Checks a solid rectangular steel bar under a moment about its major or minor axis and an axial force in compression or tension, to SIA 263:2013: the plastic section, flexural buckling, lateral-torsional buckling and their interaction. Every step shows its formula and the clause it comes from.

What it checks

Section values: A, W_pl = b · h²/4 with h in the plane of bending, N_pl,Rd and M_pl,Rd (clause 4.1.3), and n = N_Ed/N_pl,Rd.
Section resistance: a central strip of depth n · h carries N_Ed and the rest of the section carries the moment, so M_N,Rd = M_pl,Rd · (1 − n²). The check is M_Ed ≤ M_N,Rd.
Lateral-torsional buckling, clause 4.5.2: the torsion constant K = β · h · t³, the elastic critical moment for a uniform moment M_crD = (π/l_D) · √(E · I_z · G · K), with no warping resistance, then λ̄_D, χ_D on curve c (α_D = 0.49) and M_D,Rd,min = χ_D · M_R/γ_M1. For the actual moment diagram, M_D,Rd follows from C₁ and the load position z_a.
Flexural buckling, clause 4.5.1, about both axes: λ̄_K = (l_K/i)/λ_E, with λ_E = π · √(E/f_y,k), N_K,Rd = χ_K · A · f_y,k/γ_M1, and N_cr = π² · E · I/l_K².
Stability, bending and compression: N_Ed/N_K,Rd + ω · M_Ed/((1 − N_Ed/N_cr) · M_D,Rd,min) ≤ 1, the form SIA 263 gives for any constant section. N_K,Rd is the smaller buckling resistance, N_cr is taken in the plane of bending, and ω = 0.6 + 0.4 · ψ, at least 0.4.
Lateral-torsional buckling with the actual diagram, clause 4.5.2: M_Ed ≤ M_D,Rd, checked on its own.

What you enter and what you get

You enter f_y,k from table 1 of your copy of SIA 263, for the grade and the plate thickness, E and G (clause 2.2), the axis of bending, the width b and depth h, γ_M1 (clause 4.1.3), the sense and magnitude of N_Ed, the largest moment M_Ed in the plane of bending, the moment diagram with M_min and the signs of the end moments. For the member: the buckling lengths, entered or derived from the length l, the end conditions and the intermediate supports n_y and n_z in each plane. For lateral-torsional buckling, prevented by construction measures or checked: l_D, the moment diagram between the lateral supports and the point of load application.

You get the section values, the buckling and lateral-torsional buckling resistances 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

A solid rectangle is class 1 whatever the forces: it has no slender plate to buckle locally, so its plastic resistance is used.
Compression brings in the member checks. Tension leaves the section check and lateral-torsional buckling of the bending alone.
A solid rectangle resists lateral-torsional buckling by uniform torsion and lateral bending only. It has no flanges, so no warping resistance.
Intermediate supports are equally spaced, each a fixed point for buckling in its plane: the member buckles over l/(n + 1).
With end moments only, ω allows for the moment being smaller over most of the member. A transverse load between the ends, or an end not held laterally, takes ω = 1.0.
Shear is not included.

Worked example

A bar b × h = 40 × 200 mm in f_y,k = 235 N/mm², bent about its major axis, under a compression N_Ed = 150 kN and end moments of 25 and 10 kNm of the same sign, 2,000 mm long and pinned at both ends in both planes, with lateral supports 2,000 mm apart, the moment diagram M and M/2 for lateral-torsional buckling and the load at the centroid. 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,k	235 N/mm²	§ t.1
Modulus of elasticity E	210000 N/mm²	§ 2.2
Bending about	Major axis (y–y)
Width b	40 mm
Depth h	200 mm
Resistance factor γ_M1	1.05	§ 4.1.3
The axial force is	Compression
Design axial force N_Ed	150 kN
Largest moment in the plane of bending along the member M_Ed	25 kN·m
Moment diagram	End moments only — the diagram is linear
Smaller end moment \|M_min\|	10 kN·m
The two end moments have	The same sign at both ends
Buckling lengths	From the end conditions and intermediate supports
Length of the member l	2000 mm
Ends, buckling about the strong axis y	Held against sway — both ends pinned
Intermediate supports against buckling about y n_y	0
Ends, buckling about the weak axis z	Held against sway — both ends pinned
Intermediate supports against buckling about z n_z	0
Lateral-torsional buckling	Checked	§ 4.5.2
Distance between lateral supports l_D	2000 mm
Moment diagram between the lateral supports	End moments M and M/2, same sense
Load applied at	Centroid
Shear modulus G	81000 N/mm²	§ 2.2

