Steel truss of constant depth to SIA 263
Analyses a steel truss of constant depth at the ultimate limit state and verifies every member to SIA 263:2013: the chords, diagonals and verticals in tension or in compression with buckling, and the deflection. Every step shows its formula and the clause it comes from.
Expert SIA 263:2013
Part of the Expert plan
This module is included in the Expert plan and above, with its calculation note.
About this check
clauses, worked example and questions
Analyses a steel truss of constant depth at the ultimate limit state and verifies every member to SIA 263:2013: the chords, diagonals and verticals in tension or in compression with buckling, and the deflection. Every step shows its formula and the clause it comes from.
What it checks
The truss as a whole
- Geometry: the panel length a = l/n, the ratio h/l (about 1/10 to 1/15 for a first sizing) and the inclination α of the first diagonal, economical between 35° and 55°.
- Determinacy: m + 3 = 2j, so the forces do not depend on the sections.
- Loads and reactions: γG · G + γQ · Q at the nodes, the self-weight Ga = Σ A · l · 78.5 kN/m³, and the reactions Av, Ah and Bv.
- Axial force in every member, in tension or compression.
Each member group (top chord, bottom chord, diagonals, verticals)
- Section class, tables 5a and 5b, in compression, from the least favourable wall: flange and web of a rolled I or H, or the walls (h − 3t)/t of a hollow section.
- Buckling curves: for a rolled I, b in the plane and c out of it; a for a hot-finished hollow section, c for a cold-formed one.
- Tension, clause 4.1.3: NEd ≤ NRd = A · fy/γM1 on the gross section.
- Flexural buckling, clause 4.5.1: NEd ≤ χ · A · fy/γM1, from λ̄in and λ̄out, the smaller χ governing. The governing member of each group is reported.
- With rigid nodes, or loads along a chord, the members are checked for compression or tension with bending.
Deflection: the largest w, counting the elongation and shortening of every member, against wlim = l divided by the limit you enter.
What you enter and what you get
You enter fy,k from table 1 of your copy of SIA 263, E and G (clause 2.2), the pattern (Pratt, Howe or Warren), l, the depth h between chord axes, n, support at the bottom or top chord, and pinned or rigid nodes. For the loads: γG, γQ, the characteristic gtop, qtop, gbot and qbot per metre of span, at the nodes or along the chord, the self-weight and up to six nodal forces. For each group: the section (rolled I or H, rectangular or circular hollow section, two angles or two channels back to back), the buckling length in the plane and, for the chords, the distance lout between lateral supports. Then γM1 (clause 4.1.3) and the deflection limit.
You get the member forces, every check with its utilisation and a pass or fail, and the deflection. With the Expert plan, it prints as an A4 calculation note.
Assumptions and limits
- The truss is ideal: pinned nodes, member axes meeting at the nodes, straight members, loads at the nodes.
- The in-plane buckling length is the system length, always safe, or 0.9 l for a chord and 0.8 l for a diagonal or vertical.
- Out of its plane the compressed chord buckles sideways between lateral supports; lout is also the length for lateral-torsional buckling of an I chord.
- At bolted joints the net section must also be checked; it is not done here.
- One section serves every member of a group. The connections are not designed.
Worked example
A Pratt truss of 18 m span and 2 m depth in six panels, pinned, supported at its bottom chord and loaded there at the nodes with gbot = 15 kN/m and qbot = 18 kN/m plus self-weight, in fy,k = 235 N/mm², with rolled I or H chords, hot-finished hollow sections 150 × 150 × 6.3 mm and 120 × 120 × 6.3 mm for the diagonals and verticals, lateral supports every 3,000 mm (top) and 6,000 mm (bottom), and a limit of l/300 under the variable load. These are the module's own results for its worked example, computed when this page was built.
