Checks an encased or concrete-filled composite column to SIA 264:2014: the plastic resistance of the section, buckling under the normal force, compression with bending on the computed interaction curve, and the introduction of the force into the concrete. Every step shows its formula and the clause it comes from.

What it checks

Plastic resistance, equation (14): N_pl,Rd = A_a · f_y/γ_a + A_s · f_sk/γ_s + 0.85 · A_c · f_ck/γ_c, with 1.0 instead of 0.85 in a filled tube, and N_Ed ≤ N_pl,Rd. The bars count up to 6 % of the concrete area, clause 5.3.2.2.
Conditions of the method: δ, equation (13), between 0.2 and 0.9; local buckling, clause 5.3.1.7; the concrete over the profile, c ≥ max(40 mm; b/6), clause 7.3.1.2.
Stiffness: E_c,eff = E_cm/(1 + (N_G,Ed/N_Ed) · φ), equation (19). For the slenderness, E_a · I_a + E_s · I_s + 0.6 · E_c,eff · I_c, equation (17); for the second-order effects, 0.9 · (E_a · I_a + E_s · I_s + 0.5 · E_c,eff · I_c), equation (22).
Buckling: N_cr by equation (20), λ̄_y = √(N_pl,Rk/N_cr) on the characteristic resistance, equation (16), and χ_K by SIA 263 on the curve of table 7, equation (15).
Second-order moment, clause 5.3.4.1: N_cr,eff over the column length, equation (23), the imperfection e₀ of table 7, the end moments amplified by k = ω/(1 − N_Ed/N_cr,eff) ≥ 1.0, equation (24), with ω from SIA 263 clause 5.1.9.1, and the imperfection moment amplified with ω = 1. Together they give M_y,Ed,II, equation (25).
Interaction curve, clause 5.3.4.3: points A to D, and M_pl,N,Rd read off the computed curve at N_Ed. μ_d, clause 5.3.4.6, is capped at 1.0 unless the moment and the force are dependent.
Compression and bending, equation (25): M_Ed,II/(μ_d · M_pl,Rd) ≤ 0.9.
Introduction of the force, clause 5.3.1.3: the share crossing the interface over l_v ≤ min(2 · d; L/3), against bond with τ_Rd from EN 1994-1-1 table 6.6. Where bond is exceeded, headed studs carry the whole force: P_Rd by equations (30) and (31), with their spacing (clause 7.1.2) and the web thickness against the stud (clause 7.1.4).

What you enter and what you get

You enter the materials and resistance factors, what the column carries (centred compression or bending about one or both axes) and the scope: the column, or the section only. You choose a section form of figure 6 and enter the profile, the concrete outline and the bars. You enter the length, the lateral restraint, the buckling lengths, the creep data, N_Ed, the largest first-order moment, whether moment and force are dependent, any shear in the profile, the moment distribution, how the force enters and the studs.

You get the resistances, the interaction curve, the second-order moment, the studs required and the utilisation of each check with a pass or fail. With the Basic plan, it prints as an A4 calculation note.

Assumptions and limits

The simplified method of clause 5.3.4.1 is for a column held laterally at its ends. In a sway frame the forces come from a second-order analysis of the frame with stiffness (22), clause 5.3.3.3.
Clause 5.3.4.1 does not say how k and the imperfection combine; they are added, as in EN 1994-1-1 clause 6.7.3.4(5).
Not computed: the interface shear from a moment varying along the column, and the friction of studs pressing the concrete.

Worked example

A fully encased rolled I, 171 mm deep with a 180 mm flange, in 320 × 320 mm of concrete with eight 20 mm bars, f_y = 235 N/mm², f_ck = 30 N/mm² and f_sk = 500 N/mm², 4,000 mm long and held about z, under N_Ed = 2,500 kN through the steel profile and M_y,Ed = 70 kNm, with 16 mm studs four per row. It does not pass: compression and bending about y gives η = 1.027, and the stud spacing along the column, the spacing across the web and the web thickness against the stud fail. These are the module's own results for its worked example, computed when this page was built.

