EN 1993-1-1 Beam Design Worked Example — IPE 300 (Clause 6.2 & 6.3, χLT)
Quick Reference: This worked example demonstrates a complete EN 1993-1-1 beam design check for a simply supported IPE 300 beam in S275 steel spanning 7.0 m. The verification covers section classification (Cl. 5.5), bending moment resistance Mc,Rd (Cl. 6.2.5), shear resistance Vc,Rd (Cl. 6.2.6), lateral-torsional buckling resistance Mb,Rd with full χLT calculation (Cl. 6.3.2), and deflection serviceability. All partial factors and buckling curves follow EN 1993-1-1:2005 + AC:2009 with recommended values. Use the free beam capacity calculator for instant EN 1993 checks.
PRELIMINARY — NOT FOR CONSTRUCTION. All calculations are illustrative educational examples. Results must be verified by a licensed Professional Engineer before use in any design project.
Design Data
| Parameter | Value | Reference |
|---|---|---|
| Beam section | IPE 300 | EN 10365 |
| Steel grade | S275 JR | EN 10025-2 |
| Yield strength fy | 275 MPa | EN 1993-1-1 Table 3.1 (t ≤ 40 mm) |
| Tensile strength fu | 430 MPa | EN 10025-2 |
| Modulus of elasticity E | 210,000 MPa | EN 1993-1-1 §3.2.6 |
| Span L | 7.0 m | Simply supported |
| Permanent load gk | 8.5 kN/m | Self-weight included |
| Variable load qk | 12.0 kN/m | Imposed floor load |
| Partial factors | γG = 1.35, γQ = 1.50 | EN 1990 Eq. 6.10 |
IPE 300 Section Properties
| Property | Symbol | Value | Units |
|---|---|---|---|
| Depth | h | 300.0 | mm |
| Flange width | b | 150.0 | mm |
| Flange thickness | tf | 10.7 | mm |
| Web thickness | tw | 7.1 | mm |
| Root radius | r | 15.0 | mm |
| Cross-sectional area | A | 5380 | mm² |
| Plastic section modulus y-y | Wpl,y | 628,000 | mm³ |
| Elastic section modulus y-y | Wel,y | 557,000 | mm³ |
| Second moment of area y-y | Iy | 83.56 × 10⁶ | mm⁴ |
| Second moment of area z-z | Iz | 6.04 × 10⁶ | mm⁴ |
| Torsion constant | It | 202,000 | mm⁴ |
| Warping constant | Iw | 126 × 10⁹ | mm⁶ |
| Radius of gyration z-z | iz | 33.5 | mm |
| Shear area z-z | Av,z | 2568 | mm² |
Step 1: Section Classification (EN 1993-1-1 Cl. 5.5)
The epsilon factor for S275 steel:
ε = √(235/fy) = √(235/275) = 0.924
Flange Classification (outstand in compression)
The flange is an outstand element in uniform compression:
cf = (b − tw − 2r) / 2 = (150 − 7.1 − 2 × 15) / 2 = 56.45 mm
c/tf ratio = 56.45 / 10.7 = 5.28
Class 1 limit for outstand flange in compression (rolled section): 9ε = 9 × 0.924 = 8.32
5.28 < 8.32 → Flange is Class 1 ✓
Web Classification (bending)
The web is an internal element in pure bending:
cw = h − 2tf − 2r = 300 − 2 × 10.7 − 2 × 15 = 248.6 mm
c/tw ratio = 248.6 / 7.1 = 35.01
Class 1 limit for internal part in bending: 72ε = 72 × 0.924 = 66.5
35.01 < 66.5 → Web is Class 1 ✓
Conclusion: IPE 300 in S275 is Class 1 for pure bending — plastic design permitted.
