Question

The maximum stress in a section of a circular tube subject to a torque is τmax = 27 MPa . If the inner diameter is Di = 3.75 cm and the outer diameter is Do = 5.25 cm , what is the torque on the section?

Answer 1

Answer:

$T_{max} = 4.735\,kN\cdot m$

Explanation:

The shear stress due to torque can be calculed by using the following model:

$\tau_{max} = \frac{T_{max}\cdot r_{ext}}{J_{tube}}$

The maximum torque on the section is:

$T_{max} = \frac{\tau_{max}\cdot J_{tube}}{r_{ext}}$

The Torsion Constant for the circular tube is:

$J_{tube} = \frac{\pi}{32}\cdot (D_{ext}^{4}-D_{int}^{4})$

$J_{tube} = \frac{\pi}{4}\cdot [(0.053\,m)^{4}-(0.038\,m)^{4}]$

$J_{tube} = 4.560\times 10^{-6}\,m^{4}$

Now, the require output is computed:

$T_{max} = \frac{(27\times 10^{3}\,kPa)\cdot (4.560\times 10^{-6}\,m^{4})}{0.026\,m}$

$T_{max} = 4.735\,kN\cdot m$

Answer 2

Answer:

T = 4.735 KN .m

Explanation:

T max = 27 MPa = 27 * 10⁶ Pa

Di (inner diameter ) = 3.75 cm = 0.0375 m

Do ( outer diameter ) = 5.25 cm = 0.0525 m

To calculate the Torque on the section apply this formula

$\frac{T}{Jtube}$  = $\frac{Tmax}{\frac{Do}{2} }$  equation 1

J tube = torsion constant: π/2( Do^4 - Do^4 ) = π/2 (0.0525 ^4 - 0.0375^4)

= 4.560 * 10^-6 m⁴

T = torque of the section

from equation 1 :

T = (T max * J tube ) / ( Do/2 )

  = (27000000 *  4.560 * 10 ^-6 ) / 0.0525

  = 4.735 KN.m