Question 16.S-P.1: An aluminum column of length L and rectangular cross section...

Buckling in xy Plane. Referring to Fig. 16.17, we note that the effective length of the column with respect to buckling in this plane is L_{e} = 0.7L. The radius of gyration r_{z} of the cross section is obtained by writing

I_{x} = \frac{1}{12} ba^{3}                  A = ab

and, since I_{z} = Ar^{2}_{z} ,                 r^{2}_{z} = \frac{I_{z}}{A} = \frac{\frac{1}{12}ba^{3}}{ab} = \frac{a^{2}}{12}                    r_{z} = a/\sqrt{12}

The effective slenderness ratio of the column with respect to buckling in the xy plane is

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

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

\frac{L_{e}}{r_{y}} = \frac{2L}{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. (16.13^{′}), we note that this will be the case if the two values obtained above for the effective slenderness ratio are equal. We write

σ_{cr} = \frac{π^{2}E}{(L_{e}/r)^{2}}                 (16.13^{′})

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

and, solving for the ratio a/b,

\frac{a}{b} =\frac{0.7}{2}                         \frac{a}{b} = 0.35

b. Design for Given Data. Since F.S. = 2.5 is required,

P_{cr} = (F.S.)P = (2.5) (5  kips) = 12.5  kips

Using a = 0.35b, we have A = ab = 0.35b^{2} and

σ_{cr} = \frac{P_{cr}}{A} = \frac{12,500  lb}{0.35b^{2}}

Making L = 20 in. in Eq. (2), we have L_{e}/r_{y} = 138.6/b. Substituting for E, L_{e}/r,  and   σ_{cr} into Eq. (16.13^{′}), we write

σ_{cr} =\frac{π^{2}E}{(L_{e}/r)^{2}}                          \frac{12,500  lb}{0.35b^{2}} = \frac{π^{2}(10.1 × 10^{6}  psi)}{(138.6/b)^{2}}

b = 1.620  in.                   a = 0.35b = 0.567  in.