Checks a reinforced concrete column section of any shape, with its bars at their own coordinates, to SIA 262:2025: compression resistance, bending with axial force including second-order effects, detailing and service stresses. Every step shows its formula and the clause it comes from.

What it checks

Design strengths f_cd and f_yd, equations (2) and (4), with η_fc = (40/f_ck)^(1/3) ≤ 1, equation (26).
Compression resistance, check 1: N_d ≤ N_Rd = f_cd · (A_c − A_s) + f_yd · A_s.
Imperfections: α_i = 0.01/√l, between 1/300 and 1/200, clause 4.1.3.2.3, equation (17), or for a group of columns, equation (18); e_0d is the larger of α_i · l_cr/2 and d/30, clause 4.3.7.5, equation (75).
Second-order effects, clause 4.3.7: e_1d = M_1d / N_d, equation (76); e_2d = χ_d · l_cr² / c, equation (77), with c = π², equation (81); M_d = N_d · (e_0d + e_1d + e_2d), equations (73) and (74).
Bending with axial force, check 2: M_d ≤ [M_Rd], with a 0.85 · x stress block (clause 4.2.1.5).
Confinement (optional), clause 4.2.1.9: core strength k_c · f_cd, equations (29) to (31).
Detailing: ρ ≥ 0.6 % and Ø ≥ 8 mm (clause 5.5.4.2); a bar in every corner (clause 5.5.4.8); tie spacing s ≤ min(15 · Ø_min ; a_min ; 300 mm), clause 5.5.4.7, equation (117); tie diameter Ø_w ≥ Ø_sl,max/3 (clause 5.5.4.9).
Service state: [M_cr] = (f_ct,eff + N/A_i) · I_i / v_t, clause 4.4.1.3, then the stresses under N and M.

What you enter and what you get

You choose the shape (rectangle, circle, or a polygon drawn by clicking its corners) and the face compressed by the moment. You click each bar into the drawing, or place a rectangular cage or circular ring at once. You enter the ties Ø_w and s, the materials, the resistance factors, N_d and M_1d. For second-order effects: l_cr, the height l, single member or group, and the curvature method. For serviceability: the requirement level, N, M and the load case.

You get every eccentricity, resistance and check with its utilisation and clause, detailing warnings, and a drawing of the section. With the Basic plan, it prints as an A4 calculation note.

Assumptions and limits

N_d and N are positive in compression.
Each direction of bending is checked separately: change the compressed face to check the other one.
The curvature is either estimated, χ_d = 2 · f_yd / (E_s · (d − d′)), equation (78), creep and shrinkage included, or taken from the strain planes of the section under N_d (clause 4.3.7.9).
With confinement counted, the concrete outside the ties is not counted.
The tie checks are skipped if Ø_w is left blank.
When the compressed bars yield, every second bar must be held against buckling (clause 5.5.4.6).

Worked example

A 400 × 400 mm column in f_ck = 30 N/mm² with eight Ø20 bars and Ø8 ties at 240 mm, under N_d = 2,000 kN and M_1d = 60 kNm, with l_cr = 4.5 m and the estimated curvature, and a quasi-permanent service state of N = 1,300 kN and M = 60 kNm. These are the module's own results for its worked example, computed when this page was built.

