Question 10.CA.4: A column with a 2-in.-square cross section and 28-in. effect......
A column with a 2-in.-square cross section and 28-in. effective length is made of the aluminum alloy 2014-T6 (Fig. 10.35). Using the allowable-stress method, determine the maximum load P that can be safely supported with an eccentricity of 0.8 in.
Compute the radius of gyration r using the given data:
\begin{gathered} A=(2 \text { in. })^2=4 \text { in }^2 \quad I=\frac{1}{12}(2 \text { in. })^4=1.333 \text { in }^4 \\ r=\sqrt{\frac{I}{A}}=\sqrt{\frac{1.333 in ^4}{4 in ^2}}=0.5774 in. \end{gathered}
Next, compute L/r = (28 in.)∕(0.5774 in.) = 48.50.
Since L/r is between 17.0 and 52.7, use Eq. (10.47a) to determine the allowable stress for the aluminum column subjected to a centric load
17.0>L/r <52.7: \sigma_{\text {all }}=\left[39.7-0.465(L/r)+0.00121(L/r)^2\right] ksi (10.47a)
\sigma_{\text {all }}=\left[39.7-0.465(48.5)+0.00121(48.5)^2\right]=19.99 ksi
Now use Eq. (10.55) with M = Pe and c=\frac{1}{2}(2 \text { in. })=1 in. to determine the allowable load:
\frac{P}{A}+\frac{M c}{I} \leq \sigma_{\text {all }} (10.55)
\begin{gathered} \frac{P}{4 \text { in }^2}+\frac{P(0.8 \text { in. })(1 \text { in. })}{1.333 \text { in }^4} \leq 19.99 ksi \\ P \leq 23.5 kips \end{gathered}
The maximum load that can be safely applied is P = 23.5 kips.