PROFIS ENGINEERING
Tableau 12. Résistance nominale à l’ acier en tension( N sa) N sa = valeur ESR ESR-4266 Tableau 4
N sa = 41 365 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 = 5,75 po 1,5 h ef = 8,625 po 3,0 h ef = 17,25 po
A Nc =( c x- + s x1 + s x2 + s x3 + c x +)( c y- + s y12 + s y23 + s x34 + c y +) =( 8,625 + 9,0 + 9,0 + 9,0 + 8,625)( 8,0 + 6,0 + 6,0 + 6,0 + 8,625) = 1 532,16 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 = 6 po s y23 = 6 po s y34 = 6 po
A Nc0 = 9( h ef) 2 = 9( 5,75 po) 2 = 297,56 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 = 1,0 pas d’ excentricité e’ N = 0 po 1,5 h ef = 8,625 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)(( 8,0 po / 8,625 po) = 0,978 c a, min = c y- = 8,0 po 1,5 h ef = 8,625 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
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 =( 21)( 1,0)( 4 000 psi) 0, 5( 5,75 po) 1, 5 = 18 313 lb
N cbg =
A Nc A Nc0 ψ ec, N ψ ed, N ψ c, N ψ cp, N N b
N cbg =( 1 532,16 po 2 / 297,56 po 2)( 1,0)( 0,978)( 1,0)( 1,0)( 18 313 lb) = 92 221 lb
ACI 318-19 Éq.( 17.6.2.2.1) ACI 318-19 Tableau 17.2.4.1 k c, cr = 21 Consultez la norme ESR-4266 tableau 4 f’ c = 4 000 psi h ef = 5,75 po
A Nc = 1 532,16 po 2 A Nc0 = 297,56 po 2 ψ ec, N = 1,0 ψ ed, N = 0,978 ψ c, N = 1,0 ψ cp, N = 1,0 N b = 18 313 lb
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