Inputs
Input	Value	Ref.
Shape of the section	Rectangle
Width b	400 mm
Depth h	400 mm
Face compressed by the moment	Top face (+y)
Bars in the section	{"b":[[50,50,20],[200,50,20],[350,50,20],[50,350,20],[200,350,20],[350,350,20],[50,200,20],[350,200,20]],"p":[],"v":[]}
Diameter of the ties Ø_w	8 mm
Spacing of the ties s	240 mm
Characteristic compressive strength f_ck	30 N/mm²	§ 3.1.2.2.2
Ultimate compressive strain ε_c2d	3.5 ‰	§ 4.2.1.5
Mean tensile strength f_ctm	2.9 N/mm²	§ 3.1.2.2.4
Modulus of elasticity E_cm	33600 N/mm²	§ 3.1.2.3.3
Characteristic yield strength f_yk	500 N/mm²
Modulus of elasticity E_s	205000 N/mm²
Resistance factor for concrete γ_c	1.5	§ 2.4.2.6
Resistance factor for steel γ_s	1.15	§ 2.4.2.6
Duration factor η_t	1.0 — both conditions of 4.2.1.3 met	§ 4.2.1.3
Design axial force (compression) N_d	2000 kN
First-order design moment M_1d	60 kNm
Include second-order effects?	Yes — slender member (4.3.7)	§ 4.3.7.1
Buckling length l_cr	4.5 m
Height for the inclination l	3.2 m	§ 4.1.3.2.3
Inclination	Single member (eq. (17))	§ 4.1.3.2.5
Curvature of the column χ_d	Estimate (4.3.7.8)	§ 4.3.7.7
Count the confinement of the core?	No	§ 4.2.1.9
Tensile strength for cracking	Mean tensile strength (4.4.1.3)	§ 4.4.1.3
Requirement level	Normal	§ 4.4.2.2.3
Axial force in the load case (compression) N	1300 kN
Bending moment in the load case M	60 kNm
Load case	Quasi-permanent	Table 17

Calculation: Concrete section
Step	Value	Ref.
Gross concrete area	A_c = 160,000 mm²
Depth in the direction of bendingfrom the compressed face to the opposite face	h = 400 mm
Second moment of area of the concreteabout the axis of bending through the centroid	I_c = 2,133,333,333 mm⁴
Radius of gyration√(I_c/A_c)	i = 115.5 mm

Calculation: Longitudinal reinforcement
Step	Value	Ref.
Bars	n_s = 8Ø20
Total area of the barsΣ π·Ø²/4	A_s = 2,513 mm²
Reinforcement ratioA_s/A_c	ρ = 1.57 %
Smallest clear cover of the barsdistance of the bar edge to the nearest face	c = 40 mm
Effective depthcompressed face to the farthest bar	d = 350 mm
Depth of the nearest barcompressed face to the nearest bar	d′ = 50 mm

Calculation: Design strengths
Step	Value	Ref.
Factor for the brittleness of strong concrete(40/f_ck)^(1/3) ≤ 1	η_fc = 1.000	§ 4.2.1.2 (26)
Duration factor	η_t = 1.00	§ 4.2.1.3
Design compressive strength of the concreteη_fc·η_t·f_ck/γ_c	f_cd = 20.0 N/mm²	§ 2.4.2.3 (2)
Design yield strength of the steelf_yk/γ_s	f_yd = 434.8 N/mm²	§ 2.4.2.5 (4)
Design yield strain of the steelf_yd/E_s	ε_yd = 2.12 ‰

Calculation: Check 1 — compression resistance of the section: N_d ≤ N_Rd
Step	Value	Ref.
Compression resistance of the sectionf_cd·(A_c − A_s) + f_yd·A_sthe whole section uniformly compressed at ε_c2d, no eccentricity	N_Rd = 4,242 kN
Design axial force against the compression resistanceN_d/N_Rd	η = 0.471 · Pass47 % of the capacity used

Calculation: Slenderness and eccentricities
Step	Value	Ref.
Slendernessl_cr/ifor information: the larger λ, the more the deflection matters	λ = 39.0
Inclination of the column0.01/√l, between 1/300 and 1/200 (l in m)limited to 1/200	α_i = 0.00500 rad	§ 4.1.3.2.3 (17)
Eccentricity from the inclination	α·l_cr/2 = 11.3 mm	§ 4.3.7.5 (75)
Minimum eccentricity	d/30 = 11.7 mm	§ 4.3.7.5 (75)
Eccentricity due to imperfectionsthe larger of the two	e_0d = 11.7 mm	§ 4.3.7.5 (75)
First-order eccentricityM_1d/N_d	e_1d = 30.0 mm	§ 4.3.7.6 (76)
Integration constantπ²	c = 9.87	§ 4.3.7.12 (81)
Curvature of the column2·f_yd/(E_s·(d − d′))creep and shrinkage included	χ_d = 0.01414 1/m	§ 4.3.7.8 (78)
Eccentricity from the deflectionχ_d·l_cr²/c	e_2d = 29.0 mm	§ 4.3.7.7 (77)
Total eccentricitye_0d + e_1d + e_2d	e_d = 70.7 mm	§ 4.3.7.4 (74)
Design momentN_d·e_d	M_d = 141.4 kNm	§ 4.3.7.3 (73)