Step 2: ULS Design Moment
Design load (EN 1990 Eq. 6.10):
wEd = γG × gk + γQ × qk = 1.35 × 8.5 + 1.50 × 12.0 = 29.48 kN/m
Maximum design bending moment (simply supported, UDL):
MEd = wEd × L² / 8 = 29.48 × 7.0² / 8 = 180.6 kN·m
Design shear force at support:
VEd = wEd × L / 2 = 29.48 × 7.0 / 2 = 103.2 kN
Step 3: Bending Moment Resistance (Cl. 6.2.5)
For Class 1 cross-section, the plastic moment resistance applies:
Mc,Rd = Wpl,y × fy / γM0 = 628,000 mm³ × 275 N/mm² / 1.00
Mc,Rd = 628,000 × 275 / 10⁶ = 172.7 kN·m
Unity check (cross-section alone — does not yet account for LTB):
MEd / Mc,Rd = 180.6 / 172.7 = 1.046 → FAILS at cross-section level
The IPE 300 alone cannot resist the applied moment at the cross-section level. This indicates one of three paths: (a) increase the section size, (b) reduce the span, or (c) check if a higher steel grade helps. However, for the purpose of this worked example we will continue to illustrate the LTB check — in practice this beam would be upgraded to IPE 330 or IPE 360, or steel grade raised to S355.
Revised Design: IPE 360 in S275
Let us re-run the design with IPE 360 to demonstrate the full EN 1993 workflow:
| Property | IPE 360 Value |
|---|---|
| h | 360 mm |
| b | 170 mm |
| tf | 12.7 mm |
| tw | 8.0 mm |
| Wpl,y | 1,019,000 mm³ |
| Iy | 162.7 × 10⁶ mm⁴ |
| Iz | 10.43 × 10⁶ mm⁴ |
| Iw | 313.6 × 10⁹ mm⁶ |
| It | 373,000 mm⁴ |
| iz | 37.9 mm |
Section Classification for IPE 360
ε = 0.924 (unchanged, same S275 grade)
Flange: cf = (170 − 8.0 − 2 × 18) / 2 (wait — root radius r for IPE 360 is 18 mm)
cf = (170 − 8.0 − 2 × 18) / 2 = (170 − 8 − 36) / 2 = 63 mm
c/tf = 63 / 12.7 = 4.96 < 9ε = 8.32 → Class 1
Web: cw = 360 − 2 × 12.7 − 2 × 18 = 360 − 25.4 − 36 = 298.6 mm
c/tw = 298.6 / 8.0 = 37.33 < 72ε = 66.5 → Class 1
IPE 360 is Class 1 for bending ✓
Bending Resistance (IPE 360)
Mc,Rd = Wpl,y × fy / γM0 = 1,019,000 × 275 / 10⁶ = 280.2 kN·m
MEd / Mc,Rd = 180.6 / 280.2 = 0.645 → OK at cross-section level ✓
Step 4: Shear Resistance (Cl. 6.2.6)
Shear buckling need not be checked if:
hw / tw ≤ 72ε / η where η = 1.0 (conservative, EN 1993-1-5)
hw = h − 2tf = 360 − 2 × 12.7 = 334.6 mm
hw / tw = 334.6 / 8.0 = 41.8
72ε / η = 72 × 0.924 / 1.0 = 66.5
41.8 < 66.5 → Shear buckling check not required ✓
Plastic shear resistance:
Vpl,Rd = Av,z × (fy / √3) / γM0
Av,z = A − 2btf + (tw + 2r)tf for rolled I-sections loaded parallel to web
= 7273 − 2 × 170 × 12.7 + (8.0 + 2 × 18) × 12.7 = 7273 − 4318 + 558.8 = 3514 mm²
Vpl,Rd = 3514 × (275 / 1.732) / 1.00 = 3514 × 158.8 / 1000 = 558.0 kN
VEd / Vpl,Rd = 103.2 / 558.0 = 0.185 → OK ✓
Check bending-shear interaction (Cl. 6.2.8):
VEd = 103.2 kN < 0.5 × Vpl,Rd = 279 kN
No reduction in moment resistance due to shear ✓
Step 5: Lateral-Torsional Buckling Resistance Mb,Rd (Cl. 6.3.2)
The compression flange is assumed laterally unrestrained. For a simply supported beam with fork supports (lateral restraint at supports, free to warp), the effective length for LTB equals the span: Lcr = 7000 mm.