| Input | Value | Ref. |
|---|---|---|
| Characteristic yield strength fy,k | 235 N/mm² | § t.1 |
| Modulus of elasticity E | 210000 N/mm² | § 2.2 |
| Shear modulus G | 81000 N/mm² | § 2.2 |
| Pattern of the web | Pratt — verticals, diagonals in tension under gravity load | |
| Span between the supports l | 18000 mm | |
| Depth between the chord axes h | 2000 mm | |
| Number of panels n | 6 | |
| Supported | At the ends of the bottom chord | |
| Nodes | Articulated — pinned, the members carry axial force only | |
| Load factor, permanent actions γG | 1.35 | |
| Load factor, variable actions γQ | 1.5 | |
| Distributed loads reach the chords | At the nodes only — through purlins or cross-beams | |
| Permanent load on the top chord gtop | 0 kN/m | |
| Variable load on the top chord qtop | 0 kN/m | |
| Permanent load on the bottom chord gbot | 15 kN/m | |
| Variable load on the bottom chord qbot | 18 kN/m | |
| Self-weight of the members | Included — computed from the sections | |
| Number of nodal forces | 0 | |
| Resistance factor γM1 | 1.05 | § 4.1.3 |
| Deflection | Considered | |
| Deflection under | Variable load only | |
| Deflection limit, span divided by | 300 | |
| Type of section | Rolled I or H | |
| Overall depth h | 210 mm | |
| Flange width b | 220 mm | |
| Web thickness tw | 7 mm | |
| Flange thickness tf | 11 mm | |
| Root radius r | 18 mm | |
| Buckling length in the plane of the truss | System length between the nodes | |
| Distance between lateral supports of the chord lout | 3000 mm | |
| Type of section | Rolled I or H | |
| Overall depth h | 171 mm | |
| Flange width b | 180 mm | |
| Web thickness tw | 6 mm | |
| Flange thickness tf | 9.5 mm | |
| Root radius r | 15 mm | |
| Buckling length in the plane of the truss | System length between the nodes | |
| Distance between lateral supports of the chord lout | 6000 mm | |
| Type of section | Rectangular hollow section | |
| Depth, in the plane of the truss h | 150 mm | |
| Width, across the plane of the truss b | 150 mm | |
| Wall thickness t | 6.3 mm | |
| Manufactured | Hot-finished | |
| Buckling length in the plane of the truss | System length between the nodes | |
| Type of section | Rectangular hollow section | |
| Depth, in the plane of the truss h | 120 mm | |
| Width, across the plane of the truss b | 120 mm | |
| Wall thickness t | 6.3 mm | |
| Manufactured | Hot-finished | |
| Buckling length in the plane of the truss | System length between the nodes |
| Step | Value | Ref. |
|---|---|---|
| Characteristic yield strength | fy,k = 235 N/mm² | § t.1 |
| Panel lengtha = l/n | a = 3,000.0 mm | |
| Depth-to-span ratio1 : 9.0. Usual first sizing: about 1/10 to 1/15 for a truss of constant depth. | h/l = 0.1111 | |
| Inclination of the diagonalsOf the first diagonal. Between 35° and 55° is economical. | α = 33.7 ° | |
| Nodes and membersm + 3 = 2j: statically determinate — the forces do not depend on the sections. | j · m = 14 nodes · 25 members | |
| Self-weight of the membersΣ A·l·78,5 kN/m³ over all members — 1.38 kN/m of span. | Ga = 24.78 kN | |
| Total vertical design loadγG·G + γQ·Q = 1.35·G + 1.50·Q, all nodes. | ΣFd = 884.0 kN | |
