Beam stirrup spacing requirements per ACI 318-19

Q: What is the maximum stirrup spacing for a beam with d = 24 in?

It depends on Vs. If Vs <= 4*sqrt(f'c)*bw*d, the limit is min(d/2, 24 in) = min(12, 24) = 12 in. If Vs exceeds that threshold, the limit tightens to min(d/4, 12 in) = min(6, 12) = 6 in.

Q: When does the d/4 spacing limit apply?

When Vs > 4*sqrt(f'c)*bw*d. Under heavy shear the diagonal cracks are steeper and more closely spaced, so stirrups must be placed closer together to make sure every crack is crossed by at least one stirrup.

Q: Can I use stirrup yield strength above 60 ksi?

Table 20.2.2.4a caps fyt at 60 ksi for shear design, and at 80 ksi for welded deformed wire reinforcement. Higher-grade bars may be used for detailing, but only the capped value counts toward shear capacity.

Stirrup spacing is limited so that every diagonal shear crack is crossed by at least one stirrup. ACI 318-19 sets the limit by beam depth d and by how much shear the stirrups carry (V_s). This page covers the longitudinal spacing limits, the minimum stirrup area, and the lateral spacing of stirrup legs. For the full shear capacity check, see beam shear design.

Applicable clauses

ACI 318-19 reference	Topic
Cl 9.7.6.2.2	Maximum stirrup spacing
Table 9.7.6.2.2	Spacing limits (d/2 vs d/4 threshold)
Cl 9.6.3.1	Minimum shear reinforcement trigger
Cl 9.6.3.3	Minimum A_v formula
Table 20.2.2.4a	f_yt cap (60 ksi / 420 MPa)
Cl 9.7.6.3.1	Lateral spacing of stirrup legs

When stirrups are required (Cl 9.6.3.1)

Minimum shear reinforcement is required when the factored shear exceeds half the design concrete shear strength:

V_u > 0.5\,\phi\,V_c \implies \text{stirrups required} \quad (\phi = 0.75)

Below this threshold stirrups are not required by Chapter 9 (some exceptions exist, such as shallow members and slabs). Above it, stirrups must be provided at a spacing no larger than the limits below, and with at least the minimum area in the next section.

Maximum longitudinal spacing (Table 9.7.6.2.2)

The limit has two tiers, selected by the steel shear contribution V_s:

Condition	s_max
V_s ≤ 4√f′_c b_w d	min(d/2, 24 in)
V_s > 4√f′_c b_w d	min(d/4, 12 in)

s_{max} = \begin{cases} \min(d/2,\; 24\text{ in}) & V_s \leq 4\sqrt{f'_c}\,b_w\,d \\ \min(d/4,\; 12\text{ in}) & V_s > 4\sqrt{f'_c}\,b_w\,d \end{cases} \quad (f'_c \text{ in psi})

Precomputed limits by effective depth d:

d (in)	s_max, V_s ≤ 4√f′_c b_w d (in)	s_max, V_s > 4√f′_c b_w d (in)
12	6	3
18	9	4.5
24	12	6
30	15	7.5
36	18	9
42	21	10.5
48	24	12

The 24 in and 12 in caps start to govern at d = 48 in. For typical building beams (d between 12 and 36 in), the d/2 and d/4 terms govern. The tighter tier applies under heavy shear because diagonal cracks are steeper and more closely spaced.

Minimum stirrup area (Cl 9.6.3.3)

Where stirrups are required, the area per unit length must not be less than:

\frac{A_{v,min}}{s} = \max\!\left(\frac{0.75\,\sqrt{f'_c}\;b_w}{f_{yt}},\;\; \frac{50\,b_w}{f_{yt}}\right) \quad (f'_c,\,f_{yt} \text{ in psi})

Rearranged for a chosen stirrup (A_v = number of legs × bar area), this becomes a second spacing limit:

s \leq \frac{A_v\,f_{yt}}{\max\!\left(0.75\sqrt{f'_c},\; 50\right) b_w}

The 50 psi term governs up to f′_c of about 4,444 psi; above that, the 0.75√f′_c term governs. Example values for a two-leg No. 3 stirrup (A_v = 0.22 in²) with f_yt = 60 ksi:

f′_c (psi)	b_w = 12 in	b_w = 16 in	b_w = 20 in
4,000	22.0 in	16.5 in	13.2 in
5,000	20.7 in	15.6 in	12.4 in
6,000	18.9 in	14.2 in	11.4 in

These are the A_v,min limits only. The governing spacing is the smallest of the A_v,min limit, the Table 9.7.6.2.2 limit, and the spacing required by strength (V_s = A_v f_yt d / s). Stirrup yield strength is capped at 60 ksi (Table 20.2.2.4a).

Lateral spacing of stirrup legs (Cl 9.7.6.3.1)

Across the width of the beam, the spacing between legs of a multi-leg stirrup (or between stirrup legs and the beam face) must not exceed the lesser of d and 24 in. Wide beams, such as shallow transfer beams or wide band beams, may need four legs or a second stirrup nested inside the first to satisfy this limit.

What calcnote computes

The beam shear check compares the provided stirrup spacing against the limits on this page:

Compute V_s = A_v · f_yt · d / s, with f_yt capped at 60 ksi.
Select the tier from V_s vs 4√f′_c b_w d and compute s_max from Table 9.7.6.2.2.
Check the provided spacing s against s_max and warn when it is exceeded.
Check the minimum reinforcement trigger (Cl 9.6.3.1) and A_v/s against A_v,min/s (Cl 9.6.3.3).

Check your stirrup spacing: run a beam calculation or see a demo result.

Sources

ACI 318-19: Cl 9.7.6.2.2 and Table 9.7.6.2.2 (s_max), Cl 9.6.3.1 (A_v,min trigger), Cl 9.6.3.3 (A_v,min formula), Table 20.2.2.4a (f_yt cap), Cl 9.7.6.3.1 (lateral spacing of legs).

FAQ

What is the maximum stirrup spacing for a beam with d = 24 in?

It depends on V_s. If V_s ≤ 4√f′_c b_wd, the limit is min(d/2, 24 in) = min(12, 24) = 12 in. If V_s exceeds that threshold, the limit tightens to min(d/4, 12 in) = min(6, 12) = 6 in.

When does the d/4 spacing limit apply?

When V_s > 4√f′_c b_wd. Under heavy shear the diagonal cracks are steeper and more closely spaced, so stirrups must be placed closer together to make sure every crack is crossed by at least one stirrup.

Can I use stirrup yield strength above 60 ksi?

Table 20.2.2.4a caps f_yt at 60 ksi for shear design, and at 80 ksi for welded deformed wire reinforcement. Higher-grade bars may be used for detailing, but only the capped value counts toward shear capacity.