Calculation: Common values
Step	Value	Ref.
Characteristic yield strength	f_y,k = 235 N/mm²	§ t.1
Area	A = 8,000 mm²
Plastic section modulus about the axis of bendingb·h²/4 with h in the plane of bending = 40·200²/4.	W_pl = 400,000 mm³
Plastic axial resistance	N_pl,Rd = 1,790.5 kN	§ 4.1.3
Plastic moment resistance	M_pl,Rd = 89.52 kN·m	§ 4.1.3
Axial utilisation	n = 0.0838

Calculation: Check 1 — section resistance
Step	Value	Ref.
Depth carrying the axial forceA central strip of depth n·h carries N_Ed; the rest of the section carries the moment.	2a = 16.8 mm
Reduced plastic momentM_pl,Rd·(1 − n²): the strip taken by the axial force is the one near the neutral axis, where it contributed least, so the loss is quadratic.	M_N,Rd = 88.90 kN·m
Section, bending and axial forceM_Ed ≤ M_N,Rd	η = 0.281 · Pass28 % of the capacity used

Calculation: Common values — lateral-torsional buckling
Step	Value	Ref.
Torsion constantβ·h·t³ of the solid rectangle.	K = 3,728,854 mm⁴
Elastic critical moment, uniform moment(π/l_D)·√(E·I_z·G·K), no warping resistance.	M_crD = 408.6 kN·m
Lateral-torsional buckling moment, uniform momentχ_D·M_R/γ_M1, λ̄_D = 0.480, curve c (α_D = 0,49).	M_D,Rd,min = 76.48 kN·m	§ 4.5.2
Lateral-torsional buckling moment, actual diagramEnd moments M and M/2, same sense: C₁ = 1.32, z_a = 0 mm.	M_D,Rd = 79.50 kN·m	§ 4.5.2

Calculation: Common values — flexural buckling
Step	Value	Ref.
Buckling length, strong axisk·l with k = 1.00 — held against sway — both ends pinned.	l_K,y = 2,000 mm
Buckling length, weak axisk·l with k = 1.00 — held against sway — both ends pinned.	l_K,z = 2,000 mm
Relative slenderness, strong axis(l_K,y/i_y)/λ_E, i_y = 57.7 mm, λ_E = π·√(E/f_y,k) = 93.91.	λ̄_K,y = 0.3689
Relative slenderness, weak axis(l_K,z/i_z)/λ_E, i_z = 11.5 mm.	λ̄_K,z = 1.8443
Buckling resistance, strong axisχ_K·A·f_y,k/γ_M1, curve c, χ_K = 0.9137.	N_K,y,Rd = 1,635.9 kN	§ 4.5.1
Buckling resistance, weak axisχ_K·A·f_y,k/γ_M1, curve c, χ_K = 0.2252.	N_K,z,Rd = 403.1 kN	§ 4.5.1
Elastic critical load, strong axisπ²·E·I_y/l_K,y². It sets the amplification 1/(1 − N_Ed/N_cr) of the moment.	N_cr,y = 13,817 kN
Elastic critical load, weak axis	N_cr,z = 553 kN

Calculation: Check 2 — stability, bending and compression
Step	Value	Ref.
Moment factor0,6 + 0,4·ψ, ψ = 0.400, at least 0,4.	ω = 0.7600
AmplificationN_cr in the plane of bending (the "y" axis here).	1/(1 − N_Ed/N_cr) = 1.0110
Axial termThe smaller buckling resistance.	N_Ed/N_K,Rd = 0.3721
Moment term, amplifiedOver the lateral-torsional buckling moment for a uniform moment.	ω·M_Ed/M_D,Rd,min = 0.2512
Stability, bending and compressionN_Ed/N_K,Rd + ω·M_Ed/((1 − N_Ed/N_cr)·M_D,Rd,min) ≤ 1	η = 0.623 · Pass62 % of the capacity used

Calculation: Check 3 — lateral-torsional buckling with the actual moment diagram
Step	Value	Ref.
Bending alone, actual diagramM_Ed ≤ M_D,Rd	η = 0.314 · Pass31 % of the capacity used	§ 4.5.2

How this module is verified

Verified clause by clause against SIA 263:2013 and against independent calculations used on real projects, before release.

Questions

Why does the moment resistance fall with n² and not with n?

The strip taken by the axial force lies next to the neutral axis, where it contributed least to the moment. In the worked example n = 0.084 lowers M_pl,Rd from 89.5 to 88.9 kNm.

Why is lateral-torsional buckling checked separately?

The interaction uses M_D,Rd,min, the resistance under a uniform moment, and does not reduce to the bending check when N_Ed is zero. So M_Ed ≤ M_D,Rd with the actual diagram is checked on its own.

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 263:2013 · Page updated: 28 October 2026

Solid steel rectangle under bending and axial force to SIA 263

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