------------------- | ------------------- | --------------- | | Service load | P1 = 100 kip | P2 = 60 kip | | Factored load | Pu1 = 150 kip | Pu2 = 90 kip | | Distance from left end | x1 = 5 ft | x2 = 1 ft | | Column size | 16 in. × 16 in. | 12 in. × 12 in. |

Allowable soil bearing pressure: q_all = 3.0 ksf

Step 1: Determine resultant location

Total load: P = 100 + 60 = 160 kip

x_bar = (100 × 5 + 60 × 1) / 160 = (500 + 60) / 160 = 3.5 ft

The resultant is 3.5 ft from the left end of the footing.

Step 2: Determine footing dimensions

For uniform bearing pressure (e = 0), the footing centroid must coincide with the resultant location. Therefore:

L/2 = x_bar = 3.5 ft
L = 7.0 ft

But wait -- the columns are 4 ft apart (x1 = 5 ft, x2 = 1 ft), and we need room for the column footprint. Since the edge column is at 1 ft from the left end, and the footing length must extend beyond both columns:

Let us try L = 8 ft, so the centroid is at 4.0 ft from the left end.

Eccentricity: e = 3.5 - 4.0 = -0.5 ft

Required area (assuming uniform pressure): A = P / q_all = 160 / 3.0 = 53.3 sq ft

Required width: B = A / L = 53.3 / 8.0 = 6.67 ft, use B = 6 ft-8 in. (6.67 ft)

Actual area: A = 6.67 × 8.0 = 53.3 sq ft

Step 3: Check bearing pressure

q_avg = 160 / 53.3 = 3.00 ksf
q_max = (160 / 53.3) × (1 + 6 × 0.5 / 8.0) = 3.00 × 1.375 = 4.125 ksf
q_min = (160 / 53.3) × (1 - 6 × 0.5 / 8.0) = 3.00 × 0.625 = 1.875 ksf

Check: e = 0.5 ft < L/6 = 1.33 ft -- full soil contact (OK)

Since q_max = 4.125 ksf > q_all = 3.0 ksf, the footing width must be increased.

Try B = 9 ft-0 in. (9.0 ft), L = 8.0 ft:

A = 9.0 × 8.0 = 72.0 sq ft
q_avg = 160 / 72.0 = 2.22 ksf
q_max = 2.22 × (1 + 6 × 0.5 / 8.0) = 2.22 × 1.375 = 3.06 ksf
q_min = 2.22 × (1 - 6 × 0.5 / 8.0) = 2.22 × 0.625 = 1.39 ksf

q_max = 3.06 ksf is slightly above q_all = 3.0 ksf. Try B = 9 ft-3 in. (9.25 ft):

A = 9.25 × 8.0 = 74.0 sq ft
q_max = (160 / 74.0) × 1.375 = 2.16 × 1.375 = 2.97 ksf < 3.0 ksf  (OK)

Step 4: Factored bearing pressure (for strength checks)

Total factored load: Pu = 150 + 90 = 240 kip

x_bar_u = (150 × 5 + 90 × 1) / 240 = 840 / 240 = 3.5 ft
qu_avg = 240 / 74.0 = 3.24 ksf
qu_max = 3.24 × 1.375 = 4.46 ksf
qu_min = 3.24 × 0.625 = 2.03 ksf

Step 5: Maximum factored moment

The critical section for moment is at the face of the interior column (the heavier load). Taking moments about the column 1 face:

The factored bearing pressure at distance x from the left end (with linear variation) provides the loading for computing the moment. The maximum moment occurs at the face of the heavier column where the shear passes through zero.

This example demonstrates that combined footing design is an iterative process where the footing dimensions must be adjusted to satisfy bearing pressure, shear, and moment requirements simultaneously.

One-Way and Two-Way Shear Check Formulas

Shear checks in combined footings follow the same principles as isolated footings but must account for the continuous beam action along the footing length.

