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 x1 + s x2 + s x3 + c x +)( 1.5c a1, row 1 or h concrete) ACI 318-19 Fig. R17.7.2.1 b)
Assume 1.0V ua acts on anchor Row 1. A Vc =( c x- + s x1 + s x2 + s x3 + c x +)( 1.5c a1, row 1 or h concrete) 1.5c a1, row 1 > h concrete use h concrete to calculate A Vc A Vc =( 1.5c a1, row 1 + 9 + 9 + 9 + 1.5c a1, row 1)( h concrete) = 510 in 2
Check: c a1, row 1 = 8.0 in s y, 12 = s y, 23 = s y, 34 = 6.0 in s y, nn < c a1, row 1 CASE 3 applies for row 1 s x. n < 3c a1, row 1 anchors in each row act as a group c a1, row 1 = 8 in 1.5c a1, row 1 = 12 in 3.0c a1, row 1 = 24 in c x- = ∞ = 1.5c a1, row 1 c x + = ∞ = 1.5c a1, row 1 h concrete = 10 in
s x1 = 9.0 in s x2 = 9.0 in s x3 = 9.0 in A Vc0 = 4.5( c a1, row 1) 2 = 4.5( 8 in) 2 = 288 in 2 ACI 318-19 Eq.( 17.7.2.1.3)
ψ ec, V =
1 1 + e ′ V
1.5c a1
ACI 318-19 Eq.( 17.7.2.3.1)
ψ ec, V = 1.0 no eccentricity e cV = 0.0 in 1.5c a1, row 1 = 12 in 3.0c a1, row 1 = 24 in
ψ ed, V = 0.7 + 0.3 c a2, min 1.5c a1
ACI 318-19 Eq.( 17.7.2.4.1b)
ψ ed, V = 1.0 c a2, min = ∞ 1.5c a1, row 1 = 12 in ψ c, V = 1.0 cracked concrete, no 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
ACI 318-19 Eq.( 17.7.2.6.1)
h concrete < 1.5c a1 → ψ h, V =( 12 in / 10 in) 0. 5 = 1.10 c a1, row 1 = 8 in 1.5c a1, row 1 = 12 in h a = h concrete = 10 in
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