PROFIS ENGINEERING
Tableau 36. Résistance nominale à l’ acier en tension( N sa) N sa = valeur ESR ESR-4868 Tableau 11
N sa = 13 110 lb / ancrage
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 = 8,0 po 1,5 h ef = 12,0 po 3,0 h ef = 24,0 po
A Nc =( c x- + s x + c x +)( c y- + s y12 + s y23 + s x34 + c y +) =( 4,0 + 6,0 + 8,0)( 12,0 + 8,0 + 8,0 + 8,0 + 12,0)
. ec, N =
= 864 po 2 c x- = 4 po
s x = 6 po
ACI 318-19 Fig. R17.6.2.1 b)
c x + = 8 po c y- = ∞ = 1,5h ef
s y12 = 8 po s y23 = 8 po s y34 = 8 po A Nc0 = 9( h ef) 2 = 9( 8 po) 2 = 576 po 2 ACI 318-19 Éq.( 17.6.2.1.4)
1 1 + e ′ N
1,5h e f
= 1.0 aucune excentricité de tension ACI 318-19 Éq.( 17.6.2.3.1) e’ N = 0 po ψ ed, N = 0.7 + 0.3 c a, min 1.5h e f
= 0,7 +( 0,3)(( 4,0 po / 12,0 po) = 0,80
ACI 318-19 Eq.( 17.6.2.4.1b) c a, min = c x- = 4,0 po 1,5 h ef = 12,0 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
= 1.0 béton fissuré Consultez ACI 318-19 17.6.2.6.1. c ac c ac c y + = ∞ po = 1,5h ef
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)( 4 000 psi) 0, 5( 8,0 po) 1, 5 = 24 328 lb
N cbg =
A Nc A Nc0 ψ ec, N ψ ed, N ψ c, N ψ cp, N N b
N cbg =( 864 po 2 / 576 po 2)( 1,0)( 0,80)( 1,0)( 1,0)( 24 328 lb) = 29 194 lb
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 = 4 000 psi h ef = 8,0 po
A Nc = 864 po 2 A Nc0 = 576 po 2 ψ ec, N = 1,0 ψ ed, N = 0,80 ψ c, N = 1,0 ψ cp, N = 1,0 N b = 24 328 lb
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