Welded steel plate girder to SIA 263
Checks a doubly symmetric welded plate girder, hybrid or not, to SIA 263:2013: section class, bending, shear with the tension field, their interaction, the compressed flange and the deflection. 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.
About this check
clauses, worked example and questions
Checks a doubly symmetric welded plate girder, hybrid or not, to SIA 263:2013: section class, bending, shear with the tension field, their interaction, the compressed flange and the deflection. Every step shows its formula and the clause it comes from.
What it checks
- Section class, tables 5a and 5b: the compressed flange c/tf, with c = (b − tw)/2 − s measured from the weld, and the web in bending bw/tw, with bw = h − 2tf − 2s. The limits scale with εf and εw; on a hybrid section the web limits follow the web grade.
- Bending: MEd ≤ MR/γM1. Class 1 or 2 takes the larger of the flange couple fy,f · b · tf · (h − tf) and the elastic limit; class 3 the elastic resistance; class 4 the elastic resistance on the effective section. The elastic limit is computed at the flange fibre, fy,f · Wy,el, and at the web fibre, fy,w · Iy/(hw/2). The flange width required, breq, is shown.
- Shear: VEd ≤ VR/γM1, with VR = τy,k · Aw and Aw = (h − tf) · tw. The web thickness required by strength, tw,req, and by slenderness, tw,sl, is shown.
- Shear buckling of the web: the coefficient k (5.34 + 4.00/α² above α = 1, 4.00 + 5.34/α² below), τcr = 0.9 · k · E · (tw/hf)² and λw = √(τy/τcr). Below 0.9 the web is compact and VR = τy · Aw. Above the web slenderness hf/tw at which the band is mobilised, the tension field adds to the resistance.
- Interaction of bending and shear: none is required while MEd stays below Mpl,f/γM1, the moment the flanges carry alone.
- Compressed flange: vertical buckling into the web, hf/tw against a limit of 360 for S 235 and 240 for S 355; and rotation buckling, (b/2)/tf against 17 for S 235 and 14 for S 355.
- Lateral-torsional buckling, optional, by two routes: the critical moment McrD of the full section, giving χD and σD, or the compressed flange as a laterally buckling bar with Weff.
- Web-to-flange shear flow vEd = VEd · Sf/Iy, and the deflection against wlim.
What you enter and what you get
You enter fy,f, fy,w and τy,k from table 1 of your copy of SIA 263, with τy,k from the same row as fy,w, E (clause 2.2), h, b, tf, tw, the weld leg s, the span l, γM1 (clause 4.1.3), MEd and VEd. You choose whether to check shear buckling, with or without transverse stiffeners and for an interior or end panel, the compressed flange instabilities, lateral-torsional buckling, and the deflection (static system, service load, limit).
You get the section properties, the class, the resistance model, the weld shear flow and the utilisation of each check with a pass or fail, and a drawing of the section to scale. With the Basic plan, it prints as an A4 calculation note.
Assumptions and limits
- A doubly symmetric girder of constant section. Varying depth, unequal flanges and box sections are not covered.
- Aw is not the Av of equation (42a), which counts root fillets that a welded section does not have.
- On a hybrid section the flange couple ignores the web, so the check says nothing about whether the web yields.
- Lateral-torsional buckling is assumed prevented by construction measures unless one of the two routes is checked.
- The tension field exists only where the flanges and transverse stiffeners can anchor it.
- The panel next to a support, carrying interaction and support reaction together, is not covered.
- The weld itself is not designed.
Worked example
A hybrid girder with h = 500 mm, flanges 340 × 30 mm in fy,f = 355 N/mm², a web 8 mm thick in fy,w = 235 N/mm², s = 6 mm, under MEd = 1,560 kNm and VEd = 468 kN, without transverse stiffeners, with lateral-torsional buckling prevented, and a deflection check over three continuous 20 m spans under 12 kN/m against l/350. These are the module's own results for its worked example, computed when this page was built.
