The Reinforcement Design tool performs flexural and shear design and capacity checks for reinforced concrete beams and one-way slabs per ACI 318. It can operate in design mode (finding the required steel area for a given factored moment) or check mode (evaluating whether existing reinforcement is adequate).

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ACI 318 flexural design is based on the equivalent rectangular stress block. At the strength limit state, the nominal moment capacity is:

M_n = A_s × f_y × (d − a/2)

where a = A_s × f_y / (0.85 × f'_c × b) is the depth of the equivalent rectangular stress block, d is the effective depth to the centroid of the tension steel, and b is the beam width.

The design requirement is φM_n ≥ M_u, where φ = 0.90 for tension-controlled sections (the normal condition for under-reinforced beams).

• b — web width of beam or width of slab strip (in)
• h — total section depth (in). The effective depth d is calculated as h minus cover minus tie diameter minus half the bar diameter.
• d — effective depth (in), which can be entered directly if known.
• f'_c — concrete compressive strength (psi or ksi)
• f_y — steel yield strength (typically 60 ksi for Grade 60)
• M_u — factored bending moment demand (kip·ft or kip·in)
• V_u — factored shear demand (kips), for stirrup design

ACI 318 imposes minimum and maximum reinforcement requirements:

Minimum steel (ACI 318 Section 9.6.1.2): ensures the beam has adequate post-cracking capacity:

A_s,min = max(3√f'_c / f_y, 200 / f_y) × b_w × d

Maximum steel: ACI 318 limits reinforcement indirectly through the net tensile strain requirement. For a tension-controlled section (φ = 0.90), the net tensile strain in the extreme steel layer must satisfy ε_t ≥ 0.004. Heavily reinforced sections with ε_t < 0.004 are not permitted for beams and slabs.

Design flexural reinforcement for a rectangular beam with: b = 12 in, h = 24 in, d = 21.5 in, f'_c = 4,000 psi, f_y = 60 ksi, M_u = 120 kip·ft.

Step 1 — Compute the required reinforcement index R_n:

R_n = M_u / (φ × b × d²)
R_n = (120 × 12) / (0.90 × 12 × 21.5²)
R_n = 1,440 / 4,992 = 0.288 ksi

Step 2 — Compute the required reinforcement ratio ρ:

ρ = (0.85 × f'_c / f_y) × [1 − √(1 − 2R_n / (0.85 × f'_c))]
ρ = (0.85 × 4 / 60) × [1 − √(1 − 2 × 0.288 / 3.4)]
ρ = 0.0567 × [1 − √(0.830)]
ρ = 0.0567 × 0.0888 = 0.00503

Step 3 — Required steel area:

A_s,req = ρ × b × d = 0.00503 × 12 × 21.5 = 1.30 in²

Step 4 — Check minimum steel:

A_s,min = max(3 × √4000 / 60,000, 200 / 60,000) × 12 × 21.5
A_s,min = max(0.00316, 0.00333) × 258 = 0.00333 × 258 = 0.860 in²
A_s,req = 1.30 in² > A_s,min = 0.860 in² ✓

Step 5 — Select bars. Use 3-#6 bars: A_s,prov = 3 × 0.44 = 1.32 in² ≥ 1.30 in² ✓

Step 6 — Verify capacity with provided steel:

a = 1.32 × 60 / (0.85 × 4 × 12) = 79.2 / 40.8 = 1.94 in
φM_n = 0.90 × 1.32 × 60 × (21.5 − 1.94/2)
φM_n = 0.90 × 79.2 × 20.53 = 1,463 kip·in = 121.9 kip·ft > 120 kip·ft ✓

For shear, ACI 318 requires stirrups when V_u > 0.5 × φV_c, where the concrete shear capacity is:

V_c = 2 × √f'_c × b_w × d (simplified expression in psi units)

When stirrups are required, the required stirrup spacing for a given bar size (A_v) is:

s = A_v × f_y × d / V_s

where V_s = V_u/φ − V_c is the required steel shear contribution. The tool calculates all shear checks and reports the required stirrup spacing, subject to ACI 318 maximum spacing limits.

• For T-beams (flanged sections), the effective flange width must be established per ACI 318 Section 6.3.2 before entering it as the design width b.
• When the required ρ approaches the maximum permitted by the strain limit, consider increasing the section depth h to reduce the steel requirement and improve ductility.
• For two-way slabs, use the strip method to determine M_u in each direction, then run the tool for each strip separately.

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