Footings and foundation units transfer structural loads to the supporting soil or rock. The basic design idea is simple: spread the load enough that soil pressure is within the allowable bearing capacity, while also keeping settlement, overturning, sliding, and structural stresses acceptable.
$$q=\frac{P}{A}$$
$$P_{safe}=q_{allow}A$$
Common shallow foundation units include isolated footings, wall footings, combined footings, strap footings, mat or raft foundations, and pile caps. Deep foundations such as piles and drilled shafts are used when shallow soils cannot safely support the load.
Soil Pressure Under Footing
The actual soil pressure under a footing depends on soil type and on the relative rigidity of the footing and the supporting ground. For structural design, it is common to assume the bearing pressure varies linearly, and in many concentric-load cases it is taken as uniform.
$$A=BL$$
$$q_{avg}=\frac{P}{BL}$$
A rigid footing on elastic soil tends to have a nonuniform pressure distribution, but the simplified uniform or linear pressure block is normally adequate for hand design and board-exam problems.
Eccentrically Loaded Foundation
An eccentric load creates a moment at the base of the footing. The resultant soil reaction shifts away from the footing centerline by an eccentricity $e$.
$$e=\frac{M}{P}$$
$$M=Pe$$
If the resultant lies within the kern, the entire base remains in compression. For a rectangular footing with eccentricity along dimension $B$, the kern limit is $B/6$.
$$e<\frac{B}{6}\quad \text{full compression}$$
$$e=\frac{B}{6}\quad \text{zero pressure at one edge}$$
These equations are the axial stress plus bending stress form, written for soil pressure.
Triangular Pressure and Partial Contact
At the kern limit, the soil pressure diagram becomes triangular. Beyond the kern, soil cannot take tension, so part of the footing lifts off and only a compressed contact length remains.
Here $x$ is the compressed contact width measured from the highly compressed edge.
Presumptive Bearing Capacity
The handout uses NSCP-style presumptive bearing values for sandy deposits when exhaustive geotechnical investigation is not available. A base value is assigned for a minimum footing width and embedment, then increased by increments.
$$q_{allow}=q_0(1+0.20n)$$
$$n=\frac{\text{additional width or depth}}{300\text{ mm}}$$
$$q_{allow}\leq 3q_0$$
Use the exact wording of the code problem: some examples count width increments and depth increments separately, while still applying the stated maximum limit.
Load Components on a Footing
For soil pressure, the downward load includes the column load plus the footing self-weight when the problem says to consider it. If the self-weight acts through the footing centroid, it adds vertical load but does not add overturning moment about the footing centroid.
$$W_f=\gamma_cBLt$$
$$P_{total}=P_{column}+W_f$$
$$M=P_{column}e_{column}$$
This distinction matters in eccentric footing problems: adding footing weight increases $P$ and can reduce the resultant eccentricity.
Problem 157: NSCP Presumptive Capacity, 100 kPa Base
For sandy deposits, use presumptive load bearing capacity $q_0=100$ kPa for minimum footing width 300 mm and minimum embedment 300 mm. The value may be increased by 20% for each additional 300 mm of width and/or depth, up to three times the designated value. Evaluate a 1.2 m square footing at depths 300 mm and 900 mm, then find safe downward load for the second case.
1. Width 1.2 m, depth 300 mm:
$$\Delta B=1200-300=900\text{ mm}$$
$$n_B=\frac{900}{300}=3$$
$$q_{allow}=100+0.20(3)(100)=160\text{ kPa}$$
2. Width 1.2 m, depth 900 mm:
$$\Delta D=900-300=600\text{ mm}$$
$$n_D=\frac{600}{300}=2$$
$$\Delta q_D=0.20(2)(160)=64\text{ kPa}$$
$$q_{allow}=160+64=224\text{ kPa}$$
3. Safe downward load:
$$P=q_{allow}A=224(1.2)^2$$
$$P=322.56\text{ kN}$$
Answer: 160 kPa, 224 kPa, and 322.56 kN.
Problem 158: NSCP Presumptive Capacity, 75 kPa Base
For sandy deposits, use presumptive load bearing capacity $q_0=75$ kPa for minimum footing width 300 mm and minimum embedment 300 mm. Evaluate a 1.2 m square footing at depths 300 mm and 900 mm, then compute safe load for the second case.
The extracted solution applies three increments for the width contribution and three increments for the depth contribution under the stated cap interpretation.
3. Safe downward load:
$$P=165(1.2)(1.2)=237.6\text{ kN}$$
Answer: 120 kPa, 165 kPa, and 237.6 kN.
