Rolled steel profile with cover plates to SIA 263

Sizes symmetric welded cover plates on a rolled profile in bending to SIA 263:2013: the reinforced and the unreinforced zones, the length of the plates and the shear of the web. Every step shows its formula and the clause it comes from.

Basic SIA 263:2013

Part of the Basic plan

This module is included in the Basic plan and above, with its calculation note.

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About this check

clauses, worked example and questions

What it checks

Bare profile from the five dimensions h, b, t_w, t_f and r: A, I_y and W_y,pl, fillets included, then M_pl = f_y,k · W_y,pl and its design resistance M_pl/γ_M1.
Plate width required: the additional moment required, M_sem,pl,req ≥ γ_M1 · M_Ed − M_pl, and the plate width required at the chosen thickness, b_sem,req = M_sem,pl,req/(f_y,sem · t_sem · (h + t_sem)).
Reinforced zone: M_Ed ≤ (M_pl + M_pl,sem)/γ_M1, with F_sem = f_y,sem · b_sem · t_sem, the lever a = h + t_sem and M_pl,sem = F_sem · a.
Unreinforced zone: M_Ed,0 ≤ M_pl/γ_M1, on the bare profile outside l₀.
Length of the plates: the introduction length at each end, Δl = (F_sem/f_w − b_sem)/2, since the weld across the end of the plate counts with the two side welds, and the total length l_sem = l₀ + 2Δl.
Shear of the web, equation (41): V_Ed ≤ V_R/γ_M1, with V_R = τ_y,k · A_v and A_v = A − 2b · t_f + (t_w + 2r) · t_f, equation (42a), of the profile alone.

What you enter and what you get

You enter f_y,k of the profile and f_y,sem of the plates from table 1 of your copy of SIA 263, the profile dimensions without the plates, the plate thickness t_sem and width b_sem, and γ_M1 (clause 4.1.3). You enter M_Ed in the reinforced zone, the largest moment M_Ed,0 outside it, and the theoretical length l₀ read off the moment envelope. Δl follows from the weld resistance per unit length f_w, or is entered directly. For the shear check you enter τ_y,k and V_Ed.

You get the profile properties, the depth with the plates h + 2t_sem, the required width, the plate force, the plate length 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

Symmetric reinforcement only: one plate of the same size on each flange. The neutral axis does not move, so M_pl and M_pl,sem simply add. A plate on one flange only needs the section worked out afresh.
The reinforced section is taken as fully plastified.
l₀ is the length over which M_Ed exceeds M_pl/γ_M1. No formula gives it; it depends on the shape of the moment diagram.
The cover plates add nothing to the shear resistance. The web carries the shear alone.
When V_Ed is at or below half of V_R/γ_M1, no interaction between moment and shear is called for.
The class of the reinforced section is not re-examined; the class of the profile applies.
Lateral-torsional buckling is assumed prevented by construction measures.
The weld is not designed. It must transmit F_sem over b_sem + 2Δl and carry the longitudinal shear flow along the whole plate.
Fatigue is not covered. The end of a welded cover plate is among the worst fatigue details of a steel beam.

Worked example

A rolled profile with h = 114 mm, b = 120 mm, t_w = 5 mm, t_f = 8 mm and r = 12 mm, reinforced with two 100 × 10 mm plates, all in f_y,k = 235 N/mm², under M_Ed = 47.6 kNm in the reinforced zone and M_Ed,0 = 26.0 kNm outside it, with l₀ = 4,300 mm, f_w = 839 N/mm and V_Ed = 44.1 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
Overall depth of the profile, WITHOUT the cover plates h	114 mm
Flange width b	120 mm
Web thickness t_w	5 mm
Flange thickness t_f	8 mm
Root radius r	12 mm
Yield strength of the cover plate f_y,sem	235 N/mm²	§ t.1
Plate thickness t_sem	10 mm
Plate width b_sem	100 mm
Resistance factor γ_M1	1.05	§ 4.1.3
Design moment in the reinforced zone M_Ed	47.6 kNm
Largest design moment outside the reinforced zone M_Ed,0	26.0 kNm
Theoretical length of the plates l₀	4300 mm
Introduction length Δl obtained Δl	From the weld resistance per unit length
Weld resistance per unit length f_w	839 N/mm
Shear check	Considered
Characteristic shear yield strength τ_y,k	135 N/mm²	§ t.1
Design shear force V_Ed	44.1 kN

