3D model203 x 203 x 52 · L 4.00 m
drag to rotate · scroll to zoomLoading 3D model…
Lab / EC4 composite design
Composite Column Workbench
View calculation reference →Concrete-encased UC composite columns by the EC4 simplified method. Calculations are laid out top-to-bottom in the order taught in the notes — material → geometry → squash → stiffness → buckling → interaction → member checks.
StatusPASSSecond-order amplified
Axial check0.304PASS · load / capacity
Major bending check0.06PASS · demand / capacity
Minor bending check0.124PASS · demand / capacity
Cross sectionAc 93,257 mm² · ρs 2.70%
Step 1 · Design properties and creep reduction
CALCFollow the example sequence: reduce material strengths first, then use NG,Ed / NEd and the creep coefficient to obtain Ec,eff for stiffness.
Steel design strength355.0 N/mm²VALUE
Concrete design strength14.17 N/mm²VALUE
Creep-adjusted concrete stiffness11.07 kN/mm²VALUE
Show calculation audit
fyd = fy / γA
355 / 1.00
355.0 N/mm²
α · fcd = α · fck / γC
0.85 · 25 / 1.50
14.17 N/mm²
fsd = fsk / γS
500 / 1.15
434.8 N/mm²
Ec,eff = Ecm / (1 + (NG,Ed / NEd) · φ)
31.0 / (1 + 0.60 · 3.0)
11.07 kN/mm²
Step 2 · Areas and reinforcement ratio
PASSAs in the example, take Aa from the UC table, calculate As from the bars, then use Ac = bc hc - Aa - As and check the reinforcement ratio.
Steel section area6,630 mm²VALUE
Rebar area2,513 mm²VALUE
Concrete area93,257 mm²VALUE
Rebar ratio2.70%PASS
Show calculation audit
Aa (from section table)
203 x 203 x 52
6,630 mm²
As = n · π·dia²/4
8 · π · 20² / 4
2,513 mm²
Ac = bc · hc − Aa − As
320 · 320 − 6,630 − 2,513
93,257 mm²
ρs = As / Ac
2,513 / 93,257
2.70 %scope 0.3 % – 6 %PASS
Step 3 · Plastic axial resistance and steel contribution
PASSCalculate Npl,Rd and the steel contribution δ. This is the same early check used in the notes before moving into buckling.
Axial capacity before buckling4,768 kNVALUE
Concrete axial part1,321 kNVALUE
Steel contribution0.494PASS
Show calculation audit
Npl,Rd = Aa·fyd + α·Ac·fcd + As·fsd
6,630·355 + 14.17·93,257 + 2,513·435 (÷1 000)
4,768 kN
Npm,Rd = α · Ac · fcd
14.17 · 93,257 (÷1 000)
1,321 kN
δ = Aa · fyd / Npl,Rd
6,630·355 / (4,768·1 000)
0.494scope 0.2 – 0.9PASS
Step 4 · Uncracked stiffness and elastic critical loads
CALCCombine steel, reinforcement and concrete inertias into (EI)eff, then calculate Euler Ncr about each axis before the buckling reduction step.
Major-axis stiffness22.93 ×10¹²VALUE
Minor-axis stiffness15.85 ×10¹²VALUE
Major-axis critical load14,143 kNVALUE
Minor-axis critical load9,778 kNVALUE
Show calculation audit
(EI)eff,y = Ea·Ia + Es·Is + Ke·Ec,eff·Ic
210·52.6 + 200·33.2 + 0.60·11.07·788.0 ×10¹² N·mm²
22.93 ×10¹² N·mm²
(EI)eff,z = Ea·Ia + Es·Is + Ke·Ec,eff·Ic
210·17.8 + 200·33.2 + 0.60·11.07·822.8 ×10¹² N·mm²
15.85 ×10¹² N·mm²
Ncr,y = π² (EI)eff,y / (k·L)²
π² · 22.93e12 / (1.00·4.0·1 000)²
14,143 kN
Ncr,z = π² (EI)eff,z / (k·L)²
π² · 15.85e12 / (1.00·4.0·1 000)²
9,778 kN
(EI)eff,y,II = K0·(Ea·Ia + Es·Is + Ke,II·Ec,eff·Ic)
0.90 · (… + 0.50·Ec,eff·Ic)
19.85 ×10¹² N·mm²
(EI)eff,z,II = K0·(Ea·Ia + Es·Is + Ke,II·Ec,eff·Ic)
0.90 · (… + 0.50·Ec,eff·Ic)
13.45 ×10¹² N·mm²
Step 5 · Buckling resistance
PASSUse Ncr and the EC3 buckling curve to get χ and Nb,Rd, then check the applied axial force against the governing buckling resistance.
