Question 3.15: Compare the maximum shear stress in a circular tube (Fig. 3-...

Approximate theory. The shear stress obtained from the approximate theory for a thin-walled tube [Eq. (3-83)] is

$\tau_{1}=\frac{T}{2 \pi r^{2} t}=\frac{T}{2 \pi t^{3} \beta^{2}}$ (3-96)

in which the relation

$\beta=\frac{r}{t}$  (3-97)

is introduced.
Torsion formula. The maximum stress obtained from the more accurate torsion formula [Eq. (3-13) $\tau_{\max }=\frac{T r}{I_{P}}$ ] is

$\tau_{2}=\frac{T(r+t / 2)}{I_{P}}$  (a)

where

$I_{P}=\frac{\pi}{2}\left[\left(r+\frac{t}{2}\right)^{4}-\left(r-\frac{t}{2}\right)^{4}\right]$    (b)

After expansion, this expression simplifies to

$I_{P}=\frac{\pi r t}{2}\left(4 r^{2}+r^{2}\right)$  (3-98)

and the expression for the shear stress [Eq. (a)] becomes

$\tau_{2}=\frac{T(2 r+t)}{\pi r t\left(4 r^{2}+t^{2}\right)}=\frac{T(2 \beta+1)}{\pi t^{3} \beta\left(4 \beta^{2}+1\right)}$  (3-99)

Ratio. The ratio $\tau_{1} / \tau_{2}$ of the shear stresses is

$\frac{\tau_{1}}{\tau_{2}}=\frac{4 \beta^{2}+1}{2 \beta(2 \beta+1)}$ (3-100)

which depends only on the ratio β.
For values of β equal to 5, 10, and 20, we obtain from Eq. (3-100) the values $\tau_{1} / \tau_{2}$ = 0.92, 0.95, and 0.98, respectively. Thus, we see that the approximate formula for the shear stresses gives results that are slightly less than those obtained from the exact formula. The accuracy of the approximate formula increases as the wall of the tube becomes thinner. In the limit, as the thickness approaches zero and β approaches infinity, the ratio $\tau_{1} / \tau_{2}$ becomes 1.