Combined Footing Calculator

ACI 318-19 combined footing design for two columns with adjustable spacing. Checks bearing pressure, one-way shear, two-way (punching) shear, and flexure. Interactive diagram shows resultant location and kern limit.

What It Does

This tool designs a rectangular combined footing supporting two columns. It calculates:

The calculator provides immediate feedback on pass/fail status for each check with utilization ratios.

Inputs Explained

Column Loads

Column Dimensions

Footing Geometry

Material Properties

Other Inputs

Assumptions

  1. Full soil contact: Resultant falls within kern (e < L/6). No uplift is considered.

  2. Linear pressure variation: Soil pressure varies linearly across footing base based on eccentricity.

  3. Rectangular footing: Footing shape is rectangular with columns along the longitudinal axis.

  4. ACI 318-19 provisions: Checks follow ACI 318-19 strength design method.

  5. Reinforcement in one direction: Longitudinal flexure is checked. Transverse reinforcement may require separate design.

  6. No column moments: Columns are assumed to transmit only vertical loads. Additional moments are not considered.

How to Use

  1. Enter column loads: Input service loads (P) and factored loads (Pu) from your structural analysis.

  2. Define geometry: Enter column dimensions, spacing, and footing dimensions.

  3. Specify materials: Enter concrete strength, rebar yield, and allowable soil bearing.

  4. Review results: Check pass/fail status for each limit state. Utilization < 1.0 indicates passing.

  5. Adjust design: If a check fails, increase footing dimensions or thickness and recalculate.

  6. Export results: Use CSV export to document calculations for review.

Design Checks

Bearing Pressure

Bearing pressure is calculated using the resultant load (P1 + P2) and its eccentricity from the footing center. When eccentricity exceeds L/6 (kern limit), pressure becomes concentrated at one edge.

Formula: q = P/A × (1 ± 6e/L)

Where:

One-Way Shear

One-way (wide-beam) shear is checked at a distance d from the column face. The critical section extends across the full footing width.

Formula: Vc = 2 × lambda × sqrt(fc') × b × d

Where:

Two-Way Shear

Two-way (punching) shear is checked on a perimeter located at d/2 from the column face. Each column is checked separately.

Formula: Vc = 4 × lambda × sqrt(fc') × bo × d

Where:

Flexure

Flexure is checked in the longitudinal direction at the column face. The critical section is where maximum moment occurs.

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

Where:

When to Use a Combined Footing

A combined footing is used when two or more columns are located so close together that individual isolated footings would overlap, or when an edge column footing cannot extend beyond the property line. Common scenarios include:

Rectangular vs Trapezoidal Combined Footings

A rectangular combined footing is used when the property line constraint is the only limitation and the footing can extend equally on both sides of the columns in the transverse direction. It is simpler to form and construct.

A trapezoidal combined footing is used when the columns carry significantly different loads and the resultant does not fall near the center of a rectangular footing. The trapezoidal shape allows the footing width to vary along its length, keeping the bearing pressure more uniform. The area of a trapezoidal footing is:

A = (B1 + B2) / 2 × L

Where B1 and B2 are the widths at each end and L is the length.

Bearing Pressure Distribution Formulas

The bearing pressure distribution under a combined footing depends on the location of the resultant load relative to the footing centroid. For a rectangular footing of width B and length L subjected to a total vertical load P with eccentricity e from the centroid:

Concentric Loading (e = 0)

When the resultant of the column loads passes through the centroid of the footing:

q = P / (B × L)

The bearing pressure is uniform across the entire footing base.

Eccentric Loading in One Direction

When the resultant has an eccentricity e in the longitudinal direction (along the length L), the bearing pressure varies linearly:

q_max = P / (B × L) × (1 + 6e / L)
q_min = P / (B × L) × (1 - 6e / L)

This is the standard formula for a trapezoidal pressure distribution. The kern limit is e = L/6. Within the kern:

Outside the Kern (e > L/6)

When the eccentricity exceeds L/6, the minimum pressure becomes negative (tension), indicating uplift. Since soil cannot resist tension, the pressure redistributes:

q_max = 2P / (3 × B × (L/2 - e))

The effective contact length is 3 × (L/2 - e). This condition should generally be avoided in building foundations, as it concentrates pressure at one edge and may cause excessive settlement.

Determining the Resultant Location

For two column loads P1 and P2 at distances x1 and x2 from one end of the footing:

x_bar = (P1 × x1 + P2 × x2) / (P1 + P2)

The eccentricity from the footing center is:

e = x_bar - L/2

For the footing to have full soil contact without uplift, the resultant must fall within the middle third of the footing: |e| <= L/6.

Worked Example: Two-Column Combined Footing

Problem: Design a rectangular combined footing for two columns:

Parameter Column 1 (Interior) Column 2 (Edge)
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:

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

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

Check Critical Section Location Capacity Formula phi Factor
One-way shear d from column face (each side) 2 × lambda × sqrt(fc') × bw × d 0.75
Two-way shear d/2 from column face (perimeter) 4 × lambda × sqrt(fc') × bo × d 0.75
Moment Face of column phi × Mn = 0.9 × Mn 0.90

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.

Column Loads (P1 + P2) Column Spacing Footing Width (ft) Footing Length (ft) Footing Depth (in.)
100 kip + 50 kip 6 ft 4.0 - 5.0 8.0 - 10.0 18 - 24
150 kip + 80 kip 8 ft 5.0 - 6.5 10.0 - 12.0 20 - 28
200 kip + 100 kip 10 ft 6.0 - 8.0 12.0 - 15.0 24 - 30
250 kip + 150 kip 12 ft 7.0 - 9.0 14.0 - 17.0 24 - 36
300 kip + 200 kip 15 ft 8.0 - 10.0 16.0 - 20.0 30 - 36

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

See Also