Practice Question: Euler Critical Buckling Stress for a Wood Post

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Solution

\[P_{cr}=\frac{π^2 EI}{(KL)^2}\]
where:
\(L\) = length (given)
\(K\)= effective-length factor
\(E\) = modulus of elasticity (given)
\(I\) = Moment of Inertia

Moment of Inertia:

\[I=\frac{bh^3}{12}=\frac{(6 \ in)^4}{12}=108\ in^4\]

For a column with fixed-free end conditions, the effective-length factor is:

\[K=2.0\]
Calculate the critical buckling load:

\[P_{cr}=\frac{π^2×1600\ ksi×108 \ in^4}{(2.0×14 ft ×12)^2}\]

\[P_{cr}=15.1 \ k\]

We can use the critical buckling load to determine the critical buckling stress as follows:

\[σ_{cr}=\frac{P_{cr}}A\]

\[σ_{cr}=\frac{15.1 \ k}{(6 \ in)^2} =0.42\ ksi\]

\[σ_{cr}≈420 psi\]

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