A column with a 2-in.-square cross section and 28-in. effective length is made of the

Question

A column with a 2-in.-square cross section and 28-in. effective length is made of the aluminum alloy 2014-T6. Using the allowable-stress method, determine the maximum load [latex]P[/latex] that can be safely supported with an eccentricity of [latex]0.8 \mathrm{in}[/latex].

Accepted Answer

We first compute the radius of gyration [latex]r[/latex] using the given data

[latex] \begin{gathered} A=(2 ext { 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} [/latex]

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

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

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

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

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

We now use Eq. (10.53)

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

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

[latex] \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} [/latex]

The maximum load that can be safely applied is [latex]P=23.3[/latex] kips.

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

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