Question 11.1: 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 axial load at A. Two smooth and rounded fixed plates restrain end A from moving in one of the vertical planes of symmetry but allow it to move in the other plane. (1) Determine the ratio a/b of the two sides of the cross section corresponding to the most efficient design against buckling. (2) Design the most efficient cross section for the column, knowing that L = 50 cm, E = 70 GPa, P = 22 kN and that a safety factor of 2.5 for buckling is required.
Given: Loading and support conditions for column; safety factor; length and elastic modulus.
Find: Optimal rectangular cross section of column.
Assume: Hooke’s law applies (we have assumed constant E in derivations).
Figure on next page indicates that we must consider buckling in both the xy- and xz-planes and that due to the nature of the support at A, the critical load and the prospect of buckling will be different in the two planes. As the supports allow end A to move freely in the z-direction, for buckling in the xz-plane we have a fixed–free support combination; the supports constrain motion in the y-direction but do not provide a reaction moment so that in the xy-plane we have a fixed–pinned support.
1. For buckling in the xy-plane, due to the fixed–pinned support combination, we find from Table 11.1 that the effective length of the column with respect to buckling in this plane is L_e = 0.7L. And we have
I_z=\frac{1}{12} b a^3For buckling in the xz-plane, the column sees a fixed–free support situation, so the effective length is L_e = 2L. And
I_y=\frac{1}{12} a b^3The most efficient design is that for which the critical buckling loads corresponding to the two possible modes of buckling are equal; neither mode is preferred. This is the case if
\frac{\pi^2 E\left(1 / 12 b a^3\right)}{(0.7 L)^2}=\frac{\pi^2 E\left(1 / 12 a b^3\right)}{(2 L)^2} .From which we determine the ratio a/b = 0.35
2. A safety factor of 2.5 means that for our design with specific parameters we must have a critical load that is 2.5 times the load P that will actually be applied. So P_{cr} = 2.5P = 2.5(22 kN) = 55 kN.
Using the ratio a/b found above, and the buckling criterion for the xz-plane (xy is the same), we have
P_{\mathrm{cr}}=55 \mathrm{kN}=\frac{\pi^2 E\left((1 / 12)(0.35 b) b^3\right)}{(2 L)^2}.With the given values for E and L, we can solve for b = 4.1 cm and a = 0.35b = 1.4 cm.
table 11.1
\begin{aligned} &\text { Effective Length } L_{\mathrm{e}} \text { for Different End Supports }\\ &\begin{array}{lc} \text { End Conditions } & \text { Effective Length } \\ \hline \text { Fixed-free } & 2 L \\ \text { Pinned-pinned } & L \\ \text { Fixed-pinned } & 0.7 \mathrm{~L} \\ \text { Fixed-fixed } & 0.5 \mathrm{~L} \\ \hline \end{array} \end{aligned}