Calculation: Common values
Step	Value	Ref.
Yield strength of the profile	f_y,k = 235 N/mm²	§ t.1
Area of the profileComputed from h, b, t_w, t_f and r: the three plates plus four fillets of (1 − π/4)·r². Compare it with the table row of the profile you mean. The cover plates are not in it.	A = 2,534 mm²
Second moment of area of the profileComputed from the same five dimensions, fillets included. Reported so the profile can be recognised; not used by the plastic checks.	I_y = 6,060,890 mm⁴
Plastic section modulus of the profileW_y,pl = 2·S, twice the first moment of the half-section about the neutral axis, fillets included. Computed, not entered: of the bare profile, since the plates are added to its plastic moment rather than folded into it.	W_y,pl = 119,491 mm³
Depth with the cover platesThe overall depth of the reinforced section — the profile depth h entered, plus one plate on each flange.	h + 2·t_sem = 134.0 mm
Plastic moment of the profileM_pl = f_y,k·W_y,pl	M_pl = 28.080 kNm
Design resistance of the profile aloneThis is the level the cover plates have to make up the difference to, and it is also the line on the moment envelope that fixes l₀.	M_pl/γ_M1 = 26.74 kNm

Calculation: Check 1 — reinforced zone
Step	Value	Ref.
Additional moment requiredM_sem,pl ≥ γ_M1·M_Ed − M_pl = 1.05 × 47.60 − 28.08.	M_sem,pl,req = 21.90 kNm
Plate width required at this thicknessb_sem ≥ M_sem,pl,req/(f_y,sem·t_sem·(h + t_sem)). Thickness comes from stock, so it is the width that is cut to suit.	b_sem,req = 75.2 mm
Force in each plateF_sem = f_y,sem·b_sem·t_sem, on the assumption that the reinforced section is fully plastified. This is also the force the weld has to introduce.	F_sem = 235.0 kN
Lever between the plate centresa = h + t_sem: each plate centre sits half its own thickness outside the profile depth.	a = 124.0 mm
Plastic moment of the platesM_pl,sem = F_sem·a. The two plates form a couple, so nothing of the profile enters this term.	M_pl,sem = 29.140 kNm
Resistance of the reinforced sectionThe resistances simply add — which is the whole convenience of the method, and holds because the reinforcement is symmetric and the neutral axis has not moved.	(M_pl + M_pl,sem)/γ_M1 = 54.50 kNm
Gain over the bare profile	– = 103.8 %
Reinforced zoneM_Ed ≤ (M_pl + M_pl,sem)/γ_M1	η = 0.873 · Pass87 % of the capacity used

Calculation: Check 2 — unreinforced zone
Step	Value	Ref.
Unreinforced zoneM_Ed,0 ≤ M_pl/γ_M1	η = 0.972 · Pass97 % of the capacity used

Calculation: Length of the plates
Step	Value	Ref.
Theoretical lengthFrom the bending-moment envelope, entered.	l₀ = 4,300 mm
Weld length required at each endF_sem divided by the resistance per unit length. The weld across the end of the plate counts as well as the two along its sides, so the length available is b_sem + 2Δl and not 2Δl alone.	b_sem + 2Δl = 280 mm
Introduction length at each endΔl = (F_sem/f_w − b_sem)/2.	Δl = 90 mm
Total length of each platel_sem = l₀ + 2Δl. The extension is not there for bending — it is there so the weld can get the force into the plate before the plate is asked to carry it.	l_sem = 4,480 mm

Calculation: Check 3 — shear of the web
Step	Value	Ref.
Effective shear areaA_v = A − 2b·t_f + (t_w + 2r)·t_f, of the PROFILE, with A computed above. Cover plates lie in the plane of the flanges and contribute nothing to the shear.	A_v = 846 mm²	§ (42a)
Design shear resistanceV_R = τ_y,k·A_v	V_R/γ_M1 = 108.7 kN	§ (41)
Shear resistance of the webV_Ed ≤ V_R/γ_M1	η = 0.406 · Pass41 % 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

Why are the plates longer than l_0?

A plate cannot carry F_sem where it begins. The welds have to hand the force into it, and that takes weld length. The plate therefore starts Δl before the point where it is needed, at each end.

What if the unreinforced zone fails?

Then l₀ is too short. The plates have to reach further along the envelope; a thicker plate does not help.

Which plan includes this module?

The Basic plan, with its calculation note.

Norm: SIA 263:2013 · Page updated: 26 October 2026