Concentrically loaded isolated spread footing Determine the required dimensions and flexural reinforcement required for the isolated square spread footing, subjected to the service loads shown, in Figure 10.10. Given: Compressive strength of concrete: f′c = 4000 pounds/in² (psi) Allowable

Question

Concentrically loaded isolated spread footing
Determine the required dimensions and flexural reinforcement required for the isolated square spread footing, subjected to the service loads shown, in Figure 10.10.
Given:
Compressive strength of concrete: [latex]f_c^\prime[/latex] = 4000 pounds/in² (psi)
Allowable soil-bearing pressure: [latex]q_a[/latex] = 2.0 tons/ft² (tsf)
Soil density: [latex]γ_s[/latex] = 120 pounds/ft³ (pcf)
Concrete density: [latex]γ_c[/latex] = 145 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 = 2.0 tsf imes 2000 lbs/ton −\frac{18}{12}(145 psf)-\frac{30}{12}(120 psf)= 3483[/latex] psf
[latex]q_e = \frac{3483 psf}{1000(lbs/k)}=3.48[/latex] ksf
Step 2: The required bearing area of the footing can then be calculated by dividing the total service load by the effective soil pressure (Figure 10.11):
Required footing area = [latex]\frac{Service load}{q_e}=\frac{P_D+P_L}{q_e}=\frac{175 k+125 k}{3.48 ksf}=86.2[/latex] ft²
[latex]B =L =\sqrt{86.2} = 9.3^\prime[/latex] , round footing side up 9′–6″
Therefore, B = L = 9′–6″.
Step 3: The footing thickness must be checked for adequacy against both punching shear and beam shear. This is done by first calculating the ultimate bearing capacity, [latex]q_u[/latex], using load factors, which will be applied to the underside of the footing (Figures 10.12 and 10.13).
[latex]q_u=\frac{P_u}{B imes L}=\frac{1.2(175)+1.6(125)}{9.5 imes 9.5}=4.54[/latex] ksf
Determine the design ultimate shear, [latex]V_u[/latex], generated by the ultimate loads acting on the footing, to be used to check the footing’s punching shear adequacy.
[latex]V_u=q_u(A_u)[/latex]

[latex]V_u = 4.54 ksf\left[(9.5^\prime)(9.5^\prime)- \left(\frac{36.75^{\prime \prime}}{12^{\prime\prime}} ight) \left(\frac{36.75^{\prime \prime}}{12^{\prime\prime}} ight) ight][/latex]

[latex]V_u=[/latex] 367 kips
The punching shear capacity, φ[latex]V_n[/latex], of the footing is found by the following formula:
[latex]φV_n=φV_c = φ4λ\sqrt{f^\prime c} b_od[/latex] (10.5) (ACI Equation 11.33)
The critical shear perimeter, [latex]b_o[/latex] = 36.75 × 4 = 147″.
Hence,
φ[latex]V_n=0.75(4)(1.0)\frac{\sqrt{4,000}}{1,000}(147)14.75^{\prime \prime}[/latex]

φ[latex]V_n=[/latex] 411 kips > [latex]V_u[/latex]
Therefore, the footing is OK (adequate) to sustain punching shear.
Next, determine [latex]V_u[/latex] for beam shear or shear due to one-way bending. This is done similarly to the procedure used for a continuous footing; however, since the isolated footing has a finite width, in this case, 9′–6″, we take the whole area as shown in Figure 10.14.
So, the shear force, [latex]V_u[/latex], across the critical plane, which is located at a distance d away from the face of the concrete column, is calculated by multiplying the bearing area by the ultimate bearing pressure.
[latex]V_u = q_u imes A_{pressure}[/latex]

[latex]V_u = 4.54 ksf imes \frac{31.25^{\prime\prime}}{12} imes 9.5 [/latex] ft
[latex]V_u[/latex] = 112.3 kips
Now, the shear strength of the footing must be checked for adequacy to sustain the ultimate shear by comparing it to the capacity of the footing’s section.
The shear strength φ[latex]V_n[/latex] of the concrete, for one-way bending, is calculated as follows:
[latex]φV_n=φV_c = φ2λ\sqrt{f^\prime c} bwd[/latex] (10.6) (ACI Equation 11.33)
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 = 9.5′ × 12″/ft = 114″
d is 14.75″
hence,
φ[latex]V_c = .75(2)(1.0) \sqrt{4000 psi}(114^{\prime \prime})(14.75^{\prime\prime})= 159,521[/latex] lbs
φ[latex]V_c[/latex] = 159,521 lbs/1000 lbs/k = 159.5 k > [latex]V_u[/latex] = 112.3 k.
Therefore, the depth of the section is 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 for both forms of shear.
Step 4: Now that we have the dimensions of our footing, we can now determine the cross-sectional area of transverse reinforcing, [latex]A_s[/latex], required for flexure (Figure 10.15).
The design moment, [latex]M_u[/latex], is determined by assuming the flange of the footing is acting as a cantilever beam extending from the face of the concrete column. Therefore, the design moment can be calculated as follows:
[latex]M_u =\frac{wl^2}{2}, w=q_u, 1=\frac{46 in}{12 in/ft}=3.83 ft[/latex]
Hence, [latex]M_u =(4.54 ksf)(9.5 ft)\frac{(3.83 ft)^2}{2}=316.3[/latex] ft-kips
Since the design moment, here, is calculated for the full width of the isolated footing, the moment can be converted to a per linear foot of footing; then, the area of transverse steel,[latex]A_s[/latex], can be calculated for a 12-in unit length of footing.
[latex]M_u=\frac{316.3 ft-kips}{9.5 ft}=33.3[/latex] ft-kips/foot of footing
As with the continuous footing example, 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}{\phi bd^2}=\frac{33.3 ft -kips(12^{\prime\prime}/ft)}{.9(12)(14.75)^2}=0.1701[/latex] ksi
= 170.1 psi
Then,
[latex]ρ_{actual}=\frac{0.85(4000)}{60,000}\left[1-\sqrt{1-\frac{2(170.1)}{0.85(4000)}} ight] = 0.0029[/latex]

[latex]ρ_{actual}=0.0029\gt ρ_{minimum}= 0.0018[/latex]. Therefore, OK.
[latex]ρ_t=0.319\beta_1 \frac{f_c^\prime}{f_y}[/latex] (10.7) (ACI Equation 11.33)
[latex]\beta_1=.85[/latex] when [latex]f_c^\prime \leq[/latex] 4000 psi
[latex]ρ_t=0.319(0.85)\frac{4000}{60,000}=0.0181[/latex]

[latex]ρ_{actual}=0.0029 \lt ρ_t=0.0181.[/latex] Therefore, OK.
Then, the required [latex]A_s = ρ_{actual}(bd)=0.0029(12^{\prime \prime}) (14.75^{\prime \prime}) = 0.51[/latex] in²/ft, width of footing.
Selecting a single #6 reinforcing bar at 10″ O.C. yields 0.53 in²/ft, which is greater than the required [latex]A_s[/latex] = 0.51 in².
The spacing, s, and, [latex]A_{s(actual)}[/latex], calculations are as shown here.
Since
[latex]A_{s(required)}[/latex] = 0.51 in²
and
[latex]A_{\#6 Bar}[/latex] = 0.44 in²
spacing of a #6 bar [latex]=\frac{.44}{.51}(12 in)=10.35[/latex] in, round down to 10″ spacing.
[latex]A_{s(actual)}=\frac{.44}{10^{\prime\prime}spacing}(12 in)=0.53[/latex] in²/ft
Hence,
Use #6 @ 10″ O.C. Each way (Figure 10.16).

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