Question 3.15: Compare the maximum shear stress in a circular tube (Fig. 3-......
Compare the maximum shear stress in a circular tube (Fig. 3-56), as calculated by the approximate theory for a thin-walled tube, with the stress calculated by the exact torsion theory. (Note that the tube has constant thickness t and radius r to the median line of the cross section.)
Use a four-step problem-solving approach.
1, 2. Conceptualize, Categorize:
Approximate theory: The shear stress obtained from the approximate theory for a thin-walled tube [Eq. (3-83)]
\tau =\frac{T}{2\pi r^2t} (3-83)
is
\tau_1 =\frac{T}{2\pi r^2t}=\frac{T}{2\pi r^3\beta ^2} (3-95)
in which the relation
\beta=\frac{r}{t} (3-96)
is introduced.
Torsion formula: The maximum stress obtained from the more accurate torsion formula [Eq. (3-13)]
\tau_{max}=\frac{Tr}{I_p} (3-13)
is
\tau_{2}=\frac{T(r+t/2)}{I_p} (a)
where
I_p=\frac{\pi }{2} \left[\left\lgroup r+\frac{t}{2}\right\rgroup^4-\left\lgroup r-\frac{t}{2}\right\rgroup^4\right] (b)
After expansion, this expression simplifies to
I_p=\frac{\pi rt }{2} (4r^2+t^2) (3-97)
and the expression for the shear stress [Eq. (a)] becomes
\tau_2=\frac{T(2r+t)}{\pi rt (4r^2+t^2)} = \frac{T(2\beta +1)}{\pi t^3\beta (4\beta ^2+1)} (3-98)
3. Analyze:
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)} \hookleftarrow (3-99)
which depends only on the ratio β.
4. Finalize: Using values of b equal to 5, 10, and 20 in Eq. (3-99) results in values \tau_1/\tau_2= 0.92, 0.95, and 0.98, respectively. Thus, 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.