PROFIS ENGINEERING
Nominal
Concrete Breakout Strength in Shear( V cbg)
V cbg =
Case 1 / Case 2 for row 3: c a1 taken from row 3
Reference yellow 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 3 or h concrete) ACI 318-19 Fig. R17.7.2.1 b)
Assume V ua acts on anchor Row 3. Reference yellow shaded area. A Vc =( c x- + s x12 + s x23 + s x34 + c x +)( 1.5c a1, row 3 or h concrete) 1.5c a1, row 3 > h concrete use h concrete to calculate A Vc A Vc =( 12 in + 9 in + 9 in + 9 in + 1.5c a1, row 3)( h concrete) = 792 in 2 c a1, row 1 = 6.0 in s y12 = 4.0 in s y23 = 8.0 in s y12 < c a1, row 1 Case 3 applies for row 1 and row 2 s y23 > c a1, row 1 Case 1 / Case 2 applies for row 1 and row 3 s x, 12 = s x, 23 = s x. 34 < 3c a1, row 1: anchors in each row act as group c a1, row 3 = 18 in 1.5c a1, row 3 = 27 in 3.0c a1, row 3 = 54 in
c x- = c a2, x = 12 in 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 3) 2 = 4.5( 18 in) 2 = 1458 in 2 ACI 318-19 Eq.( 17.7.2.1.3)
ψ ec, V =
1 1 + e ′ V
1.5c a1
= 0.987 ACI 318-19 Eq.( 17.7.2.3.1) e cV, x = 0.361 in 1.5c a1, row 3 = 27 in ψ ed, V = 0.7 + 0.3 c a2, min 1.5c a1
= 0.833
ACI 318-19 Eq.( 17.7.2.4.1b) c a2, min = c a2, x < 1.5c a1, row 3 → calculate ψ ed, V 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
= 27in / 12in 0. 5 = 1.5
ACI 318-19 Eq.( 17.7.2.6.1) h a = h concrete = 12 in h concrete < 1.5c a1, row 3 → calculate ψ h, V
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