PROFIS ENGINEERING
V b = 7 l e d a
0.2 d a λ a f ′ c c a1
1.5 ACI 318-19 Eq.( 17.7.2.2.1a) l e = MIN h e f | 8d a
= 5.75 in h ef = 5.75 in CONTROLS d a = 1.0 in 8d a = 8.0 in
normal weight concrete: λ = 1.0 → λ a = 1.0 ACI 318-19 Table 17.2.4.1
V b = [( 7)( 5.75 in / 1.0 in) 0. 2( 1.0 in) 0. 5 ]( 1.0)( 4000 psi) 0. 5( 8.0 in) 1. 5
= 14,213 lb V b = 9λ a f ′ c c a1
1.5 f’ c = 4000 psi
ACI 318-19 Eq.( 17.7.2.2.1b)
c a1, row 1 = 8.0 in
V cbg =
V b =( 9)( 1.0)( 4000 psi) 0. 5( 8.0 in) 1. 5 = 12,880 lb check: design V b = MIN { 14,213 lb; 12,880 lb }
A Vc A Vc0
= 12,880 lb
ψ ec, V ψ ed, V ψ c, V ψ h, V ψ parallel, V V b
V cbg, row 1 =( 510 in 2 / 288 in 2)( 1.0)( 1.0)( 1.0)( 1.1)( 1.0)( 12,880 lb)
= 25,089 lb
f’ c = 4000 psi |
c a1, row 1 = 8.0 in |
ACI 318-19 17.7.2.2 |
A Vc = 510 in 2 |
A Vc0 = 288 in 2 |
ψ ec, V = 1.0 |
ψ ed, V = 1.0 |
ψ c, V = 1.0 |
ψ h, V = 1.1 |
ψ parallel, V = 1.0 |
V b = 12,880 lb |
Nominal Concrete Pryout Strength in Shear( V cpg)
There are 16 anchors in tension and 16 anchors in shear → calculate V cpg using N cbg for 16 anchors. reference p. 88 concrete breakout strength calculations: N cbg = 92,221 lb
V cpg = k cp N cbg ACI 318-19 Eq.( 17.7.3.1b) h ef = 5.75 in → k cp = 2.0 for h ef ≥ 2.5 in V cpg =( 2.0)( 92,221 lb) = 184,442 lb
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