Flexural Design & Analysis

Singly-reinforced rectangular sections per ACI 318-25. Design the steel from a given Mu, or check the capacity φMn of a placed section — with cross-section and strain diagrams. Metric units (m, MPa, kN·m, cm²).

TERCERO TABLADA
Civil & Structural Engineering Inc.
FLEXURAL DESIGN (singly reinforced)ACI 318-25

1 · Inputs

1.0 m for walls/slabs
Gr.60 = 420
from 1.2D + 1.6L

d = h − r = 0.470 m · β1 = 0.850

OK — provide 4Ø22 (As = 15.20 cm²). φMn = 228.99 ≥ Mu = 200.00 kN·m

2–4 · Required steel

Rn = Mu/(0.9·b·d²)4.024 MPa
ρ required0.01101 (0.85f′c/fy)(1−√(1−2Rn/0.85f′c))
ρmin = 1.4/fy0.00333 ACI 9.6.1.2
ρmin alt = (4/3)·ρ0.01467 ACI 9.6.1.3 (lesser governs)
ρmax (εt = 0.005)0.01355 keeps φ = 0.90
ρ design = max(ρ, ρmin)0.01101
As required = ρ·b·d12.93 cm²
Ductile? (ρ ≤ ρmax)Yes

5 · Pick commercial bars

As provided15.20 cm²need ≥ 12.93 cm²

6 · Verify the placed steel

As placed ≥ required?Yes
a (stress block) = As·fy/(0.85f′c·b)143.1 mm
c (neutral axis) = a/β1168.3 mm
εt = (d−c)/c · 0.0030.0054 ≥ 0.004 to accept
φ (transition, capped)0.900 Table 21.2.2
Mn = b·d²·fy·ρ(1−ρfy/1.7f′c)254.44 kN·m
φMn (capacity)228.99 kN·mvs Mu = 200.00

Section

N.A. c=168mmb = 250 mmh = 500 mm4ר22 (d = 470 mm)

Strain diagram

N.A.εcu = 0.003εt = 0.0054steel (d = 470 mm)