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.
From the principle of conservation of energy we know that the work done by the applied torque equals the strain energy of the bar; thus, W = U. The work is given by the equation
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)]^{2} d x}{2 G I_{P}(x)}.
To use Eq. (3-58), we need 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)]^{4}in which d(x) is the diameter of the bar at distance x from end A. From the geometry of the figure, we see that
d(x)=d_{A}+\frac{d_{B}-d_{A}}{L} x (b)
and therefore
I_{P}(x)=\frac{\pi}{32}\left(d_{A}+\frac{d_{B}-d_{A}}{L} x\right)^{4} (c)
Now we can substitute into Eq. (3-58), as follows:
U=\int_{0}^{L} \frac{[T(x)]^{2} d x}{2 G I_{P}(x)}=\frac{16 T^{2}}{\pi G} \int_{0}^{L} \frac{d x}{\left(d_{A}+\frac{d_{B}-d_{A}}{L} x\right)^{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{d x}{\left(d_{A}+\frac{d_{B}-d_{A}}{L} x\right)^{4}}=\frac{L}{3\left(d_{B}-d_{A}\right)}\left(\frac{1}{d_{A}^{3}}-\frac{1}{d_{B}^{3}}\right)Therefore, the strain energy of the tapered bar is
U=\frac{16 T^{2} L}{3 \pi G\left(d_{B}-d_{A}\right)}\left(\frac{1}{d_{A}^{3}}-\frac{1}{d_{B}^{3}}\right) (3-65)
Equating the strain energy to the work of the torque [Eq. (a)] and solving for \phi_{A}, we get
\phi_{A}=\frac{32 T L}{3 \pi G\left(d_{B}-d_{A}\right)}\left(\frac{1}{d_{A}^{3}}-\frac{1}{d_{B}^{3}}\right) (3-66)
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. 3.4-8(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, we must find angular displacements by the usual methods described in Sections 3.3, 3.4, and 3.8