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Two-column pile footing · ACI 318

A worked example — a two-column pier footing on piles

A footing carrying two columns on a 21-pile group is a D-region twice over: the load spreads from each column into the cap, and the pile reactions come back up as concentrated forces. Beam theory does not describe either. Here is the strut-and-tie design, end to end.

AStrutTie 2020 Technical Book, Example 02-2.2 (two-column pile foundation). Every figure below is taken from that published source.

Given

  • Footing 10,900 × 4,500 × 1,500 mm; two columns 1,500 × 1,500 mm at 6,400 mm centres
  • Piles: φ508 steel pipe, L = 10,000 mm — 7 across × 3 deep = 21 piles
  • fck = 24 MPa · fy = 300 MPa
  • Column 1: V = 5,700 kN · H = 900 kN · M = 5,300 kN·m
  • Column 2: V = 5,300 kN · H = 1,000 kN · M = 5,200 kN·m
  • Load case LC01 = 1.00 L1 + 1.00 L2

Model the footing upside down

In a pile-cap model the column reactions are the supports and the pile reactions are the applied loads — the structure is analysed the way the forces actually flow. AStrutTie builds this from the template: 22 nodes, 41 elements, 6 supports (three under each column), 11 bearing plates.

STM settingValue
Nodes under each column (A)3
Node offset from column face (B)100 mm
Top cover (C) · bottom cover (D)100 mm · 200 mm
Diagonal height/width ratio (F)0.8
Reinforcementtop 22-#7 · bottom 43-#8 · vertical shear 10-#6 @400 · horizontal 6-#6 @400

Optimize, then check the sign of every diagonal

The first optimization run left element E28 in tension at +45.6 kN — a diagonal that was supposed to be a strut. That is not a rounding artefact; it means the assumed truss does not match the flow. The direction was corrected and the model re-analysed (ordinary analysis, not optimization) to give the final result.

Ties — required steel

ACI 318 §23.7

All O.K. Shear ties: Tie 19 (1,299.05 kN) and Tie 23 (272.37 kN), both vertical, 10-#6 with weff = 950.0 mm and sh = 400.0 mm giving φFn = 1,523.1 kN.

As,req = Fu / (φ · fy)
φ = 0.75 · fy = 300 MPa
TieFu (kN)As,req (mm²)Provided
Tie 5 (top)119.41530.7322-#7 → 8,534.90 mm²
Tie 32875.603,891.5743-#8 → 21,788.53 mm²
Tie 341,147.025,097.8843-#8
Tie 38206.20916.4343-#8
Tie 401,612.497,166.6443-#8

Struts — required width

ACI 318 §23.4

Note S28: it is prismatic, so βs = 1.00 and its required width collapses to 34.6 mm — while S3 with the same order of force needs 133.3 mm at βs = 0.40. The strut condition, not the force alone, sets the width.

wreq = Fu / (φ · 0.85 · βs · βc · fck · b)
βc = 1.00 · b = 4,500 mm
StrutβsθFu (kN)wreq (mm)wprov (mm)
S30.400.0°3,671.8133.3200.0
S180.4051.6°3,317.5120.5203.0
S240.4051.6°2,695.397.9203.0
S281.0061.6°2,382.834.6183.0
S200.4036.9°2,165.178.6250.0

Nodal zones

ACI 318 §23.9

Every node is classified and checked face by face. Node 3 is CCC (βn = 1.00) carrying C-2 604.2, C-3 3,671.8 and C-15 1,032.6 kN; Node 5 is CTT (βn = 0.60) with C-4 1,612.7, T-5 119.4, T-19 1,299.1 and C-20 2,165.1 kN; Node 7 is CTT with five members framing in. All faces O.K.

Twenty-one piles, two columns, 41 truss elements — and the design comes down to the same three checks, each printed with the clause and factors it used. The tie that governs is Tie 40 at 1,612 kN; the strut that governs is S3 needing 133 mm of the 200 mm available. How we verify →