How to Calculate the Anchorage and Lap Length of Steel Reinforcements According to Eurocode 2

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 = α₁ α₂ α₃ α₄ α₅ 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 = 2.25η₁ η₂ f_ctd)
f_ctd = Design concrete tensile strength f_ctd = 0.21f_ck^(2/3) for f_ck ≤ 50 N/mm²
η₁ is a coefficient related to the quality of the bond condition and the position of the bar during concreting
η₁ = 1.0 when ‘good’ conditions are obtained and
η₁ = 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
η₂ is related to the bar diameter:
η₂ = 1.0 for φ ≤ 32 mm
η₂ = (132 – φ)/100 for φ > 32 mm

α₁ is for the effect of the form of the bars assuming adequate cover.
α₂ is for the effect of concrete minimum cover.

α₃ is for the effect of confinement by transverse reinforcement
α₄ is for the influence of one or more welded transverse bars ( φt > 0.6φ) along the design anchorage length l_bd
α₅ 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;
l₀ = α₁ α₂ α₃ α₅ α₆ l_b,rqd ≥ l_0,min ——— (3)

l_0,min = max{0.3α₆ l_b,rqd; 15ϕ; 200}

α₆ = √(ρ₁/25) but between 1.0 and 1.5
where ρ₁ is the % of reinforcement lapped within 0.65l₀ from the centre of the lap

Values of α₁, α₂, α₃ and α₅ may be taken as for the calculation of anchorage length but for the calculation of α₃, 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/mm², concrete cover = 35mm, Concrete cylinder strength f_ck = 25 N/mm²) for;

(a) When it is a straight bar
(b) When it is bent into any other shape
Assume good bond conditions

Solution
l_bd = α₁ α₂ α₃ α₄ α₅ l_b,req ≥ l_b,min
l_b,rqd = (ϕ/4) σ_sd/f_bd
f_bd =2.25η₁ η₂ f_ctd
η₁ = 1.0 ‘Good’ bond conditions
η₂ = 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/mm² (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/mm²
f_bd = 2.25 × 1.0 × 1.0 × 1.2 = 2.7 N/mm²

l_b,rqd = (ϕ/4) σ_sd/f_bd
σ_sd = 0.87 × 460 = 400.2 N/mm²>br/> l_b,rqd = (ϕ × 400.2 )/(4 × 2.7) = 37.05ϕ

Therefore;

l_bd = α₁ α₂ α₃ α₄ α₅(37.05ϕ)

(a) For straight bar
α₁ = 1.0
α₂ = 1.0 – 0.15 (C_d – ϕ)/ ϕ
α₂ = 1.0 – 0.15 (35 – 16)/16 = 0.8218
α₃ = 1.0 conservative value with K = 0
α₄ = 1.0 N/A
α₅ = 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
α₁ = 1.0 bC_d; = 35 is ≤ 3ϕ = 3 × 16 = 48
α₂ = 1.0 – 0.15 (Cd – 3ϕ)/ ϕ ≤ 1.0
α₂ = 1.0 – 0.15 (35 – 48)/16 = 1.121 ≤ 1.0
α₃ = 1.0 conservative value with K = 0
α₄ = 1.0 N/A
α₅ = 1.0 conservative value
l_bd = 1.0 × (37.05ϕ) = 37.05ϕ = 37.05 × 16 = 592 mm
Say 600mm

Compression anchorage (α₁ = α₂ = α₃ = α₄ = α₅ = 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,
α₁ α₂ α₃ α₅  = 1.0
As calculated above,  l_bd = 37.05ϕ

Let us say that over 50% of reinforcement is lapped within 0.65l₀ from the centre of the lap
Hence, we will  take α₆  = 1.5

Lap length therefore = 1.5  × 37.05ϕ = 55.57ϕ = 55.57 × 16 = 889.2mm
Say 900 mm

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