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 3 or h concrete) ACI 318-19 Fig. R17.7.2.1 b) Check: c a1, row 1 = 6.0 in s y, 12 = 4.0 in s y, 23 = 8.0 in
CASE 1 / CASE 2 applies for row 1 and row 3: calculate V cbg from row 3 using c a1, row 3. s y, 12 < c a1, row 1 CASE 3 applies for row 1 and row 2( orange shaded area)
( 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 =( 1.5c a1, row 3 + 9 + 9 + 9 + 1.5c a1, row 3)( h concrete) = 972 in 2 s y, 23 > c a1, row 1 CASE 1 / CASE 2 applies for row 1( orange shaded area) and row 3( yellow shaded area)
s x. nn < 3c a1, row 1 anchors in each row act as a group c a1, row 3 = 18 in 1.5c a1, row 3 = 27 in 3.0c a1, row 3 = 54 in c x- = ∞ = 1.5c a1, row 3 c x + = ∞ = 1.5c a1, row 3 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
= 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 3 = 27 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
= 27In / 12in 0. 5 = 1.5
ACI 318-19 Eq.( 17.7.2.6.1) c a1, row 3 = 18 in 1.5c a1, row 3 = 27 in h a = h concrete = 12 in h concrete < 1.5c a1
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