Question 8.19: A pin-ended column of length 1.0 m has the cross-section sho......
A pin-ended column of length 1.0 m has the cross-section shown in Fig. P.8.19. If the ends of the column are free to warp, determine the lowest value of axial load that will cause the column to buckle and specify the mode. Take E = 70,000 N/mm² and G = 25,000 N/mm².
Answer: 5,527 N. The column buckles in bending about an axis in the plane of its web.
The three possible buckling modes of the column are given by Eqs (8.77), i.e.,
P_{\mathrm{CR}(xx)}={\frac{\pi^{2}E I_{x x}}{L^{2}}}\qquad P_{\mathrm{CR(yy)}}={\frac{\pi^{2}E I_{y y}}{L^{2}}}\qquad P_{\mathrm{CR(\theta)}}={\frac{A}{I_{0}}}\left(G J+{\frac{\pi^{2}E I}{L^{2}}}\right) (8.77)
P_{\mathrm{{CR}}(x x)}={\frac{\pi^{2}E I_{x x}}{L^{2}}} (i)
P_{\mathrm{{CR}}(yy)}={\frac{\pi^{2}E I_{y y}}{L^{2}}} (ii)
P_{\mathrm{CR}(\theta)}=\frac{A}{I_{0}}\bigg(G J+\frac{\pi^{2}E T}{L^{2}}\bigg) (iii)
From Fig. P.8.16 and taking the x axis parallel to the flanges,
A=2\times20+40)\times1.5=120\,\mathrm{mm}^{2}I_{x x}=2\times20\times1.5\times20^{2}+1.5\times40^{3}/12=3.2\times10^{4}\,\mathrm{mm}^{4}
I_{y y}=1.5\times40^{3}/12=0.8\times10^{4}\,{\mathrm{mm}}^{4}
I_{0}=I_{x x}+I_{y y}=4.0\times10^{4}\,{\mathrm{mm}}^{4}
J=(20+40+20)\times1.5^{3}/3=90.0m{\mathrm{m}}^{4} (seeEq.(18.11))
J=\sum{\frac{st^{3}}{3} } or J=\frac{1}{3}\int_{sect}^{}{t^{3} ds} (18.11)
\Gamma={\frac{1.5\times20^{3}\times40^{2}}{12}}\left({\frac{2\times40\times20}{40\times2\times20}}\right) =2.0\times10^{6}\,{\mathrm{mm}}^{6} (see Eq.(ii) of Example 28.1)
Substituting the appropriate values in Eqs (i), (ii), and (iii) gives
\begin{array}{l}{{P_{\mathrm{CR}(xx)}=22107.9\mathrm{N}}}\\ {{P_{\mathrm{CR}(yy)}=5527.0\mathrm{N}}}\\ {{P_{\mathrm{CR}(\theta)}=10895.2\mathrm{N}}}\end{array}Thus the column will buckle in bending about the y axis at a load of 5527.0 N.