Find the maximum value of soil pressure for a square footing. The axial compression force at the bottom of the footing is 100 kips and the moment is 100 kip-ft. Find the pressure for footing widths of (a) 8 ft, (b) 6 ft, and (c) 5 ft.

Question

Accepted Answer

The first step is to determine the equivalent eccentricity and compare it to the kern limit for the footing to establish which of the cases shown in Figure 13.5 applies.

(a) For all parts, the eccentricity is

[latex]e = \frac{M}{N}= \frac{100}{100}=1 ft[/latex]

For the 8-ft-wide footing the kern limit is 8/6 = 1.33 ft; thus, Case 1 applies.

For the computation of soil pressure, the properties of the section (the 8 ft by 8 ft square) must be determined. Thus,

A=8×8=64 ft²

[latex]I = \frac{bd³}{12}=\frac{8×8³}{12}=341.3 ft^4[/latex]

and the maximum soil pressure is determined as

[latex]p=\frac{N}{A}+\frac{Mc}{I}=\frac{100}{64}+\frac{100×4}{341.3}[/latex]

=1.56 +1.17=2.73 ksf

(b) For the 6-ft-wide footing, the kern limit is 1 ft, the same as the eccentricity. Thus, the situation is stress Case 2 in Figure 13.5, with
N/A = Mc/I. Thus,

[latex]p=2(\frac{N}{A})=2(\frac{100}{36})=[/latex] 5.56 ksf

(c) The eccentricity exceeds the kern limit, and the investigation
must be done as illustrated in Figure 13.7.

[latex]a=\frac{5}{2}-e = 2.5-1=[/latex]1.5 ft

x= 3a = 3(1.5) = 4.5 ft

[latex]p=\frac{2N}{wx}=\frac{2×100}{5×4.5}=[/latex]8.89 ksf

Question 13.1: Find the maximum value of soil pressure for a square footing...

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