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

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

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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