Learn: Bolt shear resistance — coefficient form
A bolt can fail in shear, so its resistance must be checked for that mode independently. This equation evaluates the stated resistance of one bolt on one shear plane.
What the formula is saying
Multiply bolt ultimate strength by the appropriate area and the specified resistance coefficient, then divide by γM2. The area depends on whether the shear plane crosses the threaded or unthreaded part; it is not always the gross shank area.
Read the symbols in plain language
- αv
- Shear coefficient
Shear coefficient. The coefficient belongs to the selected bolt grade and failure-mode expression.
ratio / no unitA dimensionless ratio has no physical unit; 0.01 as a ratio is 1% when the percent option is selected.
- fub
- Bolt ultimate strength
Bolt ultimate strength. Multiply ultimate bolt strength by the area appropriate to this failure mode.
N/mm²One N/mm² equals one MPa.
- A
- Relevant bolt area
Bolt area at the shear plane: use the tensile-stress area if threads are in the plane when required, not automatically gross shank area.
mm²Square millimetres measure area; 1 mm² = 10⁻⁶ m².
- γM2
- 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.
- Fv,Rd
- Result to find
Bolt shear resistance — coefficient form. Apply the partial factor and report the force in kilonewtons for the stated bolt model.
kN
Sort out the units first
Bolt strength fub is N/mm² and the relevant area is mm². The product is N; dividing by 1000 reports kN. All resistance and partial coefficients are dimensionless.
Assumptions before calculating
This is one resistance component for the specified joint model. Bolt grade, hole type, connected material, geometry and partial factor must be compatible with the applicable adopted connection rules.
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 Fv,Rd and explain the result in the stated output unit.
- αv · Shear coefficient
- 0.6
- fub · Bolt ultimate strength
- 800 N/mm²
- A · Relevant bolt area
- 245 mm²
- γM2 · Partial factor
- 1.25
Find the bolt material-force scale
Multiply ultimate bolt strength by the area appropriate to this failure mode.
(800) × (245) = 196000 NApply the specified resistance coefficient
The coefficient belongs to the selected bolt grade and failure-mode expression.
(0.6) × (196000) = 117600 NFactor and convert the single-bolt resistance
Apply the partial factor and report the force in kilonewtons for the stated bolt model.
(117600) ÷ (1.25) ÷ 1000 = 94.08 kN
Does this worked answer make sense?
At fixed coefficient and grade, resistance is proportional to the relevant bolt area. The numerical resistance must be compared with the demand on that bolt, not automatically with the entire joint force.
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.
- αv · Shear coefficient
- 0.6
- fub · Bolt ultimate strength
- 800 N/mm²
- A · Relevant bolt area
- 157 mm²
- γM2 · Partial factor
- 1.25
Find the bolt material-force scale
Multiply ultimate bolt strength by the area appropriate to this failure mode.
(800) × (157) = 125600 NApply the specified resistance coefficient
The coefficient belongs to the selected bolt grade and failure-mode expression.
(0.6) × (125600) = 75360 NFactor and convert the single-bolt resistance
Apply the partial factor and report the force in kilonewtons for the stated bolt model.
(75360) ÷ (1.25) ÷ 1000 = 60.288 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.
- αv · Shear coefficient
- 0.6
- fub · Bolt ultimate strength
- 800 N/mm²
- A · Relevant bolt area
- 353 mm²
- γM2 · Partial factor
- 1.25
Find: Learn: Bolt shear resistance — coefficient form
A hint, not the answer
Multiply bolt ultimate strength by the appropriate area and the specified resistance coefficient, then divide by γM2. The area depends on whether the shear plane crosses the threaded or unthreaded part; it is not always the gross shank area.
Bolt strength fub is N/mm² and the relevant area is mm². The product is N; dividing by 1000 reports kN. All resistance and partial coefficients are dimensionless.
Show the full practice solution
Compare the steps with your work; revealing a solution does not mark the lesson complete.
Find the bolt material-force scale
Multiply ultimate bolt strength by the area appropriate to this failure mode.
(800) × (353) = 282400 NApply the specified resistance coefficient
The coefficient belongs to the selected bolt grade and failure-mode expression.
(0.6) × (282400) = 169440 NFactor and convert the single-bolt resistance
Apply the partial factor and report the force in kilonewtons for the stated bolt model.
(169440) ÷ (1.25) ÷ 1000 = 135.552 kN
Avoid the common trap
Do not use the connected plate strength instead of bolt strength. Do not count two shear planes when the bolt is actually in single shear, or use the wrong threaded-area convention.
When this method applies — and when it does not
Additional shear planes may contribute only under the applicable connection rules. Bearing, slip resistance, long-joint reductions and combined tension are not evaluated. A bolt-group resistance cannot be inferred without load distribution and geometry checks.
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: Bolt shear resistance — coefficient form. 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.