Products
Rewards
from HOLOOLY

We are determined to provide the latest solutions related to all subjects FREE of charge!

Enjoy Limited offers, deals & Discounts by signing up to Holooly Rewards Program

HOLOOLY

HOLOOLY
TABLES

All the data tables that you may search for.

HOLOOLY
ARABIA

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

HOLOOLY
TEXTBOOKS

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

HOLOOLY
HELP DESK

Need Help? We got you covered.

## 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?

## 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