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 ρminmax, 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.