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 2 or h concrete) ACI 318-19 Fig. R17.7.2.1 b)
Assume V ua acts on anchor Row 2. A Vc =( c x- + s x1 + s x2 + s x3 + c x +)( 1.5c a1, row 2 or h concrete) 1.5c a1, row 2 > h concrete use h concrete to calculate A Vc A Vc =( 1.5c a1, row 2 + 9 + 9 + 9 + 1.5c a1, row 2)( h concrete) = 1035 in 2
A Vc0 = 4.5( c a1, row 2) 2 = 4.5( 14 in) 2
Check: c a1, row 1 = 6.0 in s y, 12 = 8.0 in |
s y, 12 > c a1, row 1 CASE 1 / CASE 2 applies for row 2 |
s x. n < 3c a1, row 1 anchors in row 2 act as a group |
c a1, row 2 = 14 in |
1.5c a1, row 2 = 21 in |
3.0c a1, row 2 = 42 in |
c x- = ∞ = 1.5c a1, row 2 |
c x + = ∞ = 1.5c a1, row 2 |
h concrete = 15 in |
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s x1 = 9.0 in |
s x2 = 9.0 in |
s x3 = 9.0 in |
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ACI 318-19 Eq.( 17.7.2.1.3) |
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ψ ec, V =
= 882 in 2 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 2 = 21 in 3.0c a1, row 2 = 42 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 2 = 21 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 concrete
ACI 318-19 Eq.( 17.7.2.6.1) h concrete < 1.5c a1 c a1, row 2 = 14 in 1.5c a1, row 2 = 21 in h concrete = 15 in
→ ψ h, V =( 21 in / 15 in) 0. 5 = 1.18
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