Elasticity/Torsion of circular cylinders

Torsion of Circular Cylinders

File:Torsion of circular cylinder.png 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 $\varphi$.

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 ($\varphi$).
• Relation between torque ($T$) and shear stress ($\tau$).

Solution:

If $\gamma$ is small, then

$\text{(1)}L\gamma =r\varphi \Rightarrow \gamma =\frac{r\varphi }{L}$

Therefore,

$\text{(2)}{\gamma }_{\text{max}}=\frac{c\varphi }{L}\Rightarrow \gamma =\frac{r}{c}{\gamma }_{\text{max}}$

If the material is linearly elastic,

$\text{(3)}\tau =G\gamma \Rightarrow \tau =\frac{r\varphi G}{L}$

Therefore,

$\text{(4)}{\tau }_{\text{max}}=\frac{c\varphi G}{L}\Rightarrow \tau =\frac{r}{c}{\tau }_{\text{max}}$

The torque on each c.s. is given by

$\text{(5)}T=\underset{A}{\int }\tau rdA=\frac{\varphi G}{L}\underset{A}{\int }{r}^{2}dA=\frac{G\varphi J}{L}$

where $J$ is the polar moment of inertia of the c.s.

$\text{(6)}J=\{\begin{array}{cc}\frac{1}{2}\pi c^4& \text{solid circular c.s.}\\ \frac{1}{2}\pi (c_2^4-c_1^4)& \text{ annular circular c.s.}\end{array}$

Therefore,

$\text{(7)}\tau =\frac{Tr}{J}\Rightarrow {\tau }_{\text{max}}=\frac{Tc}{J}$

and

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