Optimal Area for a Rectangular Isolated Footing with an Eccentric Column and Partial Ground Compression

Featured Application

This study is a practical tool for Structural Engineers to obtain the minimum contact area with the ground for rectangular isolated footings with an eccentric column.

Abstract

This manuscript aims to present a novel model to find the optimal area of a rectangular isolated footing with an eccentric column, taking into account that the footing is partially supported; that is, one part of the contact surface is compressed and the other part has zero pressure. The methodology is developed by integration and can also be verified using the geometric properties of a triangular-based pyramid to determine the axial load, the moments in the X and Y axes in terms of the available allowable soil pressure, the footing sides, the greatest distance on one of its sides in the X-direction where it crosses the neutral axis, the greatest distance on one of its sides in the Y-direction where it crosses the neutral axis, and the coordinates at the base of the footing. Four types of numerical problems are shown to find the optimal area of a rectangular footing with an eccentric column subjected to biaxial bending: (1) the column in the center of the footing; (2) the column on the edge of the footing in the X-direction; (3) the column on the edge of the footing in the Y-direction; and (4) the column in the corner of the footing. A comparison is presented of the new model against a model proposed by another author. The new model presents a reduction of up to 42.37% for the column in the center of the footing and up to 40.32% for the column in the corner of the footing compared to the model by the other authors. Therefore, the new model will be of great help to professionals in foundation design.

1. Introduction

The main goal in structure design is to find the best structural solution (dimensions and reinforcing steel) to support the applied loads. The current footing sizing process is time-consuming, and the designer usually completes the process after several iterations. The dimensions are obtained from a procedure of successive approximations. The sizing of the contact surface of the footing with the ground is usually carried out by approximating the pressure transmitted by the ground as a linear distribution. This hypothesis is more applicable in very consolidated rocky or sandy and clay soils, and in both of the latter two, when they are normally consolidated, a parabolic pressure distribution may be closer to real conditions.

Figure 1 shows the contact pressure distribution under the base of the foundation subjected to a concentrated load located at the center of gravity of the foundation under clay and sandy soils for rigid foundations. Figure 1a,b show rigid foundations with uniform settlements and variable pressures in contact with the ground. Furthermore, since actual extended shoes approach perfect stiffness, the contact pressure distribution is not uniform. However, for simplicity, the contact pressure distribution is assumed to be uniform to facilitate the calculation of the bearing capacity and settlement, as shown in Figure 1c. The error due to this assumption is not significant [1].

The main studies by different researchers on reinforced concrete foundation structures published in the last ten years include a study by Cunha and Albuquerque [2], who presented a summary of the developments in foundation engineering that have taken place in Brazil since its early beginnings up to the present. Hassaan [3] proposed an optimal design for shallow foundations in sandy soils for machinery. Momeni et al. [4] developed a predictive model to determine the axial load capacity of piles and its distribution based on an artificial neural network. Camero [5] presented a methodology for the design of slabs on the ground for industrial pavements and floors using two-dimensional finite elements, including the ground in the design. Luévanos-Rojas [6] studied a comparison of rectangular isolated footings versus circular isolated footings, the latter being cheaper. Rezaei et al. [7] experimentally studied the bearing capacity of thin-walled, shallow foundations using artificial intelligence. Kassouf et al. [8] obtained the horizontal load–displacement curves and horizontal reaction coefficients for highly porous and collapsible soil for a foundation with piles transversally loaded on top. Luévanos-Rojas et al. [9] studied an optimal model under the minimum-cost criterion for the design of rectangular footings, assuming that the soil pressure is linear. Khatri et al. [10] investigated the settlement behavior and soil pressure in square and rectangular footings resting on sandy soils. Munévar-Peña et al. [11] included soil properties with random uncertainty in a geostatistical analysis applied to shallow foundations under the reliability design. Rodrigo-Garcia and Rocha-de Albuquerque [12] investigated a nonlinear behavior model applied to predict the settlement of deep foundations. Rawat and Mittal [13] presented an optimal design for isolated footings with an eccentric column. Zhang et al. [14] presented the bearing behavior of an isolated footing with a column. Liu and Jiang [15] investigated the consolidation and deformation characteristics of soft rocks and soils in foundations with a hydrological humidity environment to resolve this problem. Gnananandarao et al. [16] developed a multivariable regression analysis and artificial neural networks to estimate the bearing capacity and settlement in skirt footings with multiple edges on sandy soils. Al-Ansari and Afzal [17] developed a simplified model for the design of irregularly shaped footings supporting a square column. Al-Abbas et al. [18] experimentally studied the elastic deformation of an isolated foundation. López-Machado et al. [19] studied a comparison of two reinforced concrete buildings of six levels, considering the soil–structure interaction. Komolafe et al. [20] estimated circular and square footings on non-cohesive soils from a structural, geotechnical, and construction cost approach. Ekbote and Nainegali [21] investigated two interfering asymmetrical footings, considering various widths and the impact of embedment thickness to increase the ultimate load capacity and restrict settlement inside the working interval. Solorzano and Plevris [22] estimated the design of a rectangular footing with minimal costs through genetic algorithms using the ACI 318-19 standard of the American Concrete Institute [23]. Waheed et al. [24] proposed a parametric investigation and its application for the minimum-cost design of footings through a practical metaheuristic tool. Helis et al. [25] obtained the bearing capacity of a circular footing resting on sand through two reinforcing systems, a grid anchor, and a geogrid.

