Reinforced Concrete Design of Isolated Footing Design According to ACI 318-19 (LRFD) in RFEM 6

Model Description (LRFD – Isolated Footing)

In this example, a simple square isolated footing is modeled.
A square reinforced concrete column is placed centrally on a square footing slab.
No stepped footing or special foundation geometry is used, in order to keep the example as simple as possible.

Geometry

• Footing slab: 6.0 ft × 6.0 ft, thickness t = 1.75 ft
• Column: 1.0 ft × 1.0 ft, height 2.0 ft, centrally positioned on the footing slab

3D Model of Rectangular Single Foundation with Central Column in RFEM 6

Material Parameters

The following material parameter is used for the foundation design.

• Concrete unit weight: γ_c = 150 lb/ft³ = 0.150 kip/ft³

Load Cases

Four load cases are defined in the model.

Load Case 1: Dead Load (Self Weight)

• The self-weight of the model is automatically considered.
• Additionally, a vertical force of 60 kip is applied at the top of the column in the +z-direction (acting downward).

Load Case 2: Live Load

• A vertical force of 25 kip is applied at the top of the column in the +z-direction (acting downward).

Load Case 3: Wind Load (W+)

• A member load of w = 2 kip/ft acting in the positive x-direction is applied to the upper half of the column (1 ft of the column height).
• Additionally, a moment of -25 kipft about the y-axis is applied in the same rotational direction as the member load.

Load Case 4: Wind Load (W−)

• A member load of w = 2 kip/ft acting in the negative x-direction is applied to the upper half of the column (1 ft of the column height).
• Additionally, a moment of 25 kipft about the y-axis is applied in the same rotational direction as the member load.

Load Combinations (LRFD)

For the verification of the isolated footing, the load combinations according to the Load and Resistance Factor Design (LRFD) provisions of ASCE 7 | 2022, Section 2.3, are used.

The following load cases are defined in the model:
LC1 = D, LC2 = L, LC3 = W+, LC4 = W−

According to ASCE 7-22, Section 2.3, all basic combinations are first generated from the load cases present in the model. This complete set appears in RFEM under "Action Combinations". Combinations that are algebraically identical to a combination already covered under a different number are recognized as redundant and are not included in the actual calculation. Since no roof live load, snow load, rain load, or seismic action is defined in this model, Combinations 3 and 6 are not applicable.

The reduced set actually used for the calculation can be found under "Load Combinations." For this model, the following governing combinations result:

All Load Combinations listed above are evaluated for the foundation design checks. For each verification (flexural design, one-way shear, two-way shear), the governing load combination with the highest utilization ratio is determined.

Determination of Vertical Dead Load (Load Case 1)

Load Case 1 contains the total vertical dead load acting on the foundation.
This includes the applied column load, the self-weight of the footing and column, and the weight of the soil cover above the footing.

Self-weight of footing

W_f = 6.0 ft · 6.0 ft · 1.75 ft · 0.150 kip/ft³ = 9.45 kip

Self-weight of column

W_c = 1.0 ft · 1.0 ft · 2.0 ft · 0.150 kip/ft³ = 0.30 kip

Weight of soil cover

G_soil = (6.0 ft · 6.0 ft − 1.0 ft · 1.0 ft) · 1.0 ft · 0.110 kip/ft³ = 3.85 kip

Total vertical load in Load Case 1

P_LC1 = 60.0 kip + 9.45 kip + 0.30 kip + 3.85 kip = 73.60 kip

Concrete Design Checks

The isolated footing is verified according to the LRFD approach defined in IBC/ACI 318-19.
RFEM automatically evaluates all relevant load combinations and determines the governing combination for each design check.

Flexural Design (Bottom Reinforcement)

For this example, the governing load combination for the bottom reinforcement in x-direction is LRFD Combination 4 (1.20D + 1.0L + 1.0W+), and for the bottom reinforcement in y-direction it is LRFD Combination 2 (1.20D + 1.60L).
According to ACI 318-19 §13.2.6.4, the factored moment for the bottom longitudinal reinforcement results from the superposition of the moment from the compressive stress block (soil bearing pressure) and the moment from additional foundation loads, as shown in the design section at the column face.

