calcnote ACI 318-19
Methodology Feedback Run a calc

Beam shear design per ACI 318-19

Beam shear determines the stirrup requirements to resist factored shear Vu. ACI 318-19 Table 22.5.5.1 governs the concrete shear strength Vc, and Chapter 9 sets the member requirements for minimum reinforcement and spacing. For the overall beam design workflow, see the beam methodology hub.

Applicable clauses

ACI 318-19 reference Topic
Table 22.5.5.1Vc (simplified method, two equations)
Cl 22.5.5.1.3λs size-effect factor
Table 22.5.1.2Vs,max limit (section adequacy)
Cl 9.6.3.1Minimum shear reinforcement trigger
Cl 9.7.6.2.2Maximum stirrup spacing
Table 20.2.2.4afyt cap (60 ksi)
Cl 22.5.3.1√f′c cap (100 psi)

Concrete shear strength (Vc)

ACI 318-19 Table 22.5.5.1 provides two equations for Vc. Which one governs depends on whether the beam has at least the minimum shear reinforcement.

With minimum shear reinforcement (Av/s ≥ Av,min/s): both equations apply, and the larger Vc governs.

$$V_{c,a} = 2\,\lambda\,\sqrt{f'_c}\;b_w\,d \quad \text{(Table 22.5.5.1(a))}$$
$$V_{c,b} = 8\,\lambda\,\rho_w^{1/3}\,\sqrt{f'_c}\;b_w\,d \quad \text{(Table 22.5.5.1(b))}$$
$$V_c = \max(V_{c,a},\; V_{c,b})$$

Equation (a) is the classic simplified formula. Equation (b) gives credit for the longitudinal reinforcement ratio ρw = As / (bw · d). For lightly reinforced sections, (a) typically governs; for higher ρw, (b) can exceed (a).

Without minimum shear reinforcement: only Equation (b) applies, with the size-effect factor λs and a cap on √f′c:

$$V_c = 8\,\lambda\,\lambda_s\,\rho_w^{1/3}\,\sqrt{f'_{c,\text{cap}}}\;b_w\,d$$

where √f′c is capped at 100 psi per Cl 22.5.3.1, and λ is the lightweight concrete factor (1.0 for normal-weight concrete).

Size-effect factor (λs)

Cl 22.5.5.1.3 introduces the size-effect factor for members without minimum shear reinforcement:

$$\lambda_s = \sqrt{\frac{2}{1 + d/10}} \quad (d \text{ in inches})$$

For d = 10 in, λs = 1.0 (no reduction). As d increases, λs decreases: at d = 30 in, λs ≈ 0.71; at d = 50 in, λs ≈ 0.58. This accounts for the observation that larger unreinforced beams fail at lower nominal shear stresses. When minimum stirrups are present, they arrest diagonal crack growth and the size effect does not apply.

Steel shear contribution (Vs)

The shear resisted by stirrups:

$$V_s = \frac{A_v \cdot f_{yt} \cdot d}{s}$$

where Av is the total stirrup area crossing the shear plane (number of legs × bar area), fyt is the stirrup yield strength capped at 60 ksi (Table 20.2.2.4a), and s is the stirrup spacing along the beam length.

The total shear capacity and the design check:

$$\phi\,(V_c + V_s) \geq V_u \quad (\phi = 0.75)$$

DCR = Vu / φ(Vc + Vs). A value ≤ 1.0 means the section has adequate shear strength.

Maximum steel shear (Vs,max)

Table 22.5.1.2 sets an upper limit on the steel shear contribution:

$$V_{s,max} = 8\,\sqrt{f'_c}\;b_w\,d$$

If Vs exceeds Vs,max, the section is inadequate regardless of stirrup quantity. The beam must be made deeper or wider. calcnote rejects any input that violates this limit.

Minimum shear reinforcement (Cl 9.6.3.1)

Minimum stirrups are required when the factored shear exceeds half the concrete capacity:

$$V_u > 0.5\,\phi\,V_c \implies \text{stirrups required}$$

The minimum stirrup ratio:

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

When this threshold is met, Vc benefits from the two-equation max (higher capacity), and the size-effect factor λs no longer applies.

Maximum stirrup spacing (Cl 9.7.6.2.2)

Stirrup spacing is limited to ensure every diagonal crack crosses at least one stirrup:

Condition smax
Vs ≤ 4√f′c bw dmin(d/2, 24 in)
Vs > 4√f′c bw dmin(d/4, 12 in)

The tighter limit applies when Vs is high, because diagonal cracks are steeper and more closely spaced under heavy shear.

What calcnote computes

The beam shear check performs the following steps:

  1. Determine whether minimum shear reinforcement is provided (Av/s vs Av,min/s).
  2. Compute Vc from the applicable branch of Table 22.5.5.1, applying λs and the √f′c cap when no minimum reinforcement.
  3. Compute Vs = Av · fyt · d / s.
  4. Check Vs ≤ Vs,max (reject if inadequate).
  5. Compute φ(Vc + Vs) and DCR.
  6. Check minimum reinforcement trigger (Cl 9.6.3.1) and spacing limit (Cl 9.7.6.2.2).
  7. Report PASS or FAIL with warnings for each violated condition.

Check your beam shear design: run a beam calculation or see a demo result.

Sources

FAQ

Why does ACI 318-19 have two equations for Vc?

Table 22.5.5.1 provides Eq (a) = 2λ√f′c bwd (the classic simplified formula) and Eq (b) = 8λρw1/3√f′c bwd (which accounts for the longitudinal reinforcement ratio). When minimum shear reinforcement is present, the larger value governs, giving credit for the confining effect of stirrups. Without minimum reinforcement, only Eq (b) with the size-effect factor λs applies.

When does the size-effect factor λs reduce Vc?

Only when the member lacks minimum shear reinforcement. For beams with d > 10 in and no stirrups (or stirrups below the Av,min threshold), λs is less than 1.0, reducing Vc. This accounts for the observation that larger unreinforced beams fail at lower nominal shear stresses than smaller ones.

Why is fyt capped at 60 ksi for stirrups?

Table 20.2.2.4a limits stirrup yield strength to 60 ksi to control crack widths under service loads and ensure the stirrups engage before the concrete shear failure progresses. Even if higher-grade bars are used for stirrups, only 60 ksi is counted toward shear capacity.

What does it mean when Vs exceeds Vs,max?

Table 22.5.1.2 limits Vs to 8√f′c bwd. If the required Vs exceeds this, the concrete section is inadequate: no amount of stirrups can compensate. The beam must be made deeper or wider to increase the concrete contribution and bring Vs below the limit.