How do you size a spread footing?

From service loads and allowable bearing pressure: A_required = P_service / q_a. Round up to a convenient square or rectangular dimension. The allowable bearing capacity (q_a) comes from the geotechnical investigation or IBC Table 1806.2 for presumptive values.

What checks does ACI 318-19 require for an isolated footing?

Six checks: (1) service-level bearing pressure against net allowable soil capacity (13.2.1, 13.2.2), (2) flexure at the face of the column (13.2.7.1), (3) one-way shear at d from the column face (13.2.7.2, Table 22.5.5.1), (4) two-way (punching) shear at d/2 from the column face (22.6.4.1, Table 22.6.5.2), (5) development length of flexural reinforcement (25.4.2.4), and (6) column-to-footing interface bearing and dowels (22.8.3.2, 16.3.4.1).

What is the minimum depth for a concrete footing?

ACI 318-19 Section 13.3.1.2 requires an effective depth of at least 6 inches for footings on soil. Clear cover for concrete cast against and permanently exposed to earth is 3 inches (Table 20.6.1.3.1). In practice, footing thickness is usually governed by punching shear or one-way shear, not by the code minimum.

How to design a spread footing: ACI 318-19 worked example

This page walks through a complete isolated spread footing design per ACI 318-19, from sizing under service loads through all six strength checks. Every intermediate value is pinned by the same regression tests that guard the live footing calculator. For the theory behind each individual check, see the footing methodology hub.

Problem statement

Design a reinforced concrete isolated spread footing for an interior column.

Column	18 × 18 in, interior location
Service load	P_service = 150 kip
Factored load	P_u = 210 kip (no lateral, no moment)
Concrete	f′c = 4,000 psi, normal-weight (λ = 1.0)
Steel	Grade 60 (f_y = 60 ksi), #5 bars, uncoated
Soil	q_a = 4.0 ksf, γ_soil = 120 pcf, depth = 2.0 ft
Cover	3.0 in (cast against soil, Table 20.6.1.3.1)

Step 1: Size the footing from service loads

Footing proportioning uses service loads (ACI 318-19 §13.2.1). The required area is:

A_{\text{req}} = \frac{P_{\text{service}}}{q_a} = \frac{150}{4.0} = 37.5\ \text{ft}^2

Try a 7 ft × 7 ft square footing (84 × 84 in), giving A = 49 ft². Choose h = 18 in as a trial thickness. Actual service-level soil pressure:

q_{\text{actual}} = \frac{P_{\text{service}}}{A} = \frac{150}{49} = 3.061\ \text{ksf} < q_a = 4.0\ \text{ksf}\ \checkmark

Step 2: Check bearing pressure (ACI 13.2.1, 13.2.2)

The net allowable bearing deducts the overburden and footing self-weight that the soil already supports before the column load is applied (see the bearing pressure methodology):

q_{\text{net}} = q_a - \gamma_{\text{soil}} \cdot D_f - \gamma_c \cdot \frac{h}{12} = 4.0 - 0.120 \times 2.0 - 0.150 \times 1.5 = 3.535\ \text{ksf}

Service bearing check:

q_{\max} = \frac{P_{\text{service}}}{A} = \frac{150}{49} = 3.061\ \text{ksf} < q_{\text{net}} = 3.535\ \text{ksf}\ \checkmark

Eccentricity is zero (concentric load), within the kern (L/6 = 14 in). DCR = 3.061 / 3.535 = 0.866: PASS.

Factored soil pressure for strength design (used in Steps 3 through 6):

q_u = \frac{P_u}{A} = \frac{210}{49} = 4.286\ \text{ksf}

Step 3: Effective depth and flexure (ACI 13.2.7.1)

Bars are placed in both directions. The bottom layer (L-direction bars, placed first) has the larger effective depth; the top layer (B-direction bars) sits on top of it (see the flexure methodology):

d_L = h - \text{cover} - \tfrac{d_b}{2} = 18 - 3.0 - 0.3125 = 14.6875\ \text{in (bottom layer)}

d_B = h - \text{cover} - d_b - \tfrac{d_b}{2} = 18 - 3.0 - 0.625 - 0.3125 = 14.0625\ \text{in (top layer)}

The critical section for flexure is at the face of the column (§13.2.7.1). The cantilever length on each side:

\ell = \frac{L - c_{\text{col}}}{2} = \frac{84 - 18}{2} = 33\ \text{in} = 2.75\ \text{ft}

Factored moment per direction (symmetric for a square footing with concentric load):

M_u = q_u \cdot B_{\text{ft}} \cdot \frac{\ell_{\text{ft}}^2}{2} = 4.286 \times 7.0 \times \frac{2.75^2}{2} = 113.4375\ \text{k-ft} = 1{,}361.25\ \text{k-in}

Minimum reinforcement (§13.3.3.2 referencing Table 24.4.3.2):

A_{s,\min} = 0.0018 \cdot b \cdot h = 0.0018 \times 84 \times 18 = 2.722\ \text{in}^2

Required A_s from flexure = 1.738 in², so the minimum governs. Use 9-#5 bars each way (A_s,prov = 9 × 0.31 = 2.79 in² > 2.722 in²).

