UNDERSTAND IT. WORK IT OUT.

Learn: Basic required anchorage length

A reinforcing bar transfers its tensile force to surrounding concrete through bond. Basic required anchorage length estimates how much straight bonded length is needed before additional design modifiers and minimum-length rules are considered.

Beginner-friendlyFree · No accountTwo worked examples + separate practice
01

What the formula is saying

Bar force grows with area πφ²/4, while bond force per length grows with perimeter πφ. Cancelling those terms gives the diameter factor φ/4 multiplied by the steel-stress-to-bond-stress ratio.

lb,rqd = (φ/4)(σsd/fbd)

Read the symbols in plain language

φ
Bar diameter

Bar diameter. A circular bar’s area divided by its perimeter equals one quarter of its diameter.

mm

Millimetres measure length; 1000 mm = 1 m.

σsd
Design bar stress

Design bar stress. The ratio represents how much bar stress must be transferred relative to available bond stress.

N/mm²

One N/mm² equals one MPa.

fbd
Design bond strength

Design bond strength. The ratio represents how much bar stress must be transferred relative to available bond stress.

N/mm²

One N/mm² equals one MPa.

lb,rqd
Result to find

Basic required anchorage length. Multiply the geometric factor by the stress ratio before applying any later design modifiers.

mm

Sort out the units first

Use bar diameter φ in mm and both σsd and fbd in N/mm². The stress ratio is dimensionless, leaving an anchorage length in mm. σsd is the design stress to be developed, not necessarily fyd.

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.

02

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 lb,rqd and explain the result in the stated output unit.

φ · Bar diameter
20 mm
σsd · Design bar stress
435 N/mm²
fbd · Design bond strength
2.5 N/mm²
  1. Compare steel stress with bond stress

    The ratio represents how much bar stress must be transferred relative to available bond stress.

    (435) ÷ (2.5) = 174
  2. Find the area-to-perimeter diameter factor

    A circular bar’s area divided by its perimeter equals one quarter of its diameter.

    (20) ÷ 4 = 5 mm
  3. Calculate basic required anchorage

    Multiply the geometric factor by the stress ratio before applying any later design modifiers.

    (5) × (174) = 870 mm
Answer870 mm

Does this worked answer make sense?

Doubling diameter doubles basic length at the same stresses. Improving bond strength lowers the required basic length, while higher steel stress increases it.

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.

φ · Bar diameter
16 mm
σsd · Design bar stress
400 N/mm²
fbd · Design bond strength
3 N/mm²
  1. Compare steel stress with bond stress

    The ratio represents how much bar stress must be transferred relative to available bond stress.

    (400) ÷ (3) ≈ 133.3333333
  2. Find the area-to-perimeter diameter factor

    A circular bar’s area divided by its perimeter equals one quarter of its diameter.

    (16) ÷ 4 = 4 mm
  3. Calculate basic required anchorage

    Multiply the geometric factor by the stress ratio before applying any later design modifiers.

    (4) × (133.3333333) ≈ 533.3333333 mm
Answer533.3333333 mm
03

Now try your own values

Change a value or its unit. The same method will show your calculation, step by step.

Bar diameter. A circular bar’s area divided by its perimeter equals one quarter of its diameter.

Design bar stress. The ratio represents how much bar stress must be transferred relative to available bond stress.

Design bond strength. The ratio represents how much bar stress must be transferred relative to available bond stress.

English, Arabic and Persian digits are supported. The steps convert inputs to the formula’s base units.

Results update only when you calculate. The lesson example above stays unchanged.

04

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.

φ · Bar diameter
25 mm
σsd · Design bar stress
435 N/mm²
fbd · Design bond strength
3.2 N/mm²

Find: Learn: Basic required anchorage length

For repeating decimals, use at least four significant figures. Accepted rounding tolerance: 0.05% of the expected value; zero uses an absolute tolerance of 10⁻¹².

A hint, not the answer

Bar force grows with area πφ²/4, while bond force per length grows with perimeter πφ. Cancelling those terms gives the diameter factor φ/4 multiplied by the steel-stress-to-bond-stress ratio.

Use bar diameter φ in mm and both σsd and fbd in N/mm². The stress ratio is dimensionless, leaving an anchorage length in mm. σsd is the design stress to be developed, not necessarily fyd.

Show the full practice solution

Compare the steps with your work; revealing a solution does not mark the lesson complete.

  1. Compare steel stress with bond stress

    The ratio represents how much bar stress must be transferred relative to available bond stress.

    (435) ÷ (3.2) = 135.9375
  2. Find the area-to-perimeter diameter factor

    A circular bar’s area divided by its perimeter equals one quarter of its diameter.

    (25) ÷ 4 = 6.25 mm
  3. Calculate basic required anchorage

    Multiply the geometric factor by the stress ratio before applying any later design modifiers.

    (6.25) × (135.9375) = 849.609375 mm
Answer849.609375 mm

Avoid the common trap

Use diameter rather than radius. Do not substitute concrete compressive strength for bond strength, or label the basic length as the final length to detail on a drawing.

When this method applies — and when it does not

The model assumes the supplied design bond strength is applicable to the bar and concrete conditions. This is lb,rqd only; bends, cover, confinement, transverse pressure, lap rules and required minimum design anchorage are not included.

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: Basic required anchorage 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.

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