The axial force in the column supporting the timber beam shown is P = 20 kips. Determine the smallest allowable length L of the bearing plate if the bearing stress in the timber is not to exceed 400 psi.
Fig. P8.18
Since the bearing stress in the beam, should not exceed 400 psi,
\(\sigma_{\max } \leq 400\) psi
Substitute \(\frac{P}{A}\) for \(\sigma_{\max }\).
\(\frac{P}{A} \leq 400 \mathrm{psi} \ldots \ldots\) (1)
Here, \(P\) is the applied axial force, and \(A\) is the area of the bearing plate.
Calculate the area of the bearing plate.
\(A=L \times b\)
Here, \(L\) is the smallest allowable length of the bearing plate and \(b\) is the width of the plate.
Substitute 6 in. for \(b\).
\(A=L \times 6\)
\(=6 L\) in. \(^{2}\)
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 \(8.33 \mathrm{in}\).