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^{\prime } = 0, Φ^{\prime } = 35°, γ= 118 lb/ft^{3}, and N_{60} = 18. 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?
Chapter 17
Q. 17.8
Step-by-Step
Verified Solution
Bearing capacity analysis
Perform the bearing capacity analysis on the 50 k column load, because it will produce the smallest value of q_{a}.
Per Table 17.1:
| \Phi ^{\prime } (deg) | N_{c} | N_{q} | N_{\gamma } |
| 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 |
N_{q}=41.4, N_{\gamma }=47.3
σ^{\prime }_{D}=\gamma D – u= (118 lb/ft^{3})(2 ft) – 0=236 lb/ft^{3}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
q_{a}=\frac{q_{ult}}{F} =\frac{9770 + 2233B}{2} =4885 + 1120B
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}
Setting q_{a}=q and solving produces q_{a}=7780 lb/ft^{2}
Settlement analysis
Perform the settlement analysis will be performed on the 300 k column using the modifed Meyerhof’s method.
For purposes of computing K_{d} assume B = 6 ft.
K_{d} = 1 + 0.33D/B = 1 + 0.33(2/6)= 1.11\delta =\frac{0.0040\left(q – \sigma ^{\prime }_{D}\right) }{\bar{N} _{60}K_{d}} \left\lgroup\frac{B}{B + 1} \right\rgroup ^{2}
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}
This equation solves to B = 7 ft 0 in, which corresponds to q = 6422 lb/ft ^{2}. There is no need to go back and recompute K_{d}.
Synthesis and Conclusion
The settlement analysis produced a lower q, so it governs the analysis. Round off to a multiple of 500 lb/ft^{2}:
Design all footings using q_{A} = 6500 lb/ft^{2}
Sample Application
For a column carrying 100 k:
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 ftRounding to a multiple of 3 in gives B = 4 ft 0 in