The current scientific studies closest to the topic studied here include the following: Vela-Moreno et al. [26] investigated the minimum partially supported ground contact area for rectangular isolated footings with a column centered on the footing. Luévanos-Rojas [27] proposed an optimal model for rectangular isolated footings with an eccentric column, considering that the contact area with the ground is completely supported.

This study presents the optimal area for rectangular isolated footings with an eccentric column, taking into account that the footing is partially supported (one part of the contact surface is compressed and the other part has zero pressure). The methodology is developed by integration and can also be verified using the geometric properties of a pyramid with a triangular base to determine P (vertical axial load), M_xT (moment in the X-axis), and M_yT (moment in the Y-axis) in terms of σ_max (available allowable soil pressure), L_x (footing side in X-direction), L_y (footing side in Y-direction), L_x1 (greatest distance on one of its sides in the X-direction where it crosses the neutral axis), L_y1 (greatest distance on one of its sides in the Y-direction where it crosses the neutral axis), and x and y (coordinates at the base of the footing). Four types of numerical problems are shown to find the optimal area of a rectangular isolated footing with an eccentric column under biaxial bending: Problem 1: the column centered on the footing; Problem 2: the column on the edge of the footing in the X-direction; Problem 3: the column on the edge of the footing in the Y-direction; and Problem 4: the column at the corner of the footing. A comparison is presented of the new model against a model proposed by another author to observe the differences.

2. Formulation of the Model

The loads and moments on each structural member (beams, columns, and foundations) can be estimated from a structural analysis using well-known methods including dead, live, and accidental loads.

Figure 2 shows a rectangular isolated footing with an eccentric column under biaxial bending at the base of the column; the soil pressure distribution is linear and the footing rests on elastic soils.

The biaxial bending equation is

$$\sigma ={\displaystyle \frac{P}{A}}+{\displaystyle \frac{M_{xT}y}{I_x}}+{\displaystyle \frac{M_{yT}x}{I_y}},$$

(1)

where σ = the stress provided by the soil in any part of the footing (kN/m²); P = the vertical axial load (kN); A = the soil contact surface at the bottom of the footing (m²); M_xT = the total moment on the X-axis (kN-m); M_yT = the total moment on the Y-axis (kN-m); I_x = the moment of inertia on the X-axis (m⁴); I_y = the moment of inertia on the Y-axis (m⁴); and x and y = the coordinates at the base of the footing, where the ground pressure on the footing can be obtained (m).

By substituting A = L_xL_y, I_x = L_xL_y³/12, I_y = L_yL_x³/12, M_xT = M_x + Py_c, M_yT = M_y + Px_c, and the corresponding coordinates at each corner of the footing into Equation (1), the stresses are obtained as follows:

$${\sigma }_1={\displaystyle \frac{P}{L_x{L}_y}}+{\displaystyle \frac{6\left(M_x+P{y}_c\right)}{L_x{L_y}^2}}+{\displaystyle \frac{6\left(M_y+P{x}_c\right)}{{L_x}^2{L}_y}},$$

(2)

$${\sigma }_2={\displaystyle \frac{P}{L_x{L}_y}}+{\displaystyle \frac{6\left(M_x+P{y}_c\right)}{L_x{L_y}^2}}-{\displaystyle \frac{6\left(M_y+P{x}_c\right)}{{L_x}^2{L}_y}},$$

(3)

$${\sigma }_3={\displaystyle \frac{P}{L_x{L}_y}}-{\displaystyle \frac{6\left(M_x+P{y}_c\right)}{L_x{L_y}^2}}-{\displaystyle \frac{6\left(M_y+P{x}_c\right)}{{L_x}^2{L}_y}},$$

(4)

$${\sigma }_4={\displaystyle \frac{P}{L_x{L}_y}}-{\displaystyle \frac{6\left(M_x+P{y}_c\right)}{L_x{L_y}^2}}+{\displaystyle \frac{6\left(M_y+P{x}_c\right)}{{L_x}^2{L}_y}}.$$

(5)

Equation (1) assumes that the contact area of the soil on the footing works completely in compression and is not valid when P (the resultant force) is located outside of the central nucleus, a well-known and delimited area designated as an uncompressed zone.

Figure 3 shows the complete eccentricity zone resulting from the entire base of the rectangular isolated footing.

Figure 4 presents five possible cases for a rectangular isolated footing subjected to biaxial bending.

Case I: The contact area of the footing with the ground is completely supported. The stresses generated by the soil on the footing are obtained using the biaxial bending equation.

Cases II, III, IV, and V: The contact area of the footing with the ground is partially supported by the geometric properties of a pyramid with a triangular base. P (the vertical axial load), M_xT (the moment in the X-axis), and M_yT (the moment in the Y-axis) are obtained. The stresses generated by the soil on the footing are obtained by means of the general equation of the stress plane from three known points and by solving the determinant.

The three known points in the plane are

$$P_1\left({\displaystyle \frac{L_x}{2}}-L_{x1}, {\displaystyle \frac{L_y}{2}}, 0\right); P_2\left({\displaystyle \frac{L_x}{2}}, {\displaystyle \frac{L_y}{2}}-L_{y1}, 0\right); P_3\left({\displaystyle \frac{L_x}{2}},{\displaystyle \frac{L_y}{2}}, {\sigma }_{max}\right),$$

(6)

where ${\sigma }_{max}$ is the allowable bearing capacity of the soil.

