Respuesta :
Answer:
Max Normal Stress, [tex]\sigma_{m} = 63.6\ MPa[/tex]
Max Shear Stress, [tex]\tau_{m} = 34.3\ MPa[/tex]
Solution:
As per the question:
Torque, T = 11 kN.m = 11000 Nm
The inner diameter of the tank, [tex]d_{i} = 180\ mm = 0.18\ m[/tex]
Thickness of the tank, t = 12 mm = 0.012 m
Pressure, P = 8 MPa = [tex]8\times 10^{6}\ Pa[/tex]
Now,
Inner radius of the tank, [tex]R_{i} = \frac{d_{i}}{2} = 0.09\ m[/tex]
Outer radius of the tank, [tex]R_{o} = R_{i} + t = 0.9 + 0.012 = 0.102\ m[/tex]
Now,
To calculate the polar moment of inertia, J:
[tex]J = \frac{\pi}{2}(R_{o}^{4} - R_{i}^{4})[/tex]
[tex]J = \frac{\pi}{2}(0.102^{4} - 0.09^{4})[/tex]
[tex]J = \frac{\pi}{2}(0.102^{4} - 0.09^{4}) = 6.7\times 10^{- 5}\ m^{4}[/tex]
Now,
To calculate the shear stress at the surface end:
[tex]\sigma = \frac{P\times R_{i}}{t}[/tex]
[tex]\sigma = \frac{8\times 10^{6}\times 0.09}{0.012} = 60\times 10^{6}\ Pa = 60\ MPa[/tex]
Now,
To calculate the stress at the outer end:
[tex]\sigma' = \frac{P\times R_{i}}{2t}[/tex]
[tex]\sigma' = \frac{8\times 10^{6}\times 0.09}{2\times 0.012} = 30\times 10^{6}\ Pa = 30\ MPa[/tex]
Now,
To calculate the value of the shear stress,
[tex]\tau = \frac{TR_{o}}{J}[/tex]
[tex]\tau = \frac{11000\times 0.102}{6.7\times 10^{- 5}} = 18.27\times 10^{6}\ Pa[/tex]
[tex]\sigma = \sigma_{x}[/tex]
[tex]\sigma' = \sigma_{y}[/tex]
[tex]\tau_{xy} = \tau[/tex]
The resultant stress, [tex]\sigma_{R} = 0.5(\sigma + sigma') = 0.5(90) = 45\ MPa[/tex]
The overall resultant of all the stresses (all in MPa):
[tex]R = \sqrt{(\frac{\sigma - sigma'}{2})^{2} + \tau^{2}}[/tex]
[tex]R = \sqrt{(\frac{60 - 30}{2})^{2} + 18.27^{2}} = 23.638\ MPa[/tex]
Now,
Max stress, [tex]\sigma_{m} = R + \sigma_{R}[/tex]
[tex]\sigma_{m} = 23.638 + 45 = 68.638\ MPa[/tex]
Minimum stress, [tex]\sigma_{min} = 0\ MPa[/tex]
Now,
Maximum shear stress, [tex]\tau_{m} = \frac{1}{2}(\sigma_{m} - \sigma_{min})[/tex]
[tex]\tau_{m} = \frac{1}{2}(68.638 - 0) = 34.319\ MPa[/tex]
The maximum normal stress and the maximum shearing stress are 67.49 MPa and 22.49 MPa respectively.
Given the following data:
Torque, T = 11 kNm = 11,000 Nm.
Inner diameter = 180 mm to m = 0.18 m.
Wall thickness = 12 mm to m = 0.012 m.
How to determine maximum normal stress.
First of all, we would find the inner radius and outer radius of the tank:
Inner radius = [tex]r_{in}=\frac{d_{in}}{2} =\frac{180}{2} = 90\;mm[/tex].
Outer radius = [tex]r_{out}=r_{in}+t=90+12=102\;mm[/tex]
Next, we would calculate the hoop stress:
[tex]\sigma_h = \frac{P r_{in}}{t} \\\\\sigma_h =\frac{8 \times 90}{12} \\\\\sigma_h =60\;MPa[/tex]
For the shear stress at the outer end:
[tex]\sigma_o=\frac{\sigma_h}{2} \\\\\sigma_o=\frac{60}{2} \\\\\sigma_o=30\;MPa[/tex]
The average shear stress is given by:
[tex]\sigma_{avg}=\frac{\sigma_h +\sigma_o}{2} \\\\\sigma_{avg}=\frac{60+30}{2} \\\\\sigma_{avg}=45\;MPa[/tex]
For the torsions:
[tex]c_1=90\;mm[/tex] [tex]c_2 =12\;mm[/tex]
Next, we would calculate the polar moment of inertia:
[tex]J=\frac{\pi}{2} (c_2^4-c_1^4)\\\\J=\frac{\pi}{2} (102^4-90^4)\\\\J=1.571(108,243,216-65,610,000)\\\\J=1.571(42,633,216)\\\\J=66.98 \times 10^{-6}\;m^4[/tex]
For the shear stress:
[tex]\tau = \frac{Tc}{J} \\\\\tau =\frac{11000 \times 102 \times 10^{-3}}{66.98 \times 10^{-6}} \\\\\tau =\frac{1122}{66.98 \times 10^{-6}}\\\\\tau =16.75\;MPa[/tex]
Next, we would calculate the maximum shearing stress by using Mohr's circle:
[tex]\tau_{max}=\sqrt{\tau^2 + (\frac{\sigma_h -\sigma_o}{2})^2 } \\\\\tau_{max}=\sqrt{16.75^2 + (\frac{60 -30}{2})^2 }\\\\\tau_{max}=\sqrt{280.5625+225}\\\\\tau_{max}=\sqrt{505.5625}\\\\\tau_{max}=22.49\;MPa[/tex]
For the maximum normal stress:
[tex]\sigma_{max}=\sigma_{avg}+\tau_{max}\\\\\sigma_{max}=45+22.49\\\\\sigma_{max}=67.49\;MPa[/tex]
Read more on shear stress here: https://brainly.com/question/18793028
