Learn: Plastic bending resistance
A plastic section modulus represents the force couple when an admissible section develops its plastic stress distribution. Multiplying it by yield strength gives the plastic bending resistance before the supplied section partial factor.
What the formula is saying
The reference relationship is moment = section modulus × stress. Use Wpl and the section-yielding partial factor for a section permitted to develop plastic resistance.
Read the symbols in plain language
- Wpl
- Plastic section modulus
Plastic section modulus Wpl about the checked bending axis.
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.
- γM0
- 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.
- Mpl,Rd
- Result to find
Plastic 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 Mpl,Rd and explain the result in the stated output unit.
- Wpl · Plastic section modulus
- 800000 mm³
- fy · Yield strength
- 355 N/mm²
- γM0 · Partial factor
- 1
Find the reference section moment
Section modulus times yield stress gives a moment in newton-millimetres.
(800000) × (355) = 284000000 N·mmApply the relevant partial factor
The correct factor belongs to this section or member failure mode, not to every steel check.
(284000000) ÷ (1) = 284000000 N·mmReport bending resistance
One million newton-millimetres make one kilonewton-metre.
(284000000) ÷ 1000000 = 284 kN·m
Does this worked answer make sense?
At fixed strength and factor, doubling W doubles this resistance. The same member can have different elastic and plastic moments because Wel and Wpl differ.
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.
- Wpl · Plastic section modulus
- 600000 mm³
- fy · Yield strength
- 275 N/mm²
- γM0 · Partial factor
- 1
Find the reference section moment
Section modulus times yield stress gives a moment in newton-millimetres.
(600000) × (275) = 165000000 N·mmApply the relevant partial factor
The correct factor belongs to this section or member failure mode, not to every steel check.
(165000000) ÷ (1) = 165000000 N·mmReport bending resistance
One million newton-millimetres make one kilonewton-metre.
(165000000) ÷ 1000000 = 165 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.
- Wpl · Plastic section modulus
- 900000 mm³
- fy · Yield strength
- 355 N/mm²
- γM0 · Partial factor
- 1.1
Find: Learn: Plastic bending resistance
A hint, not the answer
The reference relationship is moment = section modulus × stress. Use Wpl and the section-yielding partial factor for a section permitted to develop plastic resistance.
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 the relevant partial factor
The correct factor belongs to this section or member failure mode, not to every steel check.
(319500000) ÷ (1.1) ≈ 290454545.5 N·mmReport bending resistance
One million newton-millimetres make one kilonewton-metre.
(290454545.5) ÷ 1000000 ≈ 290.4545455 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
Plastic resistance requires a suitable section class and rotation assumptions; elastic resistance uses the appropriate elastic extreme-fibre modulus. 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: Plastic 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.