Question : Comparison of open and closed thin-walled sections.

The torsional behaviour of thin-walled closed sections is quite different from that of open sections. For closed sections, the shear stress is uniformly distributed through the thickness of the wall (Figure 4.25), whereas a linear distribution through the wall thickness is found in open sections. The torsional stiffness \hat{D}_{t} is proportional to the square of the enclosed area for a closed section (Eq.(4.52)) in contrast with a thickness cubed proportionality for open sections (Eq.(4.142a) [BC].
Consider, for instance, a thin ring of circular shape and a thin-walled open circular tube, both of identical mean radius R_{m} and thickness t, as depicted in Figure 4.25. The torsional stiffness of the closed and open sections, denoted \hat{D}_{t}^{closed} and \hat{D}_{t}^{open} , respectively, are given by Eqs.(4.52) and (4.142a), respectively, as \hat{D}_{t}^{closed}=2\pi GR_{m}^{3}t and \hat{D}_{t}^{open}=2\pi GR_{m}t^{3}/3. Their ratio is

If the two sections are subjected to the same torque,M_{\hat{x}{}'} , the maximum shear stresses in the open and closed sections, denoted \tau _{max}^{open} and \tau _{max}^{closed}, respectively, are given by Eqs.(4.143) and (4.53), respectively, as

\tau _{max}^{open}=\frac{M_{\hat{x}{}'}t}{\hat{D}_{t}^{open}}=\frac{3M_{\hat{x}{}'}}{2\pi R_{m}t^{2}} , \tau _{max}^{closed}=\frac{M_{\hat{x}{}'}}{2\pi R_{m}t}

Their ratio can then be expressed as

For a typical thin-walled beam with R_{m}=20t. The torsional stiffness of the closed section will be 1200 times larger than that of the open section. Under the same applied torque, the maximum shear stress in the open section will be 60 times larger than that of the closed section. In other words, the closed section can carry a 60 times larger torque for an equal shear stress level [BC].

//(Eq.(4.52)):\hat{D}_{t}=\frac{4GtA^{2}}{L_{s}}=GJ with J=\frac{4tA^{2}}{L_{s}}

(Eq.(4.142a)):J\simeq \frac{1}{3}L_{s} \frac{t^{3}}{3} and \hat{D}_{t}\simeq \frac{G}{3}\sum_{i=1}^{n} L_{s}\frac{t^{3}}{3}

(Eqs.(4.143)) : \tau_{s}^{max}=\pm Gt\frac{\partial \theta _{\hat{x}{}'}}{\partial {x}'}=\pm Gt \frac{M_{\hat{x}{}'}}{\hat{D}_{t}}

(Eqs.(4.53)) : \tau _{s}=\frac{M_{\hat{x}{}'}}{2\pi R_{m}^{t}}