Question 8.18: A column of length 2l with the doubly symmetric cross-sectio......
A column of length 2l with the doubly symmetric cross-section, shown in Fig. P.8.18, is compressed between the parallel platens of a testing machine that fully prevents twisting and warping of the ends. Using the data that follows, determine the average compressive stress at which the column first buckles in torsion:
l = 500 mm, b = 25.0 mm, t = 2.5 mm, E = 70,000 N/mm², E/G = 2.6
Answer: \sigma_{\mathrm{CR}}=282\ \mathrm{N/mm^{2}}.
In this case Eqs (8.77) do not apply since the ends of the column are not free to warp. From Eq. (8.70) and since, for the cross-section of the column, x_{s} = y_{s} = 0,
E \Gamma{\frac{\mathrm{d}^{4}\theta}{\mathrm{d}z^{4}}}-\left(G J-I_{0}{\frac{P}{A}}\right){\frac{\mathrm{d}^{2}\theta}{\mathrm{d}z^{2}}}-P x_{S}{\frac{\mathrm{d}^{2}\text{v}}{\mathrm{d}z^{2}}}+P y_{S}{\frac{\mathrm{d}^{2}u}{\mathrm{d}z^{2}}}=0 (8.70)
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)
E \Gamma{\frac{\mathrm{d}^{4}\theta}{\mathrm{d}z^{4}}}+\left(I_{0}{\frac{P}{A}}-G J\right){\frac{\mathrm{d}^{2}\theta}{\mathrm{d}z^{2}}}=0 (i)
For buckling, P=P_{\mathrm{CR}}, the critical load and P_{\mathrm{CR}}/A=\sigma_{\mathrm{CR}}, the critical stress. Eq. (i) may then be written
{\frac{\mathrm{d}^{4}\theta}{\mathrm{d}z^{4}}}+\lambda^{2}{\frac{\mathrm{d}^{2}\theta}{\mathrm{d}z^{2}}}=0 (ii)
in which
\lambda^{2}{=}\frac{(I_{0}\sigma_{\mathrm{CR}}-G J)}{E \Gamma} (iii)
The solution of Eq. (ii) is
\theta=A\cos\lambda z+B\sin\lambda z+D z+F (iv)
The boundary conditions are:
\theta=0\;\;\;\mathrm{at}\;\;z=0\;\;\;\mathrm{and}\;\;z=2l{\frac{\mathrm{d}\theta}{\mathrm{d}z}}=0 \mathrm{at} z=0 \mathrm{and} z=2l (see Eq.(18.19))
w_{s}=-2A_{\mathrm{R}}\ {\frac{\mathrm{d}\theta}{\mathrm{d}z}} (18.19)
Then B = D = 0, F = –A, and Eq. (iv) becomes
\theta{=}A(\cos\lambda z-1) (v)
Since θ = 0 when z = 2l,
\cos\lambda2l=1or
\lambda2l=2n\piHence, for n = 1,
\lambda^{2}={\frac{\pi^{2}}{l^{2}}}i.e., from Eq. (iii),
{\frac{I_{0}\sigma_{\mathrm{CR}}-G J}{E \Gamma}}={\frac{\pi^{2}}{l^{2}}}so that
\sigma_{\mathrm{CR}}=\frac{1}{I_{0}}\left(G J+\frac{\pi^{2}E T}{l^{2}}\right) (iv)
For the cross-section of Fig. P.8.15,
J=\sum{\frac{s t^{3}}{3}} (see Eq.(18.11))
J=\sum{\frac{s t^{3}}{3}} or J=\frac{1}{3}\int_{sect}^{}{t^{3} ds} (18.11)
i.e.,
J={\frac{8b t^{3}}{3}}={\frac{8\times25.0\times2.5^{3}}{3}}=1041.7\,\mathrm{mm}^{4}and
I_{x x}=4b t(b\cos30^{\circ})^{2}+2{\frac{(2b)^{3}t\sin^{2}60^{\circ}}{12}} (see Section 16.4.5)
i.e.,
I_{x x}=4b^{3}t=4\times25.0^{3}\times2.5=156250.0 \mathrm{mm}^{4}Similarly,
I_{y y}=0\left({\frac{b t^{3}}{12}}+b t b^{2}\right)+2{\frac{(2b)^{3}t\cos^{2}60^{\circ}}{12}}={\frac{14b^{3}t}{3}}so that
I_{y y}=14\times25.0^{3}\times2.5/3=182291.7\,\mathrm{mm}^{4}Then
I_{0}=I_{x x}+I_{y y}=338541.7\,\mathrm{mm}^{4}The torsion-bending constant, Γ, is found by the method described in Section 28.2 and is given by
\Gamma=b^{5}{t}=25.0^{5}\times2.5=24\times10^{6}\mathrm{{mm}}^{4}Substituting these values in Eq. (vi) gives
\sigma_{\mathrm{CR}}=282.0\,\mathrm{N/mm^{2}}