Question

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.

Answer 1

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.

Step-by-step explanation:

First, we need to determine the torque experimented by the shaft (
`T`), measured in kilonewton-meters, whose formula is described:

`T = (\dot W)/(\omega)` (Eq. 1)

Where:

`\dot W` - Power, measured in kilowatts.

`\omega` - Angular velocity, measured in radians per second.

If we know that
`\dot W = 10\,kW` and
`\omega = 20.944\,(rad)/(s)`, then the torque experimented by the shaft:

`T = (10\,kW)/(20.944\,(rad)/(s) )`

`T =0.478\,kN\cdot m`

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

`\tau = (16\cdot T)/(\pi\cdot D^(3))` (Eq. 2)

Where:

`D` - Diameter of the shaft, measured in meters.

`\tau` - Torsional shear stress, measured in kilopascals.

If we know that
`D = 0.03\,m` and
`T =0.478\,kN\cdot m`, the torsional shear stress is:

`\tau = (16\cdot (0.478\,kN\cdot m))/(\pi\cdot (0.03\,m)^(3))`

`\tau \approx 90164.223\,kPa`

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

`S_(ys) = 0.5\cdot S_(ut)` (Eq. 3)

Where:

`S_(ys)` - Ultimate shear stress, measured in kilopascals.

`S_(ut)` - Ultimate tensile stress, measured in kilopascals.

If we know that
`S_(ut) = 440* 10^(3)\,kPa`, the ultimate shear stress of the material is:

`S_(ys) = 0.5\cdot (440* 10^(3)\,kPa)`

`S_(ys) = 220* 10^(3)\,kPa`

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

`n = (S_(ys))/(\tau)` (Eq. 4)

If we know that
`S_(ys) = 220* 10^(3)\,kPa` and
`\tau \approx 90164.223\,kPa`, the static factor of safety of the shaft is:

`n = (220* 10^(3)\,kPa)/(90164.223\,kPa)`

`n = 2.440`

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:

`S_(ys) = 0.577\cdot S_(ut)` (Eq. 5)

If we know that
`S_(ut) = 440* 10^(3)\,kPa`, the ultimate shear stress of the material is:

`S_(ys) = 0.577\cdot (440* 10^(3)\,kPa)`

`S_(ys) = 253.88* 10^(3)\,kPa`

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

`n = (253.88* 10^(3)\,kPa)/(90164.223\,kPa)`

`n = 2.816`

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