Question 3.9: Figure 3-12 shows a tube before [Part (b)] and after [Part (...

Objective    Compare the torsional stiffness and the strength of the closed tube of Figure 3-12(a) with those of the open-seam (split) tube shown in Figure 3-12(b).
Given        The tube shapes are shown in Figure 3-12. Both have the same length, diameter, and wall thickness, and both are made from the same material.
Analysis           Equation (3-13) gives the angle of twist for a noncircular member and shows that the angle is inversely proportional to the value of K. Similarly. Equation (3-11) shows that the an

\theta = TL/GK            (3-13)

\theta = TL/GJ             (3-11)

gle of twist for a hollow circular tube is inversely proportional to the polar moment of inertia J, All other terms in the two equations are the same for each design. Therefore, the ratio of θ_{open} to θ_{closed}  is equal to the ratio J/K. From Appendix 1, we find

J = \pi (D^4 – d^4)/32

From Figure 3-10, we find

K = 2\pi rt^3 /3

Using similar logic. Equations (3-12) and (3-8) show that the maximum

τ=Tr/J                  (3-8)

τ_{max} = T/Q                        (3-12)

torsional shear stress is inversely proportional to Q and Z_p for the open and closed tubes, respectively. Then we can compare the strengths of the two forms by computing the ratio Z_p/Q. By Equation (3-9), we find that

Z_p = J/c = J/(D/2)

The equation for Q for the split tube is listed in Figure 3-10.

Results We make the comparison of torsional stiffness by computing the ratio J/K, For the closed, hollow tube,

J = \pi (D^4 – d^4)/32

J = \pi(3.500^4 – 3.188^4)/32 = 4.592 \ in^4

For the open tube before the slit is welded, from Figure 3-10,

K = 2\pi rt^3 /3

K= [(2)(\pi)(1.672)(0.156)^3]/3 = 0.0133 \ in^4

Ratio = J/K = 4.592/0.0133 = 345

Then we make the comparison of the strengths of the two forms by computing the ratio Z_p/Q. The value of 7 has already been computed to be 4.592\ in^4. Then

Z_p = J/c = J/(D/2) = (4.592 \ in^4)/[(3.500in)/2] = 2.624 in^3

For the open tube.

Q=\frac{4\pi ^2r^2t^2}{(6\pi r + 1.80t)} =\frac{4\pi ^2( 1.672 \ in)^2(0.156 \ in)^2}{[6\pi (1.672 \ in) + 1.8(0.156 \ in)]} =0.0845 \ in^3

Then the strength comparison is

Ratio = Z_p/Q = 2.624/0.0845 = 31.1

Comments Thus, for a given applied torque, the slit tube would twist 345 times as much as the closed tube. The stress in the slit tube would be 31.1 times higher than in the closed tube. Also note that if the material for the tube is thin, it will likely buckle at a relatively low stress level, and the tube will collapse suddenly. This comparison shows the dramatic superiority of the closed form of a hollow section to an open form. A similar comparison could be made for shapes other than circular.