Answer
$$L_{MIN} = 8\frac{1}{3} \space in = 8.333 \space in$$
Work Step by Step
Solution:
1) Bearing Area (for Timber Beam)
$A = 6 * L $
2) Bearing Stress:
$\sigma_{BEARg} = \frac{P}{A} = \frac{P}{6*L} = 400 \space psi$
this allows us to solve for the unknown value "L":
$\Rightarrow L = \frac{20 kip}{6\space in * 400 \space psi} = \frac{20*10^3}{6*400} = 8.333 in$
$\boxed{L_{MIN} = 8.333 \space in}\space\leftarrow\space\space ANS$
*NOTE: The value of L can be rounded $\underline{down}$ to the nearest fraction for practical purposes:
$L_{MIN} \approx 3\frac{5}{16}\space in \space\space OR \space\space 3\frac{1}{4}\space in$
