PROFIS ENGINEERING
Nominal Concrete Breakout Strength in Shear( V cbg)
Check shear towards to the x- edge from Col 1,3.
V b = 7 l e d a
0.2 d a λ a f ′ c c a1
1.5
CASE 3 c a1 is taken from anchor 3 and V ua, col 1, 3 = 1.0V ua, x
ACI 318-19 Eq.( 17.7.2.2.1a) l e = MIN h e f | 8d a
= 4.0 in h ef = 4.0 in CONTROLS d a = 0.625 in 8d a = 5.0 in
normal weight concrete: λ = 1.0 → λ a = 1.0 ACI 318-19 Table 17.2.4.1
V b = [( 7)( 4.0 in / 0.625 in) 0. 2( 0.625 in) 0. 5 ]( 1.0)( 3000 psi) 0. 5( 10.0 in) 1. 5
= 13,894 lb V b = 9λ a f ′ c c a1
1.5 f’ c = 3000 psi
ACI 318-19 Eq.( 17.7.2.2.1b)
c a1, col 1, 3 = 10 in
V cbg =
V b =( 9)( 1.0)( 3000 psi) 0. 5( 10.0 in) 1. 5 = 15,588 lb check: design V b = MIN { 13,894 lb; 15,588 lb }
A Vc A Vc0
= 13,894 lb Shear towards x- edge
ψ ec, V ψ ed, V ψ c, V ψ h, V ψ parallel, V V b
V cbg, col 13 =( 171.6 in 2 / 450 in 2)( 0.991)( 0.82)( 1.0)( 1.58)( 1.0)( 13,894 lb)
V cbg =
A Vc A Vc0
= 6803 lb Shear parallel to x- edge
ψ ec, V ψ ed, V ψ c, V ψ h, V ψ parallel, V V b
f’ c = 3000 psi |
c a1, col 1, 3 = 10 in |
ACI 318-19 17.7.2.2 |
A Vc = 171.6 in 2 |
A Vc0 = 450 in 2 |
ψ ec, V = 0.991 |
ψ ed, V = 0.82 |
ψ c, V = 1.0 |
ψ h, V = 1.58 |
ψ parallel, V = 1.0 |
V b = 13,894 lb |
A Vc = 171.6 in 2 |
A Vc0 = 450 in 2 |
ψ ec, V = 0.991 |
ψ ed, V = 1.0 |
ψ c, V = 1.0 |
ψ h, V = 1.58 |
V cbg //, col 13 =( 171.6 in 2 / 450 in 2)( 0.991)( 1.0)( 1.0)( 1.58)( 2.0)( 13,894 lb)
= 16,592 lb ψ parallel, V = 2.0
V b = 13,894 lb
October 2025 148