Steel Truss Design to DBN V.2.6-198:2014: Member Forces, Effective Lengths, Buckling, Deflection

What is checked in a truss

For a flat truss with pinned joints the following checks are made:

• member forces from the joint loads (static analysis);
• strength of tension members (cl. 8.1.1, formula 8.1);
• buckling of compression members (cl. 8.1.3, formulas 8.3–8.5) with effective lengths from Table 13.1;
• limiting slenderness of compression and tension members (Tables 13.9 and 13.10);
• truss deflection (DSTU B V.1.2-3:2006).

Joints and welds are checked separately. Where the web members are attached directly to the chords, such a check is mandatory under cl. 17.3.5 of DBN (Annex F is informative). Joints are outside the scope of this article.

The design model has pinned joints and loads applied at the joints. If a load acts between joints (for example, the purlins do not sit at the nodes), the chord also bends and has to be checked as an eccentrically compressed member (section 10). That is a separate subject.

Input data

Six groups of data are gathered before the calculation. The table shows where each comes from and what is assumed in the example. Sections are selected from the square hollow section table and the rectangular hollow section table, and masses and section properties can be cross-checked in the metal sections handbook.

The layout, loads, sections, class CC2 and γc = 1.0 are assumed for illustration. A pinned-joint model is permitted if the ratio of section depth to member length is small (for hollow sections h/l ≤ 0.1, cl. 17.3.2; in the example h/l = 0.04 for the chords) and the member axes meet at the joints (cl. 17.3.1).

Step 1. Member forces

In a truss with pinned joints the forces follow from the equilibrium of joints or sections (the method of sections). For chords of parallel geometry, the chord force near midspan is approximately:

N chord = M / h

Here M is the moment at the section from external forces, as in a simply supported beam, and h is the truss depth. In the example the support reaction is 60 kN and the midspan moment is M = 60·6 − 20·(5 + 3 + 1) = 180 kN·m. That gives a force in the middle chord panels of 180 / 1.5 = 120 kN: the top chord is in compression, the bottom chord in tension. The forces in all members, found from joint equilibrium, are listed in the table.

“+” is tension, “−” is compression. The design forces for selecting sections are the maxima in each group: in the example 120 kN in the chords, 72.1 kN in the end diagonals and 48.1 kN in the other diagonals. For a real truss the forces come from analysis software, for all load combinations.

Step 2. Effective lengths

The effective length of a compression member in a flat truss depends on two things: whether it buckles in the plane of the truss (lef) or out of it (lef,1), and how the web is attached (Table 13.1). The coefficients below are for trusses where the web members are welded straight to the chords without gusset plates, as in welded hollow-section trusses, and the chords are closed sections.

Here l is the distance between the centres of the nearest joints in the plane of the truss, and l1 is the distance between joints held against out-of-plane movement. Such restraint comes from truss chords, special bracing, or rigid roof slabs fixed by welds or bolts. Purlins hold a chord only when they themselves bear on a bracing system or a rigid roof diaphragm.

For trusses of angles and for trusses with gusset plates the coefficients are different: they are in other rows of Table 13.1, and cl. 13.1.4 also applies to single angles. Which row suits a given joint is for the designer to decide.

Varying force along the chord

If the chord carries different compressive forces N1 and N2 (N1 > N2) between the restrained joints, the out-of-plane length is refined by formula (13.1), and buckling is checked for N1:

lef,1 = ( 0.75 + 0.25 · N2 / N1 ) · l1

Step 3. Tension members

The strength of a tension member without weakening is checked by formula (8.1):

N · γn / ( An · Ry · γc ) ≤ 1

Bottom chord 80×80×4: 120·10³ · 1.0 / ( 1175 · 240 · 1.0 ) = 0.43 ≤ 1 (area in mm²). Tension diagonals 50×50×4: 48.1·10³ / ( 695 · 240 ) = 0.29 ≤ 1.

