Determine the factored design moment for an eccentrically loaded isolated spread footing Consider the 10′ × 9′ footing discussed in both Examples 10.3 and 10.4. The soil-bearing pressure distributions for both directions are shown in Figure 10.22.

Question

Determine the factored design moment for an eccentrically loaded isolated spread footing
Consider the 10′ × 9′ footing discussed in both Examples 10.3 and 10.4. The soil-bearing pressure distributions for both directions are shown in Figure 10.22.

Accepted Answer

The applicable basic load combinations, as per the International Building Code (IBC), are as follows:

16–1	1.4D
16–2	1.2D + 1.6L
16–3	1.2D + 0.8W
16–4	1.2D +1.6W +1.0L
16–6	0.9D +1.6W

The governing load case in the long direction of the footing is load case 16-2:
U = 1.2D + 1.6L
[latex]P_u = 1.2(100^k) + 1.6(90^k) = 264^k[/latex]

[latex]e=\frac{M}{P}=\frac{264^k(1.5^\prime)}{264^k+1.2(26.1^k)}=1.34^\prime\\frac{L}{6}=\frac{10^\prime}{6}=1.67 \gt 1.34[/latex]
[latex]q_{max,min} =\frac{P_u\left[1\pm\frac{6e}{L} ight]}{(L)(B)}=\frac{264\left[1\pm\frac{6(1.34^\prime)}{10^\prime} ight]}{(10)(9)}[/latex]

[latex]q_{max} = 5.29 ksf, q_{min} = 0.57 ksf[/latex]
For the short direction of the footing, we will need to consider load cases 16-3, 16-4, and 16-6.
16-3: U = 1.2D + 0.8W
[latex]e=\frac{M_w}{P_{(1.2D)}}=\frac{0.8(180^{k\cdot ft})}{1.2(100^k+26.1^k)}=[/latex] 0.95′
[latex]\frac{L}{6}=\frac{9^\prime}{6}=1.5 \gt 0.95[/latex]

[latex]q_{max,min} =\frac{P_u\left[1\pm\frac{6e}{L} ight]}{(L)(B)}=\frac{151.3\left[1\pm\frac{6(0.95^\prime)}{9^\prime} ight]}{(10)(9)}[/latex]

[latex]q_{max} = 2.74 ksf, q_{min} = 0.62 ksf[/latex]
16-4: U = 1.2D + 1.6W + L
[latex]e=\frac{M_w}{P_{(1.2D+L)}}=\frac{1.6(180^{k\cdot ft})}{1.2(100^k+26.1^k)}=[/latex] 1.2′
[latex]\frac{L}{6}\gt e[/latex]

[latex]q_{max,min} =\frac{P_u\left[1\pm\frac{6e}{L} ight]}{(L)(B)}=\frac{241\left[1\pm\frac{6(1.2^\prime)}{9^\prime} ight]}{(10)(9)}[/latex]

[latex]q_{max} = 4.82 ksf, q_{min} = 0.54 ksf[/latex]
16-6: U = 0.9D + 1.6W
[latex]e=\frac{M_w}{P_{(0.9D)}}=\frac{1.6(180^{k\cdot ft})}{0.9(100^k+26.1^k)}=[/latex] 2.54′
[latex]\frac{L}{6}\lt e[/latex]

[latex]e^\prime=\frac{9^\prime}{2}-2.54^\prime=1.96^\prime[/latex]

[latex]q_{max}=\frac{2P}{3Be^\prime}=\frac{2\left[0.9(126.1) ight]}{3(10)(1.96)}=3.86[/latex] ksf
Consequently, load case 16-4 governs in the short direction of the footing ([latex]q_{max}[/latex] = 4.82 ksf).
However, because the offset concrete column creates an eccentric loading, in the long direction, under dead and live load, and there is also a moment, due to wind, which, according to load combination 16-4, will occur simultaneously with the dead and live loads, the footing will experience bending in both directions at the same time, and thus, the footing is subject to biaxial bending.
Biaxial bending can be calculated using stress distribution (P/A) and (M/S) as follows:
[latex]q=\frac{P}{BL}\pm\frac{6Pe}{BL^2}\pm\frac{6M}{LB^2}[/latex]
Using the factored loads from load case 16-4 (Figures 10.23 and 10.24):
[latex]P_u = 241^k[/latex]

[latex]M_u =1.6(180)=288^{k\cdot ft}[/latex]
then,
[latex]q=\frac{241}{(10)(9)}\pm\frac{6(214)1.5}{(9)(10)^2}\pm \frac{6(288)}{(10)(9)^2}[/latex]

[latex]q=2.68^{k/ft^2}\pm 2.41^{k/ft^2}\pm 2.13^{k/ft^2}[/latex]
For design purposes, the resulting factored bearing stress, applied to the bottom of the footing, is conservatively estimated as shown in Figure 10.25.
The design moment can be calculated as follows for both sides of the column.
The moment at the left side of the column is
[latex]M_u=(2.40)(6.5)\left(\frac{6.5}{2} ight)+\frac{1}{2}(6.5)(3.13)\left(\frac{6.5}{3} ight)=72.74^{k\cdot ft}[/latex]
The moment at the right side of the column is
[latex]M_u=(2.40)(2)(1)+(3.86)(2)(1)+\frac{1}{2}(2)(0.96)\left(\frac{1}{3} ight)=12.84^{k\cdot ft}[/latex]
Clearly, the moment to the left of the column produces the critical or design moment for the footing to resist in bending.
The beam shear and punching shear are calculated and checked for adequacy in the same manner as discussed in Example 10.2. Albeit, the factored bearing stress distribution is more complicated due to the footing’s eccentric loading, the procedure is exactly the same, and can be readily solved using a spread sheet in Excel.

Question 10.5: Determine the factored design moment for an eccentrically lo...

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