Problem 159: Eccentric Column on Rectangular Footing
A 2 m by 3 m footing is 0.50 m thick and supports a 0.60 m square column. The column external face is flush with the shorter edge of the footing and the column lies along the minor principal axis. Column load including column self-weight is 50 kN. Concrete unit weight is 24 kN/m3. Find total downward load, overturning moment, and maximum soil pressure.
$$W_f=24(2)(3)(0.5)=72\text{ kN}$$
$$P=50+72=122\text{ kN}$$
$$e_{column}=1.5-0.3=1.2\text{ m}$$
$$M=50(1.2)=60\text{ kN}\cdot\text{m}$$
$$q_{max}=\frac{P}{BL}+\frac{6M}{LB^2}$$
$$q_{max}=\frac{122}{3(2)}+\frac{6(60)}{2(3)^2}$$
$$q_{max}=40.33\text{ kPa}$$
Answer: $P=122$ kN, $M=60$ kN-m, and $q_{max}=40.33$ kPa.
Problem 160: Maximum Soil Pressure, Footing Weight Neglected
A rectangular footing 3 m by 2 m has an eccentric load $P=700$ kN. With the footing weight neglected, the extracted solution gives $e=0.5$ m and a triangular pressure block. Find maximum soil pressure.
$$M=Pe=700(0.5)=350\text{ kN}\cdot\text{m}$$
$$e>\frac{B}{6}\quad \text{so contact is partial}$$
$$x=1.5\text{ m}$$
$$q_{max}=\frac{2P}{xL}$$
$$q_{max}=\frac{2(700)}{1.5(3)}$$
$$q_{max}=311.11\text{ kPa}$$
Answer: $q_{max}=311.11$ kPa.
Problem 161: Maximum Soil Pressure with Footing Weight
A rectangular footing 3 m by 2 m is 0.4 m thick. An eccentric load $P=800$ kN acts with $e=0.5$ m. Determine maximum soil pressure considering footing weight. Use concrete unit weight $24$ kN/m3.
The kern of a rectangle is the diamond-shaped central zone where the resultant must land for the entire base to remain in compression. The kern boundaries are $e_B \leq B/6$ and $e_L \leq L/6$. When both conditions are satisfied, soil pressure is:
Use $+$ signs for the corner farthest from the resultant (maximum pressure) and $-$ signs for the nearest corner (minimum). When one eccentricity alone exceeds its kern limit, only a portion of the footing remains in contact and the triangular or trapezoidal block formulas or the $x=3(B/2-e)$ shortcut applies.
Problem: Biaxial Eccentricity — Kern Check and Corner Pressures
A 3.0 m by 4.0 m footing supports column load $P=1{,}200$ kN with moment $M_x=150$ kN·m about the 3.0 m axis and $M_y=200$ kN·m about the 4.0 m axis. Verify that the resultant is inside the kern and find maximum and minimum corner pressures.
A 2.5 m by 3.5 m footing carries a concentric vertical load of 900 kN plus a moment of 180 kN·m about the 2.5 m axis. Find maximum and minimum soil pressures and verify the footing is fully in contact with the soil.
Answer: Full contact; $q_{max}=152.2$ kPa, $q_{min}=53.5$ kPa.
Problem: Maximum Soil Pressure Under Partial Contact
A 2.0 m by 3.0 m footing carries column load $P=500$ kN with moment $M=220$ kN·m about the 2.0 m axis. Determine whether full or partial contact governs and calculate maximum soil pressure.
Answer: Partial contact governs; $x=1.68$ m in compression, $q_{max}=198.4$ kPa.
Problem: Soil Pressure Including Footing Self-Weight
A 3.0 m by 2.0 m footing is 0.45 m thick. A column load of 650 kN acts with an eccentricity of 0.30 m along the 2.0 m direction (moment $M=195$ kN·m). Concrete unit weight is 24 kN/m³. Compute resultant eccentricity and maximum soil pressure.
$$W_f = 24(3.0)(2.0)(0.45) = 64.8\text{ kN (acts through footing centroid, no eccentricity)}$$
Answer: $e_R=0.273$ m (within kern); $q_{max}=216.8$ kPa, $q_{min}=21.6$ kPa.
Problem: Maximum Column Load at Kern Limit
A 2.4 m by 3.6 m footing is designed so that the resultant load acts exactly at the kern boundary along the 2.4 m direction. If maximum allowable soil pressure is 200 kPa, find the maximum column load $P$ and the moment it implies.
At the kern limit $e=B/6=2.4/6=0.40$ m, the pressure diagram is triangular: $q_{min}=0$, $q_{max}=2P/(BL)$.
