Answer:
a) According to the maximum-shear-stress failure theory, the static factor of safety of the shaft is 2.440.
b) According to the distortion-energy failure theory, the static factor of safety of the shaft is 2.816.
Explanation:
First, we need to determine the torque experimented by the shaft ([tex]T[/tex]), measured in kilonewton-meters, whose formula is described:
[tex]T = \frac{\dot W}{\omega}[/tex] (Eq. 1)
Where:
[tex]\dot W[/tex] - Power, measured in kilowatts.
[tex]\omega[/tex] - Angular velocity, measured in radians per second.
If we know that [tex]\dot W = 10\,kW[/tex] and [tex]\omega = 20.944\,\frac{rad}{s}[/tex], then the torque experimented by the shaft:
[tex]T = \frac{10\,kW}{20.944\,\frac{rad}{s} }[/tex]
[tex]T =0.478\,kN\cdot m[/tex]
Let consider that shaft has a circular form, such that shear stress is determined by the following formula:
[tex]\tau = \frac{16\cdot T}{\pi\cdot D^{3}}[/tex] (Eq. 2)
Where:
[tex]D[/tex] - Diameter of the shaft, measured in meters.
[tex]\tau[/tex] - Torsional shear stress, measured in kilopascals.
If we know that [tex]D = 0.03\,m[/tex] and [tex]T =0.478\,kN\cdot m[/tex], the torsional shear stress is:
[tex]\tau = \frac{16\cdot (0.478\,kN\cdot m)}{\pi\cdot (0.03\,m)^{3}}[/tex]
[tex]\tau \approx 90164.223\,kPa[/tex]
a) According to the maximum-shear-stress failure theory, we get that maximum shear stress limit is:
[tex]S_{ys} = 0.5\cdot S_{ut}[/tex] (Eq. 3)
Where:
[tex]S_{ys}[/tex] - Ultimate shear stress, measured in kilopascals.
[tex]S_{ut}[/tex] - Ultimate tensile stress, measured in kilopascals.
If we know that [tex]S_{ut} = 440\times 10^{3}\,kPa[/tex], the ultimate shear stress of the material is:
[tex]S_{ys} = 0.5\cdot (440\times 10^{3}\,kPa)[/tex]
[tex]S_{ys} = 220\times 10^{3}\,kPa[/tex]
Lastly, the static factor of safety of the shaft ([tex]n[/tex]), dimensionless, is:
[tex]n = \frac{S_{ys}}{\tau}[/tex] (Eq. 4)
If we know that [tex]S_{ys} = 220\times 10^{3}\,kPa[/tex] and [tex]\tau \approx 90164.223\,kPa[/tex], the static factor of safety of the shaft is:
[tex]n = \frac{220\times 10^{3}\,kPa}{90164.223\,kPa}[/tex]
[tex]n = 2.440[/tex]
According to the maximum-shear-stress failure theory, the static factor of safety of the shaft is 2.440.
b) According to the distortion-energy failure theory, we get that maximum shear stress limit is:
[tex]S_{ys} = 0.577\cdot S_{ut}[/tex] (Eq. 5)
If we know that [tex]S_{ut} = 440\times 10^{3}\,kPa[/tex], the ultimate shear stress of the material is:
[tex]S_{ys} = 0.577\cdot (440\times 10^{3}\,kPa)[/tex]
[tex]S_{ys} = 253.88\times 10^{3}\,kPa[/tex]
Lastly, the static factor of safety of the shaft is:
[tex]n = \frac{253.88\times 10^{3}\,kPa}{90164.223\,kPa}[/tex]
[tex]n = 2.816[/tex]
According to the distortion-energy failure theory, the static factor of safety of the shaft is 2.816.