The general equation for the stress plane can be determined as follows:

$$\left|\begin{array}{ccc}x-\left({\displaystyle \frac{L_x}{2}}-L_{x1}\right)& y-{\displaystyle \frac{L_y}{2}}& {\sigma }_z-0\\ {\displaystyle \frac{L_x}{2}}-\left({\displaystyle \frac{L_x}{2}}-L_{x1}\right)& {\displaystyle \frac{L_y}{2}}-L_{y1}-{\displaystyle \frac{L_y}{2}}& 0-0\\ {\displaystyle \frac{L_x}{2}}-\left({\displaystyle \frac{L_x}{2}}-L_{x1}\right)& {\displaystyle \frac{L_y}{2}}-{\displaystyle \frac{L_y}{2}}& {\sigma }_{max}-0\end{array}\right|.$$

(7)

The value of ${\sigma }_z$ is obtained by solving the determinant:

$${\sigma }_z={\displaystyle \frac{{\sigma }_{max}\left[L_{y1}\left(2x-L_x\right)+L_{x1}\left(2y-L_y\right)+2L_{x1}{L}_{y1}\right]}{2L_{x1}{L}_{y1}}}.$$

(8)

The neutral axis equation (σ_z = 0) is

$$L_{y1}\left(2x-L_x\right)+L_{x1}\left(2y-L_y\right)+2L_{x1}{L}_{y1}=0.$$

(9)

• Case I

Figure 4a presents the case where “P” is inside the central nucleus.

General equations to determine the stresses of the soil on the footings subjected to biaxial bending at each corner of the footing are shown in Equations (2)–(5).

• Case II

Figure 4b shows a case where “P” is outside the central nucleus.

General equations to determine “P”, “M_xT”, and “M_yT” are obtained as follows:

$$P={\int }_{{\displaystyle \frac{L_y}{2}}-L_{y1}}^{{\displaystyle \frac{L_y}{2}}}{\int }_{{\displaystyle \frac{L_x}{2}}+{\displaystyle \frac{L_{x1}\left(L_y-2y-2L_{y1}\right)}{2L_{y1}}}}^{{\displaystyle \frac{L_x}{2}}}{\sigma }_{z}dxdy={\displaystyle \frac{{\sigma }_{max}{L}_{x1}{L}_{y1}}{6}},$$

(10)

$$M_{xT}={\int }_{{\displaystyle \frac{L_y}{2}}-L_{y1}}^{{\displaystyle \frac{L_y}{2}}}{\int }_{{\displaystyle \frac{L_x}{2}}+{\displaystyle \frac{L_{x1}\left(L_y-2y-2L_{y1}\right)}{2L_{y1}}}}^{{\displaystyle \frac{L_x}{2}}}{\sigma }_{z}ydxdy={\displaystyle \frac{{\sigma }_{max}{L}_{x1}{L}_{y1}\left(2L_y-L_{y1}\right)}{24}},$$

(11)

$$M_{yT}={\int }_{{\displaystyle \frac{L_y}{2}}-L_{y1}}^{{\displaystyle \frac{L_y}{2}}}{\int }_{{\displaystyle \frac{L_x}{2}}+{\displaystyle \frac{L_{x1}\left(L_y-2y-2L_{y1}\right)}{2L_{y1}}}}^{{\displaystyle \frac{L_x}{2}}}{\sigma }_{z}xdxdy={\displaystyle \frac{{\sigma }_{max}{L}_{x1}{L}_{y1}\left(2L_x-L_{x1}\right)}{24}}.$$

(12)

• Case III

Figure 4c shows a case where “P” is outside the central nucleus.

General equations to determine “P”, “M_xT”, and “M_yT” are obtained as follows:

$$P={\int }_{-{\displaystyle \frac{L_x}{2}}}^{{\displaystyle \frac{L_x}{2}}}{\int }_{{\displaystyle \frac{L_y}{2}}+{\displaystyle \frac{L_{y1}\left(L_x-2x-2L_{x1}\right)}{2L_{x1}}}}^{{\displaystyle \frac{L_y}{2}}}{\sigma }_{z}dydx={\displaystyle \frac{{\sigma }_{max}{L}_{y1}\left[{L_{x1}}^3-{\left(L_{x1}-L_x\right)}^3\right]}{6{L_{x1}}^2}},$$

(13)

$$M_{xT}={\int }_{-{\displaystyle \frac{L_x}{2}}}^{{\displaystyle \frac{L_x}{2}}}{\int }_{{\displaystyle \frac{L_y}{2}}+{\displaystyle \frac{L_{y1}\left(L_x-2x-2L_{x1}\right)}{2L_{x1}}}}^{{\displaystyle \frac{L_y}{2}}}{\sigma }_{z}ydydx={\displaystyle \frac{{\sigma }_{max}{L}_{y1}{\left(L_x-L_{x1}\right)}^3\left[L_{x1}\left(2L_y-L_{y1}\right)+L_x{L}_{y1}\right]}{24{L_{x1}}^3}}+{\displaystyle \frac{{\sigma }_{max}{L}_{y1}{L}_{x1}\left(2L_y-L_{y1}\right)}{24}},$$

(14)

$$M_{yT}={\int }_{-{\displaystyle \frac{L_x}{2}}}^{{\displaystyle \frac{L_x}{2}}}{\int }_{{\displaystyle \frac{L_y}{2}}+{\displaystyle \frac{L_{y1}\left(L_x-2x-2L_{x1}\right)}{2L_{x1}}}}^{{\displaystyle \frac{L_y}{2}}}{\sigma }_{z}xdydx={\displaystyle \frac{{\sigma }_{max}{L}_{y1}{L_x}^3\left(2L_{x1}-L_x\right)}{24{L_{x1}}^2}}.$$

(15)

• Case IV

Figure 4d shows a case where “P” is outside the central nucleus.

