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EN 1990 · EN 1993-1-1 · §5.5 · §6.2.5 · §6.2.6

How do you check a steel beam by hand to EN 1993-1-1?

A simply supported steel beam is the first calculation most engineers learn, and still one of the most common. This worked example checks an IPE 300 in S235 over 6.00 m, step by step: the design load, the cross-section class, bending, shear and deflection.

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Struct4U
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4 min read
Contents

The beam

A floor beam, simply supported over L = 6.00 m. The floor holds the compression flange laterally, so lateral-torsional buckling cannot occur. The loads per metre are a permanent load gk = 5.0 kN/m, including self-weight, and a variable load qk = 7.5 kN/m on a floor in category A (residential). The profile is an IPE 300 in S235.

Simply supported beam: load, deflection, bending moment and shearSpan 6.00 m, design load 18.0 kN/m. Maximum bending moment 81.0 kNm at midspan, shown highlighted because bending governs; shear force 54.0 kN at each support; deflection 12.0 mm under the characteristic load.qd = 18.0 kN/mδ = 12.0 mm (SLS)L = 6.00 mMMEd = 81.0 kNmVVEd = 54.0 kN−54.0 kN
FIG. 1The beam with its design load, bending moment and shear force, and the deflection under the characteristic load. Bending governs.EN 1993-1-1 §6.2

Step 1: the design load

For the ultimate limit state EN 1990 combines the loads with partial factors. With the recommended values of equation (6.10):

EN 1990 (6.10)
qd= 1.35 gk + 1.5 qk = 1.35 · 5.0 + 1.5 · 7.5 = 18.0 kN/m

The Dutch national annex, consequence class CC2, takes the less favourable of equations 6.10a and 6.10b instead. For a floor in category A, ψ0 = 0.4:

NEN-EN 1990 NB, CC2
(6.10a)1.35 · 5.0 + 1.5 · 0.4 · 7.5 = 11.25 kN/m
(6.10b)1.2 · 5.0 + 1.5 · 7.5 = 17.25 kN/m, governs

The example continues with 18.0 kN/m; the results with the Dutch annex follow at the end.

Step 2: material and cross-section

S235 with a thickness of at most 40 mm has fy = 235 N/mm² (Table 3.1), so ε = √(235 / fy) = 1.0. The partial factor γM0 = 1.0, both as the recommended value and in the Dutch annex. E = 210 000 N/mm².

IPE 300, hot-rolled (EN 10365)
PropertyValue
Depth h, width b300 mm, 150 mm
Web tw, flange tf, root radius r7.1 mm, 10.7 mm, 15 mm
Area A53.81 cm2
Second moment of area Iy8356 cm4
Plastic section modulus Wpl,y628.4 cm3
Shear area Av,z25.68 cm2

Step 3: the cross-section class

The class decides whether the plastic resistance may be used. Table 5.2 compares the width-to-thickness ratio c/t of each compressed part with limits that depend on ε.

IPE 300: the widths c and thicknesses t used for classificationCross-section of an IPE 300, 300 mm deep and 150 mm wide. The flange outstand c is 56.5 mm for a flange thickness of 10.7 mm; the web c is 248.6 mm for a web thickness of 7.1 mm. Both ratios are well within the class 1 limits of Table 5.2 for S235.b = 150h = 300cf = 56.5tf = 10.7cw = 248.6tw = 7.1r = 15IPE 300 · S235 · ε = 1.00dimensions in mmoutstand flange, compression:c/t = 5.28 ≤ 9ε = 9 → class 1internal web, bending:c/t = 35.0 ≤ 72ε = 72 → class 1
FIG. 2The flange outstand and the web of an IPE 300, with the widths c that Table 5.2 compares with the thicknesses t.EN 1993-1-1 Table 5.2
EN 1993-1-1 §5.5, Table 5.2
flangec = (b − tw − 2r) / 2 = (150 − 7.1 − 30) / 2 = 56.5 mm
c / tf = 56.5 / 10.7 = 5.28 ≤ 9ε = 9: class 1 (outstand in compression)
webc = h − 2tf − 2r = 300 − 21.4 − 30 = 248.6 mm
c / tw = 248.6 / 7.1 = 35.0 ≤ 72ε = 72: class 1 (internal part in bending)
sectionclass 1: the plastic resistance applies

