A 200 kN column load is to be supported on a 0.5 m deep footing in the sandy soil shown in Figure 17.11. The maximum allowable settlement is 20 mm. Determine the required footing width.

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Accepted Answer

[latex]σ^{\prime }_{D}=γD - u=(20.0 kN/m^{3})(0.5 m) - 0=10 kPa[/latex]

Try B = 1.0 m

[latex]K_{d} = 1 + 0.33D/B= 1 + 0.33 (0.5/1.0) =1.17 [/latex]

[latex]\bar{N} _{60}=\frac{19 + 22}{2} =21[/latex]

[latex]\delta =\frac{1.3\left(q - \sigma ^{\prime }_{D}\right) }{\bar{N} _{60}K_{d}} [/latex]

[latex]20=\frac{1.3\left(q - 10 kPa\right) }{\left(21\right) \left(1.17\right) } \rightarrow q=388 kPa[/latex]

[latex]q=\frac{P}{B^{2}} +\gamma _{c}D - u[/latex]
[latex]388 kPa = \frac{200 kN}{B^{2}} +\left(23.6 kN/m^{3}\right) \left(0.5 m\right) - 0[/latex]
[latex]B = 0.73 m[/latex]

The computed B of 0.73 m is close to the assumed value of 1.00 m. It is not necessary to run the computations again with a new assumed B, because doing so would produce only a small change in [latex]K_{d}[/latex] and no change in [latex]\bar{N} _{60}[/latex].
Footing widths are normally expressed as multiples of 3 in or 100 mm, so the design width is 800 mm.

Question 17.7: A 200 kN column load is to be supported on a 0.5 m deep foot...

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