PROFIS ENGINEERING
Résistance nominale à la rupture du béton en tension( N cbg)
N cbg =
A ACI 318-19 Eq.( 17.6.2.1b) Nc ψ A ec, N ψ ed, N ψ c, N ψ cp, N N b Nc0 h ef = 10,0 po 1,5 h ef = 15,0 po 3,0 h ef = 30,0 po
A Nc =( c x- + s x1 + s x2 + s x3 + c x +)( c y- + s y12 + s y23 + s x34 + c y +) =( 15,0 + 9,0 + 9,0 + 9,0 + 15,0)( 6,0 + 4,0 + 8,0 + 8,0 + 15,0) = 2 337 po 2 c x- = ∞ po = 1,5 h ef
ACI 318-19 Fig. R17.6.2.1 b) c x + = ∞ po = 1,5 h ef c y- = 6 po
s x1 = 9 po s x2 = 9 po s x3 = 9 po s y12 = 4 po s y23 = 8 po s y34 = 8 po
A Nc0 = 9( h ef) 2 = 9( 10 in) 2 = 900 po 2 ACI 318-19 Éq.( 17.6.2.1.4)
ψ ec, N =
1 1 + e ′ N
1.5h e f
ACI 318-19 Éq.( 17.6.2.3.1) c y + = ∞ po = 1,5 h ef ψ ec, N = 0,938 excentricité dans la direction y e’ N = 1 po 1,5 h ef = 15,0 po ψ ed, N = 0.7 + 0.3 c a, min 1.5h e f
ACI 318-19 Eq.( 17.6.2.4.1b) ψ ed, N = 0,7 +( 0,3)(( 6,0 po / 15,0 po) = 0,82 c a, min = c y- = 6,0 po 1,5 h ef = 15,0 po ψ c, N = 1,0 béton fissuré ou non fissuré Consultez ACI 318-19 17.6.2.5.1
ψ cp, N = MAX c a, min
| 1. 5h e f c ac c ac
ψ cp, N = 1,0 béton fissuré
N b = k c, xxxx λ a f ′ c h e f
1.5
Béton de poids normal: λ = 1,0 → λ a = 1,0 N b =( 17)( 1,0)( 5000 psi) 0, 5( 10,0 po) 1, 5 = 38 013 lb
N cbg =
A Nc A Nc0 ψ ec, N ψ ed, N ψ c, N ψ cp, N N b
N cbg =( 2 337 po 2 / 900 po 2)( 0,938)( 0,82)( 1,0)( 1,0)( 38 013 lb) = 75 922 lb
Consultez ACI 318-19 17.6.2.6.1.
ACI 318-19 Éq.( 17.6.2.2.1) ACI 318-19 Tableau 17.2.4.1 k c, cr = 17 Consultez la norme ESR-4868 tableau 12 f’ c = 5 000 psi h ef = 10,0 po
A Nc = 2 337 po 2 A Nc0 = 900 po 2 ψ ec, N = 0,938 ψ ed, N = 0,82 ψ c, N = 1,0 ψ cp, N = 1,0 N b = 38 013 lb
[ en ] October 2025 94