Inputs
Input	Value	Ref.
Characteristic yield strength f_y	235 N/mm²	SIA 263 tab. 1
Modulus of elasticity E_a	210000 N/mm²
Resistance factor, steel γ_a	1.05
Characteristic cylinder strength f_ck	30 N/mm²	SIA 262 tab. 3
Aggregate coefficient k_E	10700
Resistance factor, concrete γ_c	1.5
Yield strength of the reinforcement f_sk	500 N/mm²
Resistance factor, reinforcement γ_s	1.15
The column carries	Compression and bending about the strong axis y	§ 5.3.1.3
Scope of the calculation	Column — section, buckling and second-order effects	§ 5.3.1.3
Section form	Fully encased I profile	§ 5.3.1.1
Profile	Rolled I or H profile
Overall depth h	171 mm
Flange width b	180 mm
Flange thickness t_f	9.5 mm
Web thickness t_w	6 mm
Root radius r	15 mm
Width of the concrete, across the flanges b_c	320 mm	§ 7.3.1.2
Depth of the concrete, along the web h_c	320 mm	§ 7.3.1.2
Reinforcement	Longitudinal bars	§ 5.3.2.2
Bar diameter ⌀	20 mm
Bars along each face across y, corners included n_y	3
Bars along each face down z, corners included n_z	3
Distance from the concrete face to the bar centre a	40 mm	§ 7.3.1.3
Length of the column L	4000 mm	§ 5.3.4.1
Lateral restraint	Held laterally about the weak axis z	§ 5.3.2.6
Buckling length about the strong axis y L_K,y	4000 mm	§ (20)
Effective modulus of the concrete	(19) — from φ and the permanent share of N_Ed	§ 5.3.2.9
Creep coefficient φ(t,t₀)	2.0	SIA 262 §3.1.2.6
Permanent share of the normal force N_G,Ed/N_Ed	0.5	§ (19)
Design normal force, compression N_Ed	2500 kN	§ (15)
Largest first-order moment about y M_y,Ed	70 kNm	§ (25)
The moment and the normal force are	Independent — μ_d capped at 1,0	§ 5.3.4.6
Design shear force in the steel profile, if any V_a,Ed	0 kN	§ 5.3.4.5
Moment along the column	Linear moment diagram — ω from SIA 263	SIA 263 §5.1.9.1
Smaller end moment, absolute value M_min	0 kNm	SIA 263 §5.1.9.1
The two end moments have	The same sign	SIA 263 §5.1.9.1
The normal force enters the column	Through the steel profile — beams connected to it	§ 5.3.1.3
Force introduced at this level	All of N_Ed — a single-storey column or the top storey
Length over which it is introduced	The longest permitted — min(2·d; L/3)	EN 1994-1-1 §6.7.4.2
Stud diameter d	16 mm	§ 6.1.2.1
Stud length, after welding h_D	75 mm	§ 6.1.1.3
Tensile strength of the stud steel f_u	450 N/mm²	§ 6.1.2.1
Resistance factor of the connection γ_v	1.25	§ 6.1.2.1
Studs per row, on the two faces of the web n	4	§ 7.1.2
Elastic resistance of the section	Not computed

Calculation: Areas
Step	Value	Ref.
Area of the steel profileTwo flanges, the web and the four root fillets.	A_a = 4,525 mm²	§ (14)
Area of the reinforcement8 bars of 20 mm	A_s = 2,513 mm²	§ (14)
Net area of the concreteThe outline less the steel and the bars — they occupy real space.	A_c = 95,362 mm²	§ (14)
Reinforcement ratio	ρ_s = A_s/A_c = 2.64 %	§ 5.3.2.2

Calculation: Second moments of area
Step	Value	Ref.
Steel profile, about y	I_a,y = 25,102,104 mm⁴	§ (17)
Reinforcement, about y	I_s,y = 27,206,192 mm⁴	§ (17)
Concrete, uncracked, about yThe gross concrete less what the steel and the bars occupy. §5.3.2.8 takes the concrete UNCRACKED here; the 0,6 of (17) is what allows for the cracking.	I_c,y = 821,505,037 mm⁴	§ (17)
Steel profile, about z	I_a,z = 9,245,304 mm⁴	§ (17)
Reinforcement, about z	I_s,z = 27,206,192 mm⁴	§ (17)
Concrete, uncracked, about z	I_c,z = 837,361,837 mm⁴	§ (17)

Calculation: Stiffness of the section
Step	Value	Ref.
Modulus of the concretek_E·∛f_cm, with f_cm = f_ck + 8	E_cm = 35,973 N/mm²	SIA 262 §3.1.2.3.3
Effective modulus, long termE_cm/(1 + (N_G,Ed/N_Ed)·φ). Only the permanent part of the force creeps.	E_c,eff = 17,987 N/mm²	§ (19)
Stiffness for the slenderness, about yE_a·I_a + E_s·I_s + 0,6·E_c,eff·I_c — the 0,6 on the concrete alone, for its cracking.	(EI)_eff,λ,y = 19,850 kNm²	§ (17)
Stiffness for the second-order effects, about y0,9·(E_a·I_a + E_s·I_s + 0,5·E_c,eff·I_c) — lower than the one of the slenderness.	(EI)_eff,d,y = 16,535 kNm²	§ (22)
Stiffness for the slenderness, about z	(EI)_eff,λ,z = 16,692 kNm²	§ (17)
Stiffness for the second-order effects, about z	(EI)_eff,d,z = 13,667 kNm²	§ (22)