Step 4. Compression members

For each compression member the slenderness is found in two planes, λ = lef / i and λef,1 = lef,1 / i, along with the relative slenderness λ̄ = λ · √( Ry / E ) (E = 2.06·10⁵ N/mm², Table B.1) and the buckling coefficient φ by formulas (8.4) and (8.5):

φ = 0.5 / λ̄² · ( δ − √( δ² − 39.48 · λ̄² ) ); δ = 9.87 · ( 1 − α + β·λ̄ ) + λ̄²

For hollow sections and closed rectangular sections, buckling curve a applies (Table 8.1): α = 0.03; β = 0.06. Here α is the curve parameter, not to be confused with α in Table 13.9. For λ̄ > 3.8, φ is not taken greater than 7.6 / λ̄² for curve a. The smaller of the two φ values governs.

The check itself is formula (8.3):

N · γn / ( φ · A · Ry · γc ) ≤ 1

Formula (8.3) holds when the requirements of 8.3.2–8.3.8 on local buckling of walls are met. They are checked separately against Table 8.3. In the example below these requirements are taken as met for the chosen sections.

Step 5. Limiting slenderness

Buckling may pass and a member can still be too slender. According to Table 13.9, for compression members of flat trusses of hollow sections or paired angles the limiting slenderness is:

• chords and end diagonals that carry the support reactions (up to 50 m high): λu = 180 − 60α;
• other members of hollow-section trusses: λu = 210 − 60α;
• top chords not restrained during erection: λu = 220 (after erection, the first row applies).

Here α = N / ( φ · A · Ry · γc ) (without γn, per the note to Table 13.9), but not less than 0.5. The table speaks of tubes. Whether the row covers cold-formed closed rectangular sections the standard does not say outright, so the example treats them as tubes; the designer makes the final call.

For members of category B in CC3 and CC2 structures, and for all members of CC1 structures, the limiting slenderness may be increased by 10 % (cl. 13.4.2).

For tension members of statically loaded trusses (Table 13.10) the limiting slenderness is 400 (rows 1 and 2). Where the sign of the force may change, it is taken as for compression members (note 4). In the example the diagonals work in tension or compression depending on the load. For structures without dynamic actions the slenderness of tension members is checked in vertical planes only (note 1 to Table 13.10).

Step 6. Deflection

Truss deflection is found for the loads that DSTU B V.1.2-3 prescribes for the relevant requirement. For aesthetic and psychological requirements these are permanent and long-term variable loads (cl. 4.3 and 5.1); short-term loads such as snow are not included. Deflection is measured from the line joining the supports (cl. 4.4).

The convenient method is the unit load method (the Mohr integral). A unit force is applied where the deflection is wanted (for a truss, at midspan of the bottom chord), the forces Ni from the load and ni from the unit force are found, and the products are summed over all members:

f = Σ ( Ni,n · ni · li ) / ( E · Ai )

The limiting deflection comes from Table 1 of DSTU B V.1.2-3:2006 (beams, trusses and roof girders open to view, row 2, a). In the example, with a span l = 12 m and a room height up to 6 m (note 3 to the table), fu = l/250 = 48 mm is adopted; for other spans the values are interpolated linearly (note 2). If the service conditions call for other limits (technological, structural), they are taken from the corresponding rows of the table.

Example: 12 m truss of hollow sections

Top chord 80×80×4, N = 120 kN, purlins at every joint. Here lef = l = 2000 mm and lef,1 = l1 = 2000 mm. λ = 2000 / 30.7 = 65.1; λ̄ = 65.1 · √(240 / 206 000) = 2.22. By (8.4)–(8.5), φ = 0.848 (Table Zh.1: 0.851 at 2.2 and 0.820 at 2.4, interpolation gives 0.848). Check (8.3): 120·10³ / ( 0.848 · 1175 · 240 ) = 0.50 ≤ 1. Limiting slenderness: α = 0.50, λu = 180 − 60·0.50 = 150; λ = 65.1 < 150.