General equations to determine “P”, “M_xT”, and “M_yT” are obtained as follows:

$$P={\int }_{-{\displaystyle \frac{L_y}{2}}}^{{\displaystyle \frac{L_y}{2}}}{\int }_{{\displaystyle \frac{L_x}{2}}+{\displaystyle \frac{L_{x1}\left(L_y-2y-2L_{y1}\right)}{2L_{y1}}}}^{{\displaystyle \frac{L_x}{2}}}{\sigma }_{z}dxdy={\displaystyle \frac{{\sigma }_{max}{L}_{x1}\left[{L_{y1}}^3-{\left(L_{y1}-L_y\right)}^3\right]}{6{L_{y1}}^2}},$$

(16)

$$M_{xT}={\int }_{-{\displaystyle \frac{L_y}{2}}}^{{\displaystyle \frac{L_y}{2}}}{\int }_{{\displaystyle \frac{L_x}{2}}+{\displaystyle \frac{L_{x1}\left(L_y-2y-2L_{y1}\right)}{2L_{y1}}}}^{{\displaystyle \frac{L_x}{2}}}{\sigma }_{z}ydxdy={\displaystyle \frac{{\sigma }_{max}{L}_{x1}{L_y}^3\left(2L_{y1}-L_y\right)}{24{L_{y1}}^2}},$$

(17)

$$M_{yT}={\int }_{-{\displaystyle \frac{L_y}{2}}}^{{\displaystyle \frac{L_y}{2}}}{\int }_{{\displaystyle \frac{L_x}{2}}+{\displaystyle \frac{L_{x1}\left(L_y-2y-2L_{y1}\right)}{2L_{y1}}}}^{{\displaystyle \frac{L_x}{2}}}{\sigma }_{z}xdxdy={\displaystyle \frac{{\sigma }_{max}{L}_{x1}{\left(L_y-L_{y1}\right)}^3\left[L_{y1}\left(2L_x-L_{x1}\right)+L_y{L}_{x1}\right]}{24{L_{y1}}^3}}+{\displaystyle \frac{{\sigma }_{max}{L}_{x1}{L}_{y1}\left(2L_x-L_{x1}\right)}{24}}.$$

(18)

Case V

Figure 4e shows a case where “P” is outside the central nucleus.

General equations to determine “P”, “M_xT”, and “M_yT” are obtained as follows:

$$\begin{gathered} P={\int }_{{\displaystyle \frac{L_x}{2}}+{\displaystyle \frac{L_{x1}\left(L_y-L_{y1}\right)}{L_{y1}}}}^{{\displaystyle \frac{L_x}{2}}}{\int }_{-{\displaystyle \frac{L_y}{2}}}^{{\displaystyle \frac{L_y}{2}}+{\displaystyle \frac{L_{y1}\left(L_x-2x-2L_{x1}\right)}{2L_{x1}}}}{\sigma }_{z}dydx+{\int }_{-{\displaystyle \frac{L_x}{2}}}^{{\displaystyle \frac{L_x}{2}}}{\int }_{{\displaystyle \frac{L_y}{2}}+{\displaystyle \frac{L_{y1}\left(L_x-2x-2L_{x1}\right)}{2L_{x1}}}}^{{\displaystyle \frac{L_y}{2}}}{\sigma }_{z}dydx \\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}={\displaystyle \frac{{\sigma }_{max}\left[{L_{y1}}^3{\left(L_x-L_{x1}\right)}^3+{L_{x1}}^3{\left(L_y-L_{y1}\right)}^3\right]}{6{L_{x1}}^2{L_{y1}}^2}}+{\displaystyle \frac{{\sigma }_{max}{L}_{x1}{L}_{y1}}{6}}, \end{gathered}$$

(19)

$$\begin{gathered} M_{xT}={\int }_{{\displaystyle \frac{L_x}{2}}+{\displaystyle \frac{L_{x1}\left(L_y-L_{y1}\right)}{L_{y1}}}}^{{\displaystyle \frac{L_x}{2}}}{\int }_{-{\displaystyle \frac{L_y}{2}}}^{{\displaystyle \frac{L_y}{2}}+{\displaystyle \frac{L_{y1}\left(L_x-2x-2L_{x1}\right)}{2L_{x1}}}}{\sigma }_{z}ydydx+{\int }_{-{\displaystyle \frac{L_x}{2}}}^{{\displaystyle \frac{L_x}{2}}}{\int }_{{\displaystyle \frac{L_y}{2}}+{\displaystyle \frac{L_{y1}\left(L_x-2x-2L_{x1}\right)}{2L_{x1}}}}^{{\displaystyle \frac{L_y}{2}}}{\sigma }_{z}ydydx \\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}={\displaystyle \frac{{\sigma }_{max}{L}_{y1}{\left(L_x-L_{x1}\right)}^3\left[L_{x1}\left(2L_y-L_{y1}\right)+L_x{L}_{y1}\right]}{24{L_{x1}}^3}}+{\displaystyle \frac{{\sigma }_{max}{L}_{x1}{L_y}^3\left(2L_{y1}-L_y\right)}{24{L_{y1}}^2}}, \end{gathered}$$

