Checks a solid rectangular steel plate or bar under a moment and a shear force acting together, to SIA 263:2013: the shear force alone, then the moment on the lower of two plastic models. Every step shows its formula and the clause it comes from.

What it checks

Section: A = b · h and W_pl = b · h²/4, with the plastic moment without shear M_pl,Rd = f_y,k · W_pl/γ_M1.
Class and resistance model, table 4: class 1 by construction, because a solid section has no plate to classify. The plastic resistance is taken.
The shear force alone, check 1: τ_Ed = V_Ed/A ≤ τ_y,k/γ_M1. A section that fails it has nothing left for the moment.
Model 1, reduced yield strength: f_yr = √(f_y,k² − 3τ²), with τ = V_Ed/A, is what is left for a direct stress at every point. The whole rectangle works at that strength: M_R,1 = f_yr · b · h²/4/γ_M1.
Model 2, central shear strip: a strip of depth 2a at τ_y,k carries all the shear and nothing else, from τ = V_Ed/(2a · b) = τ_y,k. The two blocks outside it carry the moment at full yield strength: M_R,2 = f_y,k · b · (h²/4 − a²)/γ_M1.
The moment with the shear acting, check 2: M_y,Ed ≤ M_R, the lower of M_R,1 and M_R,2. The module names the model that governs.
Elastic comparison stress σ_g,Ed, reported for reference beside the plastic check.

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, the thickness b of the plate or bar, its depth h in the plane of bending, and γ_M1 (clause 4.1.3). You enter the design moment M_y,Ed and the design shear force V_Ed, both as magnitudes and in the same plane.

You get the area, the plastic modulus, the plastic moment without shear, the average shear stress, both plastic models with their moment resistances, the model that governs, 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. Bending is in the plane of the depth h, which is also the direction of the shear force.
Both plastic models are statically admissible, so both are safe. The check is made on the lower of the two, and both are reported.
The shear stress of check 1 and of model 1 is the average V_Ed/A.
τ_y,k is entered from table 1, not derived as f_y,k/√3. The tabulated values are not exactly f_y,k/√3: over the eight cells of table 1 the division gives the higher, unconservative value in six.
A section check only. Lateral-torsional buckling, the introduction of concentrated forces and torsion are not covered. An axial force acting with the moment and the shear is the subject of the bending, axial force and shear modules.

Worked example

A plate b = 20 mm thick and h = 300 mm deep, in f_y,k = 235 N/mm² and τ_y,k = 135 N/mm², under M_y,Ed = 40 kN·m and V_Ed = 200 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
Thickness b	20 mm
Depth in the plane of bending h	300 mm
Resistance factor γ_M1	1.05	§ 4.1.3
Design bending moment M_y,Ed	40 kN·m
Design shear force V_Ed	200 kN

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
Resistance factor	γ_M1 = 1.05	§ 4.1.3
Section classClass 1 by construction: a solid section has no plate to classify.	– = 1	§ t.4
Resistance model	PLASTIC. Table 4 allows a plastic resistance to a class 1 section.	§ t.4
AreaA = b·h	A = 6,000 mm²
Plastic section modulusW_pl = b·h²/4	W_pl = 450,000 mm³
Plastic moment resistance, no shearf_y,k·W_pl/γ_M1.	M_pl,Rd = 100.71 kN·m	§ 4.1.3

Calculation: Check 1 — the shear force alone
Step	Value	Ref.
Average shear stress	τ_Ed = V_Ed/A = 33.33 N/mm²
Shear alone, before any interactionτ_Ed ≤ τ_y,k/γ_M1	η = 0.259 · Pass26 % of the capacity used

Calculation: Check 2 — the moment with the shear acting
Step	Value	Ref.
Model 1 — reduced yield strengthf_yr = √(f_y,k² − 3τ²), τ = V_Ed/A: what is left for a direct stress at every point.	f_yr = 227.80 N/mm²
Model 1 — moment resistancef_yr·b·h²/4/γ_M1: the whole rectangle at the reduced yield strength.	M_R,1 = 97.63 kN·m
Model 2 — depth of the shear strip at τ_y,kFrom τ = V_Ed/(2a·b) = τ_y,k. The strip carries all the shear and nothing else.	2a = 74.1 mm
Model 2 — moment resistancef_y,k·b·(h²/4 − a²)/γ_M1: the two blocks outside the strip at full f_y.	M_R,2 = 94.57 kN·m
Which model is verified	The LOWER of the two, here model 2, the central shear strip at 94.57 kN·m against 97.63 kN·m for the other.
Moment with the shear force actingM_y,Ed ≤ M_R	η = 0.423 · Pass42 % of the capacity used
Elastic comparison stress, for referenceWhat the elastic check would give, against 223.8 N/mm². Not the check for a class 1 section.not used	σ_g,Ed = 133.33 N/mm²

How this module is verified

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

Questions

Which of the two plastic models governs?

The one with the lower moment resistance, and it depends on the forces. In the worked example model 2, the central shear strip, gives 94.57 kN·m against 97.63 kN·m for the reduced yield strength. Check 2 is made on model 2, with η = 0.423.

Why is the plate or bar not classified?

A solid rectangle has no slender plate to classify. It is class 1 by construction, so the plastic models are available without condition.

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: 5 November 2026

Solid steel rectangle in bending and shear to SIA 263

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