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Chapter 17

Q. 17.8

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?

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  ft

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