Anchorage of longitudinal reinforcement
RESULT
Design anchorage length l_{bd} (cm)  
Bar diameter Φ (mm)  Tension  Compression  
Good bond  Poor bond  Good bond  Poor bond  
12  35  50  49  70 
Notes (see Figure 8.2 of EC2):

DETAILS OF CALCULATION
Notation and methodology according to clause 8.4 of EC2
Good Bond conditions
1) Tensile bar
l_{bd} (design anchorage length) = 345 mm
l_{bd} = α_{1} · α_{2} · α_{3} · α_{5} · l_{b,rqd} ≥ l_{b,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
(It is not considered)  α_{5}(effect of the pressure transverse) = 1
α_{5} = 1  0.04p = 1 – 0.04·0 = 1 (≥ 0.7 and ≤ 1.0)  α_{2} · α_{3} · α_{5} = 0.71 (≥ 0.7)
 l_{b,min} = max(145 ; 120 ; 100) = 145 mm
l_{b,min} = max{0.3·l_{b,rqd}; 10·φ; 100 mm}
2) Compression bar
l_{bd} (design anchorage length) = 484 mm
l_{bd} = α_{1} · α_{2} · α_{3} · l_{b,rqd} ≥ l_{b,min}
where:
 l_{b,rqd} (basic required anchorage length) = 484 mm
 α_{1} = 1, α_{2} = 1, α_{3} = 1
 l_{b,min} = max(291 ; 120 ; 100) = 291 mm
l_{b,min} = max{0.6·l_{b,rqd}; 10·φ; 100 mm}
Poor Bond conditions
1) Tensile bar
l_{bd} (design anchorage length) = 493 mm
l_{bd} = α_{1} · α_{2} · α_{3} · α_{5} · l_{b,rqd} ≥ l_{b,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
(It is not considered)  α_{5}(effect of the pressure transverse) = 1
α_{5} = 1  0.04p = 1 – 0.04·0 = 1 (≥ 0.7 and ≤ 1.0)  α_{2} · α_{3} · α_{5} = 0.71 (≥ 0.7)
 l_{b,min} = max(208 ; 120 ; 100) = 208 mm
l_{b,min} = max{0.3·l_{b,rqd}; 10·φ; 100 mm}
2) Compression bar
l_{bd} (design anchorage length) = 692 mm
l_{bd} = α_{1} · α_{2} · α_{3} · l_{b,rqd} ≥ l_{b,min}
where:
 l_{b,rqd} (basic required anchorage length) = 692 mm
 α_{1} = 1, α_{2} = 1, α_{3} = 1
 l_{b,min} = max(415 ; 120 ; 100) = 415 mm
l_{b,min} = max{0.6·l_{b,rqd}; 10·φ; 100 mm}