Question

A torque of magnitude T = 11 kN·m is applied to the end of a tank containing compressed air under a pressure of 8 MPa. Knowing that the tank has a 180-mm inner diameter and a 12-mm wall thickness, determine the maximum normal stress and the maximum shearing stress in the tank. (Round the final answers to one decimal place.)

Answer 1

Answer:

Max Normal Stress, $\sigma_{m} = 63.6\ MPa$

Max Shear Stress, $\tau_{m} = 34.3\ MPa$

Solution:

As per the question:

Torque, T = 11 kN.m = 11000 Nm

The inner diameter of the tank, $d_{i} = 180\ mm = 0.18\ m$

Thickness of the tank, t = 12 mm = 0.012 m

Pressure, P = 8 MPa = $8\times 10^{6}\ Pa$

Now,

Inner radius of the tank, $R_{i} = \frac{d_{i}}{2} = 0.09\ m$

Outer radius of the tank, $R_{o} = R_{i} + t = 0.9 + 0.012 = 0.102\ m$

Now,

To calculate the polar moment of inertia, J:

$J = \frac{\pi}{2}(R_{o}^{4} - R_{i}^{4})$

$J = \frac{\pi}{2}(0.102^{4} - 0.09^{4})$

$J = \frac{\pi}{2}(0.102^{4} - 0.09^{4}) = 6.7\times 10^{- 5}\ m^{4}$

Now,

To calculate the shear stress at the surface end:

$\sigma = \frac{P\times R_{i}}{t}$

$\sigma = \frac{8\times 10^{6}\times 0.09}{0.012} = 60\times 10^{6}\ Pa = 60\ MPa$

Now,

To calculate the stress at the outer end:

$\sigma' = \frac{P\times R_{i}}{2t}$

$\sigma' = \frac{8\times 10^{6}\times 0.09}{2\times 0.012} = 30\times 10^{6}\ Pa = 30\ MPa$

Now,

To calculate the value of the shear stress,

$\tau = \frac{TR_{o}}{J}$

$\tau = \frac{11000\times 0.102}{6.7\times 10^{- 5}} = 18.27\times 10^{6}\ Pa$

$\sigma = \sigma_{x}$

$\sigma' = \sigma_{y}$

$\tau_{xy} = \tau$

The resultant stress, $\sigma_{R} = 0.5(\sigma + sigma') = 0.5(90) = 45\ MPa$

The overall resultant of all the stresses (all in MPa):

$R = \sqrt{(\frac{\sigma - sigma'}{2})^{2} + \tau^{2}}$

$R = \sqrt{(\frac{60 - 30}{2})^{2} + 18.27^{2}} = 23.638\ MPa$

Now,

Max stress, $\sigma_{m} = R + \sigma_{R}$

$\sigma_{m} = 23.638 + 45 = 68.638\ MPa$

Minimum stress, $\sigma_{min} = 0\ MPa$

Now,

Maximum shear stress, $\tau_{m} = \frac{1}{2}(\sigma_{m} - \sigma_{min})$

$\tau_{m} = \frac{1}{2}(68.638 - 0) = 34.319\ MPa$