A thin-walled column has the cross-section shown in Fig. 8.19, is of length L, and is subjected to an axial load through its shear center S. If the ends of the column are prevented from warping and twisting, determine the value of direct stress when failure occurs due to torsional buckling.
Question 8.6: A thin-walled column has the cross-section shown in Fig. 8.1...
The torsion bending constant \Gamma is found using the method described in Section 27.2 of Ref. 3. The position of the shear center is given but is obvious by inspection. The swept area 2 \Gamma A_{R, 0} is determined as a function of s, and its distribution is shown in Fig. 8.20. The center of gravity of the “wire: is found by taking moments about the s axis. Then,
2 A_{R}^{\prime} 5 t d=t d\left(\frac{d^{2}}{2}+\frac{5 d^{2}}{4}+\frac{3 d^{2}}{2}+\frac{5 d^{2}}{4}+\frac{d^{2}}{2}\right)
which gives
2 A_{R}^{\prime}=d^{2}
The torsion bending constant is then the “moment of inertia” of the “wire” and is
\Gamma=2 t d \frac{1}{3}\left(d^{2}\right)^{2}+\frac{t d}{3}\left(\frac{d^{2}}{2}\right)^{2} \times 2+t d\left(\frac{d^{2}}{2}\right)^{2}
from which
\Gamma=\frac{13}{12} t d^{5}
Also, the torsion constant J is given by (see Section 3.4)
J=\sum \frac{s t^{3}}{3}=\frac{5 d t^{3}}{3}
The shear center of the section and the centroid of area coincide, so that the torsional buckling load is given by Eq. (8.76). Rewriting this equation,
E \Gamma \frac{ d ^{4} \theta}{ d z^{4}}-\left(G J-I_{0} \frac{P}{A}\right) \frac{ d ^{2} \theta}{ d z^{2}}=0 (8.76)
\frac{ d ^{4} \theta}{ d z^{4}}+\mu^{2} \frac{ d ^{2} \theta}{ d z^{2}}=0 (i)
where
\mu^{2}=\left(\sigma I_{0}-G J\right) / E \Gamma, \quad(\sigma=P / A)
The solution of Eq. (i) is
\theta=A \cos \mu z+B \sin \mu z+C z+D (ii)
The boundary conditions are \theta=0 when z = 0 and z = L, and since the warping is suppressed at the ends of the beam,
\left.\frac{ d \theta}{ dz }=0, \text { when } z=0 \text { and } z=L \quad \text { (see Eq. }(17.19)\right)
w_{s}=-2 A_{ R } \frac{ d \theta}{ d z} (17.19)
Putting \theta=0 at z = 0 in Eq. (ii)
0=A+D
or
A=-D
Also,
\frac{ d \theta}{ d z}=-\mu A \sin \mu z+\mu B \cos \mu z+C
and, since ( d \theta / d z)=0 at z = 0,
C=-\mu B
When z=L, \theta=0, so that, from Eq. (ii),
0=A \cos \mu L+B \sin \mu L+C L+D
which may be rewritten
0=B(\sin \mu L-\mu L)+ A (\cos \mu L-1) (iii)
Then, for ( d \theta / d z)=0 \text { at } z=L,
0=\mu B \cos \mu L-\mu A \sin \mu L-\mu B
or
0=B(\cos \mu L-1)-A \sin \mu L (iv)
Eliminating A from Eqs. (iii) and (iv),
0=B[2(1-\cos \mu L)-\mu L \sin \mu L] (v)
Similarly, in terms of the constant C,
0=-C[2(1-\cos \mu L)-\mu L \sin \mu L] (vi)
or
B=-C
But B=-C / \mu, so that to satisfy both equations, B = C = 0 and
\theta=A \cos \mu z-A=A(\cos \mu z-1) (vii)
Since \theta=0 \text { at } z=l,
\cos \mu L=1
or
\mu L=2 n \pi
Therefore,
\mu^{2} L^{2}=4 n^{2} \pi^{2}
or
\frac{\sigma I_{0}-G J}{E \Gamma}=\frac{4 n^{2} \pi^{2}}{L^{2}}
The lowest value of torsional buckling load corresponds to n = 1, so that, rearranging the preceding,
\sigma=\frac{1}{I_{0}}\left(G J+\frac{4 \pi^{2} E \Gamma}{L^{2}}\right) (viii)
The polar second moment of area I_{0} is given by
\left.I_{0}=I_{x x}+I_{y y} \quad \text { (see Megson }^{2}\right)
that is,
I_{0}=2\left(t d d^{2}+\frac{t d^{3}}{3}\right)+\frac{3 t d^{3}}{12}+2 t d \frac{d^{2}}{4}
which gives
I_{0}=\frac{41}{12} t d^{3}
Substituting for I_{0}, J, and \Gamma in Eq. (viii),
\sigma=\frac{4}{41 d^{2}}\left(5 G t^{2}+\frac{13 \pi^{2} E d^{4}}{L^{2}}\right)
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