Checks the resistance to a shear force from a shear area entered directly, to SIA 263:2013, for a section the rolled, welded and rectangle modules do not describe. Every step shows its formula and the clause it comes from.

What it checks

Section class, table 4: entered, not derived. A shear area carries no information about the slenderness of the plates that make it up, so take the class from the classification module or from the bending check on the same section.
Stability of the web, equation (42), if you enter the web slenderness: (h − t_f)/t_w ≤ √(4E/f_y,k), 59.8 in S 235 and 48.6 in S 355. Clause 5.1.4.1 grants equation (41) only where this holds; beyond it clause 4.5.4 takes over and the module stops.
Shear area: A_v as entered, clause 5.1.4.1. For a class 3 section, clause 5.2.4 puts the web area A_w in its place.
Shear resistance, equation (41): V_Ed ≤ V_Rd = V_R/γ_M1, with V_R = τ_y,k · A_v, the stress taken as uniform over the area.
Shear area required, A_v,req = γ_M1 · V_Ed/τ_y,k, for sizing.

What you enter and what you get

You enter f_y,k and τ_y,k from table 1 of your copy of SIA 263, for the grade and the plate thickness, E (clause 2.2), the shear area A_v, the web area A_w if class 3 may govern, the web slenderness (h − t_f)/t_w if the section has a web, the section class (1 to 4), γ_M1 (clause 4.1.3) and the design shear force V_Ed as a magnitude.

You get the resistance model, the shear resistance and the utilisation of each check with a pass or fail, each with its clause. With the Basic plan, it prints as an A4 calculation note.

Assumptions and limits

Nothing here knows what section you have, so nothing checks what you enter. A_v, the class and, for a class 4 section, S, I_y and the thickness at the neutral axis are taken on trust.
Where the rolled, welded or rectangle module fits your section, use it: those derive these values from the dimensions and draw the section.
No figure of the section is drawn: an area says nothing about the shape it came from.
Leave the web slenderness empty for a section with no web, such as a solid or tubular one. Equation (42) is written for I and U sections.
τ_y,k is entered from table 1, not derived as f_y,k/√3 (clause 1.2.3).
The shear force acts alone; once V_Ed passes half of V_Rd the interaction modules apply. Shear buckling between stiffeners, connections and shear in a bolted connection (clause 5.1.4.3, referring to clause 6.2.4) are not covered.

Worked example

A class 1 section with A_v = 2,874 mm², A_w = 1,781 mm² and a web slenderness of 31.67 in f_y,k = 235 N/mm² and τ_y,k = 135 N/mm², under V_Ed = 350 kN. 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
Characteristic shear yield strength τ_y,k	135 N/mm²	§ t.1
Modulus of elasticity E	210000 N/mm²	§ 2.2
Shear area A_v	2874 mm²	§ 5.1.4.1
Web area, if class 3 may govern A_w	1781 mm²	§ 5.2.4
Web slenderness, if the section has a web (h − t_f)/t_w	31.67	§ (42)
Section class	1	§ t.4
Resistance factor γ_M1	1.05	§ 4.1.3
Design shear force V_Ed	350 kN	§ 5.1.4

Calculation: Common values
Step	Value	Ref.
Characteristic yield strength	f_y,k = 235 N/mm²	§ t.1
Characteristic shear yield strengthEntered from table 1, not derived.	τ_y,k = 135 N/mm²	§ t.1
The definition in §1.2.3§1.2.3 defines τ_y as f_y/√3 and table 1 then tabulates values that disagree with it in all eight cells. The table governs.not used	f_y,k/√3 = 135.7 N/mm²	§ 1.2.3
Resistance factor	γ_M1 = 1.05	§ 4.1.3
Section classEntered. Nothing here can derive it: a shear area carries no information about the slenderness of the plates it is made of.	– = 1	§ t.4
Resistance model	PLASTIC. §5.1.4.1 (41) takes the whole shear area at τ_y,k: the section is stocky enough to yield right through A_v before anything buckles, so the stress is taken as uniform over it.	§ 5.1.4.1

Calculation: Check 1 — stability of the web
Step	Value	Ref.
Web slendernessThe ratio (42) is written against. It is NOT the flat width of table 5a: this one runs between the flange mid-planes and takes no account of the fillets.	(h − t_f)/t_w = 31.67	§ (42)
Limit√(4·210,000/235). 59,8 in S 235 and 48,6 in S 355 — the stronger the steel, the sooner its web buckles in shear rather than yielding.	√(4E/f_y,k) = 59.79	§ (42)
Web stability — the condition (41) is granted under(h − t_f)/t_w ≤ √(4E/f_y,k)	η = 0.530 · Pass53 % of the capacity used	§ (42)

Calculation: Check 2 — shear resistance
Step	Value	Ref.
Shear areaEntered directly.	A_v = 2,874 mm²	§ 5.1.4.2
A_w, were the section class 3The two differ by a factor of 1.61 here. That factor is what the class is worth in shear, and it is larger than what the class is worth in bending.not used	A_w = 1,781 mm²	§ 5.2.4
Shear resistanceV_R = τ_y,k·A_v, the plastic model: the stress is taken as uniform over the area at τ_y,k.	V_R = 388.0 kN	§ (41)
Design shear resistanceV_Rd = V_R/γ_M1	V_Rd = 369.5 kN	§ (41)
Shear area required(41) turned round: A_v,req = γ_M1·V_Ed/τ_y,k. The sizing form, the one used to CHOOSE a section rather than to check one.	A_v,req = 2,722 mm²	§ (41)
Shear resistance of the sectionV_Ed ≤ V_Rd	η = 0.947 · Pass95 % of the capacity used	§ (41)

How this module is verified

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

Questions

When should I use this module instead of the others?

When your section is not a rolled I or H, a welded I or H or a solid rectangle, or when you want to impose a shear area of your own. The price is that the area and the class are yours to justify.

What is the class worth in shear?

In the worked example A_v and A_w differ by a factor of 1.61. That factor is what the class is worth in shear: a class 3 section is checked on A_w, not A_v.

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: 27 October 2026

Steel shear resistance from an entered shear area to SIA 263

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