Elastic Critical Moment Mcr
Using the general formula from NCCI SN003a-EN-EU for a beam with uniform moment:
Mcr = C1 × (π²EIz / Lcr²) × √(Iw/Iz + Lcr²GIt / π²EIz)
Where:
- C1 = 1.132 for simply supported beam with UDL (NCCI Table 3.2 — conservative for this check; exact value depends on load height position)
- G = E / [2(1+ν)] = 210,000 / [2(1+0.3)] = 80,769 MPa
Compute component terms:
π²EIz / Lcr² = π² × 210,000 × 10.43 × 10⁶ / 7000² = 441,000 N
Iw / Iz = 313.6 × 10⁹ / 10.43 × 10⁶ = 30,070 mm²
Lcr²GIt / π²EIz = 7000² × 80,769 × 373,000 / (π² × 210,000 × 10.43 × 10⁶) = 219,100 mm²
Mcr = 1.132 × 441,000 × √(30,070 + 219,100) / 10⁶
Mcr = 1.132 × 441,000 × √(249,170) / 10⁶
Mcr = 1.132 × 441,000 × 499.2 / 10⁶ = 249.3 kN·m
Non-Dimensional Slenderness λ̄LT
λ̄LT = √(Wpl,y × fy / Mcr) = √(1,019,000 × 275 / 249.3 × 10⁶)
λ̄LT = √(280.2 × 10⁶ / 249.3 × 10⁶) = √(1.124) = 1.060
Reduction Factor χLT — General Case (Cl. 6.3.2.2)
For rolled I-sections with h/b = 360/170 = 2.12 > 2.0, use buckling curve b:
αLT = 0.34 (EN 1993-1-1 Table 6.3, buckling curve b)
However, the UK NA recommends always using buckling curve 'a' for hot-rolled I-sections. For this generic EU example we will use the code default: h/b > 2.0 → curve b.
ΦLT = 0.5 [1 + αLT (λ̄LT − 0.2) + λ̄LT²]
ΦLT = 0.5 [1 + 0.34 × (1.060 − 0.2) + 1.060²]
ΦLT = 0.5 [1 + 0.34 × 0.860 + 1.124]
ΦLT = 0.5 [1 + 0.292 + 1.124] = 0.5 × 2.416 = 1.208
χLT = 1 / [ΦLT + √(ΦLT² − λ̄LT²)]
χLT = 1 / [1.208 + √(1.208² − 1.060²)]
χLT = 1 / [1.208 + √(1.458 − 1.124)]
χLT = 1 / [1.208 + √(0.334)] = 1 / [1.208 + 0.578] = 1 / 1.786
χLT = 0.560
Design Buckling Resistance Moment
Mb,Rd = χLT × Wpl,y × fy / γM1 = 0.560 × 1,019,000 × 275 / 10⁶
Mb,Rd = 0.560 × 280.2 = 156.9 kN·m
LTB unity check:
MEd / Mb,Rd = 180.6 / 156.9 = 1.151 → FAILS for LTB ✗
Simplified Method (Cl. 6.3.2.3) — Rolled Sections
Since h/b = 2.12 is close to the 2.0 limit, let us also check the simplified method which is commonly used in practice for IPE sections:
λ̄LT,0 = 0.4, β = 0.75
ΦLT = 0.5 [1 + αLT (λ̄LT − λ̄LT,0) + β × λ̄LT²]
ΦLT = 0.5 [1 + 0.34 × (1.060 − 0.4) + 0.75 × 1.060²]
ΦLT = 0.5 [1 + 0.224 + 0.843] = 0.5 × 2.067 = 1.034
χLT,mod = 1 / [1.034 + √(1.034² − 0.75 × 1.060²)]
χLT,mod = 1 / [1.034 + √(1.069 − 0.843)] = 1 / [1.034 + √(0.226)]
χLT,mod = 1 / [1.034 + 0.475] = 1 / 1.509 = 0.663
But χLT,mod ≤ 1/λ̄LT² = 1/1.124 = 0.890 and χLT,mod ≤ 1.0 — both satisfied.
However, χLT,mod must not exceed χLT from the general case when applying the correction factor f:
f = 1 − 0.5(1 − kc)[1 − 2.0(λ̄LT − 0.8)²] (Annex B)
For a UDL on a simply supported beam, kc = 0.94 (Table 6.6).
