Question 10.04: A column with a 2-in.-square cross section and 28-in. effect...

We first compute the radius of gyration r using the given data

\begin{gathered} A=(2 \text { in. })^{2}=4  \mathrm{in}^{2} \quad I=\frac{1}{12}(2  \mathrm{in} .)^{4}=1.333  \mathrm{in}^{4} \\ r=\sqrt{\frac{I}{A}}=\sqrt{\frac{1.333  \mathrm{in}^{4}}{4  \mathrm{in}^{2}}}=0.5774  \mathrm{in} . \end{gathered}

We next compute L / r=(28  \mathrm{in}) /.(0.5774 in. )=48.50.

Since L / r<55, we use Eq. (10.48)

C_P=\frac{1+\left(\sigma_{C E} / \sigma_C\right)}{2 c}-\sqrt{\left[\frac{1+\left(\sigma_{C E} / \sigma_C\right)}{2 c}\right]^2-\frac{\sigma_{C E} / \sigma_C}{c}}       (10.48)

to determine the allowable stress for the aluminum column subjected to a centric load. We have

\sigma_{\text {all }}=[30.9-0.229(48.50)]=19.79  \mathrm{ksi}

We now use Eq. (10.53)

{\frac{P}{A}}+{\frac{M c}{I}}\leq\sigma_{\mathrm{all}}     (10.53)

with M=P e and c=\frac{1}{2}(2  \mathrm{in})=.1 in. to determine the allowable load:

\begin{gathered} \frac{P}{4  \mathrm{in}^{2}}+\frac{P(0.8  \mathrm{in} .)(1  \mathrm{in} .)}{1.333  \mathrm{in}^{4}} \leq 19.79  \mathrm{ksi} \\ P \leq 23.3  \mathrm{kips} \end{gathered}

The maximum load that can be safely applied is P=23.3 kips.