PROFIS ENGINEERING
Nominal Concrete Breakout Strength in Shear( V cbg)
Check shear towards to the x- edge from Col 3,4.
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 4 and V ua, col 34 = 1.0V ua, x
ACI 318-19 Eq.( 17.7.2.2.1a) l e = MIN h e f | 8d a
= 3.25 in h ef = 3.25 in CONTROLS d a = 0.5 in 8d a = 4.0 in
normal weight concrete: λ = 1.0 → λ a = 1.0 ACI 318-19 Table 17.2.4.1
V b = [( 7)( 3.25 in / 0.5 in) 0. 2( 0.5 in) 0. 5 ]( 1.0)( 3000 psi) 0. 5( 5.0 in) 1. 5
= 4407 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 3, 4 = 5 in
V cbg =
V b =( 9)( 1.0)( 3000 psi) 0. 5( 5.0 in) 1. 5 = 5511 lb check: design V b = MIN { 4407 lb; 5511 lb }
A Vc A Vc0
= 4407 lb
ψ ec, V ψ ed, V ψ c, V ψ h, V ψ parallel, V V b
V cbg, col 13 =( 112.2 in 2 / 112.5 in 2)( 0.976)( 0.98)( 1.0)( 1.118)( 1.0)( 4407 lb)
V cbg =
A Vc A Vc0
= 4700 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 3, 4 = 5 in |
ACI 318-19 17.7.2.2 |
A Vc = 112.2 in 2 |
A Vc0 = 112.5 in 2 |
ψ ec, V = 0.976 |
ψ ed, V = 0.98 |
ψ c, V = 1.0 |
ψ h, V = 1.118 |
ψ parallel, V = 1.0 |
V b = 4407 lb |
A Vc = 112.2 in 2 |
A Vc0 = 112.5 in 2 |
ψ ec, V = 0.976 |
ψ ed, V = 1.0 |
ψ c, V = 1.0 |
ψ h, V = 1.118 |
V cbg //, col 13 =( 112.2 in 2 / 112.5 in 2)( 0.976)( 1.0)( 1.0)( 1.118)( 2.0)( 4407 lb)
= 9592 lb ψ parallel, V = 2.0
V b = 4407 lb
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