Lap length of reinforcements
RESULT
Lap length l_{0} (cm)  
Bar diameter Φ (mm)  Tension  Compression  
Good bond  Poor bond  Good bond  Poor bond  
12  49  70  69  98 
Notes:

DETAILS OF CALCULATION
Notation and methodology according to clause 8.7 of EC2
Good Bond conditions
1) Tensile bar
l_{0} (Lap length) = 488 mm
l_{0} = α_{1} · α_{2} · α_{3} · α_{5} · α_{6} · l_{b,rqd} ≥ l_{0,min}
where:
 l_{b,rqd} (basic required anchorage length) = 484 mm
l_{b,rqd} = (Φ / 4) (σ_{sd} / f_{bd})
with: σ_{sd} = r · f_{yk}/γ_{s} = 1·500/1.15 = 434.78 MPa
 f_{bd} (ultimate bond stress) = 2.69 MPa
f_{bd} = 2.25 · η_{1} · η_{2} · f_{ctd} η_{1} = 1, η_{2} = 1.0
 f_{ctd} = α_{ct}·f_{ctk,0.05}/γ_{c} = 1·1.8/1.5 = 1.2 MPa
 f_{ctk,0.05} = 0.21· f_{ck}^{(2/3)} = 0.21·25^{(2/3)} = 1.8 MPa
 α_{1}(effect of the form of the bars) = 1
 α_{2}(effect of concrete minimum cover) = 0.71
α_{2} = 10.15 (Cd  φ)/ φ = 10.15·(3512)/12 = 0.71
(≥ 0.7 and ≤ 1.0)  α_{3}(effect of transverse reinforcement) = 1
α_{3} = 1  Kλ = 1.05 (≥ 0.7 and ≤ 1.0)
with: λ = (∑A_{st}  ∑A_{st,min}) / A_{s} = (0.57  1.13) / 1.13 = 0.5
∑A_{st,min} = 1.0·As(σ_{sd} / f_{yd}) = 1.13 · 1  K = 0.1 (Form transverse bars Links or U bars)
 λ = (∑A_{st}  ∑A_{st,min}) / A_{s} = (0.57  1.13) / 1.13 = 0.5
 α_{5}(effect of the pressure transverse) = 1
α_{5} = 1  0.04p = 1 – 0.04·0 = 1 (≥ 0.7 and ≤ 1.0)  α_{6} = 1.41
α_{6} = (ρ_{1}/25)^{0.5} = (50/25)^{0.5} = 1.41 (≥ 1.0 and ≤ 1.5)  l_{0,min} = max(205 ; 180 ; 200) = 205 mm
l_{0,min} = max{0.3·α_{6}·l_{b,rqd}; 15·φ; 200 mm}
2) Compression bar
l_{0} (Lap length) = 685 mm
l_{0} = α_{1} · α_{2} · α_{3} · α_{6} · l_{b,rqd} ≥ l_{0,min}
where:
 l_{b,rqd} (basic required anchorage length) = 484 mm
 α_{1} = 1, α_{2} = 1, α_{3} = 1, α_{6} = 1.41
 l_{0,min} = 205 mm
Poor Bond conditions
1) Tensile bar
l_{0} (Lap length) = 697 mm
l_{0} = α_{1} · α_{2} · α_{3} · α_{5} · α_{6} · l_{b,rqd} ≥ l_{0,min}
where:
 l_{b,rqd} (basic required anchorage length) = 692 mm
l_{b,rqd} = (Φ / 4) (σ_{sd} / f_{bd})
with: σ_{sd} = r · f_{yk}/γ_{s} = 1·500/1.15 = 434.78 MPa
 f_{bd} (ultimate bond stress) = 1.89 MPa
f_{bd} = 2.25 · η_{1} · η_{2} · f_{ctd} η_{1} = 0.7, η_{2} = 1.0
 f_{ctd} = α_{ct}·f_{ctk,0.05}/γ_{c} = 1·1.8/1.5 = 1.2 MPa
 f_{ctk,0.05} = 0.21· f_{ck}^{(2/3)} = 0.21·25^{(2/3)} = 1.8 MPa
 α_{1}(effect of the form of the bars) = 1
 α_{2}(effect of concrete minimum cover) = 0.71
α_{2} = 10.15 (Cd  φ)/ φ = 10.15·(3512)/12 = 0.71
(≥ 0.7 and ≤ 1.0)  α_{3}(effect of transverse reinforcement) = 1
α_{3} = 1  Kλ = 1.05 (≥ 0.7 and ≤ 1.0)
with: λ = (∑A_{st}  ∑A_{st,min}) / A_{s} = (0.57  1.13) / 1.13 = 0.5
∑A_{st,min} = 1.0·As(σ_{sd} / f_{yd}) = 1.13 · 1  K = 0.1 (Form transverse bars Links or U bars)
 λ = (∑A_{st}  ∑A_{st,min}) / A_{s} = (0.57  1.13) / 1.13 = 0.5
 α_{5}(effect of the pressure transverse) = 1
α_{5} = 1  0.04p = 1 – 0.04·0 = 1 (≥ 0.7 and ≤ 1.0)  α_{6} = 1.41
α_{6} = (ρ_{1}/25)^{0.5} = (50/25)^{0.5} = 1.41 (≥ 1.0 and ≤ 1.5)  l_{0,min} = max(294 ; 180 ; 200) = 294 mm
l_{0,min} = max{0.3·α_{6}·l_{b,rqd}; 15·φ; 200 mm}
2) Compression bar
l_{0} (Lap length) = 978 mm
l_{0} = α_{1} · α_{2} · α_{3} · α_{6} · l_{b,rqd} ≥ l_{0,min}
where:
 l_{b,rqd} (basic required anchorage length) = 692 mm
 α_{1} = 1, α_{2} = 1, α_{3} = 1, α_{6} = 1.41
 l_{0,min} = 294 mm