Elasticity/Torsion of circular cylinders

About the problem:

 * Circular Cylinder.
 * Centroidal axis thru the center of each cross section (c.s.)
 * Length $$L$$, Outer radius $$c$$.
 * Applied torque $$T$$.
 * Angle of twist $$\phi$$.

Assumptions:

 * Each c.s. remains plane and undistorted.
 * Each c.s. rotates through the same angle.
 * No warping or change in shape.
 * Amount of displacement of each c.s. is proportional to  distance from end.

Find:

 * Shear strains in the cylinder ($$\gamma$$).
 * Shear stress in the cylinder ($$\tau$$).
 * Relation between torque ($$T$$) and angle of twist ($$\phi$$).
 * Relation between torque ($$T$$) and shear stress ($$\tau$$).

Solution:
If $$\gamma$$ is small, then
 * $$\text{(1)} \qquad

L\gamma = r\phi \Rightarrow {\gamma = \frac{r\phi}{L}} $$ Therefore,
 * $$\text{(2)} \qquad

\gamma_{\text{max}} = \frac{c\phi}{L} \Rightarrow \gamma = \frac{r}{c} \gamma_{\text{max}} $$ If the material is linearly elastic,
 * $$\text{(3)} \qquad

\tau = G\gamma \Rightarrow {\tau = \frac{r\phi G}{L}} $$ Therefore,
 * $$\text{(4)} \qquad

\tau_{\text{max}} = \frac{c\phi G}{L} \Rightarrow \tau = \frac{r}{c} \tau_{\text{max}} $$ The torque on each c.s. is given by
 * $$\text{(5)} \qquad

T = \int_A \tau r dA = \frac{\phi G}{L}\int_A r^2 dA = \frac{G\phi J}{L} $$ where $$J$$ is the polar moment of inertia of the c.s.
 * $$\text{(6)} \qquad

J = \begin{cases} \frac{1}{2} \pi c^4 & \text{solid circular c.s.} \\ \frac{1}{2} \pi (c_2^{~4}-c_1^{~4}) & \text{ annular circular c.s.} \end{cases} $$ Therefore,
 * $$\text{(7)} \qquad

{\tau = \frac{Tr}{J}} \Rightarrow \tau_{\text{max}} = \frac{Tc}{J} $$ and
 * $$\text{(8)} \qquad

{\phi = \frac{TL}{JG}} $$