Learn: LTB bending resistance
Lateral-torsional buckling can reduce a beam’s bending resistance before its cross-section reaches the reference moment capacity. This equation applies an already established χLT reduction to the appropriate section-modulus resistance.
What the formula is saying
The reference relationship is moment = section modulus × stress. Use the section modulus appropriate to the section class, multiply by χLT and divide by the applicable member factor.
Read the symbols in plain language
- χLT
- LTB reduction factor
LTB reduction factor. Use the χLT already derived for the unrestrained length, loading and restraint model.
ratio / no unitA dimensionless ratio has no physical unit; 0.01 as a ratio is 1% when the percent option is selected.
- Wy
- Section modulus
Appropriate section modulus for the stated class and LTB model.
mm³Cubic millimetres here describe a section modulus; they are a length-cubed unit.
- fy
- Yield strength
Specified yield stress of the relevant steel grade and thickness, before the material partial factor unless explicitly stated otherwise.
N/mm²One N/mm² equals one MPa.
- γM1
- Partial factor
Supplied material/resistance partial factor in the divisor. Select its code clause and National Annex before real design.
ratio / no unitA dimensionless ratio has no physical unit; 0.01 as a ratio is 1% when the percent option is selected.
- Mb,Rd
- Result to find
LTB bending resistance. One million newton-millimetres make one kilonewton-metre.
kN·m
Sort out the units first
W is in mm³ and fy in N/mm², producing N·mm. Divide by 1,000,000 for kN·m. W is a section modulus, not second moment of area I in mm⁴.
Assumptions before calculating
Use the stated first-generation teaching equation with compatible section properties, material strengths and supplied partial factors. The required section class, buckling curve, National Annex values and design situation must be established separately.
This is a first-generation Eurocode teaching relationship or an explicitly simplified coefficient calculation. The numbers supplied here are exercise data, not a recommendation for any country. Check the adopted edition, relevant clause, National Annex, applicability conditions and all other limit states before any real design.
Let’s solve one together
Read the given values, follow each operation, then check what the result means.
Read the supplied values as one complete study case. Find Mb,Rd and explain the result in the stated output unit.
- χLT · LTB reduction factor
- 0.75
- Wy · Section modulus
- 800000 mm³
- fy · Yield strength
- 355 N/mm²
- γM1 · Partial factor
- 1
Find the reference section moment
Section modulus times yield stress gives a moment in newton-millimetres.
(800000) × (355) = 284000000 N·mmApply lateral-torsional reduction
Use the χLT already derived for the unrestrained length, loading and restraint model.
(0.75) × (284000000) = 213000000 N·mmApply the relevant partial factor
The correct factor belongs to this section or member failure mode, not to every steel check.
(213000000) ÷ (1) = 213000000 N·mmReport bending resistance
One million newton-millimetres make one kilonewton-metre.
(213000000) ÷ 1000000 = 213 kN·m
Does this worked answer make sense?
At fixed strength and factor, doubling W doubles this resistance. A χLT less than 1 must reduce the reference moment, never increase it.
A second worked example — different values
A second case uses different data. Predict which way the answer will change, then calculate it without reusing the first answer.
- χLT · LTB reduction factor
- 0.6
- Wy · Section modulus
- 600000 mm³
- fy · Yield strength
- 275 N/mm²
- γM1 · Partial factor
- 1
Find the reference section moment
Section modulus times yield stress gives a moment in newton-millimetres.
(600000) × (275) = 165000000 N·mmApply lateral-torsional reduction
Use the χLT already derived for the unrestrained length, loading and restraint model.
(0.6) × (165000000) = 99000000 N·mmApply the relevant partial factor
The correct factor belongs to this section or member failure mode, not to every steel check.
(99000000) ÷ (1) = 99000000 N·mmReport bending resistance
One million newton-millimetres make one kilonewton-metre.
(99000000) ÷ 1000000 = 99 kN·m
Now try your own values
Change a value or its unit. The same method will show your calculation, step by step.
Results update only when you calculate. The lesson example above stays unchanged.
Your turn — check your understanding
Solve this separate case yourself. Use only the values below; the two worked examples use different data. Give the requested result in the selected unit.
- χLT · LTB reduction factor
- 0.8
- Wy · Section modulus
- 900000 mm³
- fy · Yield strength
- 355 N/mm²
- γM1 · Partial factor
- 1.1
Find: Learn: LTB bending resistance
A hint, not the answer
The reference relationship is moment = section modulus × stress. Use the section modulus appropriate to the section class, multiply by χLT and divide by the applicable member factor.
W is in mm³ and fy in N/mm², producing N·mm. Divide by 1,000,000 for kN·m. W is a section modulus, not second moment of area I in mm⁴.
Show the full practice solution
Compare the steps with your work; revealing a solution does not mark the lesson complete.
Find the reference section moment
Section modulus times yield stress gives a moment in newton-millimetres.
(900000) × (355) = 319500000 N·mmApply lateral-torsional reduction
Use the χLT already derived for the unrestrained length, loading and restraint model.
(0.8) × (319500000) = 255600000 N·mmApply the relevant partial factor
The correct factor belongs to this section or member failure mode, not to every steel check.
(255600000) ÷ (1.1) ≈ 232363636.4 N·mmReport bending resistance
One million newton-millimetres make one kilonewton-metre.
(232363636.4) ÷ 1000000 ≈ 232.3636364 kN·m
Avoid the common trap
Do not swap Wel and Wpl or use I instead of W. Do not call a section resistance a complete member resistance when lateral restraint is absent, and keep N·mm distinct from kN·m.
When this method applies — and when it does not
The χLT must belong to the relevant lateral-torsional mode and the chosen section property. Shear interaction, axial force, local instability, holes, fatigue and other member checks are not included.
For study and understanding, not approval of a real structure, site operation or design. Apply the correct standard, National Annex and professional review to actual engineering work.
One idea understood. Keep going.
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Sources & further reading
References open in a new tab and explain the underlying principles. The teaching text and examples here are SimpleFlick’s own; the source organisations have not endorsed this calculator.
Reading focus: LTB bending resistance. First-generation EN 1993/EN 1994 teaching: cross-section resistance, stability, connections or composite action as relevant. Member classification and other limit states remain separate checks.
Lesson updated: · Both examples and the separate practice case are checked against an independent high-precision numerical implementation. This verifies arithmetic for the stated model, not engineering certification.