One-Way (Wide-Beam) Shear

One-way shear is checked at a distance d from the face of each column, where d is the effective depth of the footing. The critical section extends across the full footing width B.

Vc = 2 × lambda × sqrt(fc') × bw × d

Where:

• lambda = 1.0 for normal-weight concrete
• fc' = concrete compressive strength (psi)
• bw = footing width (inches)
• d = effective depth (inches)

For the combined footing, one-way shear must be checked at:

• Distance d from the face of Column 1 (both sides)
• Distance d from the face of Column 2 (both sides)
• Midspan between columns (where the bending moment may reverse sign)

The factored shear Vu at each critical section is computed from the bearing pressure distribution acting on the portion of the footing beyond (or between) the critical sections. If Vu > phi × Vc, shear reinforcement must be provided or the footing depth increased.

Two-Way (Punching) Shear

Two-way shear is checked on a critical perimeter bo located at d/2 from the column face:

Vc = 4 × lambda × sqrt(fc') × bo × d

Where bo is the perimeter of the critical section:

bo = 2 × (c1 + d) + 2 × (c2 + d) = 2 × (c1 + c2 + 2d)

For a rectangular column of dimensions c1 x c2.

Each column in the combined footing must be checked independently for two-way shear. The factored shear force Vu at each column is the column factored load minus the bearing pressure over the area inside the critical perimeter:

Vu = Pu - qu × (c1 + d) × (c2 + d)

If Vu > phi × Vc (where phi = 0.75 for shear), the footing depth must be increased or shear reinforcement (stirrups or shear studs) must be provided.

Combined Footing Shear Comparison

Combined Footing Proportions

The following table provides preliminary sizing guidance for combined footings based on typical column loads and spacing. These are starting points for design iteration.

Notes: Assumes q_all = 2.0 - 3.0 ksf, fc' = 4000 psi, fy = 60,000 psi. Actual dimensions depend on soil bearing capacity, column loads, eccentricity, and shear requirements.

Structural Design of Combined Footing

The structural design of a combined footing treats it as a beam loaded from below by the soil bearing pressure and supported at discrete points by the columns. This is the inverse of a typical continuous beam problem.

Longitudinal Reinforcement (Bottom Steel)

Bottom reinforcement resists positive moments that occur between columns where the bearing pressure pushes the footing upward against the column supports. The critical sections are at the face of each column.

Required reinforcement at each critical section:

Mu = factored moment at column face
a = As × fy / (0.85 × fc' × b)
As = Mu / (phi × fy × (d - a/2))
phi = 0.90 for flexure

The maximum moment and corresponding reinforcement must be determined at:

1. Face of Column 1 (interior)
2. Face of Column 2 (edge)
3. Point of zero shear between columns (if applicable)

Longitudinal Reinforcement (Top Steel)

Top reinforcement resists negative moments that occur in the cantilever portion beyond the edge column. When the footing extends beyond the edge column, the bearing pressure on the overhang creates a cantilever moment:

Mu(overhang) = qu × B × L_overhang^2 / 2

Where L_overhang is the distance from the edge column center to the footing edge.

Top steel is also required at the midspan between columns if the moment diagram reverses sign (which occurs when the footing acts as a continuous beam with the bearing pressure load).

Transverse Reinforcement

Transverse (width-direction) reinforcement distributes the column loads across the footing width. The footing is conceptually divided into strips, with each column strip designed as a short cantilever beam extending from the column face to the footing edge.

For each column, the effective transverse strip width is typically taken as the column width plus d on each side:

b_strip = c + 2d

The factored bearing pressure under the strip creates a cantilever moment:

Mu(transverse) = qu × (B - c) / 2 × b_strip × [(B - c) / 4]

Required transverse reinforcement per strip:

As(transverse) = Mu(transverse) / (phi × fy × (d - a/2))

Transverse bars are placed below the longitudinal bars (since d is measured to the lower steel layer).

