Question 3.16: A circular tube and a square tube (Fig. 3-57) are constructe......

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.