Steel Beam Design to DBN V.2.6-198:2014: Strength, Stability, Deflection

What is checked in a beam

For a simply supported beam made of a rolled I-section, the checks are:

• strength under normal stresses from bending (cl. 9.2.1, formula 9.1);
• strength under shear stresses from the transverse force (formula 9.2);
• overall stability: whether the compression flange could buckle out of the plane of bending (cl. 9.4);
• local stresses in the web under a concentrated load or over a support, where there is one (cl. 9.2.2, formulas 9.6–9.7);
• deflection. This is the second limit state group, and it has its own standard (DSTU B V.1.2-3:2006).

Local stability of the web and flanges of rolled I-beams from the standard range is checked under cl. 9.5. The example below does not cover it.

Input data

First the input data are collected. The table shows where each quantity comes from and what the example assumes. The beam geometry comes from the I-beam sortament.

The consequence class CC2 is assumed for illustration. In a real project the designer sets it, and the category, under DBN V.1.2-14 and Annex A of DBN V.2.6-198.

Two different «categories»: do not mix them up

The «importance category» (A, B, C) that drives γn under DBN V.1.2-14 and the «category by purpose» under cl. 5.3.5 and Table A.1 of DBN V.2.6-198 are not the same thing. For beams the latter is A or B, depending on the purpose.

Where the design standards do not divide structures into importance categories, note 1 to Table 5 of DBN V.1.2-14 says to take category B. For CC2 in a persistent design situation that is γn = 1.050 for the first limit state group and 0.975 for the second (in the second group the CC2 value is the same for all categories).

Step 1. Internal forces

For a simply supported beam under uniform load:

M = q·l² / 8; Q = q·l / 2

In the example: M = 6.2 · 6² / 8 = 27.9 kN·m; Q = 6.2 · 6 / 2 = 18.6 kN.

Step 2. Design strengths of the steel

The design strength in tension, compression and bending Ry comes from Table G.2 by strength grade and thickness; for shaped rolled sections the flange thickness governs. The flange of I-beam No. 20 is 8.4 mm thick and falls in the «4 to 10 mm» range. For S255 the design strength of shaped rolled sections is Ry = 250 N/mm² (characteristic Ryn = 255 N/mm²).

The shear strength comes from Table 7.1:

Rs = 0.58 · Ryn / γm

The material factor is γm = 1.025 (Table G.2, footnote 3, and Table 7.2), so Rs = 0.58 · 255 / 1.025 ≈ 144 N/mm².

Note 2 to Table G.2 lets the strength Ryw of webs of hot-rolled I-beams and channels be raised by 10 % over Ry. The example does not use this reserve.

Step 3. Strength under normal stresses

For beams in class 1 cross-sections (elastic behaviour), formula (9.1) applies:

M · γn / ( Wn,min · Ry · γc ) ≤ 1

Here Wn,min is the smallest section modulus of the net section. For a rolled I-beam without holes it is the tabulated Wx; for I-beam No. 20, Wx = 184 cm³ = 184 000 mm³.

For the example: 27.9·10⁶ · 1.05 / ( 184 000 · 250 · 1.0 ) = 0.64 ≤ 1. Bending strength is satisfied with a 36 % margin.

Design with limited plastic deformations (class 2 sections, formula 9.10 with the coefficient сх from Annex M) is allowed for static loads under the conditions of cl. 9.2.3. The section class is set under cl. 5.3.6; for simplicity the example uses elastic behaviour (class 1), which gives a margin compared with the design to (9.10).

Step 4. Strength under shear stresses

Formula (9.2):

Q · Sx · γn / ( Ix · tw · Rs · γc ) ≤ 1

I-beam No. 20: Sx = 104 cm³, Ix = 1840 cm⁴, tw = 5.2 mm. Shear stress τ = 18.6·10³ · 104·10³ / ( 1840·10⁴ · 5.2 ) ≈ 20.2 N/mm². Check: 20.2 · 1.05 / 144 = 0.15 ≤ 1.

On spans like this, shear is almost never the governing check. It is checked at the support sections and wherever large concentrated forces act.

Step 5. Overall stability

An I-beam in class 1 is checked with formula (9.28), where φb is the bending stability factor from Annex N:

M · γn / ( φb · Wcx · Ry · γc ) ≤ 1

However, cl. 9.4.4 waives this check in two cases:

• the load is transferred through a continuous rigid deck (reinforced concrete slabs, profiled metal decking and the like) that bears continuously on the compression flange and is reliably connected to it (cl. 9.4.4, a);
• the relative slenderness of the compression flange λ̄b = ( lef / bf ) · √( Ryf / E ) does not exceed the limit λ̄ub from Table 9.1 (cl. 9.4.4, b).

lef is the distance between the points where the compression flange is restrained against lateral displacement (cl. 9.4.2). Where there are no restraints, lef = l.

