A bearing wall for a proposed building is to be supported on a 24 in deep continuous footing founded in an unsaturated clayey sand (SC). The load from this wall will be 4.0 k/ft, and the soil has c' = 100 lb/ft^2, Φ' = 28°, and γ = 119 lb/ft^3. The groundwater table is at a very great depth.

Question

A bearing wall for a proposed building is to be supported on a 24 in deep continuous footing founded in an unsaturated clayey sand (SC). The load from this wall will be 4.0 k/ft, and the soil has [latex]c^{\prime } = 100 lb/ft^{2}[/latex], [latex]Φ^{\prime } = 28°[/latex], and [latex]γ = 119 lb/ft^{3}[/latex]. The groundwater table is at a very great depth. Compute the required footing width to maintain a factor of safety of 3.0 against a bearing capacity failure.

Accepted Answer

Per Table 17.1,

[latex]\Phi ^{\prime }[/latex] (deg)	[latex]N_{c}[/latex]	[latex]N_{q}[/latex]	[latex]N_{\gamma }[/latex]
0	5.7	1.0	0.0
1	6.0	1.1	0.1
2	6.3	1.2	0.1
3	6.6	1.3	0.2
4	7.0	1.5	0.3
5	7.3	1.6	0.4
6	7.7	1.8	0.5
7	8.2	2.0	0.6
8	8.6	2.2	0.7
9	9.1	2.4	0.9
10	9.6	2.7	1.0
11	10.2	3.0	1.2
12	10.8	3.3	1.4
13	11.4	3.6	1.6
14	12.1	4.0	1.9
15	12.9	4.4	2.2
16	13.7	4.9	2.5
17	14.6	5.5	2.9
18	15.5	6.0	3.3
19	16.6	6.7	3.8
20	17.7	7.4	4.4
21	18.9	8.3	5.1
22	20.3	9.2	5.9
23	21.7	10.2	6.8
24	23.4	11.4	7.9
25	25.1	12.7	9.2
26	27.1	14.2	10.7
27	29.2	15.9	12.5
28	31.6	17.8	14.6
29	34.2	20.0	17.1
30	37.2	22.5	20.1
31	40.4	25.3	23.7
32	44.0	28.5	28.0
33	48.1	32.2	33.3
34	52.6	36.5	39.6
35	57.8	41.4	47.3
36	63.5	47.2	56.7
37	70.1	53.8	68.1
38	77.5	61.5	82.3
39	86.0	70.6	99.8

[latex]N_{c} = 31.6[/latex], [latex]N_{q}= 17.8[/latex], [latex]N _{γ}= 14.6[/latex] for [latex]Φ=28°[/latex]

[latex]σ^{\prime }_{D}=γD - u=(119 lb/ft^{3})(2 ft) - 0=238 lb/ft^{2}[/latex]

[latex]q_{ult}=c^{\prime }N_{c} + \sigma ^{\prime }_{D}N_{q}+ 0.5\gamma ^{\prime }BN_{\gamma }=\left(100 lb/ft^{3}\right) \left(31.6\right) +\left(238 lb/ft^{3}\right) \left(17.8\right) +0.4\left(119 lb/ft^{3}\right) B\left(14.6\right) =7396 lb/ft^{2}+695 B[/latex]

[latex]q_{a}=\frac{q_{ult}}{F} =\frac{7396 lb/ft^{2} + 695 B}{3.0} =2465 lb/ft^{2}+232 B[/latex]

[latex]q=\frac{P/b}{B} +\gamma _{c}D - u=\frac{4000 lb/ft}{B} +\left(150 lb/ft^{3}\right) \left(2 ft\right) - 0[/latex]

Setting [latex]q=q_{a}[/latex] and solving for B gives B = 1 ft 9 in

Question 17.4: A bearing wall for a proposed building is to be supported on...

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