Question 3.12: A tapered bar AB of solid circular cross section is supporte......
A tapered bar AB of solid circular cross section is supported at the right-hand end and loaded by a torque T at the other end (Fig. 3-43). The diameter of the bar varies linearly from d_A at the left-hand end to d_B at the right-hand end. Determine the angle of rotation \phi_A at end A of the bar by equating the strain energy to the work done by the load.
Use a four-step problem-solving approach. Combine steps as needed for an efficient solution.
1, 2. Conceptualize, Categorize: From the principle of conservation of energy, the work done by the applied torque equals the strain energy of the bar; thus, W = U . The work is given by
W=\frac{T\phi _A}{2} (a)
and the strain energy U can be found from Eq. (3-58).
U=\int_{0}^{L}{\frac{[T(x)]^2dx}{2GI_p(x)} } (3-58)
3. Analyze: To use Eq. (3-58), first find expressions for the torque T(x) and the polar moment of inertia I_p(x). The torque is constant along the axis of the bar and equal to the load T, and the polar moment of inertia is
I_p(x)=\frac{\pi }{32} [d(x)]^4in which d(x) is the diameter of the bar at distance x from end A. From the geometry of the figure, diameter d(x) is
d(x)=d_A+\frac{d_B-d_A}{L}x (b)
and therefore
I_p(x)=\frac{\pi }{32}\left\lgroup d_A+\frac{d_B-d_A}{L}x\right\rgroup ^4 (c)
Now substitute into Eq. (3-58)
U=\int_{0}^{L}{\frac{[T(x)]^2dx}{2GI_p(x)} } =\frac{16T^2}{\pi G}\int_{0}^{L}\frac{dx}{\left\lgroup d_A+\frac{d_B-d_A}{L}x\right\rgroup ^4}The integral in this expression can be integrated with the aid of a table of integrals (see Appendix C) with the result:
\int_{0}^{L}\frac{dx}{\left\lgroup d_A+\frac{d_B-d_A}{L}x\right\rgroup ^4} =\frac{L}{3(d_B-d_A)}\left\lgroup\frac{1}{d^3_A} -\frac{1}{d^3_B}\right\rgroupTherefore, the strain energy of the tapered bar is
U=\frac{16T^2L}{3\pi G(d_B-d_A)}\left\lgroup\frac{1}{d^3_A} -\frac{1}{d^3_B}\right\rgroup (3-65)
Equating the strain energy to the work of the torque [Eq. (a)] and solving for \phi_A gives
\phi _A=\frac{32TL}{3\pi G(d_B-d_A)}\left\lgroup\frac{1}{d^3_A} -\frac{1}{d^3_B}\right\rgroup \hookleftarrow (3-66)
4. Finalize: This equation gives the angle of rotation at end A of the tapered bar. [Note: This is the same angle of twist expression obtained in the solution of Prob. 3.4-10(a).]
Note especially that the method used in this example for finding the angle of rotation is suitable only when the bar is subjected to a single load, and then only when the desired angle corresponds to that load. Otherwise, angular displacements must be obtained by the usual methods described in Sections 3.3, 3.4, and 3.8.