Step 4: bending

EN 1993-1-1 §6.2.5
MEd= qd L2 / 8 = 18.0 · 6.002 / 8 = 81.0 kNm
Mc,Rd= Wpl,y fy / γM0 = 628.4 · 103 · 235 / 1.0 = 147.7 kNm(6.13)
UC= MEd / Mc,Rd = 81.0 / 147.7 = 0.55 ≤ 1.0(6.12)

Step 5: shear

For a rolled I-section loaded parallel to the web, §6.2.6(3)a gives the shear area.

EN 1993-1-1 §6.2.6
Av= A − 2b tf + (tw + 2r) tf = 5381 − 3210 + 397 = 2568 mm², at least η hw tw = 2374 mm²
VEd= qd L / 2 = 18.0 · 6.00 / 2 = 54.0 kN
Vpl,Rd= Av (fy / √3) / γM0 = 2568 · (235 / √3) / 1.0 = 348.4 kN(6.18)
UC= 54.0 / 348.4 = 0.16 ≤ 1.0(6.17)

Two more conditions. The web needs no shear buckling check, because hw / tw = 278.6 / 7.1 = 39.2 ≤ 72ε / η = 60 (§6.2.6(6), with η = 1.2 from EN 1993-1-5; η = 1.0 is allowed as a conservative value). And the shear force does not reduce the bending resistance, because VEd = 54.0 kN ≤ 0.5 Vpl,Rd = 174.2 kN (§6.2.8); at midspan, where the moment is largest, the shear force is zero anyway.

Step 6: deflection

Deflection is a serviceability check, with the characteristic loads and without partial factors.

EN 1990 Annex A1.4
δ= 5 (gk + qk) L4 / (384 E Iy) = 5 · 12.5 · 60004 / (384 · 210 000 · 8356 · 104) = 12.0 mm
δlim= L / 250 = 24.0 mm (example limit)
UC= 12.0 / 24.0 = 0.50

EN 1990 leaves the deflection limits to the national annex and the client, so L/250 is only an example here.

The result

CheckClauseUnity check
BendingEN 1993-1-1 §6.2.50.55
ShearEN 1993-1-1 §6.2.60.16
Deflection, L/250EN 1990 A1.40.50

Bending governs, with a unity check of 0.55: the IPE 300 is far from fully used. With the Dutch annex the design load is 17.25 kN/m, MEd = 77.6 kNm and the bending check 0.53; shear becomes 0.15.

An IPE 270, also class 1 (Wpl,y = 484.0 cm3, Iy = 5790 cm4), gives a bending check of 81.0 / 113.7 = 0.71 and a deflection of 17.3 mm, 0.72 of L/250.

What this example leaves out

  • Lateral-torsional buckling (§6.3.2). Without lateral restraint of the compression flange, the buckling resistance moment Mb,Rd = χLT Wy fy / γM1 (6.55) takes the place of the cross-section resistance.
  • Supports and concentrated loads. The web has to carry the reactions without crippling (EN 1993-1-5 §6).
  • Vibration of the floor, a separate serviceability criterion.
  • Fire resistance (EN 1993-1-2).

The same check in XBeam2D

XBeam2D does this calculation for beams on any number of supports, in steel, concrete and timber, and checks steel members including lateral-torsional buckling. The report shows each check the way this article does: the clause, the formula, the values that went in and the unity check. A reviewer can follow it, and you can recheck any number by hand.

Sources
  1. EN 1990:2002+A1:2005, Basis of structural design: equation (6.10) and Annex A1 (in the Netherlands NEN-EN 1990+A1+A1/C2:2011 with its national annex: 6.10a and 6.10b, consequence classes).
  2. EN 1993-1-1:2005, Eurocode 3, Part 1-1: Table 3.1, §5.5, Table 5.2, §6.2.5, §6.2.6, §6.2.8, §6.3.2.
  3. EN 1993-1-5:2006, Plated structural elements: §5 (shear buckling) and §6 (resistance to transverse forces).
  4. Profile data: hot-rolled IPE sections to EN 10365.

Related

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Every check names its clause and shows the values that went in.

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