Bottom Reinforcement, x-Direction

M_{f,x,(bottom),u} = max(M_{C,+x,(bottom)} + M_{A,+x,(bottom)} ; M_{C,−x,(bottom)} + M_{A,−x,(bottom)} ; 0) / Φ
= max(70.87 kipft − 8.38 kipft ; 47.17 kipft − 8.38 kipft ; 0) / 0.900 = 69.43 kipft
M_2w,u = √((0.60 · −28.00 kipft)² + (0.60 · 0.00 kipft)²) / 0.900 = 18.67 kipft
M_{f,x,(bottom),u} = max(69.43 kipft ; 18.67 kipft) = 69.43 kipft
A_{s,stat,x,(bottom)} = max(0.72 in² ; 0.00 in²) = 0.72 in²

Utilization ratio

η = A_{s,stat,x,(bottom)} / A_{s,prov,x,(bottom)} = 0.72 in² / 3.16 in² = 0.227 [-]
Since η < 1.0, the flexural check for the bottom reinforcement in the x-direction is satisfied.

RFEM 6 Design Check Details SD0206.01 – Flexural Design of Foundation Plate in x-Direction

Bottom Reinforcement, y-Direction

M_{f,y,(bottom),u} = max(66.83 kipft − 8.38 kipft ; 66.83 kipft − 8.38 kipft ; 0) / 0.900 = 64.94 kipft
Since LRFD Combination 2 contains no wind action, there is no moment transfer at the critical punching shear perimeter: M_2w,u = 0.00 kipft
M_y,(bottom),u = max(64.94 kipft ; 0.00 kipft) = 64.94 kipft
A_{s,stat,y,(bottom)} = max(0.69 in² ; 0.00 in²) = 0.69 in²

Utilization ratio

η = A_{s,stat,y,(bottom)} / A_{s,prov,y,(bottom)} = 0.69 in² / 3.16 in² = 0.220 [-]
Since η < 1.0, the flexural check for the bottom reinforcement in y-direction is satisfied.

RFEM 6 Design Check Details SD0206.02 – Flexural Design of Foundation Plate in y-Direction

Top Reinforcement (x- and y-Direction)

For this example, the top reinforcement utilization ratios are η = 0.000 in both directions.
The governing combinations produce compression-controlled sections (Φ_compr = 0.65) with a resulting factored top moment of 0 kipft.

One-Way Shear

The one-way shear strength is verified according to ACI 318-19 §22.5 at the critical section located at a distance d from the column face.
For this example, the governing load combination is LRFD Combination 2 (1.20D + 1.60L).

Applied Shear Force

V_u = max(0 ; V_C,red+ − V_A,red+ ; V_C,red− − V_A,red−) = max(0 ; 19.815 kip − 2.485 kip ; 19.815 kip − 2.485 kip) = 17.330 kip

Shear Resistance

V_c = V_c(c) = min(5 · λ · √f'_c · b_w · d, (8 · λ_s · λ · (ρ_w)^1/3 · √f'_c + min(0.05 · f'_c, N_u / (6 · A_g))) · b_w · d)
= min(5 · 1 · 63.2 psi · 72 in · 18.881 in, (8 · 1 · 1 · (0.23 %)^1/3 · 63.2 psi + min(0.05 · 576 ksf, 0)) · 72.000 in · 18.881 in)
= 91.093 kip
V_max = Φ_V,T · (V_c + 8 · √f'_c · b_w · d) = 0.750 · (91.093 kip + 8 · 63.2 psi · 72.000 in · 18.881 in) = 584.310 kip

Utilization ratio

η = max(V_u / (Φ_V,T · V_n) ; V_u / V_max) = max(17.330 kip / (0.750 · 91.093 kip) ; 17.330 kip / 584.310 kip) = 0.254 [-]
Since η < 1.0, the one-way shear check is satisfied for LRFD Combination 2.

RFEM 6 Design Check Details SD0207 – Shear Design of Foundation Plate

Two-Way (Punching) Shear

The punching shear strength is verified according to ACI 318-19 §22.6 at the critical perimeter located at d/2 from the column face.
For this example, the governing load combination is LRFD Combination 4 (1.20D + 1.0L + 1.0W+).

