Learn: Design bond strength — classic EC2 form
Bond strength describes the design stress that transfers force between reinforcement and concrete along the bar surface. This teaching relationship derives it from concrete design tensile strength and two supplied bond-condition factors.
What the formula is saying
The factor η1 represents bond conditions and η2 represents the bar-diameter effect in this model. Multiply both by 2.25 and by fctd; the factors do not replace the requirement for an appropriate design tensile strength.
Read the symbols in plain language
- η1
- Bond-condition coefficient
Bond-condition coefficient. The supplied coefficients scale the tensile-strength contribution for the stated bond model.
ratio / no unitA dimensionless ratio has no physical unit; 0.01 as a ratio is 1% when the percent option is selected.
- η2
- Bar-size coefficient
Bar-size coefficient. The supplied coefficients scale the tensile-strength contribution for the stated bond model.
ratio / no unitA dimensionless ratio has no physical unit; 0.01 as a ratio is 1% when the percent option is selected.
- fctd
- Design tensile strength
Design tensile strength. Multiply by the already established design tensile strength, preserving its stress unit.
N/mm²One N/mm² equals one MPa.
- fbd
- Result to find
Design bond strength — classic EC2 form. Multiply by the already established design tensile strength, preserving its stress unit.
N/mm²
Sort out the units first
fctd and fbd are in N/mm². The coefficient 2.25 and the η factors are dimensionless. Do not use mean tensile strength fctm when the required input is design tensile strength fctd.
Assumptions before calculating
Use the first-generation EC2 teaching model and the supplied design coefficients. Material strengths, geometry, load situation and coefficients must be mutually compatible; selecting them from the adopted code and National Annex is outside this calculation.
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 fbd and explain the result in the stated output unit.
- η1 · Bond-condition coefficient
- 1
- η2 · Bar-size coefficient
- 1
- fctd · Design tensile strength
- 1.3 N/mm²
Combine the bond-condition factors
The supplied coefficients scale the tensile-strength contribution for the stated bond model.
2.25 × (1) × (1) = 2.25Find design bond strength
Multiply by the already established design tensile strength, preserving its stress unit.
(2.25) × (1.3) = 2.925 N/mm²
Does this worked answer make sense?
Reducing either η factor reduces bond strength and generally increases a later required anchorage length. At η1 = η2 = 1, fbd is 2.25 fctd.
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.
- η1 · Bond-condition coefficient
- 0.7
- η2 · Bar-size coefficient
- 1
- fctd · Design tensile strength
- 1.5 N/mm²
Combine the bond-condition factors
The supplied coefficients scale the tensile-strength contribution for the stated bond model.
2.25 × (0.7) × (1) = 1.575Find design bond strength
Multiply by the already established design tensile strength, preserving its stress unit.
(1.575) × (1.5) = 2.3625 N/mm²
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.
- η1 · Bond-condition coefficient
- 1
- η2 · Bar-size coefficient
- 0.92
- fctd · Design tensile strength
- 1.4 N/mm²
Find: Learn: Design bond strength — classic EC2 form
A hint, not the answer
The factor η1 represents bond conditions and η2 represents the bar-diameter effect in this model. Multiply both by 2.25 and by fctd; the factors do not replace the requirement for an appropriate design tensile strength.
fctd and fbd are in N/mm². The coefficient 2.25 and the η factors are dimensionless. Do not use mean tensile strength fctm when the required input is design tensile strength fctd.
Show the full practice solution
Compare the steps with your work; revealing a solution does not mark the lesson complete.
Combine the bond-condition factors
The supplied coefficients scale the tensile-strength contribution for the stated bond model.
2.25 × (1) × (0.92) = 2.07Find design bond strength
Multiply by the already established design tensile strength, preserving its stress unit.
(2.07) × (1.4) = 2.898 N/mm²
Avoid the common trap
Do not interpret η1 and η2 as angles or percentages entered as whole numbers. Do not use compressive strength in place of tensile strength or apply a material partial factor twice.
When this method applies — and when it does not
The factors must match the applicable bar size, bond condition and concrete provisions. This expression does not establish good bond automatically, select casting conditions, calculate lap length or verify splitting resistance.
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: Design bond strength — classic EC2 form. First-generation EN 1992 teaching: material properties and the relevant bending, shear, serviceability, detailing or prestress relationship. Read the applicability conditions as well as the expression.
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.