Free Web Tool

RC Beam Design

Singly-reinforced rectangular concrete beam — ACI 318 / NSCP 2015. Get required steel area, ρ vs ρmin/ρmax check, suggested bar combination and φMn.

Inputs

mm
mm
mm
MPa
MPa (Grade 60 ≈ 420)
kN·m
mm
Notes & assumptions
  • Singly-reinforced rectangular section, tension-controlled (εt ≥ 0.005).
  • Whitney equivalent rectangular stress block, β₁ per ACI 318 22.2.2.4.3.
  • ρmin = max(1.4/fy, √f'c/(4·fy)) (MPa form).
  • ρmax for tension-controlled: εcu=0.003, εt=0.005 → c/d=0.375.
  • Bar suggestions assume ASTM/PNS standard sizes; verify spacing & cover.

Design

StatusOK
Rn = Mu/(φbd²)—
ρ required—
ρmin / ρmax—
As,req—
a (stress block)—
c (neutral axis)—
εt (tension steel)—
φMn provided—

Suggested Bar Combinations

CombinationAs providedStatus

About this calculator

This tool sizes the tension reinforcement of a singly-reinforced rectangular concrete beam for a given factored moment, using strength design per ACI 318 / NSCP 2015. It reports the required steel ratio, checks it against ρmin/ρmax, suggests standard bar combinations, and returns the resulting design moment capacity φMn. It's meant for preliminary sizing of beams, joists, and slab strips by students and practicing engineers — treat it as a design aid, not a substitute for a full code check.

Method & formulas

The section uses the Whitney equivalent rectangular stress block with equilibrium between the concrete compression block and the tension steel:

a = As·fy / (0.85·f'c·b)
β₁ = 0.85 for f'c ≤ 28 MPa; otherwise 0.85 − 0.05·(f'c−28)/7, not less than 0.65

The required steel ratio is solved directly from the flexural resistance Rn:

Rn = Mu / (φ·b·d²)
ρ = (0.85·f'c/fy)·[1 − √(1 − 2Rn/(0.85·f'c))]
As = ρ·b·d

with limits and the resulting capacity:

ρmin = max(1.4/fy, √f'c /(4·fy))
ρmax (tension-controlled, εt=0.005) = 0.85·β₁·(f'c/fy)·[0.003/(0.003+0.005)]
c = a/β₁ , εt = 0.003·(d−c)/c
Mn = As·fy·(d − a/2) , φMn = φ·Mn (φ = 0.90 tension-controlled)

Worked example

Using the tool's default inputs — b=300 mm, h=500 mm, d=440 mm, f'c=28 MPa, fy=420 MPa, φ=0.90, Mu=200 kN·m:

Rn = 200×10⁶ / (0.9·300·440²) = 3.83 MPa ρ = (0.85·28/420)·[1 − √(1 − 2·3.83/23.8)] ≈ 0.00999 ρmin = 0.00333 , ρmax = 0.0181 → ρmin < ρ < ρmax, OK As = 0.00999·300·440 ≈ 1319 mm² a = 1319·420/(0.85·28·300) ≈ 77.6 mm , c = a/0.85 ≈ 91.3 mm εt = 0.003·(440−91.3)/91.3 ≈ 0.0115 ≥ 0.005 → tension-controlled φMn = 0.9·1319·420·(440−38.8) ≈ 200 kN·m ✓

Reasonable bar picks: 3-25 mm bars (≈1473 mm²) or 5-20 mm bars (≈1571 mm²), both above the 1319 mm² required.

Assumptions & limitations

FAQ

Q: What does exceeding ρmax mean?

A: The section can't reach a tension-controlled failure mode with tension steel alone at that size — the tool flags it so you can enlarge the section, raise f'c, or add compression reinforcement rather than silently under-designing.

Q: Why is φ always 0.90 in the results?

A: The calculator assumes the tension-controlled limit (εt ≥ 0.005) governs. If your actual strain falls between 0.002 and 0.005 (transition zone), ACI 318 §21.2 requires a lower, interpolated φ that this simplified tool does not compute automatically.

Q: Does this replace a full beam design?

A: No — it covers flexural sizing only. Shear, torsion, deflection, crack control, and detailing still need to be checked separately, ideally with a licensed engineer reviewing the complete design.