An aluminum column of length L and rectangular cross section has a fixed end B and

Question

An aluminum column of length [latex]L[/latex] and rectangular cross section has a fixed end [latex]B[/latex] and supports a centric load at [latex]A[/latex]. Two smooth and rounded fixed plates restrain end [latex]A[/latex] from moving in one of the vertical planes of symmetry of the column, but allow it to move in the other plane. (a) Determine the ratio [latex]a / b[/latex] of the two sides of the cross section corresponding to the most efficient design against buckling. (b) Design the most efficient cross section for the column, knowing that [latex]L=20[/latex] in., [latex]E=10.1 imes 10^{6} \mathrm{psi}, P=5[/latex] kips, and that a factor of safety of 2.5 is required.

Accepted Answer

Buckling in [latex]x y[/latex] Plane. Referring to Fig. 10.17, we note that the effective length of the column with respect to buckling in this plane is [latex]L_{e}=0.7 \mathrm{~L}[/latex]. The radius of gyration [latex]r_{z}[/latex] of the cross section is obtained by writing

[latex] \begin{gathered} I_{x}=\frac{1}{12} b a^{3} \quad A=a b \ ext { and, since } I_{z}=A r_{z}^{2}, \quad r_{z}^{2}=\frac{I_{z}}{A}=\frac{\frac{1}{12} b a^{3}}{a b}=\frac{a^{2}}{12} \quad r_{z}=a / \sqrt{12} \end{gathered} [/latex]

The effective slenderness ratio of the column with respect to buckling in the [latex]x y[/latex] plane is

[latex]\frac{L_{e}}{r_{z}}=\frac{0.7 L}{a / \sqrt{12}}[/latex] (1)

Buckling in [latex]\mathbf{x z}[/latex] Plane. The effective length of the column with respect to buckling in this plane is [latex]L_{e}=2 L[/latex], and the corresponding radius of gyration is [latex]r_{y}=b / \sqrt{12}[/latex]. Thus,

[latex]\frac{L_{e}}{r_{y}}=\frac{2 L}{b / \sqrt{12}}[/latex] (2)

a. Most Efficient Design. The most efficient design is that for which the critical stresses corresponding to the two possible modes of buckling are equal. Referring to Eq. [latex]\left(10.13^{\prime} ight)[/latex],

[latex]\sigma_{\mathrm{cr}}={\frac{\pi^{2}E}{(L_{\mathrm{c}}/r)^{2}}}[/latex] (10.13′)

we note that this will be the case if the two values obtained above for the effective slenderness ratio are equal. We write

[latex]\frac{0.7 L}{a / \sqrt{12}}=\frac{2 L}{b / \sqrt{12}}[/latex]

and, solving for the ratio [latex]a / b[/latex],

[latex]\frac{a}{b}=\frac{0.7}{2} \quad \frac{a}{b}=0.35[/latex]

b. Design for Given Data. Since F.S. [latex]=2.5[/latex] is required,

[latex]P_{ ext {cr }}=(F . S .) P=(2.5)(5 ext { kips })=12.5 \mathrm{kips}[/latex]

Using [latex]a=0.35 b[/latex], we have [latex]A=a b=0.35 b^{2}[/latex] and

[latex]\sigma_{\mathrm{cr}}=\frac{P_{\mathrm{cr}}}{A}=\frac{12,500 \mathrm{lb}}{0.35 b^{2}}[/latex]

Making [latex]L=20[/latex] in. in Eq. (2), we have [latex]L_{e} / r_{y}=138.6 / b[/latex]. Substituting for [latex]E[/latex], [latex]L_{e} / r[/latex], and [latex]\sigma_{ ext {cr }}[/latex] into Eq. [latex]\left(10.13^{\prime} ight)[/latex], we write

[latex] \begin{aligned} \sigma_{\mathrm{cr}}=\frac{\pi^{2} E}{\left(L_{e} / r ight)^{2}} \quad \frac{12,500 \mathrm{lb}}{0.35 b^{2}} & =\frac{\pi^{2}\left(10.1 imes 10^{6} \mathrm{psi} ight)}{(138.6 / b)^{2}} \ b & =1.620 \mathrm{in} . \quad a=0.35 b=0.567 \mathrm{in} . \end{aligned} [/latex]

Question 10.SP.1: An aluminum column of length L and rectangular cross section...

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