PROFIS ENGINEERING
Nominal
Concrete Breakout Strength in Shear( V cbg)
V cbg =
Shear parallel to x- edge.
Case 1 / Case 2 for col 2: c a1 taken from col 2
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 y- + s y12 + s y23 + s y34 + c y +)( 1.5c a1, col 2 or h concrete) ACI 318-19 Fig. R17.7.2.1 b)
Assume 1.0V ua acts on col 2. Reference yellow shaded area. A Vc =( c y- + s y12 + s y23 + s y34 + c y +)( 1.5c a1, col 2 or h concrete) 1.5c a1, col 2 < h concrete use 1.5c a1, col 2 to calculate A Vc A Vc =( 1.5c a1, col 2 + 8in + 8in + 8in + 1.5c a1, col 2)( 1.5c a1, col 2) = 810 in 2 c a1, col 1 = 4.0 in s x, 12 = 6.0 in s x, 12 > c a1, col 1 Case 1 / Case 2 applies for col 1 and col 2 s y, 12 = s y, 23 = s y. 34 < 3c a1, col 1: anchors in each row act as group c a1, col 2 = 10 in 1.5c a1, col 2 = 15 in 3.0c a1, col 2 = 30 in c y- = ∞ → = 1.5c a1, col 2 c y + = ∞ → = 1.5c a1, col 2 h concrete = 24 in
s y, 12 = 8.0 in s y, 12 = 8.0 in s y, 12 = 8.0 in A Vc0 = 4.5( c a1, col 2) 2 = 4.5( 10 in) 2 = 450 in 2 ACI 318-19 Eq.( 17.7.2.1.3)
ψ ec, V =
1 1 + e ′ V
1.5c a1
= 1.0 ACI 318-19 Eq.( 17.7.2.3.1) e cV = 0 in ψ ed, V = 0.7 + 0.3 c a2, min 1.5c a1 shear ∣∣ edge → ψ ed, V = 1.0 ACI 318-19 17.7.2.1c)
ψ c, V = 1.0 no edge reinforcement Reference ACI 318-19 17.7.2.5.1 ψ parallel, V = 2.0 shear load acts parallel to x- edge Reference ACI 318-19 17.7.2.1( c).
ψ h, V =
1. 5c a1 h a h concrete > 1.5c a1, col 2 ψ h, V = 1.00
ACI 318-19 Eq.( 17.7.2.6.1) c a1, col 2 = 10 in 1.5c a1, col 2 = 15 in h a = h concrete = 24 in
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