Learn: Steel gross-section yielding resistance
A steel tension member can yield across its gross cross-section. This equation estimates that gross-section design resistance by multiplying the area by yield strength and applying the supplied material partial factor.
What the formula is saying
Area times yield stress gives the characteristic gross yielding force. Dividing by γM0 gives the design value for this failure mode; a separate net-section fracture check may produce a smaller resistance.
Read the symbols in plain language
- A
- Gross area
Gross area. Multiply the whole gross section area by the characteristic yield stress.
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.
- Npl,Rd
- Result to find
Steel gross-section yielding resistance. Convert to kilonewtons and keep the gross-yielding failure mode attached to the result.
kN
Sort out the units first
A is gross area in mm² and fy is N/mm², giving N. Divide by 1000 to report kN. γm is dimensionless and represents the applicable gross-yielding partial factor.
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 Npl,Rd and explain the result in the stated output unit.
- A · Gross area
- 5000 mm²
- fy · Yield strength
- 355 N/mm²
- γM0 · Partial factor
- 1
Find the gross yielding force
Multiply the whole gross section area by the characteristic yield stress.
(5000) × (355) = 1775000 NApply the gross-yielding partial factor
The supplied factor converts this characteristic resistance component to a design value.
(1775000) ÷ (1) = 1775000 NReport the gross-section resistance
Convert to kilonewtons and keep the gross-yielding failure mode attached to the result.
(1775000) ÷ 1000 = 1775 kN
Does this worked answer make sense?
Doubling gross area doubles this resistance. The final tension resistance cannot exceed a smaller governing fracture or connection resistance found in the other checks.
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.
- A · Gross area
- 3200 mm²
- fy · Yield strength
- 275 N/mm²
- γM0 · Partial factor
- 1
Find the gross yielding force
Multiply the whole gross section area by the characteristic yield stress.
(3200) × (275) = 880000 NApply the gross-yielding partial factor
The supplied factor converts this characteristic resistance component to a design value.
(880000) ÷ (1) = 880000 NReport the gross-section resistance
Convert to kilonewtons and keep the gross-yielding failure mode attached to the result.
(880000) ÷ 1000 = 880 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.
- A · Gross area
- 4500 mm²
- fy · Yield strength
- 355 N/mm²
- γM0 · Partial factor
- 1.1
Find: Learn: Steel gross-section yielding resistance
A hint, not the answer
Area times yield stress gives the characteristic gross yielding force. Dividing by γM0 gives the design value for this failure mode; a separate net-section fracture check may produce a smaller resistance.
A is gross area in mm² and fy is N/mm², giving N. Divide by 1000 to report kN. γm is dimensionless and represents the applicable gross-yielding partial factor.
Show the full practice solution
Compare the steps with your work; revealing a solution does not mark the lesson complete.
Find the gross yielding force
Multiply the whole gross section area by the characteristic yield stress.
(4500) × (355) = 1597500 NApply the gross-yielding partial factor
The supplied factor converts this characteristic resistance component to a design value.
(1597500) ÷ (1.1) ≈ 1452272.727 NReport the gross-section resistance
Convert to kilonewtons and keep the gross-yielding failure mode attached to the result.
(1452272.727) ÷ 1000 ≈ 1452.272727 kN
Avoid the common trap
Do not use net area in a formula intended for gross yielding, or ultimate tensile strength fu in place of fy. Do not forget that mm² × N/mm² gives N, not kN.
When this method applies — and when it does not
This checks gross yielding only. Bolt-hole deductions, net fracture, block tearing, eccentric connection effects, fatigue and connection resistance are not included. A positive result is a resistance component, not a complete member approval.
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: Steel gross-section yielding 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.