| Reaction at the left support, verticalPin at node b0. | Av = 442.0 kN | |
| Reaction at the left support, horizontal | Ah = 0.0 kN | |
| Reaction at the right supportRoller at node b6. | Bv = 442.0 kN |
| Step | Value | Ref. |
|---|---|---|
| TC1 — top chord, t0–t1Compression, length 3,000 mm. | NTC1 = -552.1 kN | |
| TC2 — top chord, t1–t2Compression, length 3,000 mm. | NTC2 = -883.4 kN | |
| TC3 — top chord, t2–t3Compression, length 3,000 mm. | NTC3 = -993.8 kN | |
| TC4 — top chord, t3–t4Compression, length 3,000 mm. | NTC4 = -993.8 kN | |
| TC5 — top chord, t4–t5Compression, length 3,000 mm. | NTC5 = -883.4 kN | |
| TC6 — top chord, t5–t6Compression, length 3,000 mm. | NTC6 = -552.1 kN | |
| BC1 — bottom chord, b0–b1Zero-force member under this loading. | NBC1 = 0.0 kN | |
| BC2 — bottom chord, b1–b2Tension, length 3,000 mm. | NBC2 = 552.1 kN | |
| BC3 — bottom chord, b2–b3Tension, length 3,000 mm. | NBC3 = 883.4 kN | |
| BC4 — bottom chord, b3–b4Tension, length 3,000 mm. | NBC4 = 883.4 kN | |
| BC5 — bottom chord, b4–b5Tension, length 3,000 mm. | NBC5 = 552.1 kN | |
| BC6 — bottom chord, b5–b6Zero-force member under this loading. | NBC6 = 0.0 kN | |
| D1 — diagonal, t0–b1Tension, length 3,606 mm. | ND1 = 663.5 kN | |
| D2 — diagonal, t1–b2Tension, length 3,606 mm. | ND2 = 398.1 kN | |
| D3 — diagonal, t2–b3Tension, length 3,606 mm. | ND3 = 132.7 kN | |
| D4 — diagonal, b3–t4Tension, length 3,606 mm. | ND4 = 132.7 kN | |
| D5 — diagonal, b4–t5Tension, length 3,606 mm. | ND5 = 398.1 kN | |
| D6 — diagonal, b5–t6Tension, length 3,606 mm. | ND6 = 663.5 kN | |
| V1 — vertical, b0–t0Compression, length 2,000 mm. | NV1 = -370.1 kN | |
| V2 — vertical, b1–t1Compression, length 2,000 mm. | NV2 = -223.9 kN | |
| V3 — vertical, b2–t2Compression, length 2,000 mm. | NV3 = -76.7 kN | |
| V4 — vertical, b3–t3Compression, length 2,000 mm. | NV4 = -2.3 kN | |
| V5 — vertical, b4–t4Compression, length 2,000 mm. | NV5 = -76.7 kN | |
| V6 — vertical, b5–t5Compression, length 2,000 mm. | NV6 = -223.9 kN | |
| V7 — vertical, b6–t6Compression, length 2,000 mm. | NV7 = -370.1 kN |
| Step | Value | Ref. |
|---|---|---|
| Top chord — area | A = 6,434 mm² | |
| Top chord — second moment of area in the plane | Iin = 54,093,844 mm⁴ | |
| Top chord — second moment of area out of the plane | Iout = 19,542,438 mm⁴ | |
| Top chord — radii of gyration | iin · iout = 91.7 · 55.1 mm | |
| Top chord — flange c/tfLimits 9.0 · 10.0 · 14.0 for class 1, 2, 3 in compression. | – = 8.0 | § t.5b |
| Top chord — web c/twLimits 33.0 · 38.0 · 42.0 for class 1, 2, 3 in compression. | – = 21.7 | § t.5a |
| Top chord — section classIn compression, the least favourable wall. | – = Class 1 | § t.5a |
| Top chord — buckling curvesRolled I: b about the strong axis (in the plane), c about the weak one. | – = in plane b · out of plane c | |
| Top chord — buckling length factor in the planeThe system length. | βin = 1.00 | |
| Top chord — buckling length out of the planeThe distance between lateral supports. | lK,out = 3,000 mm |
| Step | Value | Ref. |
|---|---|---|
| TC1 — flexural bucklingNEd ≤ χ·A·fy/γM1 | η = 0.481 · Pass48 % of the capacity used | § 4.5.1 |
