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Spread footing · AASHTO LRFD 2024

A worked example — the PCA square spread footing, rerun under AASHTO LRFD 2024

PCA EB231 Example 2 designs a 9 × 9 ft spread footing, 3 ft thick, under a 30 in square column, with a strut-and-tie model under AASHTO LRFD. AStrutTie has no template for that truss, so it was drawn node for node as a free model, with two zero-force members added to close it. Checked under AASHTO LRFD 2024, the forces match the source, every strength check passes with the bottom tie governing, and the hooked-bar anchorage is the one check whose verdict changes.

Problem data (geometry, materials, factored load) and the strut-and-tie model are from PCA EB231, AASHTO LRFD Strut-and-Tie Model Design Examples, Example 2: Design of Footing (Mitchell, Collins, Bhide, Rabbat; Portland Cement Association, 2004), pp. 2-1–2-7. Figures and text are not reproduced; the model and every check below were run in AStrutTie under AASHTO LRFD 2024.

Given

  • f′c = 4,000 psi (27.6 MPa) · fy = 60,000 psi (414 MPa) · φ = 0.70 for compression, 0.90 for tension
  • Footing 9 × 9 ft, 3 ft thick · column 2 ft 6 in square at the center
  • Factored column load 714 kip, including the column's own weight · column moment neglected, as in the source
  • Uniform soil pressure: nine reactions of 79.3 kip at 1 ft spacing · bottom bars 8-#8 each way, 3 in cover, 180° hooks
①

The problem

A square spread footing carries a column load into the ground. Footings are usually designed for flexure and shear at critical sections; the source uses a strut-and-tie model to follow the flow of forces instead. No AStrutTie template builds this truss, so the outline — 108 × 36 in — was drawn as a free model, in kip and inch as in the source, with the full 108 in width of the footing as the member thickness.

Spread footing outline and material in AStrutTieSpread footing outline and material in AStrutTie
Footing outline and material — PCA EB231 Example 2 drawn as a free model in AStrutTie.
②

Build the model — the PCA truss

The source puts the tie 4 in above the bottom and the top strut 2 in below the top, a lever arm of 30 in. The soil pressure is uniform, so it enters as nine reactions of 79.3 kip at nodes A–I, 1 ft apart. The column load is split into three forces of 238 kip at J, K and L, each the resultant of a 10 in third of the column, and each top node fans out to three bottom nodes. That truss is a mechanism — the panels J–C–D–K and K–F–G–L have no diagonal — and it stands only because every top node carries exactly three reactions. The app would solve it as a lower-bound equilibrium and flag it, so two members, CK and GK, were added to close it; under this load they carry nothing. The supports at A and I only hold the model in place: with the loads in equilibrium they take no reaction. The analysis gives AJ −128.0 · BJ −105.0 · CJ −87.5 · DK −85.4 · EK −79.3 kip in the struts and 100.5 · 169.2 · 206.3 · 238.0 kip along the tie, the source's forces to within its rounding.

PCA EB231 Example 2 strut-and-tie model of the spread footing in AStrutTiePCA EB231 Example 2 strut-and-tie model of the spread footing in AStrutTie
The PCA truss on the AStrutTie canvas — node and member names as in the source, member forces in kip.
③

Analyze & check — every member, every node

31 of 31 strength checks pass. The minimum safety factor is 1.43, on the bottom tie DE–EF under the column. Every strut and node keeps a wide margin; the tightest is the top strut JK at 2.64. Anchorage is reported separately from the strength verdict, and both hooked ends of the tie fail it; step ⑦ explains why.

Spread footing analysis results with member sections and overall verdictSpread footing analysis results with member sections and overall verdict
Analysis · Check — required and maximum sections, overall verdict.
④

Tie check — required reinforcement

AASHTO LRFD 2024

Ties DE and EF carry 238 kip and need 4.41 in², the source's value. The source then raises the steel to 1.33 times that, 5.87 in², because the minimum flexural reinforcement for 1.2 Mcr (8.30 in²) would otherwise govern, and chooses 8-#8 each way. The app takes a bar's area as π·d²/4 = 0.785 in², so the eight bars give 6.28 in² rather than the source's 6.32 in²: a safety factor of 1.43. The flexural minimum is a separate check from the strut-and-tie checks, and the app does not run it.

