Question 16.SP.1: An aluminum column with a length of L and a rectangular cros...
An aluminum column with a length of L and a 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 × 10^6 psi, P = 5 kips, and a factor of safety of 2.5 is required.
STRATEGY: The most efficient design is that for which the critical stresses corresponding to the two possible buckling modes are equal. This occurs if the two critical stresses obtained from Eq. (16.13b) are the same. Thus, for this problem, the two effective slenderness ratios in this equation must be equal to solve part a. Use Fig. 16.18 to determine the effective lengths. The design data can then be used with Eq. (16.13b) to size the cross section for part b.
\sigma_{cr}=\frac{\pi^2 E}{\left(L_e / r\right)^2} (16.13b)
MODELING:
Buckling in xy Plane. Referring to Fig. 16.18c, the effective length of the column with respect to buckling in this plane is L_e=0.7 L. The radius of gyration r_z of the cross section is obtained by
I_z=\frac{1}{12} b a^3 \quad A=a band because I_z=A r_z^2,
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}The effective slenderness ratio of the column with respect to buckling in the xy plane is
\frac{L_e}{r_z}=\frac{0.7 L}{a / \sqrt{12}} (1)
Buckling in xz Plane. Referring to Fig. 16.18a, 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)
ANALYSIS:
a. Most Efficient Design. The most efficient design is when the critical stresses corresponding to the two possible modes of buckling are equal. Referring to Eq. (16.13b), this is the case if the two values obtained earlier for the effective slenderness ratio are equal.
\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} \frac{a}{b}=0.35
b. Design for Given Data. Because F . S. = 2.5 is required,
P_{cr}=(F . S .) P=(2.5)(5 \text{ kips})=12.5 \text{ kips}Using a = 0.35b,
A = ab = 0.35b² and \sigma_{cr}=\frac{P_{cr}}{A}=\frac{12500 lb}{0.35 b^2}
Making L = 20 in. in Eq. (2), L_e / r_y=138.6 / b. Substituting for E, L_e / r, and \sigma_{cr} into Eq. (16.13b) gives
\sigma_{cr}=\frac{\pi^2 E}{\left(L_e / r\right)^2} \quad \frac{12,500 lb}{0.35 b^2}=\frac{\pi^2\left(10.1 \times 10^6 psi\right)}{(138.6 / b)^2}b = 1.620 in. a = 0.35b = 0.567 in.
REFLECT and THINK: The calculated critical Euler buckling stress can never be taken to exceed the yield strength of the material. In this problem, you can readily determine that the critical stress \sigma_{cr}=13.6 ksi; though the specific alloy was not given, this stress is less than the tensile yield strength \sigma_Y values for all aluminum alloys listed in Appendix C.