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

Example 1 - Square Concentrically Loaded Footing

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

Inputs

Parameter Value Column dimensions b b b = 18" and ℓ \ell ℓ =18" Footing dimensions B B B = 11'-2", L L L = 11'-2", H H H = 32" Concrete strength f c ′ f'_c fc′​ = 3,000 psi Allowable Bearing Pressure q a q_a qa​ = 6,000 psf Depth of soil over footing h s o i l h_{soil} hsoil​ = 1 ft * Soil unit weight γ s \gamma_s γs​ = 135 pcf * Reinforcement 11-#8 each way, f y f_y fy​ =60 ksi, 3" cover Dowels 4-#6 Loads D = 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

Outputs

Result Example ClearCalcs Bearing stress q s q_s qs​ * 123 ft 2 124.7 ft 2 = 98.6 % \frac{123 \text{ ft}^2}{124.7 \text{ ft}^2} =98.6\% 124.7 ft2123 ft2​=98.6% 5910 psf 6000 psf = 98.5 % \frac{5910 \text{ psf}}{6000 \text{ psf}} =98.5\% 6000 psf5910 psf​=98.5% X and Y axis moment demand M u M_{u} Mu​ = 954 kip-ft M u x M_{ux} Mux​ = 954 kip-ft, M u y M_{uy} Muy​ = 954 kip-ft X-axis moment resistance** ϕ M n \phi M_n ϕMn​ = 1070 kip-ft ϕ M n x \phi M_{nx} ϕMnx​ = 1080 kip-ft Y-axis moment resistance** ϕ M n \phi M_n ϕMn​ = 1070 kip-ft ϕ M n y \phi M_{ny} ϕMny​ = 1050 kip-ft X-axis shear demand** V u V_{u} Vu​ = 204 kip V u x V_{ux} Vux​ = 201 kip X-axis shear resistance** Y-axis shear demand** V u V_{u} Vu​ = 204 kip V u y V_{uy} Vuy​ = 208 kip Y-axis shear resistance** ϕ V c \phi V_{c} ϕVc​ = 308 kip ϕ V n y \phi V_{ny} ϕVny​ = 303 kip Two-way shear demand v u v_u vu​ = 156 psi v u v_u vu​ = 156 psi Two-way shear resistance ϕ v c \phi v_{c} ϕvc​ = 164 psi ϕ v n \phi v_{n} ϕvn​ = 164 psi Development length *** ℓ d \ell_d ℓd​ = 54.8 in ℓ d \ell_d ℓd​ = 27.5 in Concrete Bearing Strength ϕ B c \phi B_c ϕBc​ = 1070 kip ϕ B c \phi B_c ϕBc​ = 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.

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 ϕ \phi ϕ was increased from 0.55 to 0.60 in the ACI 318-14 standard.

Inputs

Parameter Value Column dimensions b b b = 16" and ℓ \ell ℓ =16" Footing dimensions B B B = 9'-10", L L L = 9'-10", H H H = 42.6" Concrete strength f c ′ f'_c fc′​ = 4,000 psi Allowable bearing pressure q a q_a qa​ = 6,000 psf Reinforcement Plain concrete Dowels 4-#8 bars Axial loads P u P_u Pu​ = 512 kip

Outputs

Result Example ClearCalcs Required effective thickness* h h h = 42.6 in h h h = 39.1 in Concrete Bearing Resistance** ϕ B c \phi B_c ϕBc​ = 479 kip ϕ B c \phi B_c ϕBc​ = 1030 kip

*The total footing thickness minus two inches, per the code. The decrease in required thickness is entirely attributable to the increase in ϕ \phi ϕ : 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 A 2 / A 1 \sqrt{A_2/A_1} A2​/A1​ ​ factor which doubles the capacity in this case to 958 kip. Then we consider the difference in ϕ \phi ϕ: 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.

Example 3 - Rectangular Eccentrically Loaded Footing (Uniaxial)

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

Inputs

Parameter Value Column dimensions b b b = 16" and ℓ \ell ℓ =16" Footing dimensions B B B = 10', L L L = 12', H H H = 26" Concrete strength f c ′ f'_c fc′​ = 3,500 psi Allowable bearing pressure q a q_a qa​ = 4,000 psf Depth of soil over footing h s o i l h_{soil} hsoil​ = 1 ft * Soil unit weight γ s \gamma_s γs​ = 235 pcf * Reinforcement 10-#7 for X-axis bending, f y f_y fy​ =60 ksi, 3.625" cover** Axial loads P D P_D PD​ = 180 kip and P L P_L PL​ = 120 kip Moment loads M D M_D MD​ = 80 kip-ft and M L M_L ML​ = 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 d d d of 22 inches, which corresponds to a cover of 3.625 inches with #7 bars.

Outputs

Result Example ClearCalcs Bearing stress q s q_s qs​ * 11 ft 12 ft = 91.7 % \frac{11 \text{ ft}}{12 \text{ ft}} =91.7\% 12 ft11 ft​=91.7% 3640 psf 4000 psf = 91.0 % \frac{3640 \text{ psf}}{4000 \text{ psf}} =91.0\% 4000 psf3640 psf​=91.0% X axis moment demand M u M_{u} Mu​ = 563 kip-ft M u x M_{ux} Mux​ = 564 kip-ft X-axis moment resistance ϕ M n \phi M_n ϕMn​ = 580 kip-ft ϕ M n x \phi M_{nx} ϕMnx​ = 579 kip-ft X-axis shear demand** V u V_{u} Vu​ = 147 kip V u x V_{ux} Vux​ = 139 kip X-axis shear resistance ϕ V c \phi V_{c} ϕVc​ = 234 kip ϕ V n x \phi V_{nx} ϕVnx​ = 234 kip Two-way shear demand*** v u v_u vu​ = 132 psi v u v_u vu​ = 138 psi Two-way shear resistance

*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 u v_u vu​ = 132 psi.

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 ootings 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 L L L, B B B 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.

Inputs

For all examples:

Parameter Value Concrete density w c w_c wc​ = 0 pcf (set using "custom" weight classification) Soil density γ s \gamma_s γs​ = 0 pcf

Per example:

Parameter | Gurfinkel (1970) | Bowles (1982)** | Köseoğlu (1975) | Wilson (1997) | Chokshi (2009) Ex 2 | Chokshi (2009) Ex 3

Result Gurfinkel (1970) Bowles (1982) Köseoğlu (1975) Wilson (1997) Chokshi (2009) Ex 2 Chokshi (2009) Ex 3 B B B (ft)* 20 6 8.2 22 16.4 19.7 L L L (ft)* 10 6 4.92 6 8.2 16.4 P D P_D PD​ (kip) 400 60 89.9 111 292 281 M D x M_{Dx} MDx​ (kip*ft) 1000 120 88.5 69.1 120 2070 M D y M_{Dy} MDy​ (kip*ft) 400 120 111 407 1330 553

* B B B and L L L 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.

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.

Result Gurfinkel (1970) Bowles (1982) Köseoğlu (1975) Wilson (1997) Chokshi (2009) Ex 2 Chokshi (2009) Ex 3 Max Bearing stress q s q_s qs​ (psf) 1749/1750 22500/22500 7797/7796 2264/2281 7523/7528 15662/15661 Loading Zone* 2 / 2 5/5 2/2 2/2 3/3 4/4

*The paper uses different numbers to identify eccentricity zones. To match from ClearCalcs zones to the paper: 1 → 2 1 \rightarrow 2 1→2, 2 → 5 2 \rightarrow 5 2→5, 5 → 1 5 \rightarrow 1 5→1 (ClearCalcs zone numbers are on the left).