9.2 Effective Footing Area, Eccentricity & Combined Footings

Foundations Subjected to Eccentric Loading

Eccentric loads arise when structural columns transfer vertical downward forces $P$ accompanied by bending moments $M$ caused by wind, seismic shear, or lateral soil pressures ($e = M / P$). In spread footings, eccentricity redistributes soil contact pressure non-uniformly across the base.

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Contact Stress Distribution & Middle-Third Rule (One-Way Eccentricity)

For a rectangular footing of width $B$ and length $L$ subjected to a load $P$ with eccentricity $e = M / P$ along axis $B$:

                   Soil Stress Profiles Under Eccentric Load P
                   
       e <= B/6 (Trapezoidal)      e = B/6 (Triangular)        e > B/6 (Base Detachment)
             P                           P                           P
             │                           │                           │
       ┌─────┴─────┐               ┌─────┴─────┐               ┌─────┴─────┐
       └─────┬─────┘               └─────┬─────┘               └─────┬─────┘
       ▲▲▲▲▲▲▲▲▲▲▲▲▲               ▲▲▲▲▲▲▲▲▲▲▲▲▲                     ▲▲▲▲▲▲▲
       │           │               │           │                     │     │
     q_min       q_max           q_min=0     q_max                 q_min=0 q_max
                                                                 ◄─── L_eff ──►

1. Eccentricity Within the Middle Third ($e \le B/6$)

The entire base remains in compression with a trapezoidal contact stress distribution: $q_{max} = \frac{P}{B L} \left( 1 + \frac{6 e}{B} \right)$ $q_{min} = \frac{P}{B L} \left( 1 - \frac{6 e}{B} \right)$ Where $q_{min} \ge 0$.

2. Eccentricity at the Limit of Middle Third ($e = B/6$)

The stress distribution becomes strictly triangular across the entire footprint: $q_{min} = 0, \quad q_{max} = \frac{2 P}{B L}$

3. Eccentricity Exceeding Middle Third ($e > B/6$)

Soil cannot sustain tension. Tensile separation (lift-off) occurs along the heel of the footing. The effective contact length reduces to $L_{eff} = 3 \left(\frac{B}{2} - e\right)$, and maximum compressive edge stress becomes: $q_{max} = \frac{2 P}{3 L \left(\frac{B}{2} - e\right)}$ $q_{min} = 0 \text{ across detached zone of length } \left[ B - 3\left(\frac{B}{2} - e\right) \right]$

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Meyerhof's Effective Area Concept

Meyerhof (1953) established that an eccentrically loaded footing can be evaluated as an equivalent concentrically loaded footing with reduced effective dimensions ($B', L'$):

$B' = B - 2 e_B$ $L' = L - 2 e_L$

Where:

• $e_B = M_L / P =$ Eccentricity along width $B$
• $e_L = M_B / P =$ Eccentricity along length $L$
• $A' = B' \cdot L' =$ Effective footing area

Application Rules for Bearing Capacity Equations

1. In all general bearing capacity equations (Meyerhof, Vesic, Hansen), use effective dimensions $B'$ and $L'$ to compute shape factors $s_c, s_q, s_gamma$ (e.g., $s_q = 1 + \frac{B'}{L'} \tan\phi'$).
2. The third term of the bearing capacity equation ($0.5 \gamma B' N_gamma s_gamma$) uses $B'$ (always the smaller of $B'$ and $L'$).
3. Ultimate load capacity $P_{ult}$ is calculated by multiplying ultimate effective bearing capacity $q_{ult}'$ by the effective area $A'$: $P_{ult} = q_{ult}' \cdot A' = q_{ult}' \cdot (B' L')$

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Bi-Directional Eccentricity and the Kern Region

When eccentricity occurs simultaneously along both axes ($e_B$ and $e_L$), base tension is avoided if the load resultant falls within the Kern region of the footing: $\frac{e_B}{B} + \frac{e_L}{L} \le \frac{1}{6}$

                         Kern Zone of Rectangular Footing
                         
                                  ┌───────────┐ +L/2
                                  │     ▲     │
                                  │    ╱│╲    │
                                  │   ╱ │ ╲   │
                           ───────┼──◄──┼──►──┼───────
                                  │   ╲ │ ╱   │ -B/6 to +B/6
                                  │    ╲│╱    │
                                  │     ▼     │
                                  └───────────┘ -L/2
                               -B/2           +B/2

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Combined Footings Design

When column loads are heavy, space is restricted, or an exterior column sits directly against a property boundary, individual spread footings may overlap or experience severe load eccentricity. Combined footings support two or more columns on a single structural slab.

1. Rectangular Combined Footings

Used when the exterior column load $P_1$ is smaller than or comparable to interior load $P_2$, allowing the footing to extend past column $P_2$.

