Practice Question: Bearing Capacity Factor of Safety

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Solution

Answer: C) 3.33

Calculate the eccentricity, e:

$$ \ e=\frac{M_O}{P}=\frac{700\ kip\ast ft}{\left(300\ kip\right)\ \ }=2.33\ ft>\frac{B}{6}=2\ ft $$

$$ If\ e<\frac{B}{6}\ then: $$

$$ q_{max}=\frac{P}{\left.\ \left(B\ast L\right.\right)\ }+\frac{\left.\ \left(6\ast M\right.\right)}{\left.\ \left(B\ast L^{2\ }\right.\right)} $$

$$ If\ e>\frac{B}{6}\ then: $$

$$ q_{max}= \frac{2 \ast P}{3 \ast B \ast (\frac{L}{2}-e)} $$

Calculate maximum bearing pressure:
$$ q_{max}=\frac{2\ast300\ kip}{\left(3\ast12\ ft\ast\left(\frac{12\ ft}{2\ \ }-\ 2.33ft\right)\ \right)\ \ }=4.5\ ksf $$

Calculate the Factory of Safety:

$$ FS=\frac{q_{net}}{q_{max\ }}=\frac{15\ ksf}{4.5\ ksf}=3.33 $$
 

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