Introduction to Elasticity/Torsion of circular cylinders

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[edit] Torsion of Circular Cylinders

Torsion of a cylinder with a circular cross section

[edit] 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 φ.

[edit] 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.

[edit] Find:

  • Shear strains in the cylinder (γ).
  • Shear stress in the cylinder (τ).
  • Relation between torque (T) and angle of twist (φ).
  • Relation between torque (T) and shear stress (τ).

[edit] Solution:

If γ 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 deformation is 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}}
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