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

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

$T = \frac{\dot W}{\omega}$ (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 $\omega = 20.944\,\frac{rad}{s}$, then the torque experimented by the shaft:

$T = \frac{10\,kW}{20.944\,\frac{rad}{s} }$

$T =0.478\,kN\cdot m$

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

$\tau = \frac{16\cdot T}{\pi\cdot D^{3}}$ (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:

$\tau = \frac{16\cdot (0.478\,kN\cdot m)}{\pi\cdot (0.03\,m)^{3}}$

$\tau \approx 90164.223\,kPa$

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 $S_{ut} = 440\times 10^{3}\,kPa$, the ultimate shear stress of the material is:

$S_{ys} = 0.5\cdot (440\times 10^{3}\,kPa)$

$S_{ys} = 220\times 10^{3}\,kPa$

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

$n = \frac{S_{ys}}{\tau}$ (Eq. 4)

If we know that $S_{ys} = 220\times 10^{3}\,kPa$ and $\tau \approx 90164.223\,kPa$, the static factor of safety of the shaft is:

$n = \frac{220\times 10^{3}\,kPa}{90164.223\,kPa}$

$n = 2.440$

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 $S_{ut} = 440\times 10^{3}\,kPa$, the ultimate shear stress of the material is:

$S_{ys} = 0.577\cdot (440\times 10^{3}\,kPa)$

$S_{ys} = 253.88\times 10^{3}\,kPa$

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

$n = \frac{253.88\times 10^{3}\,kPa}{90164.223\,kPa}$

$n = 2.816$

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