Major-axis slenderness0.63PASS
Minor-axis slenderness0.76PASS
Governing axial capacity3,288 kNVALUE
Axial utilisation0.304PASS
Show calculation audit
Npl,Rk = Aa·fy + α·Ac·fck + As·fsk
6,630·355 + 0.85·93,257·25 + 2,513·500 (÷1 000)
5,592 kN
Y
Ncr,y = π² (EI)eff,y / (k·L)²
π² · 22.93e12 / (1.00·4.0·1 000)²
14,143 kN
λ̄y = √(Npl,Rk / Ncr,y)
√(5,592 / 14,143)
0.63
Φy = 0.5·(1 + α(λ̄ − 0.2) + λ̄²)
0.5·(1 + 0.49·(0.63 − 0.2) + 0.63²)
0.80
χy = 1 / (Φ + √(Φ² − λ̄²))
1 / (0.80 + √(0.80² − 0.63²))
0.77
Nb,Rd,y = χ · Npl,Rd
0.77 · 4,768
3,662 kN
Z
Ncr,z = π² (EI)eff,z / (k·L)²
π² · 15.85e12 / (1.00·4.0·1 000)²
9,778 kN
λ̄z = √(Npl,Rk / Ncr,z)
√(5,592 / 9,778)
0.76
Φz = 0.5·(1 + α(λ̄ − 0.2) + λ̄²)
0.5·(1 + 0.49·(0.76 − 0.2) + 0.76²)
0.92
χz = 1 / (Φ + √(Φ² − λ̄²))
1 / (0.92 + √(0.92² − 0.76²))
0.69
Nb,Rd,z = χ · Npl,Rd
0.69 · 4,768
3,288 kN
Axial check NEd / Nb,Rd ≤ 1
1,000 / 3,288
0.304PASS
Step 6 · Column length and second-order moments
REVIEWUse the column length and the reduced second-order stiffness to get Ncr,eff. M1,Ed comes from the entered top/base moments; Simple mode exposes the base moments for the bottom column.
Minimum Ncr,eff / NEd8.29×REVIEW
First-order end moments0.0 / 0.0 kNmVALUE
Major-axis design moment20.0 kNmVALUE
Minor-axis design moment30.3 kNmVALUE
Show calculation audit
Y
Ncr,eff,y = π² (EI)eff,y,II / (k·L)²
π² · 19.85e12 / (k·L)²
12,245 kN
Ncr,eff,y ≥ 10 NEd?
12,245 ≥ 10·1,000
skip 2nd-order
β = 0.66 + 0.44·r
r = M₂ / M₁
1.00
k_end = β / (1 − NEd / Ncr,eff)
1.00 / (1 − 1,000 / 12,245)
1.00
k_imp = 1 / (1 − NEd / Ncr,eff)
1 / (1 − 1,000 / 12,245)
1.00
e0,y = L / (L/e0)
4,000 / 200 (applied)
20.0 mm
M_imp,y = NEd · e0,y
1,000 · 0.020 m
20.0 kNm
My,Ed (uniaxial) = max(M₁, k_end·M₁ + k_imp·M_imp)
max(0.0, 1.00·0.0 + 1.00·20.0)
20.0 kNm
My,Ed (for biaxial)
imperfection dropped (other axis governs)
0.0 kNm
Z
Ncr,eff,z = π² (EI)eff,z,II / (k·L)²
π² · 13.45e12 / (k·L)²
8,294 kN
Ncr,eff,z ≥ 10 NEd?
