Question 10.05: Use the interaction method to determine the maximum load P t...
Use the interaction method to determine the maximum load P that can be safely supported by the column of Example 10.04 with an eccentricity of 0.8 \mathrm{in}. The allowable stress in bending is 24 \mathrm{ksi}.
The value of \left(\sigma_{\text {all }}\right)_{\text {centric }} has already been determined in Example 10.04. We have
\left(\sigma_{\text {all }}\right)_{\text {centric }}=19.79 \mathrm{ksi} \quad\left(\sigma_{\text {all }}\right)_{\text {bending }}=24 \mathrm{ksi}
Substituting these values into Eq. (10.55),
\frac{P / A}{\left(\sigma_{\text {all }}\right)_{\text {centric }}}+\frac{M c / I}{\left(\sigma_{\text {all }}\right)_{\text {bending }}} \leq 1 (10.55)
we write
\frac{P / A}{19.79 \mathrm{ksi}}+\frac{M c / I}{24 \mathrm{ksi}} \leq 1.0
Using the numerical data from Example 10.04, we write
\begin{gathered} \frac{P / 4}{19.79 \mathrm{ksi}}+\frac{P(0.8)(1.0) / 1.333}{24 \mathrm{ksi}} \leq 1.0 \\ P \leq 26.6 \mathrm{kips} \end{gathered}
The maximum load that can be safely applied is thus P=26.6 kips.