Bearing Pressure for an Eccentrically Loaded Foundation

Updated solution to consider the correct eccentricity (0.75 ft) in calculating bearing pressure.

The square foundation shown supports a vertical load and lateral load at the base of a column. 

Determine the maximum bearing pressure imposed on the underlying soil.

Assume all loads have appropriate factors.
Assume vertical load provided includes foundation self-weight.


Vertical Load: \(P=100 \ k\)
Lateral Load: \(V=15 \ k\)
 

A) 1.9 ksf


B) 2.6 ksf


C) 3.4 ksf


D) 4.0 ksf


Solution

Answer: A) 1.9 ksf

Width of Footing:
\[B=7 \ ft\]
Length of Footing: 
\[L=7 \ ft\]
Vertical Load:
\[P=100 \ k\]
Lateral Load:
\[V=15\ k\]
Distance from Base of Footing:
\[h=5 \ ft\]
Moment at Base of Footing:
\[M=V×h\]
\[M=15 \ k*5 \ ft=75 \ k⋅ft\]
Eccentricity:
\[e=M/P=(75 \ k⋅ft)/(100 \ k)\]
\[e=0.75 \ ft\]
Check that eccentricity is less than B/6:
\[e<B/6=\frac{7 \ ft}6=1.12 \ ft\]

Since eccentricity is less than B/6, minimum bearing pressure can be calculated as follows:
\[q_{max}= \frac{P}{BL} (1 + \frac{6e}{B})\]
\[q_{max}= \frac{100 \ k}{7 \ ft × 7 \ ft} (1+ \frac{6×\sout{1.12 \ ft}}{7 \ ft})\]\[q_{max}= \frac{100 \ k}{7 \ ft × 7 \ ft} (1+ \frac{6×0.75 \ ft}{7 \ ft})\]

\[\sout{q_{max}=1.9 \ ksf}\]
\[q_{max}=3.35 \ ksf\]