8,294 ≥ 10·1,000
amplification required
β = 0.66 + 0.44·r
r = M₂ / M₁
1.00
k_end = β / (1 − NEd / Ncr,eff)
1.00 / (1 − 1,000 / 8,294)
1.14
k_imp = 1 / (1 − NEd / Ncr,eff)
1 / (1 − 1,000 / 8,294)
1.14
e0,z = L / (L/e0)
4,000 / 150 (applied)
26.7 mm
M_imp,z = NEd · e0,z
1,000 · 0.027 m
26.7 kNm
Mz,Ed (uniaxial) = max(M₁, k_end·M₁ + k_imp·M_imp)
max(0.0, 1.14·0.0 + 1.14·26.7)
30.3 kNm
Mz,Ed (for biaxial)
imperfection applied
30.3 kNm
Step 7 · Interaction diagram read-off
CALCThis is the graph used in the example: draw the horizontal line at NEd and read Mpl,N,Rd where it intersects the A-C-D-B polygon.
Axial load line1,000 kNVALUE
Major bending capacity367.7 kNmVALUE
Minor bending capacity272.7 kNmVALUE
Capacity reduction1.03 / 1.00VALUE
Show calculation audit
Y
hn,y = Npm,Rd / (2·bc·αfcd + 2·tw·(2fyd − αfcd))
see EC4 Appendix B/C
65.86 mm
Wpa,y
section table
567,000 mm³
Wps,y ≈ As · offset
2,513 · 115
289,027 mm³
Wpc,y = bc·hc²/4 − Wpa − Wps
uncracked concrete
7.34 ×10⁶ mm³
Mmax,Rd = Wpa·fyd + Wps·fsd + 0.5·Wpc·αfcd
full polygon top
378.9 kNm
Mpl,Rd = Mmax,Rd − Wpa,n·fyd − 0.5·Wpc,n·αfcd
subtract neutral region
357.2 kNm
Mpl,N,Rd (polygon at NEd = 1,000 kN)
read off A-C-D-B
367.7 kNm
μ_d = Mpl,N,Rd / Mpl,Rd
367.7 / 357.2
1.03
Z
hn,z = (Npm,Rd + tw·(2tf − h)·(2fyd − αfcd)) / (2·hc·αfcd + 4·tf·(2fyd − αfcd))
see EC4 Appendix B/C
7.41 mm
Wpa,z
section table
264,000 mm³
Wps,z ≈ As · offset
2,513 · 115
289,027 mm³
Wpc,z = bc·hc²/4 − Wpa − Wps
uncracked concrete
7.64 ×10⁶ mm³
Mmax,Rd = Wpa·fyd + Wps·fsd + 0.5·Wpc·αfcd
full polygon top
273.5 kNm
Mpl,Rd = Mmax,Rd − Wpa,n·fyd − 0.5·Wpc,n·αfcd
subtract neutral region
271.9 kNm
Mpl,N,Rd (polygon at NEd = 1,000 kN)
read off A-C-D-B
272.7 kNm
μ_d = Mpl,N,Rd / Mpl,Rd
272.7 / 271.9
1.00
Interaction polygon (A-C-D-B)
The blue read-off line is the notes example method: start from NEd, meet the polygon, and drop to the M axis to get Mpl,N,Rd. The green or red point is the actual design moment demand.
Step 8 · Member checks (axial · uniaxial · biaxial)
PASSFinish with plain pass/fail checks: axial buckling, major bending, minor bending, and the combined biaxial interaction sum.
Axial check0.304PASS
Major bending check0.060PASS
Minor bending check0.124PASS
Combined bending check0.111PASS
Show calculation audit
My,Ed / Mpl,N,Rd,y ≤ αM
20.0 / 367.7 ≤ 0.90
0.054PASS
Mz,Ed / Mpl,N,Rd,z ≤ αM
30.3 / 272.7 ≤ 0.90
0.111PASS
My,biax / (μ_dy · Mpl,Rd,y) + Mz,biax / (μ_dz · Mpl,Rd,z) ≤ 1
0.000 (My 0.0) + 0.111 (Mz 30.3)
0.111PASS
Scope of EC4 simplified method
Outside these bounds the simplified method does not apply — use the general method.
0.3 % ≤ ρs ≤ 6 %2.70 %PASS
0.2 ≤ δ ≤ 0.90.494PASS
λ̄ ≤ 2.00.63 / 0.76PASS
Ncr,eff ≥ 10 NEd8.29×REVIEW
Axial NEd / Nb,Rd ≤ 10.304PASS
Biaxial Σ ≤ 10.111PASS