Now the same chord, but with purlins at every second joint (l1 = 4000 mm). λ = 4000 / 30.7 = 130.3; λ̄ = 4.45; φ = 7.6 / λ̄² = 0.384 (Table Zh.1: 0.393 at 4.4 and 0.359 at 4.6). Check (8.3): 120·10³ / ( 0.384 · 1175 · 240 ) = 1.11 > 1, so buckling is not satisfied. In addition λ = 130 exceeds the limit λu = 180 − 60·1.11 = 113.4. Refining the length by formula (13.1) for such a layout lowers the utilisation slightly, but the conclusion is the same.

End diagonal 60×60×4, N = −72.1 kN. lef = l = 1803 mm, lef,1 = l1 = 1803 mm; λ = 1803 / 22.6 = 79.8; λ̄ = 2.72; φ = 0.762. Check (8.3): 72.1·10³ / ( 0.762 · 855 · 240 ) = 0.46 ≤ 1; λu = 180 − 60·0.50 = 150 > 79.8.

Diagonal 50×50×4, N = −48.1 kN. For the other web members lef = 0.9 l = 1623 mm (in plane) and lef,1 = 0.9 l1 = 1623 mm (out of plane); λ = 1623 / 18.5 = 87.7; λ̄ = 2.99; φ = 0.706. Check (8.3): 48.1·10³ / ( 0.706 · 695 · 240 ) = 0.41 ≤ 1; λu = 210 − 60·0.50 = 180 > 87.7.

Tension members (step 3): bottom chord 0.43, tension diagonals 0.29. Limiting slenderness. The bottom chord is in tension without a change of sign: in the vertical plane λ = 2000 / 30.7 = 65, well below 400. The 50×50×4 diagonals work in tension and compression in turn, so under note 4 to Table 13.10 their limit is taken as for compression members: λu = 210 − 60α = 180. The diagonal slenderness λ = 1623 / 18.5 = 87.7 (as in the buckling check) is below 180.

Deflection (step 6) with characteristic Pn = 16 kN: Σ( Ni,n · ni · li ) / ( E · Ai ) = 10.6 mm. With γn = 0.975 (DBN V.1.2-14, Table 5, persistent design situation, second group of limit states, CC2) this gives 10.3 mm ≤ fu = 48 mm, a utilisation of 21 %.

Summary of checks

The results are in the table. The truss with 80×80×4 chords and 60×60×4 and 50×50×4 web members passes every check if the purlins restrain the top chord at every joint. When the purlins stand at every second joint only, the top chord does not hold buckling out of the truss plane.

Common mistakes

• taking lef,1 as the chord length between supports instead of the distance between joints restrained out of the truss plane;
• using the coefficients of Table 13.1 from the wrong row: trusses with web members attached straight to the chords and trusses with gusset plates have different ones;
• using buckling curve b for closed sections, when curve a applies to tubes and closed rectangular sections;
• forgetting limiting slenderness: a compression member can have spare strength and buckling resistance and still exceed λu;
• not checking the top chord for the erection stage, when the purlins are not yet fitted (Table 13.9, third row: λu = 220);
• not comparing the deflection with the right row of DSTU B V.1.2-3: the limits depend on the requirements (aesthetic and psychological, structural, technological).

Standards used for this article

• DBN V.2.6-198:2014 “Steel structures. Design standards” (cl. 7.3, 8.1, 13.1, 13.4, Tables 5.1, 8.1, 13.1, 13.9, 13.10, B.1, G.2, Zh.1);
• DBN V.1.2-14:2018 “General principles of ensuring the reliability and structural safety of buildings and structures” (Table 5, values of γn);
• DSTU B V.1.2-3:2006 “Deflections and displacements. Design requirements” (Table 1, notes 2 and 3).

When you need help with KM and KMD

The load capacity of a truss depends as much on the joints and welds as on the members. The POLYGONAL company prepares KM and KMD drawings (see KM and KMD design) and fabricates steel structures to the customer’s drawings or to its own design, with hot-dip galvanising and powder coating.

If you already have a layout, send it, and we will clarify what is needed before fabrication. Installation is done by a partner construction company.