Calculation: Plastic resistance to the normal force
Step	Value	Ref.
Carried by the steel	A_a·f_y/γ_a = 1,013 kN	§ (14)
Carried by the reinforcement	A_s·f_sk/γ_s = 1,093 kN	§ (14)
Carried by the concreteThe 0,85 is on the concrete alone.	0.85·A_c·f_ck/γ_c = 1,621 kN	§ (14)
Plastic resistance of the section	N_pl,Rd = 3,727 kN	§ (14)
The same with all resistance factors at 1,0What the slenderness (16) needs. Putting the design value there would understate λ̄ and overstate the resistance.	N_pl,Rk = 4,752 kN	§ (16)

Calculation: Conditions of the method
Step	Value	Ref.
Section parameter(A_a·f_y/γ_a)/N_pl,Rd — the steel's share of the resistance. A test on the method, not a verification.between 0,2 and 0,9	δ = 0.272	§ (13)
Depth to width of the section	h/b = 1.00	§ 5.3.2.3
Local buckling of the steelA fully encased profile is held by the concrete all round.	– = may be neglected	§ 5.3.1.7
Concrete over the profilec ≥ max(40 mm; b/6)	η = 0.571 · Pass57 % of the capacity used	§ 7.3.1.2
Yield strain of the barsThe plastic resistance assumes the bars yield before the concrete crushes at 3 ‰.	ε_ys = f_sk/E_s = 2.38 ‰

Calculation: Introduction of the force — shear at the interface
Step	Value	Ref.
The force entersThe concrete and the bars carry their share only once bond has passed it to them from the steel.	– = through the steel	§ 5.3.1.3
Share that has to cross the interfaceThe more severe of the plastic shares of (14) and the elastic ones — by the axial stiffnesses, with E_cm and with the creep-reduced E_c,eff.the elastic, E_cm distribution, the most severe of 0.728 / 0.806 / 0.702	(N_c + N_s)/N = 0.806	EN 1994-1-1 §6.7.4.2
Introduction lengthThe length over which the force is assumed to pass by bond. It may not exceed twice the smaller dimension of the section nor a third of the column length.min(2·d; L/3), d = 320 mm	l_v = 640 mm	EN 1994-1-1 §6.7.4.2
Bonded perimeter, the whole profileτ_Rd = 0.30 N/mm²	u = 1,050 mm	EN 1994-1-1 §6.7.4.3
Force bond can carryτ_Rd from table 6.6 of SN EN 1994-1-1 — SIA 264 §5.3.1.3 requires the check and gives no values of its own.	Σ τ_Rd·u·l_v = 202 kN	EN 1994-1-1 §6.7.4.3
Force to carry across the interface0.806 × 2,500 kN, all of N_Ed	F_v,Ed = 2,016 kN	EN 1994-1-1 §6.7.4.2
Mean shear stress on the bonded perimeterF_v,Ed/(u·l_v), over the faces that can carry bond.above τ_Rd = 0.30 N/mm² — bond is not enough	τ_Ed = 3.000 N/mm²	EN 1994-1-1 §6.7.4.2

Calculation: Check — introduction of the force
Step	Value	Ref.
Resistance of one studThe smaller of (30) 0,29·d²·√(f_ck·E_cm)/γ_v and (31) 0,8·f_u·(π·d²/4)/γ_v — the same studs as in a beam.the shank governs	P_Rd = 57.9 kN	§ 6.1.2.1
Studs needed for the whole forcerounded up to 35, rows of 4 → 9 rows	N = F_v,Ed/P_Rd = 34.82	§ 5.3.1.3
Introduction of the force — by the studsF_v,Ed ≤ N·P_Rd	η = 0.967 · Pass97 % of the capacity used	§ 6.1.2.1

Calculation: Detailing of the studs over l_v
Step	Value	Ref.
Spacing of the rows along the column9 rows over 640 mm	e = l_v/rows = 71 mm	§ 7.1.2
Spacing along the column, more than 5·de > 5·d	η = 1.125 · Fail113 % of the capacity used	§ 7.1.2
Spacing across, up the web, on each face, more than 4·de_t > 4·d	η = 1.049 · Fail105 % of the capacity used	§ 7.1.2
Web against the studt/d ≥ 0,5	η = 1.333 · Fail133 % of the capacity used	§ 7.1.4

Calculation: Check — the section under the normal force
Step	Value	Ref.
Plastic resistance of the sectionN_Ed ≤ N_pl,Rd	η = 0.671 · Pass67 % of the capacity used	§ (14)