| Input | Value | Ref. |
|---|---|---|
| Characteristic shear yield strength τy,k | 135 N/mm² | § t.1 |
| Modulus of elasticity E | 210000 N/mm² | § 2.2 |
| Yield strength of the flanges fy,f | 355 N/mm² | § t.1 |
| Yield strength of the web fy,w | 235 N/mm² | § t.1 |
| Overall depth h | 500 mm | |
| Flange width b | 340 mm | |
| Flange thickness tf | 30 mm | |
| Web thickness tw | 8 mm | |
| Weld leg at the web-flange joint s | 6 mm | § t.5b |
| Span l | 20000 mm | |
| Resistance factor γM1 | 1.05 | § 4.1.3 |
| Design bending moment MEd | 1560 kNm | |
| Design shear force VEd | 468 kN | |
| Web voilement in shear | Checked | |
| Transverse stiffeners | No transverse stiffeners | |
| Panel considered | Interior panel | |
| Compressed flange instabilities | Checked | |
| Critical moment of the full section | Prevented by construction measures | |
| Compressed flange as a laterally buckling bar | Prevented by construction measures | |
| Deflection | Considered | |
| Static system | Continuous over equal spans — uniformly distributed load | |
| Number of equal spans n | 3 | |
| Service load, uniformly distributed q | 12 kN/m | |
| Deflection limit, span divided by | 350 |
| Step | Value | Ref. |
|---|---|---|
| Yield strength of the flanges | fy,f = 355 N/mm² | § t.1 |
| Yield strength of the web | fy,w = 235 N/mm² | § t.1 |
| Reduction factor, flangesε = √(235/fy). | εf = 0.8136 | § t.5a |
| Reduction factor, web | εw = 1.0000 | § t.5a |
| Resistance factor | γM1 = 1.05 | § 4.1.3 |
| Area | A = 23,920 mm² | |
| Second moment of areaIy = [b·h³ − (b − tw)·hw³]/12, the difference of two rectangles. Exact for a welded section, which has no fillet. | Iy = 1,184.909 10⁶ mm⁴ | |
| Elastic section modulusWel = 2Iy/h | Wy,el = 4,739,637 mm³ | |
| First moment of one flangeSf = b·tf·(h − tf)/2, about the neutral axis of the whole section. It is what the weld shear flow needs. | Sf = 2,397,000 mm³ | |
| Shear area of the webAw = (h − tf)·tw, for a doubly symmetric girder. It is NOT the Av of SIA 263 (42a) — that expression counts root fillets, which a welded section does not have. | Aw = 3,760 mm² | |
| Depth-to-span ratio1 : 40.0. Usual first sizing: 1/7 to 1/12 for a heavy crane runway, 1/12 to 1/20 for a rail bridge, 1/15 to 1/25 for a hall or building frame and 1/20 to 1/30 for a road bridge. | h/l = 0.02500 | |
| Compressed flange — flat widthc = (b − tw)/2 − s, measured from the weld. | c = 160.00 mm | § t.5b |
| Compressed flange — slendernessLimits 7.32 · 8.14 · 11.39 at εf = 0.814. | c/tf = 5.33 | § t.5b |
| Web in bending — flat widthh − 2tf − 2s | bw = 428.0 mm | § t.5a |
| Web in bending — slendernessLimits 72.0 · 83.0 · 124.0 at εw = 1.000. On a hybrid section the web limits follow the WEB grade, not the flange grade. | bw/tw = 53.50 | § t.5a |
| Section classThe least favourable of the two compressed parts. A plate girder usually cannot reach full plastification at all, so PP and EP are generally out and EE is the normal case — but a shallow girder can reach class 2. | – = Class 1 | § t.5a |
| Step | Value | Ref. |
|---|---|---|
| Resistance modelDecided from the section class. A class 1 or 2 section may take the flange couple — the two flanges at fy forming a couple, the web left to the shear. The elastic limit is also a lower bound of the plastic moment, so whichever of the two is larger is used. | – = Class 1 — plastic admitted | § t.5a |
| Flange width requiredb ≥ γM1·MEd/(fy,f·tf·(h − tf)), the sizing form of the flange couple. Depth and flange thickness are settled first, so it is the width that follows. | breq = 327.2 mm | |
| Flange coupleThe two flanges at fy on the lever between their centres, with the web ignored. It is a plastic model, so it belongs to a class 1 or 2 section, and it is conservative there because the web is left out of a resistance it does contribute to. | fy,f·b·tf·(h − tf) = 1,701.9 kNm | |
| Elastic limit at the flange fibre | fy,f·Wy,el = 1,682.6 kNm | |
| Elastic limit at the web fibreOn a hybrid section the web reaches its own, lower, yield strength at its own extreme fibre — a shorter lever but a smaller fy. Which of the two fibres governs an elastic calculation is not obvious and is worth computing. | fy,w·Iy/(hw/2) = 1,265.7 kNm | |
| Governing resistanceThe flange couple, larger than the elastic limit. | MR/γM1 = 1,620.8 kNm | |
| Bending resistanceMEd ≤ MR/γM1 | η = 0.962 · Pass96 % of the capacity used |
| Step | Value | Ref. |
|---|---|---|
