Learn: Plastic shear resistance
A steel cross-section has a shear area that carries the shear force in the chosen direction. This plastic shear expression combines that effective shear area with the yield-based shear stress and the section partial factor.
What the formula is saying
The shear-yield scale fy/√3 follows the von Mises relationship used in this model. Multiply that stress by Av and divide by the supplied factor; Av is not automatically the entire cross-sectional area.
Read the symbols in plain language
- Av
- Shear area
Shear area. Multiply the shear stress by the area that actually belongs to the shear direction.
mm²Square millimetres measure area; 1 mm² = 10⁻⁶ m².
- 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.
- Vpl,Rd
- Result to find
Plastic shear resistance. Use the supplied section factor and convert the force from newtons to kilonewtons.
kN
Sort out the units first
Av is mm² and fy is N/mm². √3 and γm are dimensionless. The raw result is N; divide by 1000 for kN.
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 Vpl,Rd and explain the result in the stated output unit.
- Av · Shear area
- 3000 mm²
- fy · Yield strength
- 355 N/mm²
- γM0 · Partial factor
- 1
Find the yield-based shear stress
The von Mises relationship introduces the square-root-of-three divisor.
(355) ÷ √(3) ≈ 204.9593456 N/mm²Apply the effective shear area
Multiply the shear stress by the area that actually belongs to the shear direction.
(3000) × (204.9593456) ≈ 614878.0367 NFactor and convert the shear resistance
Use the supplied section factor and convert the force from newtons to kilonewtons.
(614878.0367) ÷ (1) ÷ 1000 ≈ 614.8780367 kN
Does this worked answer make sense?
Doubling Av doubles the computed resistance. The shear-yield stress fy/√3 is below fy, as expected for this multiaxial yielding model.
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.
- Av · Shear area
- 2400 mm²
- fy · Yield strength
- 275 N/mm²
- γM0 · Partial factor
- 1
Find the yield-based shear stress
The von Mises relationship introduces the square-root-of-three divisor.
(275) ÷ √(3) ≈ 158.771324 N/mm²Apply the effective shear area
Multiply the shear stress by the area that actually belongs to the shear direction.
(2400) × (158.771324) ≈ 381051.1777 NFactor and convert the shear resistance
Use the supplied section factor and convert the force from newtons to kilonewtons.
(381051.1777) ÷ (1) ÷ 1000 ≈ 381.0511777 kN
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.
- Av · Shear area
- 3500 mm²
- fy · Yield strength
- 355 N/mm²
- γM0 · Partial factor
- 1.1
Find: Learn: Plastic shear resistance
A hint, not the answer
The shear-yield scale fy/√3 follows the von Mises relationship used in this model. Multiply that stress by Av and divide by the supplied factor; Av is not automatically the entire cross-sectional area.
Av is mm² and fy is N/mm². √3 and γm are dimensionless. The raw result is N; divide by 1000 for kN.
Show the full practice solution
Compare the steps with your work; revealing a solution does not mark the lesson complete.
Find the yield-based shear stress
The von Mises relationship introduces the square-root-of-three divisor.
(355) ÷ √(3) ≈ 204.9593456 N/mm²Apply the effective shear area
Multiply the shear stress by the area that actually belongs to the shear direction.
(3500) × (204.9593456) ≈ 717357.7095 NFactor and convert the shear resistance
Use the supplied section factor and convert the force from newtons to kilonewtons.
(717357.7095) ÷ (1.1) ÷ 1000 ≈ 652.1433722 kN
Avoid the common trap
Do not omit √3, use gross area without checking Av, or treat a web susceptible to shear buckling as automatically having full plastic shear resistance.
When this method applies — and when it does not
The shear area must suit the section and force direction. Shear buckling of slender webs, torsion, openings, bending–shear interaction and connection transfer can govern instead and are not assessed here.
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.
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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 shear 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.