Problem

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

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

Step-by-Step Solution

Solution 1

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 \(8.33 \mathrm{in}\).

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