                       Rectangular Combined Footing Layout
                       
        Property Line
             │   P1 = Exterior              P2 = Interior
             │   Column                     Column
             ▼     │                          │
             ┌─────┴──────────────────────────┴─────┐
             │              Footing                 │
             └──────────────────┬───────────────────┘
             ◄────── x_bar ─────►
             ◄───────────────────── L ──────────────►

Design Equations for Rectangular Footings:

• Resultant load magnitude: $R = P_1 + P_2$
• Resultant location $\bar{x}$ measured from center of exterior column $P_1$ (spaced distance $S$ from $P_2$): $\bar{x} = \frac{P_2 \cdot S}{P_1 + P_2}$
• Distance from property line to resultant: $X_{cg} = L_{margin} + \bar{x}$ (where $L_{margin}$ is distance from property line to column $P_1$ center).
• To achieve uniform base pressure ($e = 0$), footing length $L$ must be centered on the resultant: $L = 2 \cdot X_{cg} = 2 (L_{margin} + \bar{x})$
• Required footing width $B$ based on net allowable soil pressure $q_{net,all}$: $B = \frac{P_1 + P_2}{L \cdot q_{net,all}}$

2. Trapezoidal Combined Footings

Used when property line constraints limit the extension of $L$ ($L < 2 X_{cg}$), or when exterior load $P_1 > P_2$.

                      Trapezoidal Combined Footing Footprint
                      
                      Property Line
                           │
                           ▼  ┌───────────────────────┐ Width B1 (Larger)
                           │  │   P1            P2    │
                           │  └───────────────────────┘ Width B2 (Smaller)
                           ◄───────────── L ──────────►

Design Equations for Trapezoidal Footings:

$\text{Area } A = \left( \frac{B_1 + B_2}{2} \right) L = \frac{P_1 + P_2}{q_{net,all}}$ $X_{cg} = \frac{L}{3} \left( \frac{B_1 + 2 B_2}{B_1 + B_2} \right)$ Solving simultaneously for width dimensions $B_1$ (exterior edge) and $B_2$ (interior edge): $B_1 = \frac{2 A}{L} \left( 3 \frac{X_{cg}}{L} - 1 \right)$ $B_2 = \frac{2 A}{L} \left( 2 - 3 \frac{X_{cg}}{L} \right)$

3. Strap (Cantilever) Footings

Strap footings connect an eccentrically loaded exterior column footing to an interior footing using a rigid concrete beam (strap). The strap beam is elevated above the soil so it transfers zero vertical soil bearing pressure, serving purely as a rigid moment transfer tie.

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Comprehensive Worked Calculation Example

Problem Statement

An exterior column $P_1 = 650\text{ kN}$ is located at a property boundary ($L_{margin} = 0.3\text{ m}$ from property line to column center). An interior column $P_2 = 1100\text{ kN}$ is located $4.2\text{ m}$ center-to-center from column $P_1$. The net allowable bearing pressure is $q_{net,all} = 140\text{ kPa}$.

Design a rectangular combined footing to achieve uniform contact pressure ($e = 0$).

1. Determine required footing length $L$
2. Determine required footing width $B$
3. Calculate uniform ground pressure under service loads

Solution Steps

Step 1: Calculate Resultant Force and Location $\bar{x}$

$R = P_1 + P_2 = 650 + 1100 = 1750\text{ kN}$ Taking moments about the center of column $P_1$: $\bar{x} = \frac{P_2 \cdot S}{P_1 + P_2} = \frac{1100 \times 4.2}{1750} = \frac{4620}{1750} = 2.64\text{ m}$

Step 2: Calculate Distance from Property Line to Resultant $X_{cg}$

$X_{cg} = L_{margin} + \bar{x} = 0.3 + 2.64 = 2.94\text{ m}$

Step 3: Compute Required Footing Length $L$

To ensure $e = 0$, the resultant must coincide with the midpoint of $L$: $L = 2 \cdot X_{cg} = 2 \times 2.94 = 5.88\text{ m}$

Step 4: Compute Required Footing Width $B$

$A_{req} = \frac{R}{q_{net,all}} = \frac{1750\text{ kN}}{140\text{ kPa}} = 12.50\text{ m}^2$ $B = \frac{A_{req}}{L} = \frac{12.50}{5.88} = 2.126\text{ m} \approx 2.15\text{ m}$

Step 5: Verify Final Uniform Contact Stress

Using $B = 2.15\text{ m}$ and $L = 5.88\text{ m}$ ($A_{actual} = 12.642\text{ m}^2$): $q_{actual} = \frac{1750}{12.642} = 138.4\text{ kPa} \le 140\text{ kPa} \quad \text{(OK)}$