*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. 7-22This 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 was increased from 0.55 to 0.60 in the ACI 318-14 standard.
*The total footing thickness minus two inches, per the code. The decrease in required thickness is entirely attributable to the increase in 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 A2/A1 factor which doubles the capacity in this case to 958 kip. Then we consider the difference in 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.
10-#7 for X-axis bending, fy =60 ksi, 3.625” cover**
Axial loads
PD = 180 kip and PL = 120 kip
Moment loads
MD = 80 kip-ft and 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 of 22 inches, which corresponds to a cover of 3.625 inches with #7 bars.
*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 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 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 L, 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.
wc = 0 pcf (set using “custom” weight classification)
Soil density
γs = 0 pcf
Per example:
Parameter
Gurfinkel (1970)
Bowles (1982)**
Köseoğlu (1975)
Wilson (1997)
Chokshi (2009) Ex 2
Chokshi (2009) Ex 3
B (ft)*
20
6
8.2
22
16.4
19.7
L (ft)*
10
6
4.92
6
8.2
16.4
PD (kip)
400
60
89.9
111
292
281
MDx (kip·ft)
1000
120
88.5
69.1
120
2070
MDy (kip·ft)
400
120
111
407
1330
553
*B and 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 are given in the format: (Example result / Calcs.com 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 qs (psf)
17501749
2250022500
77967797
22812264
75287523
1566115662
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 Calcs.com zones to the paper: 1→2, 2→5, 5→1 (Calcs.com zone numbers are on the left).