(20)

$$\begin{gathered} M_{yT}={\int }_{{\displaystyle \frac{L_x}{2}}+{\displaystyle \frac{L_{x1}\left(L_y-L_{y1}\right)}{L_{y1}}}}^{{\displaystyle \frac{L_x}{2}}}{\int }_{-{\displaystyle \frac{L_y}{2}}}^{{\displaystyle \frac{L_y}{2}}+{\displaystyle \frac{L_{y1}\left(L_x-2x-2L_{x1}\right)}{2L_{x1}}}}{\sigma }_{z}xdydx+{\int }_{-{\displaystyle \frac{L_x}{2}}}^{{\displaystyle \frac{L_x}{2}}}{\int }_{{\displaystyle \frac{L_y}{2}}+{\displaystyle \frac{L_{y1}\left(L_x-2x-2L_{x1}\right)}{2L_{x1}}}}^{{\displaystyle \frac{L_y}{2}}}{\sigma }_{z}xdydx, \\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}={\displaystyle \frac{{\sigma }_{max}{L}_{x1}{\left(L_y-L_{y1}\right)}^3\left[L_{y1}\left(2L_x-L_{x1}\right)+L_y{L}_{x1}\right]}{24{L_{y1}}^3}}+{\displaystyle \frac{{\sigma }_{max}{L}_{y1}{L_x}^3\left(2L_{x1}-L_x\right)}{24{L_{x1}}^2}}. \end{gathered}$$

(21)

The optimal area, “A_min” (objective function), is obtained as follows:

$$A_{min}=L_x{L}_y.$$

(22)

Table 1. Biaxial bending.

Case	Constraint Functions
I	Equations (2)–(5) and 0 ≤ σ1, σ2, σ3, σ4 ≤ σmax
II	Equations (10)–(12), Lx ≥ Lx1 and Ly ≥ Ly1
III	Equations (13)–(15), Lx ≤ Lx1 and Ly ≥ Ly1
IV	Equations (16)–(18), Lx ≥ Lx1 and Ly ≤ Ly1
V	Equations (19)–(21), Lx ≤ Lx1 and Ly ≤ Ly1

L_x, L_y, L_x1, and L_y1 are limited to satisfy the conditions of each case.

shows the constraint functions for the resultant force and the resultant moments for each case.

Figure 5 shows the flowchart of the procedure for finding the optimal area of a rectangular footing; this procedure is performed for all five cases, and the smallest area is taken.

Figure 6 presents the flowchart, obtained using Maple 15 software, for the optimal area of a rectangular footing, considering each case separately.

3. Numerical Problems

Four types of numerical problems are shown to find the optimal area of a rectangular footing with an eccentric column under biaxial bending. Problem 1: the column at the center of the footing is presented in

Table A1. Problem 1.

Case	Lx (m)	Ly (m)	Lx1 (m)	Ly1 (m)	σmax-act (kN/m2)	Amin (m2)
Problem 1.A: P = 1200 kN, Mx = 800 kN-m, My = 400 kN-m, xc = 0, and yc = 0
I	4.00	8.00	-	-	75.00	32.00
II	No solution available
III	5.36	3.13	16.86	3.13	200.00	16.76
IV	1.56	10.71	1.56	33.71	200.00	16.76
V	3.01	5.00	5.46	7.58	200.00	15.04
FS-V	3.05	5.00	5.59	7.56	196.43	15.25
Problem 1.B: P = 800 kN, Mx = 800 kN-m, My = 400 kN-m, xc = 0, and yc = 0
I	6.00	12.00	-	-	22.22	72.00
II	No solution available
III	3.70	4.17	6.14	4.17	200.00	15.42
IV	2.08	7.40	2.08	12.29	200.00	15.42
V	2.72	5.44	3.50	7.01	200.00	14.80
FS-V	2.75	5.45	3.57	7.02	196.06	14.99
Problem 1.C: P = 400 kN, Mx = 800 kN-m, My = 400 kN-m, xc = 0, and yc = 0
I	12.00	24.00	-	-	2.78	288.00
II	3.22	6.45	2.45	4.90	200.00	20.80
III	4.00	5.50	4.00	3.00	200.00	22.00
IV	2.75	8.00	1.50	8.00	200.00	22.00
V	No solution available
FS-II	3.25	6.45	2.50	4.90	195.92	20.96

FS is the final solution; σ_max-act is the maximum pressure acting on the footing due to the soil.

(see Appendix A). Problem 2: the column at the edge of the footing in the X-direction is shown in

Table A2. Problem 2.