Calculation: Check 2 — bending with axial force: M_d ≤ M_Rd(N_d)
Step	Value	Ref.
Depth of the neutral axisfrom N = N_d, 0.85·x block at f_cd, ε_c2d at the compressed face	x = 258.6 mm	§ 4.2.1.5
Concrete compression forcef_cd·A_c,comp	F_c = 1,758 kN
Force in the compressed barsΣ A_s,i·(σ_s,i − f_cd), bars inside the block	F_s′ = 481 kN
Force in the stretched barsΣ A_s,i·σ_s,i	F_s = 239 kN
Bending resistance under N_dmoment of the internal forces about the axis of the column	M_Rd(N_d) = 252.9 kNm
Design moment against the bending resistance under N_dM_d/M_Rd	η = 0.559 · Pass56 % of the capacity used

Calculation: Check 3 — minimum longitudinal reinforcement
Step	Value	Ref.
Reinforcement ratio of the column0.6 %/ρ	η = 0.382 · Pass38 % of the capacity used	§ 5.5.4.2
Smallest bar diameter8/Ø_min	η = 0.400 · Pass40 % of the capacity used	§ 5.5.4.2

Calculation: A bar in every corner
Step	Value	Ref.
Corners of the section with a longitudinal barthe rule is met	n_corner = all 4	§ 5.5.4.8

Calculation: Check 4 — ties
Step	Value	Ref.
Smallest dimension of the section	a_min = 400 mm
Spacing of the tiess/s_max	η = 0.800 · Pass80 % of the capacity used	§ 5.5.4.7 (117)
Diameter of the ties(Ø_max/3)/Ø_w	η = 0.833 · Pass83 % of the capacity used	§ 5.5.4.9

Calculation: Idealised section
Step	Value	Ref.
Modular ratio of the barsE_s/E_cm	α_s = 6.10
Area of the idealised sectionA_c + (α_s − 1)·A_s	A_i = 172,821 mm²
Second moment of area, uncracked	I_i = 2,349,682,477 mm⁴
Distance from its centroid to the stretched face	v_t = 200.0 mm
Dimension for the size factorh/3, member in bending	t = 133 mm	§ 4.4.1.3
Size factor1/(1 + 0.5·t), t in m	k_t = 0.938	§ 4.4.1.3 (100)
Effective tensile strength of the concretek_t·f_ctm	f_ct,eff = 2.72 N/mm²	§ 4.4.1.3 (99)
Cracking moment under N(f_ct,eff + N/A_i)·I_i/v_tthe compression of N delays cracking	M_cr(N) = 120.3 kNm	§ 4.4.1.3

Calculation: State under N = 1,300 kN and M = 60.0 kNm
Step	Value	Ref.
State of the section	uncracked (M ≤ M_cr)
Concrete stress at the compressed face−N/A_i − M_i·v_c/I_icompression is negative; M_i about the centroid of the idealised section	σ_c,c = -12.63 N/mm²
Concrete stress at the opposite face−N/A_i + M_i·v_t/I_istill in compression	σ_c,t = -2.42 N/mm²
Stress of the bar nearest the opposite facein compression	σ_s = -22.5 N/mm²
Stress of the bar nearest the compressed facecompression negative	σ_s′ = -69.3 N/mm²

How this module is verified

Verified clause by clause against SIA 262:2025 and against independent calculations used on real projects, before release.

Questions

When can second-order effects be left out?

Only if the column is not significantly affected, or if M_1d comes from an analysis of the deformed structure that already includes imperfections and deformations (clause 4.3.7.2).

Which curvature should I choose?

The estimate of equation (78) is simpler. The strain-plane search is usually less conservative.

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 262:2025 · Page updated: 24 October 2026

Reinforced concrete column to SIA 262

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