Reinforced concrete structures rely on the effective transfer of stress between steel reinforcement bars. Lap splices, where two reinforcement bars overlap within the concrete element, are a common method to achieve this continuity. Lapping of reinforcement is a frequent occurrence in construction due to supply length limitations, construction joints, stage construction, and ease of handling of rebars.

Therefore, determining the appropriate lap length is important for ensuring structural integrity and preventing premature failure. This article discusses how to calculate the lap length for reinforcements in concrete structures.

**Factors Influencing Lap Length**

Several factors influence the required lap length:

**Reinforcement diameter (**Larger diameter bars require longer lap lengths for effective stress transfer.*Φ*):**Concrete grade (f**Higher concrete strength allows for shorter lap lengths due to improved bond characteristics._{ck}):**Reinforcement yield strength (f**Higher yield strength steel requires longer lap lengths to develop its full capacity._{yk}):**Concrete cover:**The thickness of the concrete cover influences the bond transfer and hence, the anchorage length.**Bond condition:**The orientation of the reinforcement (straight or bent) and the concrete cover can lead to a good or poor bond condition which can influence the anchorage length of reinforcements.**Confinement conditions:**The presence of stirrups or ties around the lap splice can significantly reduce the required lap length.**Type of Lap Splice:**Standard lap splices require longer lengths compared to hooked or welded lap splices.

**Methods** **of Calculating Lap Length**

Different design codes and standards provide methods for calculating lap length. Here’s a breakdown of two common approaches:

**1. Basic Lap Length Formula:**

This simplified method often serves as a preliminary estimate:

**Lap Length (L) = 50d**

Where:

d = Reinforcement diameter (mm)

This formula assumes standard lap splices in normal concrete conditions.

**2. Code-Based Calculations:**

Detailed calculations are typically based on specific design codes like Eurocode 2 (EC2) or ACI 318. These codes provide equations and tables that consider various factors influencing lap length:

**EC2 Method:**

EC2 employs a formula that incorporates the bar diameter, concrete grade, steel yield strength, and a factor accounting for the percentage of bars lapped in one section.

**ACI 318 Method:**

ACI 318 provides tables with tabulated lap lengths based on bar diameter, concrete strength, and type of splice (standard, hooked, or welded).

**Note:**

**Minimum Lap Length:**Codes often specify minimum lap lengths regardless of calculations.**Development Length:**Lap length should be greater than or equal to the development length of the bar, which is the minimum length required for the bar to develop its full yield strength in the surrounding concrete.**Longitudinal Spacing of Bars:**The spacing between lapped bars within the overlap zone can affect the required lap length.**Lightweight Concrete:**Adjustments are often necessary for lap lengths in lightweight concrete due to its lower bond strength.

**Anchorage Length**

Apart from the lapping of steel reinforcements, reinforcing bars should be well anchored so that the bond forces are safely transmitted to the concrete to avoid longitudinal cracking or spalling. The length of rebar required to achieve adequate transfer of forces is known as the anchorage length. Transverse reinforcement shall be provided if necessary.

Apart from straight bars, other shapes that are specified in the code are;

(a) standard bend,

(b) standard hook, and

(c) standard loop.

The detailing rules and the equivalent anchorage length for each of these standard shapes are defined in EN1992-1-1 Figure 8.1. 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 10ϕ or 100mm

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

‘Good’ bond conditions are applicable when any of the following conditions are fulfilled:

(a) Vertical bars or almost vertical bars inclined at an angle 45° ≤ α ≤ 90° from the horizontal,

(b) bars that are located up to 250 mm from the bottom of the formwork for elements with height *h* ≤ 600 mm, or

(c) bars that are located at least 300 mm from the free surface during concreting for elements with height *h* > 600 mm.

‘Poor’ bond conditions are applicable for all other cases and also for bars in structural elements built with slip-forms, unless it can be shown that ‘good’ bond conditions exist. The different bond regions are shown in the figure above that is reproduced from EN1992-1-1 Figure 8.2.

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;

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 T16mm bar (f_{yk} = 460 N/mm^{2}, concrete cover = 35 mm, 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 500 mm

**(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

Good bond conditions | Poor bond conditions | |||

Concrete class | • Straight bars in tension / compression• Other-than-straight bars in compression | • Other-than-straight bars in tension and adequate cover c_{d} > 3Φ | • Straight bars in tension/compression• Other-than-straight bars in compression | • Other-than-straight bars in tension and adequate cover c_{d} > 3Φ |

C16/20 | 55Φ | 39Φ | 78Φ | 55Φ |

C20/25 | 47Φ | 33Φ | 67Φ | 47Φ |

C25/30 | 41Φ | 29Φ | 58Φ | 41Φ |

C30/37 | 36Φ | 26Φ | 52Φ | 36Φ |

C35/45 | 33Φ | 23Φ | 47Φ | 33Φ |

C40/50 | 30Φ | 21Φ | 43Φ | 30Φ |

**Example on the 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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