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.

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 (), measured in kilonewton-meters, whose formula is described:

(Eq. 1)

Where:

- Power, measured in kilowatts.

- Angular velocity, measured in radians per second.

If we know that and , then the torque experimented by the shaft:

Let consider that shaft has a circular form, such that shear stress is determined by the following formula:

(Eq. 2)

Where:

- Diameter of the shaft, measured in meters.

- Torsional shear stress, measured in kilopascals.

If we know that and , the torsional shear stress is:

a) According to the maximum-shear-stress failure theory, we get that maximum shear stress limit is:

(Eq. 3)

Where:

- Ultimate shear stress, measured in kilopascals.

- Ultimate tensile stress, measured in kilopascals.

If we know that , the ultimate shear stress of the material is:

Lastly, the static factor of safety of the shaft (), dimensionless, is:

(Eq. 4)

If we know that and , the static factor of safety of the shaft is:

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:

(Eq. 5)

If we know that , the ultimate shear stress of the material is:

Lastly, the static factor of safety of the shaft is:

According to the distortion-energy failure theory, the static factor of safety of the shaft is 2.816.

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