As,req = Fu / (φ · fy)
Required reinforcement table for the footing tieRequired reinforcement table for the footing tie
Tie checks — required vs provided A_s per member.
⑤

Strut check — effective strength & width

AASHTO LRFD 2024 Table 5.8.2.5.3a-1

A footing needs no crack control reinforcement, and this one has none, so every strut-to-node interface takes ν = 0.45: fce = 1.80 ksi with φ = 0.70. The source used the strain-based strut limit of its edition: for strut AJ, with the tie strain at node A, fcu = 2.73 ksi on a width of 12 in · sin 38.3° = 7.4 in, so φPn = 1,527 kip against 128 kip. The 2024 edition drops that limit and checks a strut where it meets a node. AJ needs 0.94 in; the app finds 13.71 in at node A, where the tie's 8 in height adds to the bearing length the source used, and 9.34 in at node J, which governs: a safety factor of 9.92. The top struts JK and KL, 4 in deep, carry 206.3 kip and are the tightest at 2.64.

fce = m · ν · f′c
wreq = Fu / (φ · fce · b) ≤ wprov
Strut strength verification of the footingStrut strength verification of the footing
Strut checks — f_ce, required width and safety factor per member.
⑥

Nodal-zone check — by face and node type

AASHTO LRFD 2024 Table 5.8.2.5.3a-1

ν by face, for f′c = 4 ksi. Without crack control reinforcement every face takes 0.45. Nodes J, K and L, where the column frames in, are CCC; the source calls them not critical and does not check them. The app checks them: the 4 in face of the top strut needs 1.52 in (2.64) and the column bearing 1.75 of 10 in (5.72). At node A, a CCT node, the source checks the tie's anchorage face, 0.12 ksi against 2.1 ksi. The app has the same face at 0.74 of 8 in (10.83) — a smaller margin, because ν drops from 0.75 to 0.45. Along the tie, nodes B–H are CTT; the most loaded, D–F, have 4.57.

fce = m · ν · f′c
wreq = Fu / (φ · fce · b) ≤ wprov
FaceWith crack control reinforcementWithout
CCC — bearing · back face0.850.45
CCT — bearing · back face0.700.45
CTT — bearing · back face0.650.45
Strut-to-node interface0.650.45
Nodal-zone strength verification of the footingNodal-zone strength verification of the footing
Nodal-zone checks — each face against its node type.
⑦

Anchorage check — the 2024 development equation

AASHTO LRFD 2024 §5.8.2.4.2 · §5.10.8.2.4a

The tie ends in 180° hooks, #8 with 3 in cover. The source looks at the critical section 12 in from the edge, finds only 15.9 ksi in the bars there (100.5 kip on 8 × 0.79 in²), and judges the 9 in left to the end of the hook enough to develop it. The app measures the available length along the tie from the extended nodal zone, 14.1 in, and asks for the 2024 development length of a #8 hook, 20.0 in, without the excess-reinforcement reduction: a safety factor of 0.70 at A and at I. The NG is shown as the app reports it.

Tie anchorage verification of the footingTie anchorage verification of the footing
Anchorage check — available ℓanc vs required development ℓd.
⑧

What changes from the source

PCA EB231 (2004) → AASHTO LRFD 2024

Same footing, same truss, same forces — with two zero-force members added so the model is stable. Every strength check passes under the 2024 edition, as it did in the source. The numbers move for known reasons: the strain-based strut limit is gone, a footing without crack control reinforcement takes ν = 0.45, the app checks the CCC nodes the source leaves out, the bar-area convention, and the development equation, which turns the hooked anchorage into the one NG.

CheckPCA EB231 (2004)This example — AASHTO LRFD 2024Why
Member forcesAJ −128 · BJ −105 · CJ −88 · DK −85 · EK −79.3 · DE 238 kip …AJ −128.0 · BJ −105.0 · CJ −87.5 · DK −85.4 · EK −79.3 · DE 238.0 kip …Same truss and statics
Model19 members, a mechanism that stands under this load+ CK · GK at 0 kip — statically determinateThe app flags a mechanism and solves it only as a lower bound
Tie DE, 8-#84.41 in² required · 6.32 in² provided4.407 in² required · 6.283 provided (1.43)Same φ = 0.90 · bar area π·d²/4 in the app
Minimum flexural reinforcement1.33 × 4.41 = 5.87 in² ≤ 6.32 in²Not part of the STM checksA hand check outside the app
Strut AJfcu 2.73 ksi · w 7.4 in · φPn 1,527 ≥ 128 kipfce 1.80 ksi · 9.34 in at J (9.92)No strain-based strut limit in 2024 · ν 0.45 · width from the nodal zone
Nodes J · K · L (CCC)Not critical — not checkedTop-strut face 2.64 · column bearing 5.72The app checks every node
Node A (CCT), tie face0.12 ≤ 2.1 ksi0.74 of 8 in (10.83)ν 0.75 → 0.45 without crack control reinforcement
Crack control reinforcementFootings exemptTaken as not provided — ν = 0.45 on every faceTable 5.8.2.5.3a-1
Hooked #8 anchorage at A15.9 ksi in the bars · 9 in to the hook end enough20.0 in required > 14.1 in (0.70) — NG2024 development equation; no excess-reinforcement reduction in the app

These are the checks AStrutTie runs on every model — here against a published example anyone can open, with the forces matching the source and every difference traced to a clause, a convention or an edition. How we verify →

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