PROFIS ENGINEERING
Nominal Concrete Breakout Strength in 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 = 12.0 in 1.5h ef = 18.0 in 3.0h ef = 36.0 in
A Nc =( c x- + s x1 + s x2 + s x3 + c x +)( c y- + s y12 + s y23 + s x34 + c y +) =( 18.0 + 9.0 + 9.0 + 9.0 + 18.0)( 6.0 + 8.0 + 8.0 + 8.0 + 18.0) = 3024 in 2
A Nc0 = 9( h ef) 2 c x- = ∞ in = 1.5h ef
ACI 318-19 Fig. R17.6.2.1 b) c x + = ∞ in = 1.5h ef c y- = 6 in
s x1 = 9 in s x2 = 9 in s x3 = 9 in s y12 = 8 in s y23 = 8 in s y34 = 8 in c y + = ∞ in = 1.5h ef
= 9( 12 in) 2 ACI 318-19 Eq.( 17.6.2.1.4) ψ ec, N =
= 1296 in 2 1 1 + e ′ N
1.5h e f
ACI 318-19 Eq.( 17.6.2.3.1) ψ ec, N = 1.0 no eccentricity e’ N = 0 in 1.5h ef = 18.0 in 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 in / 18.0 in) = 0.8 c a, min = c y- = 6.0 in 1.5h ef = 18.0 in
ψ c, N = 1.0 cracked or uncracked concrete Reference 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 cracked concrete N b = k c, xxxx λ a f ′ c h e f
1.5
normal weight concrete: λ = 1.0 → λ a = 1.0
N cbg =
N b =( 17)( 1.0)( 4500 psi) 0. 5( 12.0 in) 1. 5
A Nc A Nc0
= 47,405 lb
ψ ec, N ψ ed, N ψ c, N ψ cp, N N b
N cbg =( 3024 in 2 / 1296 in 2)( 1.0)( 0.8)( 1.0)( 1.0)( 47,405 lb) = 88,489 lb
Reference ACI 318-19 17.6.2.6.1.
ACI 318-19 Eq.( 17.6.2.2.1) ACI 318-19 Table 17.2.4.1 k c, cr = 17 Reference ESR-3814 Table 7 f ' c = 4500 psi h ef = 12.0 in
A Nc = 3024 in 2 A Nc0 = 1296 in 2 ψ ec, N = 1.0 ψ ed, N = 0.8 ψ c, N = 1.0 ψ cp, N = 1.0 N b = 47,405 lb
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