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)
8d a = 6.0 in l e = MIN h e f | 8d a = 6.0in h ef = 10.0 in d a = 0.75 in CONTROLS normal weight concrete: λ = 1.0 → λ a = 1.0 ACI 318-19 Table 17.2.4.1 V b = [( 7)( 6.0 in / 0.75 in) 0. 2( 0.75 in) 0. 5 ]( 1.0)( 5000 psi) 0. 5( 12.0 in) 1. 5 = 27,009 lb f’ c = 5000 psi c a1, col 1 = 12.0 in
1.5 V b = 9λ a f ′ c c a1 ACI 318-19 Eq.( 17.7.2.2.1b)
V b =( 9)( 1.0)( 5000 psi) 0. 5( 12.0 in) 1. 5 = 26,455 lb f’ c = 5000 psi c a1, col 1 = 12.0 in check: design V b = MIN { 27,009 lb; 26,455 lb } = 26,455 lb ACI 318-19 17.7.2.2
V cbg =
A Vc A Vc0 ψ ec, V ψ ed, V ψ c, V ψ h, V ψ parallel, V V b
V cbg, col 1 =( 528 in 2 / 648 in 2)( 0.989)( 0.8)( 1.2)( 1.23)( 1.0)( 26,455 lb) = 25,173 lb
A Vc = 528 in 2 ψ ec, V = 0.989 ψ c, V = 1.2 ψ parallel, V = 1.0
A Vc0 = 648 in 2 ψ ed, V = 0.8 ψ h, V = 1.23 V b = 26,455 lb
Nominal Concrete Pryout Strength in Shear( V cpg) ACI 318-19 Section 17.7.3
There are 16 anchors in tension and 16 anchors in shear → calculate V cpg using N cbg and N ag for 16 anchors.
Pryout is a shear load. Calculate N cbg and N ag for pryout using shear eccentricity values.
V cpg = k cp MIN { N cbg: N ag } h ef = 10 in → k cp = 2.0 for h ef ≥ 2.5 in
V cpg =( 2.0)( 72,818 lb) = 145,636 lb
N cbg = 71,926 lb ψ ec, N = 0.938 N cbg, modified = N cbg / ψ ec, N = 76,680 lb
N ag = 93,973 lb ψ ec, Na = 0.917 N ag, modified = N ag / ψ ec, Na = 102,479 lb e c, x = 0.469 in ψ ec, Vx = 0.97 e c, y = 0.326 in ψ ec, Vy = 0.979 N cbg, pryout =( N cbg, modified)( ψ ec, Vx)( ψ ec, Vy) = 72,818 lb CONTROLS e c, x = 0.469 in ψ ec, Vx = 0.959 e c, y = 0.326 in ψ ec, Vy = 0.971 N ag, pryout =( N ag, modified)( ψ ec, Vx)( ψ ec, Vy) = 95,427 lb
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