\phi M_n = 2{,}168.67\ \text{k-in (L dir)},\quad 2{,}074.50\ \text{k-in (B dir)}

DCR_L = 0.628, DCR_B = 0.656: PASS.

Step 4: One-way shear (ACI 13.2.7.2, Table 22.5.5.1)

The critical section is at d from the face of the column (§13.2.7.2). The shear force is the factored soil pressure acting on the area beyond the critical section (see the one-way shear methodology):

Direction	V_u (kip)	φV_c (kip)	DCR	Status
L	45.78	61.45	0.745	PASS
B	47.34	59.70	0.793	PASS

Step 5: Two-way (punching) shear (ACI 22.6.4.1, Table 22.6.5.2)

The critical perimeter forms at d/2 from the column face, where d is the average effective depth of both reinforcement layers (see the punching shear methodology):

d_{\text{avg}} = \frac{d_L + d_B}{2} = \frac{14.6875 + 14.0625}{2} = 14.375\ \text{in}

b_o = 4 \times (c_{\text{col}} + d_{\text{avg}}) = 4 \times (18 + 14.375) = 129.5\ \text{in}

The factored shear on the critical section is the column load minus the soil reaction inside the perimeter:

V_u = P_u - q_u \cdot A_{\text{inside}} = 210 - 4.286 \times \frac{(18 + 14.375)^2}{144} = 178.81\ \text{kip}

Table 22.6.5.2 gives three expressions for v_c (the least governs). With β_col = 1.0 (square column) and α_s = 40 (interior):

v_{c,a} = 4\,\lambda\,\sqrt{f'_c} = 252.98\ \text{psi}\quad \text{(governs)}

v_{c,b} = \left(2 + \frac{4}{\beta}\right)\lambda\,\sqrt{f'_c} = 379.47\ \text{psi}

v_{c,c} = \left(2 + \frac{\alpha_s\,d}{b_o}\right)\lambda\,\sqrt{f'_c} = 407.31\ \text{psi}

\phi V_c = 0.75 \times v_{c,a} \times b_o \times d_{\text{avg}} = 0.75 \times 0.25298 \times 129.5 \times 14.375 = 353.21\ \text{kip}

DCR = 178.81 / 353.21 = 0.506: PASS.

Step 6: Development length (ACI 25.4.2.4)

The flexural bars must be developed between the critical section (face of column) and the end of the footing. Using the general equation (Eq. 25.4.2.4a) with ψ factors (see the development length methodology):

\ell_d = 14.23\ \text{in}

Adjusted for excess reinforcement (A_s,req / A_s,prov), with the 12-in code minimum governing:

\ell_d \times \frac{A_{s,\text{req}}}{A_{s,\text{prov}}} = 14.23 \times \frac{1.738}{2.79} = 8.86\ \text{in}

\ell_{d,\text{adj}} = \max(8.86,\; 12.0) = 12.0\ \text{in}\quad \text{(12-in minimum governs)}

Available development length = cantilever − cover = 33 − 3 = 30 in.

\text{DCR} = \frac{12.0}{30.0} = 0.40

Both directions are identical for this symmetric case: PASS.

Step 7: Column-footing interface (ACI 22.8.3.2, 16.3.4.1)

The concrete bearing strength at the column-footing interface depends on the confinement ratio √(A₂ / A₁), capped at 2.0 (§22.8.3.2):

A_1 = 18 \times 18 = 324\ \text{in}^2

A_2 = (c_{\text{col}} + 2h)^2 = (18 + 36)^2 = 54^2 = 2{,}916\ \text{in}^2

\sqrt{A_2 / A_1} = \sqrt{2916 / 324} = 3.0 \;\Rightarrow\; \text{capped at } 2.0

\phi B_n = 0.65 \times 0.85 \times f'_c \times A_1 \times 2.0 = 0.65 \times 0.85 \times 4 \times 324 \times 2.0 = 1{,}432.08\ \text{kip}

DCR = 210 / 1,432 = 0.147: PASS. Minimum dowel area = 0.005 × A₁ = 1.62 in² (§16.3.4.1). Feasible arrangements: 9-#4, 6-#5, or 4-#6.

Summary

All checks pass. The 84 × 84 × 18 in footing with 9-#5 bars each way is adequate for the 18 × 18 in interior column carrying P_u = 210 kip.

Check	ACI reference	DCR	Status
Service bearing	13.2.1	0.866	PASS
Flexure (L)	13.2.7.1	0.628	PASS
Flexure (B)	13.2.7.1	0.656	PASS
One-way shear (L)	Table 22.5.5.1	0.745	PASS
One-way shear (B)	Table 22.5.5.1	0.793	PASS
Punching shear	Table 22.6.5.2	0.506	PASS
Development (L, B)	25.4.2.4	0.400	PASS
Interface bearing	22.8.3.2	0.147	PASS