Anchorage of longitudinal reinforcement
RESULT
Design anchorage length lbd (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
lbd (design anchorage length) = 345 mm
lbd = α1 · α2 · α3 · α5 · lb,rqd ≥ lb,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
(It is not considered) - α5(effect of the pressure transverse) = 1
α5 = 1 - 0.04p = 1 – 0.04·0 = 1 (≥ 0.7 and ≤ 1.0) - α2 · α3 · α5 = 0.71 (≥ 0.7)
- lb,min = max(145 ; 120 ; 100) = 145 mm
lb,min = max{0.3·lb,rqd; 10·φ; 100 mm}
2) Compression bar
lbd (design anchorage length) = 484 mm
lbd = α1 · α2 · α3 · lb,rqd ≥ lb,min
where:
- lb,rqd (basic required anchorage length) = 484 mm
- α1 = 1, α2 = 1, α3 = 1
- lb,min = max(291 ; 120 ; 100) = 291 mm
lb,min = max{0.6·lb,rqd; 10·φ; 100 mm}
Poor Bond conditions
1) Tensile bar
lbd (design anchorage length) = 493 mm
lbd = α1 · α2 · α3 · α5 · lb,rqd ≥ lb,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
(It is not considered) - α5(effect of the pressure transverse) = 1
α5 = 1 - 0.04p = 1 – 0.04·0 = 1 (≥ 0.7 and ≤ 1.0) - α2 · α3 · α5 = 0.71 (≥ 0.7)
- lb,min = max(208 ; 120 ; 100) = 208 mm
lb,min = max{0.3·lb,rqd; 10·φ; 100 mm}
2) Compression bar
lbd (design anchorage length) = 692 mm
lbd = α1 · α2 · α3 · lb,rqd ≥ lb,min
where:
- lb,rqd (basic required anchorage length) = 692 mm
- α1 = 1, α2 = 1, α3 = 1
- lb,min = max(415 ; 120 ; 100) = 415 mm
lb,min = max{0.6·lb,rqd; 10·φ; 100 mm}