PROFIS ENGINEERING
V b = 7 l e d a
0.2 d a λ a f ′ c c a1
1.5
ACI 318-19 Eq.( 17.7.2.2.1a)
l e = MIN h e f | 8d a = 5.0in h ef = 8.0 in d a = 0.625 in
normal weight concrete: λ = 1.0 → λ a = 1.0 ACI 318-19 Table 17.2.4.1
V b = [( 7)( 5.0 in / 0.625 in) 0. 2( 0.625 in) 0. 5 ]( 1.0)( 4000 psi) 0. 5( 8.0 in) 1. 5
= 12,004 lb V b = 9λ a f ′ c c a1
1.5 f’ c = 4000 psi
ACI 318-19 Eq.( 17.7.2.2.1b)
c a1, col 2 = 8.0 in
V b =( 9)( 1.0)( 4000 psi) 0. 5( 8.0 in) 1. 5 = 12,880 lb f’ c = 4000 psi c a1, col 2 = 8.0 in
8d a = 5.0 in CONTROLS
V cbg =
check: design V b = MIN { 12,004 lb; 12,880 lb }
A Vc A Vc0
= 12,004 lb
ψ ec, V ψ ed, V ψ c, V ψ h, V ψ parallel, V V b
V cbg, col 2 =( 576 in 2 / 288 in 2)( 1.0)( 1.0)( 1.0)( 1.0)( 2.0)( 12,004 lb)
= 48,016 lb
ACI 318-19 17.7.2.2 |
A Vc = 576 in 2 |
A Vc0 = 288 in 2 |
ψ ec, V = 1.0 |
ψ ed, V = 1.0 |
ψ c, V = 1.0 |
ψ h, V = 1.0 |
ψ parallel, V = 2.0 |
V b = 12,004 lb |
Summary of shear concrete breakout calculations.
October 2025 139