Reinforcing bars should be well anchored so that the bond forces are safely transmitted to the concrete to avoid longitudinal cracking or spalling. Transverse reinforcement shall be provided if necessary. Types of anchorage are shown in the figure below (Figure 8.1 EC2).

For bent bars, the basic tension anchorage length is measured along the centreline of the bar from the section in question to the end of the bar, where:

l_{bd} = α_{1} α_{2} α_{3} α_{4} α_{5} l_{b,req} ≥ l_{b,min} ———— (1)

where;

l_{b,min} is the minimum anchorage length taken as follows:

In tension, the greatest of 0.3l_{b,rqd} or 10 ϕ or 100mm

In compression, the greatest of 0.6l_{b,rqd} or 10 ϕ or 100mm

l_{b,rqd} is the basic anchorage length given by;

l_{b,rqd} = (ϕ/4) σ_{sd}/f_{bd} ————– (2)

Where;

σ_{sd} = The design strength in the bar (take 0.87f_{yk})

f_{bd} = The design ultimate bond stress (for ribbed bars = 2.25η_{1} η_{2} f_{ctd})

f_{ctd} = Design concrete tensile strength f_{ctd} = 0.21f_{ck}^{(2/3)} for f_{ck} ≤ 50 N/mm^{2}

η_{1} is a coefficient related to the quality of the bond condition and the position of the bar during concreting

η_{1} = 1.0 when ‘good’ conditions are obtained and

η_{1} = 0.7 for all other cases and for bars in structural elements built with slip-forms, unless it can be shown that ‘good’ bond conditions exist

η_{2} is related to the bar diameter:

η_{2} = 1.0 for φ ≤ 32 mm

η_{2} = (132 – φ)/100 for φ > 32 mm

α_{1} is for the effect of the form of the bars assuming adequate cover.

α_{2} is for the effect of concrete minimum cover.

α_{3} is for the effect of confinement by transverse reinforcement

α_{4} is for the influence of one or more welded transverse bars ( φt > 0.6φ) along the design anchorage length l_{bd}

α_{5} is for the effect of the pressure transverse to the plane of splitting along the design anchorage length.

The values of these coefficients can be adequately obtained by following the Table below;

**CALCULATION OF LAP LENGTH**

The design lap length of reinforcements is given by;

l_{0} = α_{1} α_{2} α_{3} α_{5} α_{6 }l_{b,rqd} ≥ l_{0,min} ——— (3)

l_{0,min} = max{0.3α_{6} l_{b,rqd}; 15ϕ; 200}

α_{6} = √(ρ_{1}/25) but between 1.0 and 1.5

where ρ_{1} is the % of reinforcement lapped within 0.65l_{0} from the centre of the lap

Values of α_{1}, α_{2}, α_{3} and α_{5} may be taken as for the calculation of anchorage length but for the calculation of α_{3}, ΣA_{st,min} should be taken as 1.0As(σ_{sd}/f_{yd}), with As = area of one lapped bar.

**SOLVED EXAMPLE FOR ANCHORAGE LENGTH**

Calculate the design tension anchorage length of X16mm bar (f_{yk} = 460 N/mm^{2}, concrete cover = 35mm, Concrete cylinder strength f_{ck} = 25 N/mm^{2}) for;

(a) When it is a straight bar

(b) When it is bent into any other shape

Assume good bond conditions

**Solution**

l_{bd} = α_{1} α_{2} α_{3} α_{4} α_{5} l_{b,req} ≥ l_{b,min}

l_{b,rqd} = (ϕ/4) σ_{sd}/f_{bd}

f_{bd} =2.25η_{1} η_{2} f_{ctd}

η_{1} = 1.0 ‘Good’ bond conditions

η_{2} = 1.0 bar size ≤ 32

f_{ctd} = (α_{ct} f_{ctk 0.05})/γ_{c} ————— (3)

where;

f_{ctk 0.05} = characteristic tensile strength of concrete at 28 days = 1.8 N/mm^{2} (Table 3.1 EC2)

γ_{c} = partial (safety) factor for concrete = 1.5

α_{ct} = coefficient taking account of long-term effects on the tensile strength, this is an NDP with a recommended value of 1.

f_{ctd} = (1.0 × 1.8)/1.5 = 1.2 N/mm^{2}

f_{bd} = 2.25 × 1.0 × 1.0 × 1.2 = 2.7 N/mm^{2}

l_{b,rqd} = (ϕ/4) σ_{sd}/f_{bd}

σ_{sd} = 0.87 × 460 = 400.2 N/mm^{2}>br/> l_{b,rqd} = (ϕ × 400.2 )/(4 × 2.7) = 37.05ϕ

Therefore;

l_{bd} = α_{1} α_{2} α_{3} α_{4} α_{5}(37.05ϕ)

**(a) For straight bar**

α_{1} = 1.0

α_{2} = 1.0 – 0.15 (C_{d} – ϕ)/ ϕ

α_{2} = 1.0 – 0.15 (35 – 16)/16 = 0.8218

α_{3} = 1.0 conservative value with K = 0

α_{4} = 1.0 N/A

α_{5} = 1.0 conservative value

l_{bd} = 0.8218 × (37.05ϕ) = 30.4ϕ = 30.4 × 16 = 486.4 mm

Say 500mm

**(b) For other shape bar**

α_{1} = 1.0 bC_{d}; = 35 is ≤ 3ϕ = 3 × 16 = 48

α_{2} = 1.0 – 0.15 (Cd – 3ϕ)/ ϕ ≤ 1.0

α_{2} = 1.0 – 0.15 (35 – 48)/16 = 1.121 ≤ 1.0

α_{3} = 1.0 conservative value with K = 0

α_{4} = 1.0 N/A

α_{5} = 1.0 conservative value

l_{bd} = 1.0 × (37.05ϕ) = 37.05ϕ = 37.05 × 16 = 592 mm

Say 600mm

**Compression anchorage** (α_{1} = α_{2} = α_{3} = α_{4} = α_{5} = 1.0)

l_{bd} = 37.05ϕ

**For poor bond conditions**

Anchorage for ‘Poor’ bond conditions = ‘Good’/0.7

**Example on calculation of lap length of 4X16mm bars of a column in a multi-storey building**

Since the bars are in compression,

α_{1} α_{2} α_{3} α_{5 }= 1.0

As calculated above, l_{bd} = 37.05ϕ

Let us say that over 50% of reinforcement is lapped within 0.65l_{0} from the centre of the lap

Hence, we will take α_{6 }= 1.5

Lap length therefore = 1.5 × 37.05ϕ = 55.57ϕ = 55.57 × 16 = 889.2mm

Say 900 mm

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The design ultimate bond stress is for ribbed bars, may I ask is there any guideline for smooth bars?

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