Question 10.SP.1: An aluminum column of length L and rectangular cross section...
An aluminum column of length L and rectangular cross section has a fixed end B and supports a centric load at A. Two smooth and rounded fixed plates restrain end A 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 a / b 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 L=20 in., E=10.1 \times 10^{6} \mathrm{psi}, P=5 kips, and that a factor of safety of 2.5 is required.
Buckling in x y Plane. Referring to Fig. 10.17, we note that the effective length of the column with respect to buckling in this plane is L_{e}=0.7 \mathrm{~L}. The radius of gyration r_{z} of the cross section is obtained by writing
\begin{gathered} I_{x}=\frac{1}{12} b a^{3} \quad A=a b \\ \text { 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}The effective slenderness ratio of the column with respect to buckling in the x y plane is
\frac{L_{e}}{r_{z}}=\frac{0.7 L}{a / \sqrt{12}} (1)
Buckling in \mathbf{x z} Plane. The effective length of the column with respect to buckling in this plane is L_{e}=2 L, and the corresponding radius of gyration is r_{y}=b / \sqrt{12}. Thus,
\frac{L_{e}}{r_{y}}=\frac{2 L}{b / \sqrt{12}} (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. \left(10.13^{\prime}\right),
\sigma_{\mathrm{cr}}={\frac{\pi^{2}E}{(L_{\mathrm{c}}/r)^{2}}} (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
\frac{0.7 L}{a / \sqrt{12}}=\frac{2 L}{b / \sqrt{12}}
and, solving for the ratio a / b,
\frac{a}{b}=\frac{0.7}{2} \quad \frac{a}{b}=0.35
b. Design for Given Data. Since F.S. =2.5 is required,
P_{\text {cr }}=(F . S .) P=(2.5)(5 \text { kips })=12.5 \mathrm{kips}
Using a=0.35 b, we have A=a b=0.35 b^{2} and
\sigma_{\mathrm{cr}}=\frac{P_{\mathrm{cr}}}{A}=\frac{12,500 \mathrm{lb}}{0.35 b^{2}}
Making L=20 in. in Eq. (2), we have L_{e} / r_{y}=138.6 / b. Substituting for E, L_{e} / r, and \sigma_{\text {cr }} into Eq. \left(10.13^{\prime}\right), we write
\begin{aligned} \sigma_{\mathrm{cr}}=\frac{\pi^{2} E}{\left(L_{e} / r\right)^{2}} \quad \frac{12,500 \mathrm{lb}}{0.35 b^{2}} & =\frac{\pi^{2}\left(10.1 \times 10^{6} \mathrm{psi}\right)}{(138.6 / b)^{2}} \\ b & =1.620 \mathrm{in} . \quad a=0.35 b=0.567 \mathrm{in} . \end{aligned}
