RC Beam Design Calculator
Governing checkFlexure As89.9%ALL CHECKS PASS
1130 mm² → 1257 mm²
All 9 checks pass
FlexurePASS
As,req
1130mm²
As,prov
1257mm²
Utilization89.9%
ShearPASS
VEd
150kN
VRd,c
90.6kN
Utilization32.2%
Crack WidthPASS
wk
0.223mm
wmax
0.4mm
Utilization55.8%
DeflectionPASS
l/d actual
10.8
l/d allow
18.9
Utilization57.0%
Summary
| Check | Required | Provided | Util. | Status |
|---|---|---|---|---|
| Flexure As | 1130 mm² | 1257 mm² | 89.9% | PASS |
| As,min | 252 mm² | 1257 mm² | 20.1% | PASS |
| As,max | ≤ 7200 mm² | 1257 mm² | 17.5% | PASS |
| x/d ratio | ≤ 0.450 | 0.240 | 53.4% | PASS |
| VRd,c (links required) | VEd = 150 kN | 90.6 kN | 165.5% | PASS |
| Stirrup Asw/s | 0.275 | 0.503 mm²/mm | 54.8% | PASS |
| VRd,max | VEd = 150 kN | 465.5 kN | 32.2% | PASS |
| Crack width wk | ≤ 0.4 mm | 0.223 mm | 55.8% | PASS |
| Span/depth l/d | ≤ 18.9 | 10.8 | 57.0% | PASS |
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Flexure [EC2 §6.1]▼
Flexural Design
EC2 §3.1.6(1)Design concrete strength
fcd = αcc · fck / γc
fcd = 0.85 × 30 / 1.5
fcd = 17.00 MPa
EC2 §3.2.7(2)Design steel strength
fyd = fyk / γs
fyd = 500 / 1.15
fyd = 434.8 MPa
EC2 §6.1(4)Normalized moment coefficient
K = MEd / (b · d² · fck)
K = 250000000 / (300 × 557.0² × 30)
K = 0.0895
EC2 §6.1(4)Check: K ≤ Kbal
K = 0.0895 ≤ Kbal = 0.167
Singly reinforced design
✓ Singly reinforced
EC2 §6.1(4)Lever arm
z = d · (0.5 + √(0.25 − K/1.134)) ≤ 0.95d
z = 557.0 × (0.5 + √(0.25 − 0.0895/1.134))
z = 508.9 mm
EC2 §6.1(4)Required tension reinforcement
As,req = MEd / (fyd · z)
As,req = 250000000 / (434.8 × 508.9)
As,req = 1130 mm²
EC2 §5.5(4)Neutral axis depth ratio
x/d = (As·fyd − As2·fyd) / (0.8·b·fcd·d)
x = 133.9 mm
x/d = 0.240 ✓ (limit 0.45)
EC2 §9.2.1.1(1)Minimum reinforcement
As,min = max(0.26·fctm/fyk·b·d, 0.0013·b·d)
As,min = max(0.26×2.9/500×300×557.0, 0.0013×300×557.0)
As,min = 252 mm²
EC2 §9.2.1.1(3)Maximum reinforcement
As,max = 0.04 · b · h
As,max = 0.04 × 300 × 600
As,max = 7200 mm²
Provided tension steel
As,prov
As,prov = 1257 mm²
Shear [EC2 §6.2]▼
Shear Design
EC2 §6.2.2(1)Size effect factor
k = min(1 + √(200/d), 2.0)
k = min(1 + √(200/557.0), 2.0)
k = 1.599
EC2 §6.2.2(1)Concrete shear resistance (no reinforcement)
VRd,c = max[CRd,c·k·(100ρl·fck)^⅓·bw·d, vmin·bw·d]
ρl=0.00752, vmin=0.388
VRd,c = 90.6 kN
EC2 §6.2.3(3)Required shear reinforcement
Asw/s = VEd / (z · fyd · cotθ)
Asw/s = 150×1000 / (501.3 × 434.8 × 2.5)
Asw/s,req = 0.275 mm²/mm
Provided shear reinforcement
Asw/s = nlegs · π·φ²/4 / s
Asw/s = 2×50.3 / 200
Asw/s,prov = 0.503 mm²/mm
EC2 §6.2.3(3)Maximum shear capacity (strut)
VRd,max = ν₁·fcd·bw·z / (cotθ + tanθ)
ν₁=0.528, fcd=17.0
VRd,max = 465.5 kN
EC2 §9.2.2(5)Minimum shear reinforcement
ρw,min = 0.08·√fck/fyk
ρw,min = 0.08×√30/500
Asw/s,min = 0.263 mm²/mm
EC2 §9.2.2(6)Maximum stirrup spacing
smax = min(0.75d, 600)
smax = min(0.75×557.0, 600)
smax = 418 mm ✓
Crack Width [EC2 §7.3]▼
Crack Control
EC2 §7.3.2(3)Cracked elastic neutral axis
x from b·x²/2 = αe·As·(d−x)
αe = Es/Ecm = 6.06, As = 1257 mm², d = 557.0 mm
x = 144.7 mm
EC2 §7.3.2(3)Effective tension area
hc,eff = min(2.5(h−d), (h−x)/3, h/2)
hc,eff = min(107.5, 151.8, 300.0)
Ac,eff = 300 × 107.5 = 32250 mm², ρp,eff = 0.0390
EC2 §7.3.4(2)Steel stress under service load
σs ≈ Ms / (As · 0.87d)
σs = 170000000 / (1257 × 484.6)
σs = 279.2 MPa
EC2 Eq. 7.11Maximum crack spacing (c = cover to the longitudinal bars)
sr,max = 3.4·c + 0.425·k₁·k₂·φ/ρp,eff
sr,max = 3.4×33 + 0.425×0.8×0.5×20/0.0390
sr,max = 199.5 mm
EC2 Eq. 7.9Mean strain difference
(εsm−εcm) = max([σs − kt·fct,eff·(1+αe·ρp,eff)/ρp,eff]/Es, 0.6σs/Es)
kt=0.6, αe=6.06
(εsm−εcm) = 1.1198 × 10⁻³
EC2 Eq. 7.8Calculated crack width
wk = sr,max · (εsm − εcm)
wk = 199.5 × 0.001120
wk = 0.223 mm ✓ (limit: 0.4 mm)
Deflection [EC2 §7.4]▼
Deflection Check
EC2 §7.4.2(2)Reference reinforcement ratio
ρ₀ = 10⁻³√fck
ρ₀ = 10⁻³×√30
ρ₀ = 0.00548
EC2 Table 7.4NStructural system factor
K per support type
Support: simply_supported
K = 1
EC2 Eq. 7.16bBasic l/d ratio (ρ > ρ₀)
l/d = K·[11 + 1.5·√fck·ρ₀/(ρ−ρ') + √fck/12·√(ρ'/ρ₀)]
ρ=0.00752, ρ'=0.00000
l/d basic = 17.0
EC2 §7.4.2(2)Steel provision modification
mod = min(As,prov/As,req, 1.5)
mod = min(1257/1130, 1.5)
mod = 1.11
EC2 §7.4.2Final span/depth check
l/d allowable vs l/d actual
allowable = 17.0 × 1.11 × 1.0 × 1.00
l/d allow = 18.9, l/d actual = 10.8 ✓
Bar Schedule▼
Suggested Arrangements (As,req = 1130 mm²)
10Ø12
1131 mm²✓
6Ø16
1206 mm²✓
8Ø14
1232 mm²✓
2Ø28
1232 mm²✓
4Ø20
1257 mm²✓
3Ø25
1473 mm²✓