Profis Engineering rupture du béton en cisaillement | Page 95

PROFIS ENGINEERING
Force d’ adhérence nominale à la traction( N ag)
N ag =
A Na A Na0 ψ ec, Na ψ ed, Na ψ cp, Na N ba ACI 318-19 Eq.( 17.6.5.1b)
c Na = 10d a τ uncr 1100
ACI 318-19 Eq.( 17.6.5.1.2b) conception τ k, uncr =( ESR τ k, uncr)( f’ c / 2 500 psi) 0, 1 =( 2 220 psi)( 5 000 psi / 2 500 psi) 0, 1 = 2 379 psi c Na =( 10)( 0,75 po)( 2 379 psi / 1 100 psi) 0, 5 = 11,03 po
A Na =( c x- + s x1 + s x2 + s x3 + c x +)( c y- + s y12 + s y23 + s x34 + c y +) =( 11,03 + 9,0 + 9,0 + 9,0 + 11,03)( 6,0 + 4,0 + 8,0 + 8,0 + 11,03) = 1 816,69 po 2
Consultez la norme ESR-4868 tableau 15 pour les valeurs τ k, uncr et() xx. c x- = ∞ po = c Na
ESR τ k, uncr = 2 220 psi d a = 0,75 po() xx = 0,10 f’ c = 5000 psi
ACI 318-19 Fig. R17.6.5.1( b) c x + = ∞ po = c Na 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 Na0 =( 2c Na) 2 =( 22,06 po) 2 = 486,64 po 2 ACI 318-19 Eq.( 17.6.5.1.2a)
ψ ec, Na =
1 1 + e ′ N c Na
= excentricité de 0,917 dans la direction y c Na = 11,03 po 2c Na = 22,06 po
ACI 318-19 Éq.( 17.6.5.3.1) e’ N = 1,0 po c Na = 11,03 po c y + = ∞ po = c Na ψ ed, Na = 0.7 + 0.3 c a, min c Na
= 0,7 +( 0,3)( 6,0 po / 11,03 po) = 0,863
Consultez ACI 318-19 Eq.( 17.6.5.4.1b) c a, min = c y- = 6,0 po c Na = 11,03 po ψ cp, Na = MAX c a, min | c Na
= 1,0 béton fissuré Consultez ACI 318-19 Eq.( 17.6.5.5.1b) c ac c ac
N ba = λ a α N, seis τ xxxx πd a h e f
ACI 318-19 Éq.( 17.6.5.2.1)
Consultez la norme ESR-4868 tableau 15 pour les valeurs α N, seis. Béton de poids normal: λ = 1,0 → λ a = 1,0 ACI 318-19 Tableau 17.2.4.1 conception τ k, cr =( ESR τ k, cr)( f’ c / 2 500 psi) 0, 15 =( 1 260 psi)( 5 000 psi / 2 500 psi) 0, 1 = 1 350 psi
Consultez la norme ESR-4868 tableau 15 pour les valeurs τ k, cr et() xx.
ESR τ k, cr = 1 260 psi f’ c = 5 000 psi() xx = 0,1 N ba =( 1,0)( 1,0)( 1 350 psi)( π)( 0,75 po)( 10,0 po) = 31 809 lb d a = 0,75 po α N, seis = 1,0 h ef = 10 po
N ag =
A Na A Na0 ψ ec, Na ψ ed, Na ψ cp, Na N ba
N ag =( 1 816,69 po 2 / 486,64 po 2)( 0,917)( 0,863)( 1,0)( 31 809 lb) = 93 973 lb
A Na = 1 816,69 po 2 A Na0 = 486,64 po 2 ψ ec, Na = 0,917 ψ ed, Na = 0,863 ψ cp, Na = 1,0 N ba = 31 809 lb
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