Learn: Bolt bearing resistance — coefficient form
A bolt can press against the side of its hole and damage the connected plate. This bearing-resistance component uses the plate’s ultimate strength, bolt diameter, plate thickness and supplied geometry-dependent coefficients.
What the formula is saying
The projected bearing area is d t. Multiply it by fu, αb and k1, then divide by γM2. The coefficients summarize limits linked to end distance, spacing and material properties; they are not chosen arbitrarily.
Read the symbols in plain language
- k1
- Bearing coefficient
Bearing coefficient governed by transverse edge distances and spacing in the selected code expression.
ratio / no unitA dimensionless ratio has no physical unit; 0.01 as a ratio is 1% when the percent option is selected.
- αb
- Bearing coefficient αb
Bearing coefficient governed by longitudinal edge/spacing ratios and material-strength ratios, with the code cap.
ratio / no unitA dimensionless ratio has no physical unit; 0.01 as a ratio is 1% when the percent option is selected.
- fu
- Plate ultimate strength
Plate ultimate strength. The connected material’s strength, not the bolt’s strength, scales this bearing force.
N/mm²One N/mm² equals one MPa.
- d
- Bolt diameter
Bolt diameter. Bolt diameter times connected-plate thickness defines the projected area in this model.
mmMillimetres measure length; 1000 mm = 1 m.
- t
- Plate thickness
Plate thickness. Bolt diameter times connected-plate thickness defines the projected area in this model.
mmMillimetres measure length; 1000 mm = 1 m.
- 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.
- Fb,Rd
- Result to find
Bolt bearing resistance — coefficient form. Apply the relevant partial factor and convert newtons to kilonewtons.
kN
Sort out the units first
d and t are mm, while plate strength fu is N/mm². The product is N, converted to kN by dividing by 1000. Use plate fu, not bolt fub, as the strength input in this expression.
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 Fb,Rd and explain the result in the stated output unit.
- k1 · Bearing coefficient
- 2.5
- αb · Bearing coefficient αb
- 0.8
- fu · Plate ultimate strength
- 510 N/mm²
- d · Bolt diameter
- 20 mm
- t · Plate thickness
- 12 mm
- γM2 · Partial factor
- 1.25
Find the projected bearing area
Bolt diameter times connected-plate thickness defines the projected area in this model.
(20) × (12) = 240 mm²Apply the plate ultimate strength
The connected material’s strength, not the bolt’s strength, scales this bearing force.
(510) × (240) = 122400 NApply the supplied bearing coefficients
The coefficients represent the specified geometry and strength limits for this bolt location.
(2.5) × (0.8) × (122400) = 244800 NFactor and convert bearing resistance
Apply the relevant partial factor and convert newtons to kilonewtons.
(244800) ÷ (1.25) ÷ 1000 = 195.84 kN
Does this worked answer make sense?
At fixed coefficients, a thicker plate increases this bearing component proportionally. A larger result does not improve a separate bolt-shear limit.
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.
- k1 · Bearing coefficient
- 2.5
- αb · Bearing coefficient αb
- 0.6
- fu · Plate ultimate strength
- 430 N/mm²
- d · Bolt diameter
- 16 mm
- t · Plate thickness
- 10 mm
- γM2 · Partial factor
- 1.25
Find the projected bearing area
Bolt diameter times connected-plate thickness defines the projected area in this model.
(16) × (10) = 160 mm²Apply the plate ultimate strength
The connected material’s strength, not the bolt’s strength, scales this bearing force.
(430) × (160) = 68800 NApply the supplied bearing coefficients
The coefficients represent the specified geometry and strength limits for this bolt location.
(2.5) × (0.6) × (68800) = 103200 NFactor and convert bearing resistance
Apply the relevant partial factor and convert newtons to kilonewtons.
(103200) ÷ (1.25) ÷ 1000 = 82.56 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.
- k1 · Bearing coefficient
- 2.5
- αb · Bearing coefficient αb
- 0.75
- fu · Plate ultimate strength
- 510 N/mm²
- d · Bolt diameter
- 20 mm
- t · Plate thickness
- 8 mm
- γM2 · Partial factor
- 1.25
Find: Learn: Bolt bearing resistance — coefficient form
A hint, not the answer
The projected bearing area is d t. Multiply it by fu, αb and k1, then divide by γM2. The coefficients summarize limits linked to end distance, spacing and material properties; they are not chosen arbitrarily.
d and t are mm, while plate strength fu is N/mm². The product is N, converted to kN by dividing by 1000. Use plate fu, not bolt fub, as the strength input in this expression.
Show the full practice solution
Compare the steps with your work; revealing a solution does not mark the lesson complete.
Find the projected bearing area
Bolt diameter times connected-plate thickness defines the projected area in this model.
(20) × (8) = 160 mm²Apply the plate ultimate strength
The connected material’s strength, not the bolt’s strength, scales this bearing force.
(510) × (160) = 81600 NApply the supplied bearing coefficients
The coefficients represent the specified geometry and strength limits for this bolt location.
(2.5) × (0.75) × (81600) = 153000 NFactor and convert bearing resistance
Apply the relevant partial factor and convert newtons to kilonewtons.
(153000) ÷ (1.25) ÷ 1000 = 122.4 kN
Avoid the common trap
Do not use hole diameter in place of nominal bolt diameter without the relevant rule. Do not swap plate and bolt strengths or assume the same coefficients apply to every bolt in a group.
When this method applies — and when it does not
k1 and αb must already reflect the correct edge distances, end distances, spacings and hole conditions. Thin plates, countersunk details, plate tear-out, block tearing, bolt shear and group distribution need their applicable 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.
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: Bolt bearing 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.