Lap length of reinforcements
EUROCODE-2

Lap length of reinforcements

DATA
Coefficient αct (?)

National choice 3.1.6 (2)P: Coefficient of fatigue strength of the concrete. The recommended value is 1.0. Valid values from 0.5 to 1

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% Lapped bars ρ1 (?)

Percentage of lapped bars relative to the total cross-section area of reinforcement. Valid values from 1 to 100

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Bar diameter Φ(mm) (?)

Insert the nominal diameter of a reinforcing bar. Valid values from 5 to 40

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Steel strength fyk(MPa) (?)

Insert the characteristic tensile strength of reinforcement fyk. Valid values from 400 to 600

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Ratio (r) σsd/fyd (?)

Ratio of the design stress (at the position from where the anchorage is measured from) and the design yield strength of reinforcement.
Approx. ratio of area of steel required and area of steel provided at this section. Valid values from 0 to 1

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Concrete strength fck(MPa) (?)

Insert the characteristic compressive cylinder strength of concrete fck. Valid values from 20 to 50

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Minimum cover Cd(mm) (?)

Concrete minimum cover
(see Figure 8.3)
Valid values from 10 to 100

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Transverse pressure p(MPa) (?)

Transverse pressure at ultimate limit state along lbd. Valid values from 0 to fck

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Transverse reinforcement placed between the concrete surface and the lapped bar
Area reinforc. Ast(cm2) (?)

Cross-sectional area of the transverse reinforcement in the lap zone. Valid values from Ast,min (not less than the area As of one lapped bar if its diameter is greater than or equal to 20 mm. Otherwise 1Φ4) and Ast,max (9Φ32).
You can enter the area directly or through the following data:

Num Φ(mm)
Intro
Form transverse bars (?)

Note clause 8.7.4.1(3):
You must select "Links or U bars" if more than 50% of the reinforcement is lapped at one point, and the distance between adjacent laps is ≤ 10·Φ

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RESULT

Lap length l0 (cm)
Bar diameter Φ (mm) Tension Compression
Good bond Poor bond Good bond Poor bond
12 49 70 69 98
Notes:
  • Good bond (Good bond conditions): for the reinforcements during concreting form with the horizontal an angle between 45° and 90° or in the case of forming an angle less than 45°, are located in the lower half of the section or at a distance equal or greater than 30 cm of the upper surface of a layer of concrete.
  • Poor bond (Poor bond conditions): for the reinforcements that, during concreting, not found in any of the above cases.


DETAILS OF CALCULATION

Notation and methodology according to clause 8.7 of EC2

Good Bond conditions

1) Tensile bar

l0 (Lap length) = 488 mm
l0 = α1 · α2 · α3 · α5 · α6 · lb,rqd ≥ l0,min
where:

  • lb,rqd (basic required anchorage length) = 484 mm
    lb,rqd = (Φ / 4) (σsd / fbd)
    with:
    • σsd = r · fyk/γs = 1·500/1.15 = 434.78 MPa
    • fbd (ultimate bond stress) = 2.69 MPa
      fbd = 2.25 · η1 · η2 · fctd
      • η1 = 1, η2 = 1.0
      • fctd = αct·fctk,0.05/γc = 1·1.8/1.5 = 1.2 MPa
      • fctk,0.05 = 0.21· fck(2/3) = 0.21·25(2/3) = 1.8 MPa
  • α1(effect of the form of the bars) = 1
  • α2(effect of concrete minimum cover) = 0.71
    α2 = 1-0.15 (Cd - φ)/ φ = 1-0.15·(35-12)/12 = 0.71
    (≥ 0.7 and ≤ 1.0)
  • α3(effect of transverse reinforcement) = 1
    α3 = 1 - Kλ = 1.05 (≥ 0.7 and ≤ 1.0)
    with:
    • λ = (∑Ast - ∑Ast,min) / As = (0.57 - 1.13) / 1.13 = -0.5
      ∑Ast,min = 1.0·As(σsd / fyd) = 1.13 · 1
    • K = 0.1 (Form transverse bars Links or U bars)
  • α5(effect of the pressure transverse) = 1
    α5 = 1 - 0.04p = 1 – 0.04·0 = 1 (≥ 0.7 and ≤ 1.0)
  • α6 = 1.41
    α6 = (ρ1/25)0.5 = (50/25)0.5 = 1.41 (≥ 1.0 and ≤ 1.5)
  • l0,min = max(205 ; 180 ; 200) = 205 mm
    l0,min = max{0.3·α6·lb,rqd; 15·φ; 200 mm}

2) Compression bar

l0 (Lap length) = 685 mm
l0 = α1 · α2 · α3 · α6 · lb,rqd ≥ l0,min
where:

  • lb,rqd (basic required anchorage length) = 484 mm
  • α1 = 1, α2 = 1, α3 = 1, α6 = 1.41
  • l0,min = 205 mm
Poor Bond conditions

1) Tensile bar

l0 (Lap length) = 697 mm
l0 = α1 · α2 · α3 · α5 · α6 · lb,rqd ≥ l0,min
where:

  • lb,rqd (basic required anchorage length) = 692 mm
    lb,rqd = (Φ / 4) (σsd / fbd)
    with:
    • σsd = r · fyk/γs = 1·500/1.15 = 434.78 MPa
    • fbd (ultimate bond stress) = 1.89 MPa
      fbd = 2.25 · η1 · η2 · fctd
      • η1 = 0.7, η2 = 1.0
      • fctd = αct·fctk,0.05/γc = 1·1.8/1.5 = 1.2 MPa
      • fctk,0.05 = 0.21· fck(2/3) = 0.21·25(2/3) = 1.8 MPa
  • α1(effect of the form of the bars) = 1
  • α2(effect of concrete minimum cover) = 0.71
    α2 = 1-0.15 (Cd - φ)/ φ = 1-0.15·(35-12)/12 = 0.71
    (≥ 0.7 and ≤ 1.0)
  • α3(effect of transverse reinforcement) = 1
    α3 = 1 - Kλ = 1.05 (≥ 0.7 and ≤ 1.0)
    with:
    • λ = (∑Ast - ∑Ast,min) / As = (0.57 - 1.13) / 1.13 = -0.5
      ∑Ast,min = 1.0·As(σsd / fyd) = 1.13 · 1
    • K = 0.1 (Form transverse bars Links or U bars)
  • α5(effect of the pressure transverse) = 1
    α5 = 1 - 0.04p = 1 – 0.04·0 = 1 (≥ 0.7 and ≤ 1.0)
  • α6 = 1.41
    α6 = (ρ1/25)0.5 = (50/25)0.5 = 1.41 (≥ 1.0 and ≤ 1.5)
  • l0,min = max(294 ; 180 ; 200) = 294 mm
    l0,min = max{0.3·α6·lb,rqd; 15·φ; 200 mm}

2) Compression bar

l0 (Lap length) = 978 mm
l0 = α1 · α2 · α3 · α6 · lb,rqd ≥ l0,min
where:

  • lb,rqd (basic required anchorage length) = 692 mm
  • α1 = 1, α2 = 1, α3 = 1, α6 = 1.41
  • l0,min = 294 mm