The axial force in the column supporting the timber beam shown is P = 20 kips. Determine t...

Calculate the smallest allowable length of the bearing plate.

Substitute \(6 L\) in. \(^{2}\) for \(A\) and 20 kip for \(P\) in equation (1).

\(\frac{P}{A} \leq 400 \mathrm{psi}\)

\(\frac{20 \times 10^{3} \mathrm{lb}}{6 L \text { in. }^{2}} \leq 400 \mathrm{psi}\)

\(\frac{20 \times 10^{3}}{6 L} \leq 400\)

\(L \geq \frac{20 \times 10^{3}}{6 \times 400}\)

\(L \geq 8.33\) in.

For the beam to be under safe condition, the length of the beam should be greater than or equal to \(8.33 \mathrm{in}\). but not less than \(8.33\) in. Hence the smallest length of the beam is \(8.33\) in.

\(L=8.33\) in.

Therefore, the smallest allowable length of the timber beam is \(8.33 \mathrm{in}\).