Applied Shear Stress

v_u = v_uv − (γ_v,1 · M_sc,1,sl · e₁ / J₁) + (γ_v,2 · M_sc,2,sl · e₂ / J₂) = 4.748 ksf − (0.40 · −28.00 kipft · 15.597 in / 388401 in⁴) + (0.40 · 0.00 kipft · −15.597 in / 388401 in⁴) = 5.526 ksf

Shear Resistance

v_c = min(v_c,(a) ; v_c,(b) ; v_c,(c)) = min(36.429 ksf ; 54.644 ksf ; 74.253 ksf) = 36.429 ksf

Utilization ratio

η = |v_u| / (Φ_V,T · v_n) = |5.526 ksf| / (0.750 · 36.429 ksf) = 0.202 [-]
Since η < 1.0, the two-way (punching) shear check is satisfied for LRFD Combination 4.

RFEM 6 Design Check Details SD0208 – Punching Shear Check of Foundation Plate

Minimum Reinforcement

In addition to the flexural design checks above, the minimum longitudinal reinforcement area for the bottom reinforcement is verified for both the x- and y-direction according to ACI 318-19. The required reinforcement area results from the greater of the statically required reinforcement and the minimum reinforcement area specified by the code.

Minimum Reinforcement, x-Direction

A_{s,stat,x,(bottom)} = max(A_{s,stat,t,x,(bottom)} ; A_{s,stat,p,x,(bottom)}) = max(0.50 in² ; 0.00 in²) = 0.50 in²
M_{f,x,(bottom),u} = max(M_{C,+x,(bottom)} + M_{A,+x,(bottom)} ; M_{C,−x,(bottom)} + M_{A,−x,(bottom)} ; 0) / Φ
= max(53.66 kipft − 9.78 kipft ; 53.66 kipft − 9.78 kipft ; 0) / 0.900 = 48.76 kipft
A_{s,req,x,(bottom)} = max(A_{s,stat,x,(bottom)} ; A_{s,min,x,(bottom)}) = 2.72 in²

Utilization ratio

η = A_{s,req,x,(bottom)} / A_{s,prov,x,(bottom)} = 2.72 in² / 3.16 in² = 0.862 [-]
Since η < 1.0, the minimum reinforcement check for the bottom reinforcement in the x-direction is satisfied.

RFEM 6 Design Check Details SD0216.01 – Required Bottom Longitudinal Reinforcement in x-Direction

Minimum Reinforcement, y-Direction

A_{s,stat,y,(bottom)} = max(A_{s,stat,t,y,(bottom)} ; A_{s,stat,p,y,(bottom)}) = max(0.52 in² ; 0.00 in²) = 0.52 in²
M_{f,y,(bottom),u} = max(M_{C,+y,(bottom)} + M_{A,+y,(bottom)} ; M_{C,−y,(bottom)} + M_{A,−y,(bottom)} ; 0) / Φ
= max(53.66 kipft − 9.78 kipft ; 53.66 kipft − 9.78 kipft ; 0) / 0.900 = 48.76 kipft
A_{s,req,y,(bottom)} = max(A_{s,stat,y,(bottom)} ; A_{s,min,y,(bottom)}) = 2.72 in²

Utilization ratio

η = A_{s,req,y,(bottom)} / A_{s,prov,y,(bottom)} = 2.72 in² / 3.16 in² = 0.862 [-]
Since η < 1.0, the minimum reinforcement check for the bottom reinforcement in the y-direction is satisfied.

RFEM 6 Design Check Details SD0216.02 – Required Bottom Longitudinal Reinforcement in y-Direction

Conclusion

This example demonstrates the complete structural design workflow for an isolated footing in RFEM according to ACI 318-19, using LRFD load combinations according to ASCE 7. The workflow includes all relevant reinforced concrete design checks, including flexural design of the bottom and top reinforcement, one-way shear, and two-way (punching) shear.

The evaluation shows that all governing verification criteria are satisfied. The provided reinforcement is sufficient in both principal directions, while the shear and punching shear checks confirm adequate structural capacity without requiring shear reinforcement.

Overall, this example illustrates how RFEM provides a transparent and efficient workflow for reinforced concrete footing design, enabling engineers to evaluate design moments, shear forces, reinforcement requirements, and governing load combinations in a single design process.

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