PROFIS ENGINEERING
Nominal Concrete Breakout Strength in Shear( V cbg)
V cbg =
Case 3 for col 1, col 2 and col 3: c a1 taken from col 1
Reference red shaded area.
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) c a1, col 1 = 12.0 in s x12 = 9.0 in s x23 = 9.0 in
Assume V ua acts on anchor column 1. Reference red shaded area. s x12 < c a1, col 1 Case 3 applies for col 1 and col 2 s x23 < c a1, col 1 Case 3 applies for col 2 and col 3 s y, 12 = 4.0 in s y, 23 = s y. 34 = 8.0 in
1.5c a1, col 1 > h concrete use h concrete to calculate A Vc A Vc =( c y + + s y34 + s y23 + s y12 + c y-)( h concrete) A Vc =( 1.5c a1, col 1 + 8 in + 8 in + 4 in + 6 in)( h concrete) = 528 in 2 s y, 12, s y, 23, s y. 34 < 3c a1 col 1: anchors in each column act as group c a1, col 1 = 12 in 1.5c a1, col 1 = 18 in 3.0c a1, col 1 = 36 in c y + = ∞ → = 1.5c a1col 1 c y- = c a2, y = 6 in h concrete = 12 in
s y12 = 4.0 in s y23 = 8.0 in s y34 = 8.0 in A Vc0 = 4.5( c a1, row 1) 2 = 4.5( 12 in) 2 = 648 in 2 ACI 318-19 Eq.( 17.7.2.1.3)
ψ ec, V =
1 1 + e ′ V
1.5c a1
= 0.989 ACI 318-19 Eq.( 17.7.2.3.1) e cV, y = 0.21 in ψ ed, V = 0.7 + 0.3 c a2, min 1.5c a1
= 0.8
ACI 318-19 Eq.( 17.7.2.4.1b) c a2, min = c a2, y = 6 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 x- edge Reference ACI 318-19 17.7.2.1( c).
ψ h, V =
1. 5c a1 h a
= 18in / 12in 0. 5 = 1.23
ACI 318-19 Eq.( 17.7.2.6.1) h a = h concrete = 12 in h concrete < 1.5c a1, col 1 → calculate ψ h, V
October 2025 125