Elasticity/Torsion of circular cylinders

About the problem:

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

Solution:

If ${\displaystyle \gamma }$ is small, then

${\displaystyle {\text{(1)}}\qquad L\gamma =r\phi ~~\Rightarrow ~~{\gamma ={\frac {r\phi }{L}}}}$

Therefore,

${\displaystyle {\text{(2)}}\qquad \gamma _{\text{max}}={\frac {c\phi }{L}}~~\Rightarrow ~~\gamma ={\frac {r}{c}}\gamma _{\text{max}}}$

If the material is linearly elastic,

${\displaystyle {\text{(3)}}\qquad \tau =G\gamma ~~\Rightarrow ~~{\tau ={\frac {r\phi G}{L}}}}$

Therefore,

${\displaystyle {\text{(4)}}\qquad \tau _{\text{max}}={\frac {c\phi G}{L}}~~\Rightarrow ~~\tau ={\frac {r}{c}}\tau _{\text{max}}}$

The torque on each c.s. is given by

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

where ${\displaystyle J}$ is the polar moment of inertia of the c.s.

${\displaystyle {\text{(6)}}\qquad J={\begin{cases}{\frac {1}{2}}\pi c^{4}&{\text{solid circular c.s.}}\\{\frac {1}{2}}\pi (c_{2}^{~4}-c_{1}^{~4})&{\text{ annular circular c.s.}}\end{cases}}}$

Therefore,

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

and

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