PROFIS ENGINEERING
Nominal Bond Strength in Tension( 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
design τ k, uncr =( ESR τ k, uncr)( f ' c / 2500 psi) 0. 1 =( 2220 psi)( 4000 psi / 2500 psi) 0. 1 = 2327 psi c Na =( 10)( 0.625 in)( 2327 psi / 1100 psi) 0. 5 = 9.09 in A Na =( c x- + s x + c x +)( c y- + s y12 + s y23 + s x34 + c y +) =( 4 in + 6.0 in + 8.0 in)( 9.09 in + 8.0 in + 8.0 in + 8.0 in + 9.09 in) = 759.24 in 2
ACI 318-19 Eq.( 17.6.5.1.2b) Reference ESR-4868 Table 15 for τ k, uncr and() xx. ESR τ k, uncr = 2220 psi d a = 0.625 in() xx = 0.10 f’ c = 4000 psi
ACI 318-19 Fig. R17.6.5.1( b) c x- = 4 in c x + = 8 in c y- = c Na c y + = c Na s x = 6 in s y12 = 8 in s y23 = 8 in s y34 = 8 in ψ ec, Na =
A Na0 =( 2c Na) 2 =( 18.18 in) 2 = 330.51 in 2 ACI 318-19 Eq.( 17.6.5.1.2a)
1 c Na = 9.09 in 2c Na = 18.18 in
1 + e ′ N c Na
= 1.0 no tension eccentricity ACI 318-19 Eq.( 17.6.5.3.1) e’ N = 0.0 in ψ ed, Na = 0.7 + 0.3 c a, min c Na
= 0.7 +( 0.3)( 4.0 in / 9.09 in) = 0.832
Reference ACI 318-19 Eq.( 17.6.5.4.1b) c a, min = c x- = 4.0 in c Na = 9.09 in ψ cp, Na = MAX c a, min | c Na
= 1.0 cracked concrete Reference 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 Eq.( 17.6.5.2.1)
Reference ESR-4868 Table 15 for α N, seis. normal weight concrete: λ = 1.0 → λ a = 1.0 ACI 318-19 Table 17.2.4.1 design τ k, cr =( ESR τ k, cr)( f ' c / 2500 psi) 0. 15
Reference ESR-4868 Table 15 for τ k, cr.
=( 1170 psi)( 4000 psi / 2500 psi) 0. 1 = 1226 psi N ba =( 1.0)( 1.0)( 1226 psi)( π)( 0.625 in)( 8.0 in) = 19,258 lb
N ag =
A Na A Na0 ψ ec, Na ψ ed, Na ψ cp, Na N ba
N ag =( 759.24 in 2 / 330.51 in 2)( 1.0)( 0.832)( 1.0)( 19,258 lb) = 36,807 lb
Reference ESR-4868 Table 15 for() xx. τ k, cr = 1170 psi f ' c = 4000 psi() xx = 0.1 d a = 0.625 in α N, seis = 1.0 h ef = 8.0 in
A Na = 759.24 in 2 A Na0 = 330.51 in 2 ψ ec, Na = 1.0 ψ ed, Na = 0.832 ψ cp, Na = 1.0 N ba = 19,258 lb
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