| Web thickness required by strengthtw ≥ γM1·VEd/(τy,k·(h − tf)), on the conservative first assumption that the web takes the whole of the shear. | tw,req = 7.74 mm | |
| Web thickness required by slenderness(h − 2tf − 2s) divided by the class 2 limit of 83.0. Whichever of the two governs settles the plate ordered. | tw,sl = 5.16 mm | § t.5a |
| Ultimate shear resistanceVR = τy,k·Aw with Aw = (h − tf)·tw. | VR = 507.6 kN | |
| Design shear resistance | VR/γM1 = 483.4 kN | |
| Shear resistanceVEd ≤ VR/γM1 | η = 0.968 · Pass97 % of the capacity used |
| Step | Value | Ref. |
|---|---|---|
| Longitudinal shear flowv = VEd·Sf/Iy. It comes from differentiating the flange force N = M·Sf/Iy along the member: dN/dx = V·Sf/Iy, and the equilibrium of a flange over a length dx requires v·dx = dN. | vEd = 946.7 N/mm | |
| Force in one flange at this sectionN = MEd·Sf/Iy. The weld exists to build this up along the member, and v is the rate at which it does so. | N = 3,155.8 kN |
| Step | Value | Ref. |
|---|---|---|
| Static system | – = Continuous over equal spans — uniformly distributed load | |
| Deflection coefficientw = k·q·l⁴/(E·I), that is 1/145.26. | k = 0.0068842 | |
| Deflection | w = 53.1 mm | |
| Limitl/350, the same limits as for a rolled profile. | wlim = 57.1 mm | |
| Deflectionw ≤ wlim | η = 0.930 · Pass93 % of the capacity used |
| Step | Value | Ref. |
|---|---|---|
| Panel aspect ratioWith none, α → ∞ and k → 5,34. α = 3 may nevertheless be taken for the band: tests on unstiffened welded girders showed the tension band never inclines below the value for a fictitious panel of α ≈ 3 — PROVIDED it can be anchored at the beam ends. | α = a/hf = no transverse stiffeners | |
| Shear buckling coefficientFor a panel hinged along all four sides: 4,00 + 5,34/α² below α = 1, 5,34 + 4,00/α² above. | k = 5.340 | |
| Critical shear stress0,9·k·E·(tw/hf)². The 0,9 is π²/(12(1−ν²)) = 0.9038, the constant of plate buckling. | τcr = 292.4 N/mm² | |
| Shear yield strengthfy,w/√3. | τy = 135.7 N/mm² | |
| Panel slenderness√(τy/τcr). Below 0,9 the web is COMPACT and there is no voilement problem at all; between 0,9 and 1,12, τcr exceeds the proportionality limit 0,8·τy and is reduced; above 1,12 no reduction applies. | λw = 0.6812 | |
| Regimeλw < 0,9: VR = τy·Aw, the SAME expression as for a rolled profile. No tension field is needed. | – = compact | |
| Slenderness at which the band is mobilisedWithout vertical stiffeners it is 78 for S 235 and 63 for S 355. Below it the simple resistance suffices; above it, the tension field adds to it. | hf/tw limit = 77.6 | |
| Web slenderness | hf/tw = 58.8 | |
| Pre-critical contributionτcr·hf·tw. | Vcr = 1,099.4 kN | |
| Shear on a compact webVEd ≤ τy·Aw/γM1 | η = 0.963 · Pass96 % of the capacity used |
| Step | Value | Ref. |
|---|---|---|
| Moment the FLANGES alone carryfy·Af·hf, on the EFFECTIVE flange of 340.0 mm — not the gross 340 mm where the flange is reduced. While MEd stays below Mpl,f/γM1 there is NO interaction to check: the flanges carry the moment alone and the web is free to give all of itself to the shear. | Mpl,f = 1,702 kNm | |
| Threshold for interaction | Mpl,f/γa = 1,621 kNm | |
| InteractionMEd = 1,560 kNm stays below Mpl,f/γM1 = 1,621 kNm, so the flanges alone carry the moment and the whole web remains available for shear. | – = Not required |
| Step | Value | Ref. |
|---|---|---|
| Web slendernesshf measured between the MID-PLANES of the flanges. | hf/tw = 58.8 - | |
| Vertical buckling of the flange into the webhf/tw < 238 | η = 0.247 · Pass25 % of the capacity used |
| Step | Value | Ref. |
|---|---|---|
| Half-flange slenderness | (b/2)/tf = 5.67 - | |
| Rotation buckling of the compressed flange(b/2)/tf < 13.6 | η = 0.417 · Pass42 % of the capacity used |
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 does the tension field come into play?
When the web slenderness hf/tw exceeds the limit at which the band is mobilised: without vertical stiffeners, 78 for S 235 and 63 for S 355. Below it the simple shear resistance suffices.
Why does the rotation limit exceed the class limit?
They answer different questions. The class limit chooses the method of calculation; the rotation limit asks only whether the whole flange width works. An outstand can be class 4 and still be fully effective.
Which plan includes this module?
The Basic plan, with its calculation note.
Related
- Rolled I or H section in bending
- Castellated beam in bending
- Steel bending: all sections
- SIA 263: modules and clauses
Norm: SIA 263:2013 · Page updated: 26 October 2026