Minimum Reinforcement

ACI 318-19 requires minimum reinforcement in both directions:

Shrinkage and temperature (longitudinal):

As,min = 0.0018 × b × h  (for Grade 60 bars)

Flexural minimum (where reinforcement is required by analysis):

As,min = max(0.0018 × b × h, 3 × sqrt(fc') × b × d / fy)

These minimums must be checked against the calculated reinforcement and the larger value provided.

Notes

• Resultant location: The diagram shows the resultant location relative to footing center. If outside the kern (L/6), consider increasing footing length.

• Shear capacity: Both one-way and two-way shear must be satisfied. The lower capacity governs.

• Reinforcement design: Longitudinal reinforcement is provided based on calculated moment. Transverse reinforcement may need separate design for local strip action.

• Minimum reinforcement: ACI 318 minimum temperature and shrinkage reinforcement is not explicitly checked in this tool.

• Educational use only: This tool provides screening calculations. Final design requires qualified engineer verification per ACI 318-19 and project requirements.

Code References

• ACI 318-19 Chapter 13 -- Foundation design including combined footings, mat foundations, and pile caps
• ACI 318-19 Section 13.4 -- Distribution of soil pressure under combined footings and strap footings
• ASCE 7-22 Section 12.13 -- Foundation design requirements for seismic loading including overturning checks
• IBC Section 1808 -- Foundation walls and footings including allowable bearing pressures and minimum dimensions

Worked Example

Problem: Design a rectangular combined footing for two columns: Column A (12x12 in., PD = 120 k, PL = 80 k) and Column B (14x14 in., PD = 160 k, PL = 100 k) spaced at 16 ft center-to-center. Allowable soil bearing qa = 4.0 ksf.

Given:

• Column A: 12 in. x 12 in., DL = 120 k, LL = 80 k
• Column B: 14 in. x 14 in., DL = 160 k, LL = 100 k
• Column spacing: L = 16 ft
• qa = 4.0 ksf, fc' = 3,500 psi
• f'c = 3,500 psi, Fy = 60 ksi (rebar)

Solution:

Step 1 -- Service loads and resultant location:

P_A = 120 + 80 = 200 kips
P_B = 160 + 100 = 260 kips
P_total = 460 kips

Resultant location from Column A:
x_bar = P_B * 16 / P_total = 260 * 16 / 460 = 9.04 ft

Step 2 -- Footing dimensions:

A_required = 460 / 4.0 = 115 ft^2
Length >= 2 * x_bar + column_offset = 2 * 9.04 + 1.0 = 19.1 ft -> use L = 20 ft
Width = 115 / 20 = 5.75 ft -> use B = 6 ft

A_provided = 20 * 6 = 120 ft^2 > 115 OK

Step 3 -- Factored loads for structural design:

Pu_A = 1.2*120 + 1.6*80 = 144 + 128 = 272 kips
Pu_B = 1.2*160 + 1.6*100 = 192 + 160 = 352 kips
Pu_total = 624 kips

x_bar_u = 352 * 16 / 624 = 9.03 ft
eccentricity = 10 - 9.03 = 0.97 ft (footing centroid at 10 ft)

Step 4 -- Factored soil pressure:

q_max = (624/(20*6)) * (1 + 6*0.97/20) = 5.20 * 1.291 = 6.71 ksf
q_min = 5.20 * (1 - 6*0.97/20) = 5.20 * 0.709 = 3.69 ksf

Longitudinal reinforcement designed for trapezoidal pressure distribution per ACI 318-19 Chapter 13.

Result: Combined footing 20 ft x 6 ft x 24 in. deep, with longitudinal top reinforcement at column locations and bottom reinforcement at midspan. Soil pressure distribution is trapezoidal with all compressive stresses (e/L = 0.97/20 = 0.049 < 1/6, resultant within kern).

See Also

• Beam Capacity Calculator
• Steel Beam Deflection Check
• Steel Beam Sizes Reference
• Beam Design Guide
• Beam Span Reference