Case	Lx (m)	Ly (m)	Lx1 (m)	Ly1 (m)	σmax-act (kN/m2)	Amin (m2)
Problem 2.A: P = 1200 kN, Mx = 400 kN-m, My = −800 kN-m, xc = Lx/2 − c2/2, and yc = 0
I	1.73	4.88	-	-	200.00	8.46
II	No solution available
III	No solution available
IV	3.01	5.52	3.01	17.83	200.00	16.61
V	2.11	4.67	4.79	16.24	200.00	9.84
FS-I	1.75	4.95	-	-	198.46	8.66
Problem 2.B: P = 800 kN, Mx = 400 kN-m, My = −800 kN-m, xc = Lx/2 − c2/2, and yc = 0
I	2.40	3.22	-	-	200.00	7.73
II	No solution available
III	No solution available
IV	4.65	3.45	4.65	5.42	200.00	16.04
V	2.42	3.20	374.41	3.34	200.00	7.76
FS-I	2.40	3.25	-	-	197.24	7.80
Problem 2.C: P = 400 kN, Mx = 400 kN-m, My = −800 kN-m, xc = Lx/2 − c2/2, and yc = 0
I	4.40	6.00	-	-	30.30	26.40
II	8.80	2.68	8.80	1.36	200.00	23.60
III	4.40	2.61	5,629,531.75	0.91	200.00	11.47
IV	No solution available
V	No solution available
FS-III	4.45	2.65	134.23	0.99	187.51	11.79

(see Appendix A). Problem 3: the column on the edge of the footing in the Y-direction is presented in

Table A3. Problem 3.

Case	Lx (m)	Ly (m)	Lx1 (m)	Ly1 (m)	σmax-act (kN/m2)	Amin (m2)
Problem 3.A: P = 1200 kN, Mx = −800 kN-m, My = 400 kN-m, xc = 0, and yc = Ly/2 − c1/2
I	4.88	1.73	-	-	200.00	8.46
II	No solution available
III	5.52	3.01	17.83	3.01	200.00	16.61
IV	No solution available
V	4.67	2.11	16.24	4.79	200.00	9.84
FS-I	4.95	1.75	-	-	198.46	8.66
Problem 3.B: P = 800 kN, Mx = −800 kN-m, My = 400 kN-m, xc = 0, and yc = Ly/2 − c1/2
I	3.22	2.40	-	-	200.00	7.73
II	No solution available
III	3.45	4.65	5.42	4.65	200.00	16.04
IV	No solution available
V	3.20	2.42	3.34	374.41	200.00	7.76
FS-I	3.25	2.40	-	-	197.24	7.80
Problem 3.C: P = 400 kN, Mx = −800 kN-m, My = 400 kN-m, xc = 0, and yc = Ly/2 − c1/2
I	6.00	4.40	-	-	30.30	26.40
II	2.68	8.80	1.36	8.80	200.00	23.60
III	No solution available
IV	2.61	4.40	0.91	563,951.12	200.00	11.47
V	No solution available
FS-IV	2.65	4.45	0.99	134.23	187.51	11.79

(see Appendix A). Problem 4: the column at the corner of the footing is shown in

Table A4. Problem 4.

Case	Lx (m)	Ly (m)	Lx1 (m)	Ly1 (m)	σmax-act (kN/m2)	Amin (m2)
Problem 4.A: P = 800 kN, Mx = −800 kN-m, My = −700 kN-m, xc = Lx/2 − c2/2, and yc = Ly/2 − c1/2
I	2.08	2.32	-	-	200.00	4.82
II	No solution available
III	No solution available
IV	No solution available
V	No solution available
FS-I	2.10	2.35	-	-	184.03	4.94
Problem 4.B: P = 600 kN, Mx = −800 kN-m, My = −700 kN-m, xc = Lx/2 − c2/2, and yc = Ly/2 − c1/2
I	2.34	2.63	-	-	194.86	6.16
II	No solution available
III	2.71	2.28	92.09	2.28	200.00	6.18
IV	2.04	3.04	2.04	103.32	200.00	6.18
V	2.33	2.62	4.59	5.15	200.00	6.10
FS-V	2.35	2.65	4.71	2.65	188.94	6.23
Problem 4.C: P = 400 kN, Mx = −800 kN-m, My = −700 kN-m, xc = Lx/2 − c2/2, and yc = Ly/2 − c1/2
I	3.34	3.77	-	-	2.79	12.61
II	No solution available
III	2.93	2.95	4.17	2.95	200.00	8.66
IV	2.62	3.31	2.61	4.71	200.00	8.66
V	2.76	3.12	3.27	3.69	200.00	8.61
FS-V	2.80	3.15	3.44	3.83	184.29	8.82

(see Appendix A). The general data for all the footings are c₁ (the side of the column in the Y-direction) = c₂ (the side of the column in the X-direction) = 0.40 m, and σ_max = 200 kN/m².

4. Results

The way to check the proposed model according to the pressure under the footing is shown in

Table 2. Verification of the proposed model.

Case	Coordinates		σZ
	x	y
II	Lx/2	Ly/2	σmax
	Lx/2	Ly/2 − Ly1	0
	Lx/2 − Lx1	Ly/2	0
III	Lx/2	Ly/2	σmax
	−Lx/2	Ly/2 − Ly2 (Ly2 = Ly1(Lx1 − Lx)/Lx1)	0
	Lx/2	Ly/2 − Ly1	0
IV	Lx/2	Ly/2	σmax
	Lx/2 − Lx1	Ly/2	0
	Lx/2 − Lx2 (Lx2 = Lx1(Ly1 − Ly)/Ly1)	−Ly/2	0
V	Lx/2	Ly/2	σmax
	−Lx/2	Ly/2 − Ly2 (Ly2 = Ly1(Lx1 − Lx)/Lx1)	0
	Lx/2 − Lx2 (Lx2 = Lx1(Ly1 − Ly)/Ly1)	−Ly/2	0

. The coordinates are substituted into Equation (8).