| TC2 — flexural bucklingNEd ≤ χ·A·fy/γM1 | η = 0.769 · Pass77 % of the capacity used | § 4.5.1 |
| TC3 — flexural bucklingNEd ≤ χ·A·fy/γM1 | η = 0.865 · Pass87 % of the capacity used | § 4.5.1 |
| TC4 — flexural bucklingNEd ≤ χ·A·fy/γM1 | η = 0.865 · Pass87 % of the capacity used | § 4.5.1 |
| TC5 — flexural bucklingNEd ≤ χ·A·fy/γM1 | η = 0.769 · Pass77 % of the capacity used | § 4.5.1 |
| TC6 — flexural bucklingNEd ≤ χ·A·fy/γM1 | η = 0.481 · Pass48 % of the capacity used | § 4.5.1 |
| Top chord — governing memberl = 3,000 mm; lK,in = 3,000 mm, λ̄in = 0.348, χin = 0.946 (curve b); lK,out = 3,000 mm, λ̄out = 0.580, χout = 0.797 (curve c). | – = TC3 | § 4.5.1 |
| Step | Value | Ref. |
|---|---|---|
| Bottom chord — area | A = 4,525 mm² | |
| Bottom chord — second moment of area in the plane | Iin = 25,101,340 mm⁴ | |
| Bottom chord — second moment of area out of the plane | Iout = 9,244,525 mm⁴ | |
| Bottom chord — radii of gyration | iin · iout = 74.5 · 45.2 mm | |
| Bottom chord — flange c/tfLimits 9.0 · 10.0 · 14.0 for class 1, 2, 3 in compression. | – = 7.6 | § t.5b |
| Bottom chord — web c/twLimits 33.0 · 38.0 · 42.0 for class 1, 2, 3 in compression. | – = 20.3 | § t.5a |
| Bottom chord — section classIn compression, the least favourable wall. | – = Class 1 | § t.5a |
| Bottom chord — buckling curvesRolled I: b about the strong axis (in the plane), c about the weak one. | – = in plane b · out of plane c | |
| Bottom chord — buckling length factor in the planeThe system length. | βin = 1.00 | |
| Bottom chord — buckling length out of the planeThe distance between lateral supports. | lK,out = 6,000 mm |
| Step | Value | Ref. |
|---|---|---|
| Bottom chord — design tension resistanceA·fy/γM1 on the gross section. At bolted joints the NET section must also be checked — not done here. | NRd = 1,012.8 kN | § 4.1.3 |
| BC2 — tensionNEd ≤ NRd | η = 0.545 · Pass55 % of the capacity used | § 4.1.3 |
| BC3 — tensionNEd ≤ NRd | η = 0.872 · Pass87 % of the capacity used | § 4.1.3 |
| BC4 — tensionNEd ≤ NRd | η = 0.872 · Pass87 % of the capacity used | § 4.1.3 |
| BC5 — tensionNEd ≤ NRd | η = 0.545 · Pass55 % of the capacity used | § 4.1.3 |
| Step | Value | Ref. |
|---|---|---|
| Diagonals — area | A = 3,621 mm² | |
| Diagonals — second moment of area in the plane | Iin = 12,486,868 mm⁴ | |
| Diagonals — second moment of area out of the plane | Iout = 12,486,868 mm⁴ | |
| Diagonals — radii of gyration | iin · iout = 58.7 · 58.7 mm | |
| Diagonals — wall (h − 3t)/tLimits 33.0 · 38.0 · 42.0 for class 1, 2, 3 in compression. | – = 20.8 | § t.5a |
| Diagonals — wall (b − 3t)/tLimits 33.0 · 38.0 · 42.0 for class 1, 2, 3 in compression. | – = 20.8 | § t.5a |
| Diagonals — section classIn compression, the least favourable wall. | – = Class 1 | § t.5a |
| Diagonals — buckling curvesHot-finished hollow section: a. | – = in plane a · out of plane a | |
| Diagonals — buckling length factor in the planeThe system length. | βin = 1.00 |
| Step | Value | Ref. |
|---|---|---|
| Diagonals — design tension resistanceA·fy/γM1 on the gross section. At bolted joints the NET section must also be checked — not done here. | NRd = 810.5 kN | § 4.1.3 |
