Mechanics of Materials, 7th Edition

$$L_{MIN} = 8\frac{1}{3} \space in = 8.333 \space in$$
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: $L_{MIN} \approx 3\frac{5}{16}\space in \space\space OR \space\space 3\frac{1}{4}\space in$