Therefore, this model has been verified mathematically.

For all cases, when the load, P, decreases, the eccentricities (e_x = M_yT/P and e_y = M_xT/P) increase since the moments remain constant. Therefore, the load, P, tends to move away from the center of the footing.

Table A1 (see Appendix A) for Problem 1 shows the following: For Problems 1.A and 1.B, the minimum area appears in Case V, and for Problem 1.C, the minimum area appears in Case II. For this problem, in all cases, the load, P, is located outside the central nucleus.

Table A2 (see Appendix A) for Problem 2 shows the following: For Problems 2.A and 2.B, the minimum area appears in Case I, and for Problem 2.C, the minimum area appears in Case III. For this problem, in the first two cases, the load, P, is located inside the central nucleus (Case I), and in the last case, the load, P, is located outside the central nucleus (Case III).

Table A3 (see Appendix A) for Problem 3 shows the following: For Problems 3.A and 3.B, the minimum area appears in Case I, and for Problem 3.C, the minimum area appears in Case IV. For this problem, in the first two cases, the load, P, is located inside the central nucleus (Case I), and in the last case, the load, P, is located outside the central nucleus (Case IV).

Table A4 (see Appendix A) for Problem 4 shows the following: For Problem 4.A, the minimum area appears in Case I, and for Problems 4.B and 4.C, the minimum area appears in Case V. For this problem, in the first case, the load, P, is located inside the central nucleus (Case I), and in the last two cases, the load, P, is located outside the central nucleus (Case V).

One way to justify this study is to make some comparisons with the results of other authors.

A comparison is made between the model proposed by Luévanos-Rojas [27], which considers only Case I, and the new model, which takes into account five cases (I, II, III, IV, and V) for biaxial bending with an eccentric column.

Table 3. Example 1: Luévanos-Rojas’ model against the new model.

Case	Lx (m)	Ly (m)	Lx1 (m)	Ly1 (m)	σmax-act (kN/m2)	Amin (m2)
Problem 1.1: P = 1000 kN, Mx = 225 kN-m, My = 150 kN-m, xc = 0, and yc = 0
I	2.52	3.78	-	-	180.00	9.52
II	No solution available
III	9.77	1.24	110.85	1.24	180.00	12.15
IV	0.83	14.65	0.83	166.28	180.00	12.15
V	4.38	2.19	26.03	3.33	180.00	9.60
FS-I	2.55	3.80	-	-	176.29	9.69
Problem 1.2: P = 850 kN, Mx = 225 kN-m, My = 150 kN-m, xc = 0, and yc = 0
I	2.43	3.64	-	-	180.00	8.84
II	No solution available
III	7.63	1.41	58.74	1.41	180.00	10.79
IV	0.94	11.45	0.94	88.11	180.00	10.79
V	2.43	3.64	5.22	7.83	180.00	8.83
FS-V	2.45	3.65	6.58	9.70	156.63	8.94
Problem 1.3: P = 750 kN, Mx = 225 kN-m, My = 150 kN-m, xc = 0, and yc = 0
I	2.40	3.60	-	-	173.61	8.64
II	No solution available
III	6.41	1.55	37.37	1.55	180.00	9.94
IV	1.03	9.62	1.03	56.05	180.00	9.94
V	2.37	3.55	4.70	7.05	180.00	8.39
FS-V	2.40	3.55	4.82	7.03	176.68	8.52
Problem 1.4: P = 600 kN, Mx = 225 kN-m, My = 150 kN-m, xc = 0, and yc = 0
I	3.00	4.50	-	-	88.89	13.50
II	No solution available
III	4.87	1.81	18.11	1.81	180.00	8.83
IV	1.21	7.30	1.21	27.17	180.00	8.83
V	2.28	3.42	3.97	5.96	180.00	7.78
FS-V	2.30	3.45	4.03	6.05	175.30	7.93

shows example 1 from Luévanos-Rojas [27] for P = 1000 kN (Problem 1.1), P = 850 kN (Problem 1.2), P = 750 kN (Problem 1.3), P = 600 kN (Problem 1.4), M_x = 225 kN-m, M_y = 150 kN-m, x_c = 0, y_c = 0, c₁ = c₂ = 0.40 m, and σ_max = 180 kN/m².

Table 4. Example 4: Luévanos-Rojas’ model against the new model.

