One-Way Shear Check
Flexural Reinforcing
For bottom bars in major direction,
Mu = 1080.2 k-ft
ϕMn = ϕRbd2 ³ Mu
R ≥ Mu /ϕbd2 = (1080.2 k-ft ∙ 12 in/ft)/[0.9 ∙ (180 in) ∙ (26.63 in)2] = 112.8 psi
ρ ≥ 0.001925
As ≥ 0.001925 · 180 in · 26.63 in = 9.23 in2 ∴ (21) #6 bottom required
With regard to the minimum steel requirement, as we will see below, (11) #4 top will be provided in major direction, thus,
(21 ∙ 0.44 in2 + 11 ∙ 0.20 in2) = 11.44 in2 > As,min
For bottom bars in minor direction,
Mu = (3.85 ksf / 2) ∙ 12.57 ft ∙ 6.75 ft ∙ (6.75 ft / 2) = 551.2 k-ft
By inspection minimum reinforcing will control. Provide bottom bars such that minimum reinforcing limit is met, including top bars.
From calculation below, (11) #4 top will be provided,
As > 9.72 in2 - (11 · 0.20 in2) = 7.52 in2
Required number of #6 bars = 7.52 in2 / 0.44 in2 = 18
(18) #6 bottom required
For top bars,
Uplift occurs for Concrete Load Combination thus top bars are required. However, by inspection negative moment induced in footing is small and thus only minimum top bar requirements need be satisfied.
Use (11) #4 top each way to satisfy spacing requirements