Steel member in tension to SIA 263

The code governs which inputs appear and how every value is derived.

Member

Material properties, shared by every verification. The cross-section is not shared: SIA 263 classifies it from the internal forces, so it belongs to each verification and appears with it.

About this check

clauses, worked example and questions

Checks the resistance of a steel section to an axial tensile force at the ultimate limit state, to SIA 263:2013: the gross section, and the net section where there are holes. Every step shows its formula and the clause it comes from.

What it checks

What you enter and what you get

You enter fy,k from table 1 of your copy of SIA 263, for the grade and the plate thickness; every grade drops by 20 to 30 N/mm² once the thickness passes 40 mm. Then the gross area, directly or from the width b and thickness t of a flat bar; whether there are holes in the tension length; γM1 (clause 4.1.3, 1.05); the design tensile force NEd (clause 5.1.2) as a magnitude; and the restrained thermal action: not considered, the upper bound of a fully restrained member, or the force from the global analysis.

You get the gross area, the resistance, the area required, the utilisation with a pass or fail, and the clause of each step. With the Basic plan, it prints as an A4 calculation note.

Assumptions and limits

Worked example

A flat bar 80 × 12 mm in fy,k = 235 N/mm², with no holes in the tension length, γM1 = 1.05 and a design tensile force of 200 kN, without thermal action. These are the module's own results for its worked example, computed when this page was built.

Inputs
InputValueRef.
Characteristic yield strength fy,k235 N/mm²§ t.1
Gross area given byFlat bar — b × t§ 1.1
Width b80 mm
Thickness t12 mm
HolesNo holes in the tension length§ 5.1.2.1
Resistance factor, gross section γM11.05§ 4.1.3
Design tensile force NEd200 kN§ 5.1.2
Imposed axial force from restrained temperature changeNot considered§ 2.3.3
Calculation: Material and factors
StepValueRef.
Characteristic yield strengthfy,k = 235 N/mm²§ t.1
Resistance factor, gross sectionγM1 = 1.05§ 4.1.3
Calculation: Action
StepValueRef.
Design tensile forceNEd = 200.0 kN§ 5.1.2
Calculation: Gross section — equation (38)
StepValueRef.
Gross areaA = b·t = 80.0 × 12.0A = 960 mm²§ 1.1
Plastic axial resistanceNRd = fy,k·A/γM1 = 235 × 960 / 1.05. Every fibre reaches fy,k together under a centred axial force, so the elastic and plastic limits coincide and there is no shape factor to gain.NRd = 214.9 kN§ (38)
Gross area requiredA ≥ γM1·NEd/fy,k, the sizing form of equation (38).Areq = 894 mm²§ (38)
Calculation: Verification
StepValueRef.
Design resistanceEquation (38) alone; with no hole there is no net section to check.NRd = 214.9 kN§ 5.1.2.1
Tensile resistance of the sectionNEd ≤ NRdη = 0.931 · Pass93 % of the capacity used§ 5.1.2.1

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 is the section class not needed?

Classification guards against local buckling of compressed plates. A section in tension has no compressed part, so it carries its full resistance whatever its class in compression.

When does the net section govern?

Whenever there is a hole in the tension length. Equation (39) carries fu, γM2 and the factor 0.9, so one bolt hole can make it decisive.

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.

Related

Norm: SIA 263:2013 · Page updated: 27 October 2026