4|| |P_DP\_D (kip)|400|60|89.9|111|292|281|| |M_DxM\_{Dx} (kip*ft)|1000|120|88.5|69.1|120|2070|| |M_DyM\_{Dy} (kip*ft)|400|120|111|407|1330|553|| *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)Köseoğlu (1975)Wilson (1997)Chokshi (2009) Ex 2Chokshi (2009) Ex 3
Max Bearing stress q_sq\_s (psf)1749/175022500/225007797/77962264/22817523/752815662/15661
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: 1rightarrow21 \\rightarrow 2, 2rightarrow52 \\rightarrow 5, 5rightarrow15 \\rightarrow 1 (ClearCalcs zone numbers are on the left).