How does wall footing design differ from isolated footing design?

Wall footings are analyzed per 1-ft strip of length. They skip the punching shear check because the wall load is continuous (no closed critical perimeter). Reinforcement is uniform across the width, with no gamma_s band concentration.

What is the critical section for one-way shear in a wall footing?

Per ACI 318-19 Section 13.2.7.2, the critical section is at a distance d (effective depth) from the face of the wall. The factored shear force equals the upward soil pressure times the distance from the footing edge to the critical section.

Why is there no punching shear check for wall footings?

The wall load extends the full length of the footing. No closed critical perimeter can form around the loaded area (Section 22.6.2). Only one-way shear applies.

Wall footing design per ACI 318-19

A wall (strip) footing is a continuous reinforced-concrete footing that supports a load-bearing wall along its full length. Unlike an isolated column footing, a wall footing is analyzed per 1-ft-of-length strip. All loads, areas, and capacities are per ft. Every intermediate value is pinned by the same regression tests that guard the live wall footing calculator.

Key differences from isolated footings:

No punching shear check (the loaded area extends the full wall width; no closed critical perimeter forms)
Flexure is one-way only (cantilever from face of wall, transverse direction)
Reinforcement is uniform across the width (no γ_s band concentration)
ACI minimum steel: Table 24.4.3.2 (0.0018 · b · h for Grade 60)

Problem statement

Design a strip footing for a 12-in. concrete bearing wall.

Footing width	B = 48 in. (4.0 ft)
Footing thickness	h = 12 in.
Wall thickness	12 in.
Clear cover	3.0 in. (Table 20.6.1.3.1, cast against soil)
Concrete	f′c = 3,000 psi, normal-weight (λ = 1.0)
Steel	Grade 60 (f_y = 60 ksi)
Reinforcement	#4 bars (A_b = 0.20 in²)
Allowable bearing	q_a = 3.0 ksf
Factored load	P_u = 7.0 kip/ft
Service load	P_service = 5.0 kip/ft

Step 1: Geometry

Effective depth and cantilever overhang:

d = h - \text{cover} - \tfrac{d_b}{2} = 12 - 3 - 0.25 = 8.75\ \text{in}

B_{\text{ft}} = \frac{48}{12} = 4.0\ \text{ft}

l_c = \frac{B - t_w}{2} = \frac{48 - 12}{2} = 18.0\ \text{in} = 1.5\ \text{ft}

Step 2: Bearing check (Sections 13.2.1, 13.2.2)

The service-level soil pressure must not exceed the net allowable bearing capacity. The footing self-weight and soil overburden are deducted from the gross allowable pressure (see the bearing pressure methodology):

q_{\max} = \frac{P_{\text{service}}}{B_{\text{ft}}} = \frac{5.0}{4.0} = 1.250\ \text{ksf}

Net allowable bearing after deducting footing and soil weight:

q_{\text{net}} = 2.670\ \text{ksf}

q_max = 1.250 ksf < q_net = 2.670 ksf: PASS.

Step 3: Flexure at face of wall (Section 13.2.7.1)

The critical section for flexure is at the face of the wall. The cantilevered overhang is loaded by the factored upward soil pressure (see the flexure methodology):

q_u = \frac{P_u}{B_{\text{ft}}} = \frac{7.0}{4.0} = 1.750\ \text{ksf}

M_u = \frac{q_u \cdot l_c^2}{2} = \frac{1.750 \times 1.5^2}{2} = 1.969\ \text{k-ft/ft} = 23.625\ \text{k-in/ft}

Per 1-ft strip (b = 12 in.), with #4 bars at 6-in. spacing (2 bars per ft, A_s = 0.400 in²/ft):

a = \frac{A_s\,f_y}{0.85\,f'_c\,b} = \frac{0.400 \times 60}{0.85 \times 3 \times 12} = 0.784\ \text{in}

\phi M_n = 0.9 \times 0.400 \times 60 \times (8.75 - 0.392) = 180.5\ \text{k-in/ft}

\text{DCR} = \frac{23.625}{180.5} = 0.131

Minimum reinforcement (Table 24.4.3.2):

A_{s,\min} = 0.0018 \times b \times h = 0.0018 \times 12 \times 12 = 0.259\ \text{in}^2\text{/ft}

A_s = 0.400 in²/ft > A_s,min = 0.259 in²/ft: OK. Flexure DCR = 0.131: PASS.

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

The critical section is at a distance d from the face of the wall (see the one-way shear methodology):

\text{arm} = l_c - d = 18.0 - 8.75 = 9.25\ \text{in} = 0.771\ \text{ft}

V_u = q_u \times 1.0 \times 0.771 = 1.349\ \text{kip/ft}

Concrete shear capacity per Table 22.5.5.1 (ρ_w-dependent equation, capped at 5 λ_s λ √f′c b_w d):

\rho_w = \frac{A_s}{b_w\,d} = \frac{0.400}{12 \times 8.75} = 0.003810

V_c = \frac{\min(8\,\rho_w^{1/3},\;5)\;\lambda_s\,\lambda\,\sqrt{f'_c}\;b_w\,d}{1{,}000}

8 \times 0.003810^{1/3} = 1.249 < 5.0

V_c = \frac{1.249 \times 1.0 \times 1.0 \times 54.77 \times 12 \times 8.75}{1{,}000} = 7.186\ \text{kip/ft}

\phi V_c = 0.75 \times 7.186 = 5.389\ \text{kip/ft}

\text{DCR} = \frac{1.349}{5.389} = 0.250

One-way shear: PASS (DCR = 0.250). λ_s = 1.0 per Section 13.2.6.2 for shallow foundations.

Step 5: Development length (Section 25.4.2.4)

The bottom bars must develop their full strength within the available cantilever length (see the development length methodology):

l_d = 13.145\ \text{in (general equation, Eq. 25.4.2.4a)}

l_{d,\text{adj}} = 12.0\ \text{in (adjusted for } A_{s,\text{req}} / A_{s,\text{prov}} \text{)}

l_{\text{avail}} = l_c - \text{cover} = 18.0 - 3.0 = 15.0\ \text{in}

\text{DCR} = \frac{l_{d,\text{adj}}}{l_{\text{avail}}} = \frac{12.0}{15.0} = 0.800

Modification factors: ψ_t = 1.0 (bottom bars), ψ_e = 1.0 (uncoated), ψ_s = 0.8 (#4, d_b < 0.875 in.), ψ_g = 1.0, c_b = 2.75 in.

Development length: PASS (DCR = 0.800).

Why no punching shear?

For an isolated column footing, the factored shear force concentrates at the column perimeter, creating a closed critical section at d/2 from the column face. A wall footing supports a continuous line load along its full length. No closed perimeter can form around the loaded area, so ACI 318-19 Section 22.6.2 does not apply. Only one-way shear governs.

Summary

All checks pass. The 48-in. strip footing with #4 bars at 6-in. spacing is adequate for a 12-in. wall carrying P_u = 7.0 kip/ft on soil with q_a = 3.0 ksf.

Check	ACI 318-19 reference	DCR / Value	Status
Bearing	Sections 13.2.1, 13.2.2	1.250 < 2.670 ksf	PASS
Flexure	Section 13.2.7.1	0.131	PASS
Min. reinforcement	Table 24.4.3.2	0.400 > 0.259	OK
One-way shear	Table 22.5.5.1	0.250	PASS
Development length	Section 25.4.2.4	0.800	PASS