Question 8.20: A pin-ended column of height 3.0 m has a circular cross-sect......
A pin-ended column of height 3.0 m has a circular cross-section of diameter 80 mm, wall thickness 2.0 mm, and is converted to an open section by a narrow longitudinal slit; the ends of the column are free to warp. Determine the values of axial load that would cause the column to buckle in (a) pure bending and (b) pure torsion. Hence, determine the value of the flexural–torsional buckling load. Take E = 70,000 N/mm² and G = 22,000 N/mm². Note: the position of the shear center of the column section may be found using the method described in Chapter 17.
Answer: (a)\ 3.09\times10^{4}\,{\mathrm{N}}, ({ b})\;1.78\times10^{4}\,{ N},\,1.19\times10^{4}\,{ N}
The separate modes of buckling are obtained from Eqs (8.77), i.e.,
P_{\mathrm{CR}(x x)}={\frac{\pi^{2}E I_{x x}}{L^{2}}} P_{\mathrm{CR}(yy)}={\frac{\pi^{2}E I_{y y}}{L^{2}}} P_{\mathrm{CR}(\theta)}={\frac{A}{I_{0}}}\left(G J+{\frac{\pi^{2}E \Gamma}{L^{2}}}\right) (8.77)
P_{\mathrm{CR}(xx)}=P_{\mathrm{CR}(yy)}={\frac{\pi^{2}E I}{L^{2}}}\bigl(I_{x x}=I_{y y}=I,\mathrm{say}\bigr) (i)
and
P_{\mathrm{CR}(\theta)}={\frac{A}{I_{0}}}{\bigg(}G J+{\frac{\pi^{2}E T}{L^{2}}}{\bigg)} (ii)
In this case,
I_{x x}=I_{y y}=\pi r^{3}t=\pi\times40^{3}\times2.0=4.02\times10^{5}\,{\mathrm{mm}}^{4}A = 2\pi r t=2\pi\times40\times2.0=502.7\,\mathrm{mm}^{2}
J = 2\pi rt^{3}/3=2\pi\times40\times2.0^{3}/3=670.2\,{\mathrm{mm}}^{4}
From Eq. (8.68),
T(z)={ P}\biggl(x_{S}\frac{\mathrm{d}^{2}\nu}{\mathrm{d}z^{2}}-y_{S}\frac{\mathrm{d}^{2}u}{\mathrm{d}z^{2}}\biggr)-\frac{{ P}}{A}\frac{\mathrm{d}^{2}\theta}{\mathrm{d}z^{2}}(A y_{S}^{2}+I_{x x}+A x_{S}^{2}+I_{y y}) (8.68)
I_{0}=I_{x x}+I_{y y}+A x_{s}^{2} (note that y_{s} = 0)
in which x_{s} is the distance of the shear center of the section from its vertical diameter; it may be shown that x_{s} = 80 mm (see S.17.3). Then
I_{0}=2\times4.02\times10^{5}+502.7\times80^{2}=4.02\times10^{6}\,{\mathrm{mm}}^{4}The torsion-bending constant Γ is found in a similar manner to that for the section shown in Fig. P.28.3 and is given by
\Gamma=\pi r^{5}t\left({\frac{2}{3}}\pi^{2}-4\right)i.e.,
\Gamma=\pi\times40^{5}\times2.0{\bigg(}{\frac{2}{3}}\pi^{2}-4{\bigg)}=1.66\times10^{9}\mathrm{mm}^{6}(a) P_{\mathrm{CR}(xx)}=P_{\mathrm{CR}(yy)}={\frac{\pi^{2}\times70000\times4.02\times10^{5}}{\left(3.0\times10^{3}\right)^{2}}}=3.09\times10^{4}\mathrm{N}
(b) P_{\mathrm{CR}(\theta)}=\frac{502.7}{40.2\times10^{6}}\left(22000\times670.2+\frac{\pi^{2}\times70000\times1.66\times10^{9}}{\left(3.0\times10^{3}\right)^{2}}\right) =1.78\times10^{4}\,\mathrm{N}
The flexural-torsional buckling load is obtained by expanding Eq. (8.79). Thus,
\begin{vmatrix} P-P_{CR(xx)} & -Px_{S} \\ -Px_{S} & I_{0}(P-P_{CR(\theta)})/A) \end{vmatrix} =0 (8.79)
(P-P_{\mathrm{CR}(xx)})(P-P_{\mathrm{CR}(\theta)})I_{0}/A-P^{2}x_{\mathrm{s}}^{2}=0
from which
P^{2}\bigl(1-A x_{s}^{2}/I_{0}\bigr)-P\bigl(P_{\mathrm{CR}(xx)}+P_{\mathrm{CR}(\theta)}+P_{\mathrm{CR}(\theta)}\bigr)+P_{\mathrm{CR}(x x)}P_{\mathrm{CR}(\theta)}=0 (iii)
Substituting the appropriate values in Eq. (iii) gives
P-24.39\times10P^{4}+27.54\times10^{8}=0 (iv)
The solutions of Eq. (iv) are
P=1.19\times10^{4}\mathrm{N~~or~~23.21\times10^{4}\mathrm{N}}Therefore, the least flexural-torsional buckling load is 1.19\times10^{4}\,\mathrm{N}.
