Question 8.17: Figure P.8.17 shows the doubly symmetrical cross-section of ......
Figure P.8.17 shows the doubly symmetrical cross-section of a thin-walled column with rigidly fixed ends. Find an expression, in terms of the section dimensions and Poisson’s ratio, for the column length for which the purely flexural and the purely torsional modes of instability occur at the same axial load. In which mode does failure occur if the length is less than the value found? The possibility of local instability is to be ignored.
Answer: l=(2\pi b^{2}/t)\,\sqrt{(1+\text{v})/25}, torsion
The purely flexural instability load is given by Eq. (8.7) in which, from Table 8.1 l_{e}=0.5l where l is the actual column length. Also it is clear that the least second moment of area of the column cross-section occurs about an axis coincident with the web. Thus,
P_{\mathrm{CR}}={\frac{\pi^{2}E I}{l_{\mathrm{e}}^{2}}} (8.7)
I=2\times{\frac{2t b^{3}}{12}}={\frac{t b^{3}}{3}}
Then
P_{\mathrm{{CR}}}={\frac{\pi^{2}E I}{(0.5l)^{2}}}i.e.,
P_{\mathrm{CR}}={\frac{4\pi^{2}E t b^{3}}{3l^{2}}} (i)
The purely torsional buckling load is given by the last of 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 I}{L^{2}}}\right) (8.77)
P_{\mathrm{CR}(\theta)}=\frac{A}{I_{0}}\bigg(G J+\frac{4\pi^{2}E T}{l^{2}}\bigg) (ii)
Note: For a column with fixed ends, l_{e}=0.5l.
In Eq. (ii) A = 5bt and
i.e.,
I_{0}={\frac{17t b^{3}}{12}}Also, from Eq. (18.11),
J=\sum{\frac{s t^{3}}{3}} or J=\frac{1}{3}\int_{sect}^{}{t^{3}} ds (18.11)
J=\sum{\frac{s t^{3}}{3}}={\frac{1}{3}}(2b8t^{3}+b t^{3})={\frac{17b t^{3}}{3}}
and, referring to S.28.4,
\Gamma={\frac{t b^{5}}{12}}Then, from Eq. (ii),
P_{\mathrm{CR}(\theta)}={\frac{20}{17b}}\Biggl(17G t^{3}+{\frac{\pi^{2}E t b^{4}}{l^{2}}}\Biggr) (iii)
Now equating Eqs (i) and (iii),
\frac{4\pi^{2}E t b^{3}}{3l^{2}}=\frac{20}{17b}{\biggl(}17G t^{3}+\frac{\pi^{2}E t b^{4}}{l^{2}}{\biggr)}from which
l^{2}={\frac{2\pi^{2}E b^{4}}{255G t^{2}}}From Eq. (1.50), E/G = 2(1 + v). Hence,
\gamma={\frac{2(1+\nu)}{E}}\tau (1.50)
l={\frac{2\pi b^{2}}{t}}{\sqrt{\frac{1+\nu}{255}}}Eqs (i) and (iii) may be written, respectively, as
P_{\mathrm{CR}}={\frac{1.33C_{1}}{l^{2}}}and
P_{\mathrm{CR}(\theta)}=C_{2}+{\frac{1.175C_{1}}{l^{2}}}where C_{1} and C_{2} are constants. Thus, if l were less than the value found, the increase in the last term in the expression for P_{\mathbf{CR}(\theta)} would be less than the increase in the value of P_{\mathrm{CR}},{\mathrm{i.e.}},P_{\mathrm{CR}(\theta)}\lt P_{C R} for a decrease in l and the column would fail in torsion.
Table 8.1 Column Length Solutions
| Ends | l_{e}/l | Boundary conditions | |||
| Both pinned | 1.0 | v = 0 at z = 0 and l | |||
| Both fixed | 0.5 | v = 0 at z = 0 and z = l, dv/dz =| 0 at z = l | |||
| One fixed, the other free | 2.0 | v = 0 and dv/dz = 0 at z = 0 | |||
| One fixed, the other pinned | 0.6998 | dv/dz = 0 at z = 0, v = 0 at z = l and z = 0 |