f = 1 − 0.5 × (1 − 0.94) × [1 − 2.0 × (1.060 − 0.8)²] = 1 − 0.5 × 0.06 × [1 − 2.0 × 0.0676] = 1 − 0.03 × [1 − 0.135] = 1 − 0.03 × 0.865 = 1 − 0.026 = 0.974
χLT,mod = χLT / f = 0.560 / 0.974 = 0.575 ≤ 1.0
(Maximum value: χLT,mod ≤ 1/λ̄LT² = 0.890 ✓)
Mb,Rd,mod = 0.575 × 280.2 = 161.1 kN·m
MEd / Mb,Rd = 180.6 / 161.1 = 1.121 → STILL FAILS
Resolution: Increase Section
The IPE 360 fails the LTB check. Let us move to IPE 400 (Wpl,y = 1,308,000 mm³, Iy=231.3×10⁶ mm⁴, Iz=13.18×10⁶ mm⁴, Iw=490×10⁹ mm⁶, h/b=400/180=2.22):
Mc,Rd = 1,308,000 × 275 / 10⁶ = 359.7 kN·m
Recalculate Mcr with IPE 400 properties...
π²EIz/Lcr² = π² × 210,000 × 13.18 × 10⁶ / 7000² = 557,600 N
Iw/Iz = 490 × 10⁹ / 13.18 × 10⁶ = 37,180 mm²
It for IPE 400 ≈ 510,000 mm⁴
Lcr²GIt/π²EIz = 7000² × 80,769 × 510,000 / (π² × 210,000 × 13.18 × 10⁶) = 290,100 mm²
Mcr = 1.132 × 557,600 × √(37,180 + 290,100) / 10⁶ = 1.132 × 557,600 × 572.1 / 10⁶ = 361.1 kN·m
λ̄LT = √(359.7 × 10⁶ / 361.1 × 10⁶) = √(0.996) = 0.998
Using buckling curve b (h/b > 2.0):
ΦLT = 0.5 [1 + 0.34 × (0.998 − 0.2) + 0.998²] = 0.5 [1 + 0.271 + 0.996] = 1.134
χLT = 1 / [1.134 + √(1.134² − 0.998²)] = 1 / [1.134 + √(1.286 − 0.996)] = 1 / [1.134 + 0.539] = 0.598
Mb,Rd = 0.598 × 359.7 = 215.1 kN·m
MEd/Mb,Rd = 180.6 / 215.1 = 0.840 → OK ✓
Step 6: Deflection Serviceability (Cl. 7.2)
Calculated from the characteristic (unfactored) load combination:
wchar = gk + qk = 8.5 + 12.0 = 20.5 kN/m
Total deflection (simply supported UDL):
δtotal = 5 × wchar × L⁴ / (384 × E × Iy)
δtotal = 5 × 20.5 × 7000⁴ / (384 × 210,000 × 231.3 × 10⁶)
δtotal = 5 × 20.5 × 2.401 × 10¹⁵ / (384 × 210,000 × 231.3 × 10⁶)
δtotal = 246.1 × 10¹² / (1.866 × 10¹³) = 13.2 mm
Deflection limit (floor beam): L/250 = 7000/250 = 28.0 mm
13.2 mm < 28.0 mm ✓
Summary of Results (IPE 400 in S275 — Final Selection)
| Check | Resistance | Applied | Ratio | Result |
|---|---|---|---|---|
| Bending (Mc,Rd) | 359.7 kN·m | 180.6 kN·m | 0.502 | OK |
| Shear (Vpl,Rd) | 658 kN* | 103.2 kN | 0.157 | OK |
| LTB (Mb,Rd) | 215.1 kN·m | 180.6 kN·m | 0.840 | OK |
| Deflection | 28.0 mm | 13.2 mm | 0.471 | OK |
* Approximate shear area for IPE 400.
Final section: IPE 400 meets all EN 1993-1-1 requirements for this loading configuration.
Key Takeaways
- Section classification drives everything. A Class 1 classification permits full plastic moment capacity. Class 4 sections require effective width calculations per EN 1993-1-5, dramatically reducing capacity.
- Lateral-torsional buckling is the governing check for intermediate-to-long unbraced beam spans. The χLT factor for this example (0.598) reduces the cross-section capacity by 40%, demonstrating that a section adequate for bending alone may fail LTB.
- The elastic critical moment Mcr calculation is sensitive to end restraint, load height, and moment distribution. Use the C1 and C2 factors from NCCI SN003a for accurate results, not simplified assumptions.
- National Annex variations matter. The UK NA specifies buckling curve 'a' for rolled I-sections regardless of h/b ratio, which gives a higher χLT than the generic Eurocode curve selection. The German NA uses γM1=1.10 (vs 1.00 recommended), reducing Mb,Rd by ~9%.