Eccentrically loaded isolated spread footing The 24-in thick, 10 ft × 8 ft, isolated footing has a concrete column, which is built eccentric to the center of the footing, along the long dimension of the footing, as shown in Figure 10.18. The footing is supporting 100 kips of dead load and 90 kips

Question

Eccentrically loaded isolated spread footing
The 24-in thick, 10 ft × 8 ft, isolated footing has a concrete column, which is built eccentric to the center of the footing, along the long dimension of the footing, as shown in Figure 10.18. The footing is supporting 100 kips of dead load and 90 kips of live load imparted by upper floors. The bearing capacity of the soil is as given below.
Determine if the bearing area of the footing is adequate for the allowable bearing pressure.
Given:
Compressive strength of concrete: [latex]f_c^\prime[/latex] = 4000 pounds/in² (psi)
Allowable soil-bearing pressure: [latex]q_a[/latex] = 3.0 tons/ft² (tsf)
Concrete density: [latex]γ_c[/latex] = 145 pounds/ft³ (pcf)
Typically, and as with this example, the footing design is begun with known dimensions, selected by the designer. So, the footing design becomes, effectively, a check to determine the adequacy of the bearing area and thickness of the footing to sustain the imposed loads.

Accepted Answer

Determine the soil pressure distribution along the width of the footing.
[latex]P_{column} = P_{DL} + P_{LL}[/latex] = 100 kips + 90 kips = 190 kips (unfactored)
[latex]P_{footing} = 10 ft imes 9 ft imes 2 ft imes 145 \frac{\#}{ft^3}=26.1[/latex] kips
[latex]e=\frac{M}{P}=\frac{190(1.5)}{190+26.1}=1.32^\prime[/latex]

[latex]\frac{L}{6}=\frac{10}{6}=1.67 \gt e[/latex]
Therefore, since the calculated eccentricity (e) is less than L/6, use the full bearing equation as shown in Figure 10.17.
[latex]q_{\max , \min} = \frac{P\left[1\pm (6e/L) ight]}{(L)(B)}[/latex] (10.8)
[latex]q_{\max, \min} = \frac{(190+26.2)\left[1\pm (6(1.32)/10) ight]}{(10)(9)}[/latex]

[latex]q_{\max} = 4.3 ksf, q_{\min} = 0.5 ksf[/latex]
The effective bearing pressure is calculated by accounting for the weight of the footing
[latex]q_e = q_a[/latex]−weight of concrete footing
[latex]q_e[/latex] = 3 tons/sf (2k/ton)−2 ft (.145 k/ft³) = 5.71 ksf
Therefore,
[latex]q_{\max} = 4.3 ksf \lt q_e = 5.71 ksf,[/latex] therefore OK.
The maximum design bearing pressure is less than the effective bearing pressure (accounting for the weight of the footing) (Figure 10.19).

Question 10.3: Eccentrically loaded isolated spread footing The 24-in thick...

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