Answer:
[tex]T_{max} = 4.735\,kN\cdot m[/tex]
Explanation:
The shear stress due to torque can be calculed by using the following model:
[tex]\tau_{max} = \frac{T_{max}\cdot r_{ext}}{J_{tube}}[/tex]
The maximum torque on the section is:
[tex]T_{max} = \frac{\tau_{max}\cdot J_{tube}}{r_{ext}}[/tex]
The Torsion Constant for the circular tube is:
[tex]J_{tube} = \frac{\pi}{32}\cdot (D_{ext}^{4}-D_{int}^{4})[/tex]
[tex]J_{tube} = \frac{\pi}{4}\cdot [(0.053\,m)^{4}-(0.038\,m)^{4}][/tex]
[tex]J_{tube} = 4.560\times 10^{-6}\,m^{4}[/tex]
Now, the require output is computed:
[tex]T_{max} = \frac{(27\times 10^{3}\,kPa)\cdot (4.560\times 10^{-6}\,m^{4})}{0.026\,m}[/tex]
[tex]T_{max} = 4.735\,kN\cdot m[/tex]
