Introduction to Elasticity/Torsion of circular cylinders

Torsion of Circular Cylinders

Torsion of a cylinder with a circular cross section

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

$\text{(8)} \qquad {\phi = \frac{TL}{JG}}$