Learn: Design anchorage length from basic length
The basic required anchorage length is adjusted for specific detailing and bond-related effects through a product of design modifiers. This lesson evaluates that product only, keeping every modifier visible so none is accidentally omitted.
What the formula is saying
Multiply α1 through α5, then apply the resulting factor to lb,rqd. These factors represent different specified effects such as bar form, cover and confinement; they are not five extra lengths to add.
Read the symbols in plain language
- α1
- Coefficient α1
Anchorage modifier for bar shape; supply the applicable value and verify code bounds separately.
ratio / no unitA dimensionless ratio has no physical unit; 0.01 as a ratio is 1% when the percent option is selected.
- α2
- Coefficient α2
Anchorage modifier for concrete cover; supply the applicable value and verify code bounds separately.
ratio / no unitA dimensionless ratio has no physical unit; 0.01 as a ratio is 1% when the percent option is selected.
- α3
- Coefficient α3
Anchorage modifier for confinement by transverse reinforcement; supply the applicable value and verify code bounds separately.
ratio / no unitA dimensionless ratio has no physical unit; 0.01 as a ratio is 1% when the percent option is selected.
- α4
- Coefficient α4
Anchorage modifier for welded transverse reinforcement; supply the applicable value and verify code bounds separately.
ratio / no unitA dimensionless ratio has no physical unit; 0.01 as a ratio is 1% when the percent option is selected.
- α5
- Coefficient α5
Anchorage modifier for transverse pressure; supply the applicable value and verify code bounds separately.
ratio / no unitA dimensionless ratio has no physical unit; 0.01 as a ratio is 1% when the percent option is selected.
- lb,rqd
- Basic required length
Basic required length. Apply the combined factor; minimum anchorage and factor-combination limits must still be checked.
mmMillimetres measure length; 1000 mm = 1 m.
- lbd
- Result to find
Design anchorage length from basic length. Apply the combined factor; minimum anchorage and factor-combination limits must still be checked.
mm
Sort out the units first
All α factors are dimensionless. lb and the resulting product length are in mm. The input lb is the basic required anchorage length, not an already modified final anchorage length.
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 lbd and explain the result in the stated output unit.
- α1 · Coefficient α1
- 1
- α2 · Coefficient α2
- 1
- α3 · Coefficient α3
- 1
- α4 · Coefficient α4
- 1
- α5 · Coefficient α5
- 1
- lb,rqd · Basic required length
- 800 mm
Combine the first three modifiers
Each supplied modifier acts multiplicatively on the same basic anchorage requirement.
(1) × (1) × (1) = 1Include the remaining two modifiers
Complete the five-factor product without treating any coefficient as an extra length.
(1) × (1) × (1) = 1Calculate the modified product length
Apply the combined factor; minimum anchorage and factor-combination limits must still be checked.
(1) × (800) = 800 mm
Modified anchorage product only; apply the separate minimum-length and factor limits before detailing.
Does this worked answer make sense?
When every modifier is 1, the product equals the basic length. A smaller product does not bypass a larger required minimum anchorage length.
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.
- α1 · Coefficient α1
- 0.7
- α2 · Coefficient α2
- 1
- α3 · Coefficient α3
- 1
- α4 · Coefficient α4
- 1
- α5 · Coefficient α5
- 1
- lb,rqd · Basic required length
- 900 mm
Combine the first three modifiers
Each supplied modifier acts multiplicatively on the same basic anchorage requirement.
(0.7) × (1) × (1) = 0.7Include the remaining two modifiers
Complete the five-factor product without treating any coefficient as an extra length.
(0.7) × (1) × (1) = 0.7Calculate the modified product length
Apply the combined factor; minimum anchorage and factor-combination limits must still be checked.
(0.7) × (900) = 630 mm
Modified anchorage product only; apply the separate minimum-length and factor limits before detailing.
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.
- α1 · Coefficient α1
- 1
- α2 · Coefficient α2
- 0.8
- α3 · Coefficient α3
- 1
- α4 · Coefficient α4
- 1
- α5 · Coefficient α5
- 1
- lb,rqd · Basic required length
- 750 mm
Find: Learn: Design anchorage length from basic length
A hint, not the answer
Multiply α1 through α5, then apply the resulting factor to lb,rqd. These factors represent different specified effects such as bar form, cover and confinement; they are not five extra lengths to add.
All α factors are dimensionless. lb and the resulting product length are in mm. The input lb is the basic required anchorage length, not an already modified final anchorage length.
Show the full practice solution
Compare the steps with your work; revealing a solution does not mark the lesson complete.
Combine the first three modifiers
Each supplied modifier acts multiplicatively on the same basic anchorage requirement.
(1) × (0.8) × (1) = 0.8Include the remaining two modifiers
Complete the five-factor product without treating any coefficient as an extra length.
(0.8) × (1) × (1) = 0.8Calculate the modified product length
Apply the combined factor; minimum anchorage and factor-combination limits must still be checked.
(0.8) × (750) = 600 mm
Modified anchorage product only; apply the separate minimum-length and factor limits before detailing.
Avoid the common trap
Do not add the modifiers or automatically set all of them below 1. Do not apply them twice, and do not detail a reduced result before checking the relevant minimum anchorage rule.
When this method applies — and when it does not
This formula ID contains only α1α2α3α4α5lb,rqd and does not implement lb,min or limits on permitted factor combinations. The lesson labels the output as a modified product length, not the final code-compliant anchorage to provide.
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.
This optional checkmark is saved only in this browser. It is your own progress note, not a certificate.
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 anchorage length from basic length. 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.