Calculation: Interaction curve about y
Step	Value	Ref.
Point A — compression aloneM = 0	N_pl,Rd = 3,727 kN	§ 5.3.4.3
Point B — bending aloneThe plastic moment of the section with no normal force: steel and bars at their design strength on both sides of the neutral axis, the concrete at 0.85·f_cd on the compressed side only.N = 0	M_pl,y,Rd = 217.6 kNm	§ 5.3.4.3
Point C — the same moment, with compressionFor a doubly symmetric section it is the whole concrete at its design strength, 0.85·f_cd·A_c = 1,621 kN — a check on the integration.neutral axis mirrored from B	N_pm,Rd = 1,621 kN	§ 5.3.4.3
Point D — the largest momentat N = N_pm,Rd/2 = 811 kN	M_max,Rd = 236.0 kNm	§ 5.3.4.3
Plastic moment at the design normal forceRead off the computed curve at N_Ed — the neutral axis that gives N_Ed, and the moment that goes with it.	M_pl,N,Rd = 147.4 kNm	§ 5.3.4.3
Ratio to the plastic moment	μ_d = 0.677	§ 5.3.4.6
The same read off the polygon A–C–BThe straight-line simplification an older calculation will have used. Figure 7 of SIA 264 draws the curve, and the curve is what this module checks against; the polygon lies inside it.segment A–C	M_pl,N,Rd \| polygon = 126.7 kNm

Calculation: Buckling about y
Step	Value	Ref.
Buckling curveI profile, y–y	curve = b	§ tab. 7
Elastic critical loadπ²·(EI)_eff,λ/L_K²	N_cr,y = 12,245 kN	§ (20)
Relative slenderness√(N_pl,Rk/N_cr) — the CHARACTERISTIC plastic resistance.	λ̄_y = 0.623	§ (16)
Buckling coefficientBy SIA 263, on curve b (α = 0.34).	χ_K,y = 0.8254	§ (15)
Buckling resistance	χ_K·N_pl,Rd = 3,076 kN	§ (15)

Calculation: Buckling about z
Step	Value	Ref.
Buckling about zThe column is held laterally about this axis.	– = prevented	§ 5.3.2.6

Calculation: Second-order moment about y
Step	Value	Ref.
Critical load for the second-order effectsπ²·(EI)_eff,d/L², over the LENGTH of the column, §5.3.4.1.	N_cr,eff,y = 10,200 kN	§ (23)
Equivalent imperfectionI profile, y–y	e₀ = L/200 = 20.0 mm	§ tab. 7
Ratio to itabove 0,1	N_Ed/N_cr,eff = 0.245	§ (21)
Moment factor	ω_y = 0.600	SIA 263 §5.1.9.1
Amplification of the end momentsthe floor of 1,0 governs	k = ω/(1 − N_Ed/N_cr,eff) = 1.000	§ (24)
Moment of the imperfection, first order	N_Ed·e₀ = 50.0 kNm	§ 5.3.1.4
Its amplificationω = 1 — the imperfection is a sine	1/(1 − N_Ed/N_cr,eff) = 1.325	§ (24)
Design moment, second order and imperfection includedk·M_Ed + N_Ed·e₀/(1 − N_Ed/N_cr,eff)	M_y,Ed,II = 136.2 kNm	§ (25)

Calculation: Check — compression and bending about y
Step	Value	Ref.
Compression and bending about yM_Ed,II/(μ_d·M_pl,Rd) ≤ 0.9	η = 1.027 · Fail103 % of the capacity used	§ (25)

How this module is verified

Verified clause by clause against SIA 264:2014 and against independent calculations used on real projects, before release.

Questions

Why does the bending check fail when the section check passes?

The section carries N_Ed at η = 0.671, but the second-order moment is almost twice the first-order one. k is held at 1.0, and the imperfection moment of 50 kNm is amplified by 1.325, so M_y,Ed,II = 136.2 kNm. With μ_d = 0.677, the resistance is 147.4 kNm, and 136.2/147.4 exceeds 0.9: η = 1.027.

Why are studs needed?

The mean stress on the bonded perimeter is 3.00 N/mm² against τ_Rd = 0.30 N/mm². The studs then take all of F_v,Ed = 2,016.0 kN: 35 studs in 9 rows of 4 over 640 mm, η = 0.967, at a spacing of 71.1 mm, below 5 · d.

Can I print the calculation?

The full calculation is shown on screen, free and without an account. A print-ready A4 calculation note is part of the Basic plan.

Norm: SIA 264:2014 · Page updated: 8 November 2026

Composite steel–concrete column to SIA 264

Member

About this check