Reinforced Concrete Beam Design — Rebar, Stirrups & Cracked Section (ACI 318-19)

How the flexure check works

ACI 318 assumes plane sections remain plane, concrete crushes at a strain of 0.003, and steel is elastic–perfectly plastic. For any trial neutral-axis depth c, the strain at each bar layer follows from similar triangles, the bar stress is E_sε capped at f_y, and the concrete compression is the Whitney block 0.85f′_c over depth a = β₁c. The calculator finds the c that balances tension and compression, then takes moments. This is the same procedure as the textbook "assume compression steel yields, then check" method, but it never needs the second iteration: if the compression bars are still elastic, the strain-compatibility solution already has their true stress. Where the compression bars sit inside the stress block, the concrete they displace (0.85f′_cA′_s) is subtracted.

For a T-beam, if a ≤ h_f the beam behaves as a rectangle of width b_f. If a > h_f, the full flange plus the web down to depth a is in compression and the compression centroid moves down accordingly.

ε_ty = f_y/E_s = 0.00207 for Grade 60, so the tension-controlled limit is 0.00507. For Grade 80 it is 0.00576.

Shear and stirrups

ACI 318-19 rewrote the one-way shear equations. When the beam has at least A_v,min, you may use the familiar 2λ√f′_cb_wd or the newer 8λρ_w^1/3√f′_cb_wd, which rewards heavily reinforced beams; the calculator takes the larger. Without A_v,min, the ρ_w form must be used with the size-effect factor λ_s, which reduces V_c for beams deeper than about 10 in. The "spacing needed for V_u" output is the smallest of the spacing that supplies the required V_s, the spacing that still gives A_v,min, and s_max.

Cracked section, deflection and crack control

Once the service moment passes M_cr = f_rI_g/y_t (f_r = 7.5λ√f′_c), the tension concrete cracks and stiffness drops toward I_cr. The cracked neutral axis kd and I_cr are the inputs to the effective moment of inertia in ACI 318-19 Table 24.2.3.5, where I_e = I_cr/[1 − ((2/3)M_cr/M_a)²(1 − I_cr/I_g)] for M_a > (2/3)M_cr. The service steel stress f_s = nM_a(d − kd)/I_cr drives the crack-control limit of §24.3.2: bar spacing s ≤ 15(40,000/f_s) − 2.5c_c ≤ 12(40,000/f_s). If no service moment is entered, the code-permitted f_s = (2/3)f_y is used.

Worked example

Limits of this tool

• Positive moment with the flange (if any) in compression. For negative moment over a support, model the beam as rectangular with width b_w and the top bars as "tension bars".
• One tension layer is assumed when computing d. For two layers, enter d to the bar-group centroid; ε_t is then evaluated at that d, which is slightly unconservative compared with the extreme layer d_t.
• Vertical stirrups, no axial force (N_u = 0) and no torsion. Development length, bar cutoffs, deflection limits and side-face skin reinforcement (h > 36 in) are not checked.
• E_c = 57,000√f′_c is for normal-weight concrete. For lightweight concrete compute E_c = 33w_c^1.5√f′_c separately.

Frequently asked questions

How do you calculate rebar for a concrete beam?

Start from the factored moment Mu. Assume φ = 0.90 and solve Mu = φ·As·fy·(d − a/2) with a = As·fy/(0.85·f′c·b). In closed form for a rectangular beam, Rn = Mu/(φbd²) and ρ = (0.85f′c/fy)[1 − √(1 − 2Rn/(0.85f′c))]. Pick bars with at least As = ρbd, then check As,min, bar spacing and that εt ≥ εty + 0.003. This calculator does the search numerically, including compression steel and T-beam flanges.

What changed for beam flexure in ACI 318-19?

The tension-controlled strain limit became εt ≥ εty + 0.003 (0.00507 for Grade 60) instead of a flat 0.005, so higher-grade bars need more strain. Nonprestressed beams with small axial load must be tension-controlled (§9.3.3.1); ACI 318-14 allowed εt down to 0.004 in the transition zone.

How is Vc calculated in ACI 318-19?

Table 22.5.5.1. When at least the minimum shear reinforcement Av,min is provided, Vc may be taken as 2λ√f′c·bw·d or 8λ(ρw)^(1/3)√f′c·bw·d. Without Av,min, Vc = 8λsλ(ρw)^(1/3)√f′c·bw·d, where the size-effect factor λs = √(2/(1 + d/10)) ≤ 1 penalizes deep members. Vc is capped at 5λ√f′c·bw·d, and √f′c is limited to 100 psi.

What is the maximum stirrup spacing?

For nonprestressed beams, d/2 but not more than 24 in while Vs ≤ 4√f′c·bw·d. Above that, halve it: d/4 but not more than 12 in. Vs may never exceed 8√f′c·bw·d; beyond that the section must be enlarged.

When is minimum shear reinforcement required?

ACI 318-19 §9.6.3.1 requires Av,min wherever Vu > φλ√f′c·bw·d, with exceptions in Table 9.6.3.1 for shallow members and slabs. Av,min = max(0.75√f′c, 50)·bw·s/fyt.

What is kd in a cracked concrete section?

kd is the depth of the neutral axis of the cracked transformed section, measured from the compression face. Concrete below it is assumed cracked and ignored, and the steel is transformed to n·As with n = Es/Ec. kd comes from setting the first moment of the transformed area about the axis to zero; Icr follows from the parallel-axis theorem and feeds the effective moment of inertia used for deflection.

References: ACI 318-19, Building Code Requirements for Structural Concrete — §9.3.3.1, §9.6.1, §9.6.3, §9.7.6.2, Table 21.2.2, §22.2, Table 22.5.5.1, §24.2–24.3, §25.2. Wight, J.K., Reinforced Concrete: Mechanics and Design, 8th ed., Pearson (2022), ch. 4–6. Bar data per ASTM A615.