Concentrically loaded continuous wall footing Determine the required width of the continuous wall footing, W, shown in Figure 10.4, when subjected to the following service (not factored) loads shown. wdead load = 9.5 Kips/ft wlive load = 5.75 Kips/ft Given: Compressive strength of concrete:

Question

Concentrically loaded continuous wall footing
Determine the required width of the continuous wall footing, W, shown in Figure 10.4, when subjected to the following service (not factored) loads shown.
[latex]w_{dead load}[/latex] = 9.5 Kips/ft
[latex]w_{live load}[/latex] = 5.75 Kips/ft
Given:
Compressive strength of concrete: [latex]f_c^\prime[/latex] = 4000 pounds/in² (psi).
Allowable soil-bearing pressure: [latex]q_a[/latex] = 1.75 tons/ft² (tsf).
Soil density: [latex]γ_s[/latex] = 120 pounds/ft³ (pcf).
Concrete density: [latex]γ_c[/latex] = 150 pounds/ft³ (pcf).

Accepted Answer

Step 1: Determine the effective allowable soil pressure, [latex]q_e[/latex], by subtracting the weight of the footing and soil from the allowable soil pressure, [latex]q_a[/latex], on a per foot basis along the length of the footing.
Effective soil pressure, [latex]q_e = q_a−[/latex]weight of concrete−weight of soil
Hence, [latex]q_e = 1.75 tsf imes 2000 lbs/ton −\frac{18}{12}(150 psf)-\frac{48-18}{12}(120 psf)= 2975[/latex] psf
[latex]q_e = \frac{2975 psf}{1000(lbs/k)}=2.98[/latex] ksf
Step 2: The width of the footing can then be calculated by dividing the total service load by the effective soil pressure.
Width, w = [latex]\frac{Service load}{q_e}=\frac{w_D+w_L}{q_e}=\frac{9.5(k/ft)+5.75(k/ft)}{2.98 ksf}=5.1[/latex] ft, use w = 5′ 0′′
It is important to note that when calculating the effective bearing pressure and determining the size of the footing, always use nonfactored or service loads.
Once the width of the continuous wall footing is determined, the concrete footing itself must be checked for bending and beam shear.
Hence, to check the adequacy of the footing thickness, for the continuous wall footing in Example 10.3, the shear capacity needs to be determined.
The critical section for shear is located at distance, d, away from the stem of the wall as shown in Figure 10.5.
The effective bearing area creates a shear force across the failure or critical plane in the footing (Figure 10.6).
Step 1: Determine the strength design bearing pressure ([latex]q_u[/latex]) using appropriate load factors:
[latex]q_u=\frac{w_u}{w (length of wall)}[/latex]

[latex]=\frac{1.2(9.5k)+1.6(5.75k)}{5^\prime(1^\prime)}[/latex]
= 4.12 ksf
Step 2: Check beam shear along the critical section.
This is done by determining the ultimate shear ([latex]V_u[/latex]) at the critical section, which is equal to the bearing pressure times the effective area, [latex]V_u = q_u imes A_1. A_1[/latex] is the effective bearing pressure area, shearing across the critical section (Figure 10.7).
For Example 10.1, the ultimate shear is then calculated as follows:
[latex]V_u = 4.12 ksf\left(\frac{10.25^{\prime \prime}}{12^{\prime\prime}} ight) [/latex]( 1 ft unit width of wall) = 3.52 Kips/linear foot
Determine the shear strength of the footing and check adequacy to sustain the ultimate shear determined in the previous step.
The shear strength of the concrete must first be calculated as follows:
[latex]φV_c = φ2λ\sqrt{f^\prime c} bwd[/latex] (10.3) (ACI Equation 11.3)
where:
φ[latex]V_c[/latex] is design strength of concrete
φ is found in section ACI 9.3.2.3 = .75
λ is normal weight of concrete found in section ACI 8.6.1 = 1.0
[latex]f ^\prime c[/latex] is compressive strength of concrete = 4000 psi
bw is unit length of wall = 12″
d is 14.5″
hence,
φ[latex]V_c = .75(2)(1.0) \sqrt{4000 psi}(12^{\prime \prime})(14.5^{\prime\prime})= 16,507[/latex] plf
φ[latex]V_c[/latex] = 16,507 psf/1000 lbs/k = 16.5 klf > [latex]V_u[/latex] = 3.52 klf. Therefore, OK.
The footing thickness of 18 in. is more than adequate to sustain the ultimate shear produced by the bearing pressure acting on the bottom of the footing.
Step 3: Determine the cross-sectional area of transverse reinforcing, [latex]A_s[/latex], required for flexure (Figure 10.8).
The design moment, [latex]M_u[/latex], is determined by assuming the flange of the footing to act as a cantilever beam extending from the face of the foundation wall. Therefore, the design moment can be calculated as follows:
[latex]M_u =\frac{wl^2}{2}, w=q_u, 1=\frac{25 in}{12 in/ft}=2.083 ft[/latex]
Hence, [latex]M_u =(4.12 ksf)\frac{(2.083 ft)^2}{2}=8.94[/latex] ft-kips/linear foot of footing
= 107.26 in-kips/ft
Since the design moment is calculated per linear foot of footing, the area of transverse steel, [latex]A_s[/latex], should also be calculated for a 12-in. unit length of footing.
The actual steel ratio, [latex]ρ_{actual}[/latex], is readily calculated by the following equation:
[latex]ρ_{actual}=\frac{0.85 f^\prime c}{f_y}\left[1-\sqrt{1-\frac{2R_u}{0.85 f^\prime c}} ight],R_u=\frac{M_u}{φ bd^2}[/latex] (10.4)
where φ = .90, for flexure is found in section ACI 9.3.2.3
[latex]R_u=\frac{M_u}{φ bd^2}=\frac{107.26 in -kips}{.9(12)(14.75)^2}=0.0456[/latex] ksi
= 45.6 psi
Then,
[latex]ρ_{actual}=\frac{0.85(4,000)}{60,000}\left[1-\sqrt{1-\frac{2(45.6)}{0.85(4,000)}} ight] = 0.0008[/latex]

[latex]ρ_{actual}\lt ρ_{minimum}= 0.0018[/latex]
The actual steel ratio is less than the minimum required to satisfy temperature and shrinkage requirements, [latex]ρ_{minimum} = 0.0018,[/latex] as per ACI 318, 7.12
Therefore, the required [latex]A_s = ρ_{min}(bh)=0.018(12^{\prime \prime}) (18^{\prime \prime}) = 0.39[/latex] in², where bh = gross concrete cross-sectional area.
For design, selecting a single #6 reinforcing bar at 12″ O.C. yields [latex]A_{\#6 Bar}[/latex] = 0.44 in²/ft, which is greater than the required [latex]A_s[/latex] = 0.39 in² (Figure 10.9).
Hence,
use #6 at 12″ O.C.

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