Free Tool · EN 1992-1-1 · Reinforced Concrete Beam · §6.1 + §6.2

RC Beam Bending & Shear

ULS design per EN 1992-1-1:2004. Tab 1: §6.1 bending — rectangular stress block, neutral axis, MRd, As,req. Tab 2: §6.2 shear — VRd,c without reinforcement, VRd,s variable strut inclination.

Presets:
Section Geometry
Distance from compression face to centroid of tension steel
Materials
Bending Load
Leave 0 for required-As design mode
§6.1 Equations
fcd = αcc·fck / γC (αcc=0.85, γC=1.50)
fyd = fyk / γS (γS=1.15)
λ = 0.8, η = 1.0 (fck ≤ 50 MPa)
MEd = η·fcd·b·(λx)·(d − λx/2) → solve for x
As,req = η·fcd·b·(λx) / fyd
MRd = As·fyd·z where z = d − λx/2
ξ_lim = 0.45 (fck ≤ 50 MPa) — §5.5(4)
As,min = max(0.26·fctm/fyk·b·d, 0.0013·b·d) — §9.2.1.1
Results
M_Rd (kNm)150.0
Utilization MEd/MRd—
Neutral axis x (mm)96.7
x/d ratio0.195
ξ_lim (ductility limit)0.45
As,req (mm²)756
As,min (mm²)198
Lever arm z (mm)456.3
fcd (MPa)14.17
fyd (MPa)434.78
Governing checkbending As req
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Related tools
RC Beam Shear (§6.2)
Crack Width (§7.3)
Concrete Column (§5.8)
Pad Foundation (EN 1997-1)
Section Geometry
Materials
Shear Load
Longitudinal Steel (for V_Rd,c)
Tension reinforcement crossing the shear check section
Stirrups (optional check)
e.g. 2 × Ø10 = 157 mm². Leave 0 for required-Asw/s design
§6.2 Equations
VRd,c = [CRd,c·k·(100·ρl·fck)^(1/3)]·bw·d — §6.2.2 eq.6.2a
k = 1 + √(200/d) ≤ 2.0 | ρl = Asl/(bw·d) ≤ 0.02
VRd,min = (vmin + k1·σcp)·bw·d — eq.6.2b
VRd,s = (Asw/s)·z·fywd·cotθ — §6.2.3 eq.6.8
VRd,max = αcw·bw·z·ν1·fcd/(cotθ+tanθ) — eq.6.9
ν1 = 0.6·(1−fck/250) | z ≈ 0.9·d
Asw/s,min = 0.08·√fck·bw/fyk — §9.2.2 eq.9.5N
sl,max = 0.75·d — §9.2.2 eq.9.6N
Results
V_Rd,c (kN)85.5
Shear reinf. required?Yes
Asw/s required (mm)0.248
Asw/s,min (mm)0.24
s_l,max (mm)371.3
Utilization VEd/VRd—
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Related tools
RC Beam Shear (§6.2)
Crack Width (§7.3)
Concrete Column (§5.8)
Pad Foundation (EN 1997-1)

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Frequently Asked Questions

What does EN 1992-1-1 §6.1 cover?
Section 6.1 of Eurocode 2 covers the ULS bending capacity of cross-sections. For rectangular beams it uses a simplified rectangular stress block with effective depth factor λ (= 0.8 for fck ≤ 50 MPa) and effective strength factor η (= 1.0 for fck ≤ 50 MPa) to locate the neutral axis and compute M_Rd.
What is the ductility limit x/d ≤ 0.45?
EN 1992-1-1 §5.5(4) limits the neutral axis depth ratio x/d to ξ_lim = 0.45 (for fck ≤ 50 MPa) to ensure a ductile, under-reinforced failure mode where tension steel yields before concrete crushes.
How is V_Rd,c calculated (no shear reinforcement)?
Per §6.2.2 eq.6.2a: V_Rd,c = [C_Rd,c · k · (100·ρl·fck)^(1/3)] · bw · d, where k = 1 + √(200/d) ≤ 2.0 and ρl = Asl/(bw·d) ≤ 0.02. A minimum eq.6.2b applies.
What is the variable strut inclination method?
EN 1992-1-1 §6.2.3 allows the designer to choose a strut angle θ in the range 21.8°–45° (1.0 ≤ cot θ ≤ 2.5). A shallower strut (larger cot θ = 2.5) minimises the required stirrup area Asw/s but raises the concrete strut demand. The calculator optimises θ automatically.
What is V_Rd,max?
V_Rd,max is the maximum shear capacity governed by concrete strut crushing: V_Rd,max = αcw · bw · z · ν₁ · fcd / (cot θ + tan θ). If V_Ed exceeds V_Rd,max at any permissible θ, the section must be enlarged.
Which National Annexes are covered?
The recommended EN values are used (C_Rd,c = 0.18/γC, γC = 1.5, γS = 1.15, αcc = 0.85, k₁ = 0.15). German NA uses C_Rd,c = 0.15/γC; apply override in γC input if needed.