A proposed warehouse will have design column loads between 50 and 300 k. Each column is to be supported on a spread footing foundation underlain by sandy soils with the following engineering properties: c' = 0, Φ' = 35°, γ= 118 lb/ft^3, and N60 = 18. The minimum acceptable factor of safety against

Question

A proposed warehouse will have design column loads between 50 and 300 k. Each column is to be supported on a spread footing foundation underlain by sandy soils with the following engineering properties: [latex]c^{\prime } = 0[/latex], [latex]Φ^{\prime } = 35°[/latex], [latex]γ= 118 lb/ft^{3}[/latex], and [latex]N_{60} = 18[/latex]. The minimum acceptable factor of safety against a bearing capacity failure is 2.0 and the maximum allowable settlement is 1.0 in. The groundwater table is at a depth of 50 ft and all of the footings will be located 2 ft below the ground surface. What allowable bearing pressure should be used to design them? Using this allowable bearing pressure, what is the required footing width for a column that carries a 100 k load?

Accepted Answer

Bearing capacity analysis

Perform the bearing capacity analysis on the 50 k column load, because it will produce the smallest value of [latex]q_{a}[/latex].

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_{q}=41.4[/latex], [latex]N_{\gamma }=47.3[/latex]

[latex]σ^{\prime }_{D}=\gamma D - u= (118 lb/ft^{3})(2 ft) - 0=236 lb/ft^{3}[/latex]

[latex]q_{ult}=1.3c^{\prime }N_{c} + σ^{\prime }_{D}N_{q} + 0.4 \gamma BN_{\gamma }=0 + (236 lb/ft^{2})(41.4) + 0.4(118 lb/ft^{2})B(47.3)=9770 + 2233B[/latex]

[latex]q_{a}=\frac{q_{ult}}{F} =\frac{9770 + 2233B}{2} =4885 + 1120B[/latex]

[latex]q=\frac{P}{A} +\gamma _{c}D - u =\frac{50,000 lb}{B^{2}} +\left(150 lb/ft^{3}\right) \left(2 ft\right) - 0=\frac{50,000 lb}{B^{2}} +300 lb/ft^{2}[/latex]

Setting [latex]q_{a}=q[/latex] and solving produces [latex]q_{a}=7780 lb/ft^{2}[/latex]

Settlement analysis

Perform the settlement analysis will be performed on the 300 k column using the modifed Meyerhof's method.

For purposes of computing [latex]K_{d}[/latex] assume B = 6 ft.

[latex]K_{d} = 1 + 0.33D/B = 1 + 0.33(2/6)= 1.11[/latex]

[latex]\delta =\frac{0.0040\left(q - \sigma ^{\prime }_{D}\right) }{\bar{N} _{60}K_{d}} \left\lgroup\frac{B}{B + 1} \right\rgroup ^{2}[/latex]

[latex]1=\frac{0.0040\left[\left(300,000/B^{2} + 300 lb/ft^{2}\right) - 236 lb/ft^{2}\right] }{\left(18\right) \left(1.11\right) } \left\lgroup\frac{B}{B + 1} \right\rgroup ^{2}[/latex]

This equation solves to B = 7 ft 0 in, which corresponds to [latex]q = 6422 lb/ft ^{2}[/latex]. There is no need to go back and recompute [latex]K_{d}[/latex].

Synthesis and Conclusion

The settlement analysis produced a lower q, so it governs the analysis. Round off to a multiple of [latex]500 lb/ft^{2}[/latex]:

Design all footings using [latex]q_{A} = 6500 lb/ft^{2}[/latex]

Sample Application

For a column carrying 100 k:

[latex]A=B^{2}=\frac{P}{q_{A} - \gamma _{c}D + u} =\frac{100,000 lb}{6500 lb/ft^{2} - \left(150 lb/ft^{3}\right) \left(2 ft\right) + 0} \rightarrow B=4.02 ft[/latex]

Rounding to a multiple of 3 in gives B = 4 ft 0 in

Question 17.8: A proposed warehouse will have design column loads between 5...

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