Notes  /  ACI 318

What actually changed in the ACI 318-19 shear provisions.

One equation became a table of three, and two new terms crept into it. For beams with stirrups, almost nothing changed. For slabs, footings and mats, capacity can fall by more than half.

ACI 318 9 min read

If you learned reinforced concrete on ACI 318-14 or earlier, one number is probably burned in: 2√f'c. It was the concrete contribution to one-way shear, it applied to nearly everything, and you could estimate it in your head.

ACI 318-19 removed it as a general rule. In its place is Table 22.5.5.1, which gives three expressions for Vc and makes you choose between them based on how much shear reinforcement the member carries. Two quantities that never appeared in the old equation now appear in the new one: the longitudinal reinforcement ratio, and the effective depth itself.

The consequence is narrow but severe. Members carrying at least minimum shear reinforcement are broadly unaffected. Members carrying none — one-way slabs, pad footings, mat foundations, thick transfer slabs — can compute a design shear capacity 40 to 60% below what the same section was credited with under 318-14. Designs that passed a checker in 2018 do not necessarily pass now.

The rule that went away

ACI 318-14 gave a single expression for members without axial load, and the whole profession used it as the default:

318-14 Eq. 22.5.5.1 Vc = 2 λ √f'c · bw · d λ = 1.0 for normalweight concrete · f'c in psi · bw, d in inches. In SI the coefficient is 0.17, with f'c in MPa.

It did not care how heavily the member was reinforced in flexure. It did not care whether the member was 8 inches deep or 8 feet deep. Both omissions were known to be unconservative, and both were corrected in 318-19.

What replaced it

Table 22.5.5.1 splits on a single question: does the member carry at least the minimum area of shear reinforcement, Av,min?

If it does, you get two options and may take the more favourable:

Av ≥ Av,min (a)  Vc = [ 2 λ √f'c + Nu / (6 Ag) ] · bw · d
(b)  Vc = [ 8 λ (ρw)1/3 √f'c + Nu / (6 Ag) ] · bw · d

Option (a) is the old familiar expression, preserved intact. This is why beams with stirrups barely moved. If your member has shear reinforcement you can still use 2λ√f'c exactly as before, and only reach for option (b) when a high reinforcement ratio makes it more generous.

If the member carries less than Av,min — which includes every member exempted from minimum shear reinforcement altogether, so every ordinary slab and footing — there is no option. You get one equation, and it is the harsh one:

Av < Av,min Vc = [ 8 λs λ (ρw)1/3 √f'c + Nu / (6 Ag) ] · bw · d ρw = As / (bw d), the ratio of longitudinal tension reinforcement · λs = size effect factor · Nu positive for compression, negative for tension.

Two independent penalties are stacked in that line. Take them one at a time.

Penalty one: the reinforcement ratio

The term 8(ρw)^(1/3) replaces the constant 2. It is worth finding the crossover point where the two are equal:

8 (ρw)1/3 = 2  →  ρw = 0.253 = 0.0156  (1.56%)

Below a longitudinal reinforcement ratio of about 1.6%, the new expression gives less than the old 2√f'c — before the size effect factor is applied at all. Slabs and footings typically sit between 0.2% and 0.6%. At ρw = 0.3% the term evaluates to roughly 1.15√f'c, already only 58% of the old value.

The reasoning behind it is long established: aggregate interlock across a shear crack depends on crack width, crack width depends on the strain in the flexural steel, and a lightly reinforced section strains more. The old constant simply averaged over all of that.

Penalty two: the size effect factor

The second term, λs, is new in 318-19 and is the one most likely to catch a foundation design:

318-19 Eq. 22.5.5.1.3 λs = √( 2 / (1 + d/10) ) ≤ 1.0 d in inches. In SI, replace d/10 with d/250 for d in millimetres. λs applies only where Av < Av,min.

It reaches 1.0 at d = 10 in. and falls away steadily above that. Note that λs is a separate factor from λ, the lightweight-concrete modifier — they multiply together, and confusing the two is an easy way to be wrong by 30%.

Effective depth dλsReduction
10 in.1.000
16 in.0.87712%
20 in.0.81618%
24 in.0.76723%
30 in.0.70729%
40 in.0.63237%
60 in.0.53547%

The size effect is why "make the footing thicker" is no longer a reliable way out of a shear problem — beyond a point, added depth buys less capacity than the linear d term suggests.

The ceiling and the axial term

Two limits close out the table. Vc may not be taken greater than 5λ√f'c bw d (§22.5.5.1.1), which caps what a heavily reinforced section can claim from option (b). And the axial term Nu/(6Ag) now handles compression and tension in one expression, replacing the separate 318-14 equations — with Nu signed, so a member in net tension is penalised automatically rather than being sent to a different clause.