For I-beam No. 20 in S255 steel (E = 2.06·10⁵ N/mm² per Table B.1): bf = 100 mm, tf = 8.4 mm, h0 = h − tf = 191.6 mm. The ratio bf / tf = 11.9 is below 15, so under note 2 to Table 9.1 the formulas use bf / tf = 15. The ratio h0 / bf = 1.92 is within the permitted range of 1 to 6.

The Table 9.1 formulas for λ̄ub:

(9.30) top flange: λ̄ub = 0.35 + 0.0032·bf/tf + ( 0.76 − 0.02·bf/tf ) · bf/h0

(9.32) between braces or pure bending: λ̄ub = 0.41 + 0.0032·bf/tf + ( 0.73 − 0.016·bf/tf ) · bf/h0

The example assumes that the load is passed on by a continuous rigid deck, so overall stability is ensured under cl. 9.4.4, a and needs no separate check to (9.28). For comparison, the table gives three cases for I-beam No. 20 on a 6 m span:

Bracing of the compression flange by braces or a deck is designed for the conditional (or actual) transverse force (cl. 9.4.5); friction forces are not to be counted for a deck (cl. 9.4.4, a). The Annex N method assumes the support sections are restrained against rotation about the beam axis.

What this means for the example

Without a deck, I-beam No. 20 over 6 m does not meet cl. 9.4.4, b. Stability then has to be checked with formula (9.28), with φb from Annex N. The way out is simpler: add a deck, brace the flange, or take a larger section.

Step 6. Local stresses (where needed)

If a concentrated load bears on the flange (another beam, say), or the web at a support section is not stiffened by a rib, the web is checked for local stresses by formulas (9.6) and (9.7): σloc = F / ( lef1 · tw ), with the condition σloc · γn / ( Ry · γc ) ≤ 1.

The design length of the bearing area lef1 depends on how the load is applied (formulas 9.8–9.9).

Step 7. Deflection

Deflection belongs to the second limit state group. The loads it is found from are taken under cl. 4.7 and 4.10 of DBN V.1.2-2:2006, to which DSTU B V.1.2-3 refers. For a limit based on aesthetic and psychological requirements these are the permanent and long-term variable loads, and deflections are taken with full creep. For a simply supported beam under uniform load:

f = 5 · qn · l⁴ / ( 384 · E · Ix )

In the example: f = 5 · 5.0 · 6000⁴ / ( 384 · 2.06·10⁵ · 1840·10⁴ ) ≈ 22.3 mm.

The limit deflection fu comes from Table 1 of DSTU B V.1.2-3:2006. For floor beams and trusses that are open to view (aesthetic and psychological requirements, item 2, a), at a span l = 6 m fu = l/200 = 30 mm; for intermediate spans the value is interpolated linearly (note 2).

The reliability factor for importance in the second limit state group (DBN V.1.2-14, cl. 7.6.4, Table 5) is 0.975 for CC2 in a persistent design situation, and it multiplies the action effect, deflection included: f · γn = 22.3 · 0.975 ≈ 21.7 mm ≤ 30 mm, 72 % used. Deflection is measured from the straight line joining the supports (cl. 4.4). The load qn here is the permanent and long-term variable load (cl. 4.3); short-term loads such as snow are left out.

If there are partitions under the beam, or screeds and floors on it that crack, the limit deflections and loads for them come from other rows of Table 1 (for example l/150 for structural requirements). The row is chosen by the purpose of the floor.

Summary of the example

All the checks of the example in one place:

Common mistakes

• taking the design strength of plate for an I-beam: for shaped rolled sections Table G.2 gives separate values;
• calculating deflection from design loads instead of characteristic ones (or the reverse, strength from characteristic loads);
• forgetting the spacing of braces in the stability check: lef is not always the span;
• ignoring the weakening by bolt holes: formula (9.1) uses the net section modulus Wn,min;
• leaving off the KM drawings the category of structures under Annex A, the importance class, the γn adopted and the design life (cl. 5.1.3 of DBN V.2.6-198).

When you need help with KM and KMD

You can pick the section by this procedure yourself, but connections, welds and bracing still need a separate check and have to be put on drawings. POLYGONAL prepares KM and KMD drawings, fabricates steel structures and does hot-dip galvanizing.

If you have a drawing or a sketch, write to us or call. For the geometry of other sections see the metal sections handbook.

Standards used for this article

• DBN V.2.6-198:2014 «Steel structures. Design standards» (sections 5, 7, 9, Tables 5.1, 7.1, 7.2, G.2, B.1);
• DBN V.1.2-14:2018 «General principles of ensuring the reliability and structural safety of buildings and structures» (cl. 7.6.4, Table 5);
• DBN V.1.2-2:2006 «Loads and actions»;
• DSTU B V.1.2-3:2006 «Deflections and displacements. Design requirements» (in force from 01.01.2007; cl. 4.3, 5.1, Table 1).