Question 3.16: A circular tube and a square tube (Fig. 3-57) are constructe......
A circular tube and a square tube (Fig. 3-57) are constructed of the same material and subjected to the same torque. Both tubes have the same length, same wall thickness, and same cross-sectional area.
What are the ratios of their shear stresses and angles of twist? (Disregard the effects of stress concentrations at the corners of the square tube.)
Use a four-step problem-solving approach.
1,2. Conceptualize, Categorize:
Circular tube: For the circular tube, the area A_{m1} enclosed by the median line of the cross section is
A_{m1}=\pi r^2 (a)
where r is the radius to the median line. Also, the torsion constant
[Eq. (3-93)]
J=2\pi r^3t (3-93)
and cross-sectional area are
J_1=2\pi r^3t A_1=2\pi rt (b,c)
Square tube: For the square tube, the cross-sectional area is
A_2=4bt (d)
where b is the length of one side measured along the median line. Inasmuch as the areas of the tubes are the same, length b is b=πr/2. Also, the torsion constant [Eq. (3-94)]
J=\frac{2b^2h^2t_1t_2}{bt_1+ht_2} (3-94)
and area enclosed by the median line of the cross section are
J_2=b^3t=\frac{\pi ^3r^3t}{8} A_{m2}=b^2=\frac{\pi ^2r^2}{4} (e,f)
3. Analyze:
Ratios: The ratio \tau _{1}/\tau _{2} of the shear stress in the circular tube to the shear stress in the square tube [from Eq. (3-81)]
\tau =\frac{T}{2tA_m} (3-81)
is
\frac{\tau_1}{\tau_2} =\frac{A_{m2}}{A_{m1}} =\frac{\pi ^2r^2/4}{\pi r^2}=\frac{\pi }{4} =0.79 \hookleftarrow (g)
From the torque-displacement relation \phi =\frac{TL}{GJ} , the ratio of the angles of twist is
\frac{\phi_1}{\phi_2} =\frac{J_{2}}{J_{1}} =\frac{\pi ^3r^3t/8}{2\pi r^3t}=\frac{\pi^2 }{16} =0.62 \hookleftarrow (j)
4. Finalize: These results show that the circular tube not only has a 21% lower shear stress than does the square tube but also a greater stiffness against rotation.