A pad footing, worked both ways

Abstract percentages do not persuade anyone. Here is an ordinary isolated pad footing, checked for one-way shear on a 12-inch design strip.

InputValue
Concrete strength, f'c4,000 psi (normalweight, λ = 1.0)
Overall depth, h24 in.
Effective depth, d20 in.
Design strip, bw12 in.
Flexural steel#6 @ 12 in. → As = 0.44 in²/ft
Shear reinforcementNone (exempt from Av,min)
Axial through section, Nu0

Under ACI 318-14, one line:

Vc = 2 (1.0) √4000 × 12 × 20 = 30,360 lb = 30.4 kip
φVc = 0.75 × 30.4 = 22.8 kip

Under ACI 318-19, the footing has no stirrups, so the third row of Table 22.5.5.1 governs:

ρw = 0.44 / (12 × 20) = 0.00183  →  (ρw)1/3 = 0.1223
λs = √( 2 / (1 + 20/10) ) = √0.667 = 0.816

Vc = 8 (0.816)(1.0)(0.1223) √4000 × 12 × 20
    = 50.5 psi × 240 in² = 12,130 lb = 12.1 kip
φVc = 0.75 × 12.1 = 9.1 kip Check the ceiling: 5λ√f'c bw d = 75.9 kip, so §22.5.5.1.1 does not govern here.
The result

φVc falls from 22.8 kip to 9.1 kip — the same concrete, the same steel, the same geometry, retaining 40% of its former design capacity. A footing sized against a factored shear of 15 kip passed comfortably under 318-14 and fails outright under 318-19.

Most of that loss is the reinforcement ratio, not the size effect. The ρw term on its own takes Vc to 49% of the old value; λs then carries it down to 40%. That ordering matters, because it is the opposite of the intuition most engineers bring to the change — the headline is the new size effect factor, but the quieter cube-root term is doing roughly three quarters of the damage.

The quieter change: when Av,min is triggered

Alongside the Vc rewrite sits a second change that matters more than its column-inches suggest. Under 318-14, minimum shear reinforcement was required in nonprestressed beams wherever Vu > 0.5φVc. Under 318-19 §9.6.3.1 the general trigger for nonprestressed beams becomes:

318-19 §9.6.3.1 Av,min required where  Vu > φ λ √f'c · bw · d Except for the cases listed in Table 9.6.3.1, where the trigger remains Vu > φVc.

That is a lower threshold than before, so stirrups are now required over longer stretches of a beam than they used to be. It also interacts with the table above in a way worth thinking through, because the two provisions form a loop: whether you need Av,min depends on Vc, and which Vc equation you may use depends on whether you provided Av,min.

The design leverage

That loop cuts in your favour. Providing Av,min — often a modest amount of steel — moves the member from the third row of Table 22.5.5.1 to the first, which restores the full 2λ√f'c and switches off λs entirely.

On a deep, lightly reinforced member, the cheapest route to a passing shear check is frequently to add nominal stirrups rather than concrete. That inversion — reinforcement being more efficient than depth — is the most useful habit to take from 318-19.

What this means in practice

  • Beams with stirrups: essentially unchanged. Option (a) preserves the old expression, and option (b) is a bonus available when the reinforcement ratio is high.
  • One-way slabs and pad footings: the most exposed members on any project. No stirrups, low reinforcement ratios and generous depths put them squarely in the worst row of the table.
  • Mats and thick transfer slabs: worst affected of all. At d = 40 in., λs alone removes 37%, before the ρw term has been applied.
  • Joists and shallow members: λs = 1.0 below d = 10 in., so the size effect is silent — but the reinforcement ratio term is not, and it is usually the larger of the two penalties anyway.
  • Legacy spreadsheets: anything written before 2019 that hard-codes 2√f'c is now unconservative for every member without shear reinforcement. It will not error. It will simply pass sections that the current code fails.

That last point is the one worth acting on this week. The change is invisible inside an old workbook: no warning, no broken formula, just an answer that used to be right. If you are running a sheet you inherited from a colleague, open the shear cell and look at what it multiplies √f'c by.

Our ACI 318 pack applies the current rule.

The beam, slab and footing sheets implement Table 22.5.5.1 in full — testing Av against Av,min, computing λs from the actual effective depth, and showing the section reference beside every line. The footing sheet will flag a section that a pre-2019 calculation would have passed.

The sample workbook is a complete, unlocked beam sheet with its Parameters tab, so you can read the formulas before paying anything.

Section numbers refer to ACI 318-19, Building Code Requirements for Structural Concrete. This note is a summary written to help engineers navigate the change and is not a substitute for the code text — verify every provision against the current published edition before relying on it in design. Structural Studio is independent and not affiliated with ACI.