A 30-mm-diameter shaft, made of AISI 1018 HR steel, transmits 10 kW of power while rotating at 200 rev/min. Assume any bending moments present in the shaft to be negligibly small compared to the torque. Determine the static factor of safety based on:a) The maximum-shear-stress failure theory.b) The distortion-energy failure theory.

Answers

Answer 1

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.


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