| D1 — tensionNEd ≤ NRd | η = 0.819 · Pass82 % of the capacity used | § 4.1.3 |
| D2 — tensionNEd ≤ NRd | η = 0.491 · Pass49 % of the capacity used | § 4.1.3 |
| D3 — tensionNEd ≤ NRd | η = 0.164 · Pass16 % of the capacity used | § 4.1.3 |
| D4 — tensionNEd ≤ NRd | η = 0.164 · Pass16 % of the capacity used | § 4.1.3 |
| D5 — tensionNEd ≤ NRd | η = 0.491 · Pass49 % of the capacity used | § 4.1.3 |
| D6 — tensionNEd ≤ NRd | η = 0.819 · Pass82 % of the capacity used | § 4.1.3 |
| Step | Value | Ref. |
|---|---|---|
| Verticals — area | A = 2,865 mm² | |
| Verticals — second moment of area in the plane | Iin = 6,192,443 mm⁴ | |
| Verticals — second moment of area out of the plane | Iout = 6,192,443 mm⁴ | |
| Verticals — radii of gyration | iin · iout = 46.5 · 46.5 mm | |
| Verticals — wall (h − 3t)/tLimits 33.0 · 38.0 · 42.0 for class 1, 2, 3 in compression. | – = 16.0 | § t.5a |
| Verticals — wall (b − 3t)/tLimits 33.0 · 38.0 · 42.0 for class 1, 2, 3 in compression. | – = 16.0 | § t.5a |
| Verticals — section classIn compression, the least favourable wall. | – = Class 1 | § t.5a |
| Verticals — buckling curvesHot-finished hollow section: a. | – = in plane a · out of plane a | |
| Verticals — buckling length factor in the planeThe system length. | βin = 1.00 |
| Step | Value | Ref. |
|---|---|---|
| V1 — flexural bucklingNEd ≤ χ·A·fy/γM1 | η = 0.616 · Pass62 % of the capacity used | § 4.5.1 |
| V2 — flexural bucklingNEd ≤ χ·A·fy/γM1 | η = 0.373 · Pass37 % of the capacity used | § 4.5.1 |
| V3 — flexural bucklingNEd ≤ χ·A·fy/γM1 | η = 0.128 · Pass13 % of the capacity used | § 4.5.1 |
| V4 — flexural bucklingNEd ≤ χ·A·fy/γM1 | η = 0.004 · Pass0 % of the capacity used | § 4.5.1 |
| V5 — flexural bucklingNEd ≤ χ·A·fy/γM1 | η = 0.128 · Pass13 % of the capacity used | § 4.5.1 |
| V6 — flexural bucklingNEd ≤ χ·A·fy/γM1 | η = 0.373 · Pass37 % of the capacity used | § 4.5.1 |
| V7 — flexural bucklingNEd ≤ χ·A·fy/γM1 | η = 0.616 · Pass62 % of the capacity used | § 4.5.1 |
| Verticals — governing memberl = 2,000 mm; lK,in = 2,000 mm, λ̄in = 0.458, χin = 0.937 (curve a); lK,out = 2,000 mm, λ̄out = 0.458, χout = 0.937 (curve a). | – = V1 | § 4.5.1 |
| Step | Value | Ref. |
|---|---|---|
| Largest deflectionAt node b3, under the variable characteristic load — the elongation and shortening of every member, from the same stiffness analysis. | w = 15.05 mm | |
| Limitl/300 | wlim = 60.00 mm | |
| Deflectionw ≤ wlim | η = 0.251 · Pass25 % 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
Why does the out-of-plane direction govern the top chord?
Its lateral supports are further apart than its nodes, and a rolled I buckles about its weak axis on curve c. In the worked example λ̄out = 0.580 against λ̄in = 0.348.
Do the member forces depend on the sections?
Only through the self-weight, which is computed from the sections. The truss is statically determinate, so the forces follow from equilibrium.
Which plan includes this module?
The Expert plan, with its calculation note.
Related
- Triangular steel roof truss
- Rolled steel section in compression
- Steel trusses: all shapes
- SIA 263: modules and clauses
Norm: SIA 263:2013 · Page updated: 29 October 2026