PROFIS ENGINEERING
V cbg =
A Vc A Vc0 ψ ec, V ψ ed, V ψ c, V ψ h, V ψ parallel, V V b ACI 318-19 Eq.( 17.7.2.1b)
A Vc =( c x- + s x12 + s x23 + s x34 + c x +)( 1.5c a1, row 1 or h concrete) ACI 318-19 Fig. R17.7.2.1 b)
CASE 3 applies for row 1 and row 2: calculate V cbg from row 1 using c a1, row 1.( orange shaded area) A Vc =( c x- + s x12 + s x23 + s x34 + c x +)( 1.5c a1, row 1 or h concrete) 1.5c a1, row 1 < h concrete use 1.5c a1, row 1 to calculate A Vc A Vc =( 1.5c a1, row 1 + 9 + 9 + 9 + 1.5c a1, row 1)( 1.5c a1, row 1) = 405 in 2
Check: c a1, row 1 = 6.0 in s y, 12 = 4.0 in s y, 23 = 8.0 in s y, 12 < c a1, row 1 CASE 3 applies for row 1 and row 2( orange shaded area) s y, 23 > c a1, row 1 CASE 1 / CASE 2 applies for row 1 and row 3( yellow shaded area) s x. nn < 3c a1, row 1 anchors in each row act as a group c a1, row 1 = 6 in 1.5c a1, row 1 = 9 in 3.0c a1, row 1 = 18 in c x- = ∞ = 1.5c a1, row 1 c x + = ∞ = 1.5c a1, row 1 h concrete = 12 in
s x12 = 9.0 in s x23 = 9.0 in s x34 = 9.0 in A Vc0 = 4.5( c a1, row 1) 2 = 4.5( 6 in) 2 = 162 in 2 ACI 318-19 Eq.( 17.7.2.1.3)
ψ ec, V =
1 1 + e ′ V
1.5c a1
= 1.0 no eccentricity ACI 318-19 Eq.( 17.7.2.3.1) e cV = 0.0 in ψ ed, V = 0.7 + 0.3 c a2, min 1.5c a1
= 1.0
ACI 318-19 Eq.( 17.7.2.4.1b) c a2, min = ∞ 1.5c a1, row 1 = 9 in ψ c, V = 1.2 cracked concrete with # 5 edge bars Reference ACI 318-19 17.7.2.5.1 ψ parallel, V = 1.0 shear load acts towards y- edge Reference ACI 318-19 17.7.2.1( c).
ψ h, V =
1. 5c a1 h a
h concrete ≥ 1.5c a1, row 1 → ψ h, V = 1.00
ACI 318-19 Eq.( 17.7.2.6.1) c a1, row 1 = 6 in 1.5c a1, row 1 = 9 in h a = h concrete = 12 in
October 2025 97