PLAIN AND CIVIL: EXAMPLE: 3.3.1 TORQUE FOR A GIVEN ANGLE OF TWIST

Friday 17 April 2020

What torque should be applied to the end of the shaft to produce a twist of 5 degrees? The modulus of rigidity for steel is G=77GPa.

From the formula of angle of twist, we will be able to solve for the torque, T.

   $\phi =\frac{TL}{JG}$

   $T=\frac{JG}{L}\phi$

where:

   $G=77GPa\simeq 77,000 MPa\simeq 77,000\frac{N}{mm^{2}}$

   $L=1.50m\simeq 1500mm$

   $\phi =5^{o}\left ( \frac{2\pi rad}{360^{o}} \right )=\frac{\pi }{36}rad$

For J, the same formula for the polar moment of inertia is use. The shaft is hollow so:

$J=\frac{\pi }{2}\left ( R^{4}-r^{4} \right )$

$J=\frac{\pi }{2}\left ( 30^{4}-20^{4} \right )=325000\pi mm^{4}$

With all parameters determined, we can compute for the required torque, T:

$T=\frac{325000\pi mm^{4}\left ( 77000\frac{N}{mm^{2}} \right )}{1500mm}\left ( \frac{\pi }{36} rad\right )={\color{Red} 1,455,895.484N-m\simeq 1.46kN-m}$

PLAIN AND CIVIL

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Friday 17 April 2020

EXAMPLE: 3.3.1 TORQUE FOR A GIVEN ANGLE OF TWIST

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