Case	Lx (m)	Ly (m)	Lx1 (m)	Ly1 (m)	σmax-act (kN/m2)	Amin (m2)
Problem 4.1: P = 750 kN, Mx = −750 kN-m, My = −600 kN-m, xc = Lx/2 − c2/2, and yc = Ly/2 − c1/2
I	1.96	2.36	-	-	180.00	4.63
II	No solution available
III	No solution available
IV	No solution available
V	No solution available
FS-I	2.00	2.35	-	-	169.76	4.70
Problem 4.2: P = 605 kN, Mx = −750 kN-m, My = −600 kN-m, xc = Lx/2 − c2/2, and yc = Ly/2 − c1/2
I	2.15	2.59	-	-	180.00	5.57
II	No solution available
III	No solution available
IV	No solution available
V	No solution available
FS-I	2.15	2.60	-	-	178.37	5.59
Problem 4.3: P = 450 kN, Mx = −750 kN-m, My = −600 kN-m, xc = Lx/2 − c2/2, and yc = Ly/2 − c1/2
I	2.63	3.20	-	-	107.00	8.41
II	No solution available
III	2.71	2.61	8.20	2.61	180.00	7.09
IV	2.15	3.30	2.15	9.98	180.00	7.09
V	2.40	2.92	3.67	4.46	180.00	6.99
FS-V	2.40	2.95	3.66	4.70	173.04	7.08
Problem 4.4: P = 300 kN, Mx = −750 kN-m, My = −600 kN-m, xc = Lx/2 − c2/2, and yc = Ly/2 − c1/2
I	3.77	4.63	-	-	34.37	17.46
II	2.91	3.58	2.85	3.50	180.00	10.42
III	2.93	3.55	2.93	3.41	180.00	10.42
IV	2.89	3.60	2.78	3.60	180.00	10.42
V	No solution available
FS-II	2.90	3.60	2.80	3.60	178.57	10.44

Note: In Problems 4.1 and 4.2, a solution is not obtained in Cases II, III, IV, and V because Case I reaches the maximum pressure, which is 180 kN/m².

shows example 4 from Luévanos-Rojas [27] for P = 750 kN (Problem 4.1), P = 605 kN (Problem 4.2), P = 450 kN (Problem 4.3), P = 300 kN (Problem 4.4), M_x = −750 kN-m, M_y = −600 kN-m, x_c = L_x/2 − c₂/2, y_c = L_y/2 − c₁/2, c₁ = c₂ = 0.40 m, and σ_max = 180 kN/m².

Table 3 shows example 1 from Luévanos-Rojas [27] versus the new model (the column in the center of the footing):

• For Problem 1.1, the minimum area appears in Case I, and for Problems 1.2, 1.3, and 1.4, the minimum area appears in Case V.
• For Problem 1.2, in terms of the contact area between the footing and the ground, there is no difference (theoretical areas) when using the new model.
• For Problem 1.3, there is a reduction of 2.89% (theoretical areas) in the contact area between the footing and the ground when using the new model.
• For Problem 1.4, there is a reduction of 42.37% (theoretical areas) in the contact area between the footing and the ground when using the new model.

Table 4 shows example 4 from Luévanos-Rojas [27] versus the new model (the column in the corner of the footing):

• For Problems 4.1 and 4.2, the minimum area appears in Case I; for Problem 4.3, the minimum area appears in Case V; and for Problem 4.4, the minimum area appears in Case II.
• For Problem 4.3, there is a reduction of 16.88% in the contact area between the footing and the ground when using the new model.
• For Problem 4.4, there is a reduction of 40.32% in the contact area between the footing and the ground when using the new model.

Note: In Table A1, Table A2, Table A3 and Table A4 of Appendix A and Table 3 and Table 4, a legend appears indicating that there is no solution available; this is because it is restricted according to each case, as shown in Table 1. For example, for Case II, L_x ≥ L_x1 and L_y ≥ L_y1; for Case III, L_x ≤ L_x1 and L_y ≥ L_y1; for Case IV, L_x ≥ L_x1 and L_y ≤ L_y1; and for Case V, L_x ≤ L_x1 and L_y ≤ L_y1.

5. Conclusions

This paper shows a general model for rectangular isolated footings under biaxial bending with an eccentric column assuming that the surface in contact with the ground is partially supported to determine the minimum area. The new model considers that the soil pressure is linear.

The new optimal model is developed from a mathematical approach based on the minimum area. The independent variables (constant or known parameters) are σ_max, P, M_x, M_y, c₁, c₂, x_c, and y_c. The dependent variables (decision or unknown variables) are A_min, L_x, L_y, L_x1, and L_y1.

The main difference between the current model and the new model is that the current model considers Case I (the load, P, is located in the central nucleus of the footing), and the new model takes into account Cases I, II, II, IV, and V (the load, P, is located anywhere on the footing, that is, inside and outside the central nucleus of the footing).

The limitations of this study are as follows:

• When the columns are very close to each other, the isolated footings tend to overlap. In this case, a combined footing must be used.
• When an isolated footing is limited by space situations (property boundary), a combined footing must be used.

The main conclusions are as follows:

• Other authors propose very complex equations to find the maximum load capacity of the footing due to the soil, and other authors obtain the sides of the footing through iterations.
• It is also observed that for the minimum area of the examples considered under biaxial bending, when Case II appears, Case V is not present. This means that when the resultant force decreases, Case II appears, and, conversely, Case V does not appear. This can be seen in Figure 2; when the resultant force is very small, the resultant force tends to fall in zone II. Therefore, it is not possible to obtain a solution for both cases of the same problem.
• This paper shows simplified and precise equations for P, M_x, and M_y in each case, as well as the minimum area and its constraint functions. Furthermore, it shows a significant reduction in the minimum contact area with the ground with respect to the model proposed by the other authors; if the resultant force is outside the central nucleus, this is because the footing is partially supported.
• This model can be used to verify the allowable load capacity of the soil considering maximizing “σ_max” (the objective function), the same constraint functions, and the known parameters “L_x” and “L_y”.

Suggestions for the next investigation:

• Minimum-cost design for rectangular isolated footings with an eccentric column, assuming that the footing is partially supported.
• Minimum area for rectangular combined footings with eccentric columns, assuming that the footing is partially supported.