Chapter 3.5, Problem 74P

Three shafts and four gears are used to form a gear train that will transmit power from the motor at A to a machine tool at F. (Bearings for the shafts are omitted in the sketch.) The diameter of each shaft is as follows: d_AB = 16mm, d_CD = 20 mm, d_EF = 28 mm. Knowing that the frequency of the motor is 24 Hz and that the allowable shearing stress for each shaft is 75 MPa, determine the maximum power that can be transmitted.

To determine

The maximum power that can be transmitted by shaft.

Answer to Problem 74P

The maximum power that can be transmitted by the shaft is $\underset{\_}{7.11\text{kW}}$.

Explanation of Solution

Given information:

The frequency of the motor is 24 Hz.

The allowable shearing stress in each shaft is 75 MPa.

The diameter of the shaft AB is $d_{AB}=16\text{mm}$.

The diameter of the shaft CD is $d_{CD}=20\text{mm}$.

The diameter of the shaft EF is $d_{EF}=28\text{mm}$.

Calculation:

The maximum shear stress in the shaft $\left({\tau }_{\max }\right)$ is expressed as shown below:

${\tau }_{\max }=\frac{Tc}{J}$ (1)

Here, T is the torque transmitted by the shaft, c is the radius of the shaft, and J is the polar moment of inertia of the shaft.

The power transmitted by the shaft $\left(P\right)$ is expressed as follows:

$P=\left(2\pi f\right)T$ (2)

For shaft AB:

The polar moment of inertia of shaft AB with radius $c$ is,

$\begin{array}{c}{J}_{AB}=\frac{\pi }{2}{\left(c\right)}^4\\ =\frac{\pi }{2}{\left(8\text{mm}\right)}^4\\ =6.434\times {10}^3{\text{mm}}^4\\ =6.434\times {10}^{-9}{\text{m}}^4\end{array}$

Substitute 75 MPa for ${\tau }_{\max }$, 8 mm for $c$, and $6.434\times {10}^{-9}{\text{m}}^4$ for J in Equation (1).

$\begin{array}{l}75\text{MPa}=\frac{\left(T\right)\left(8\text{mm}\right)}{6.434\times {10}^{-9}{\text{m}}^4}\\ 75\text{MPa}\times \frac{{10}^6\text{Pa}}{1\text{MPa}}=\frac{\left(T\right)\left(8\text{mm}\times \frac{{10}^{-3}\text{m}}{1\text{mm}}\right)}{6.434\times {10}^{-9}{\text{m}}^4}\\ T=60.319\text{N}\cdot \text{m}\end{array}$

The frequency of the shaft AB is $f_{AB}=24\text{Hz}$.

Substitute $60.319\text{N}\cdot \text{m}$ for T and $24\text{Hz}$ for f in Equation (2).

$\begin{array}{c}P=2\pi \left(24\text{Hz}\right)\left(60.319\text{N}\cdot \text{m}\right)\\ =9.1\times {10}^3\text{W}\\ =9.1\text{kW}\end{array}$

For shaft CD:

The polar moment of inertia of shaft CD with radius $c$ is,

$\begin{array}{c}{J}_{CD}=\frac{\pi }{2}{\left(c\right)}^4\\ =\frac{\pi }{2}{\left(10\text{mm}\right)}^4\\ =1.5708\times {10}^4{\text{mm}}^4\\ =1.5708\times {10}^{-8}{\text{m}}^4\end{array}$

Substitute 75 MPa for ${\tau }_{\max }$, 10 mm for $c$, and $1.5708\times {10}^{-8}{\text{m}}^4$ for J in Equation (1).

$\begin{array}{l}75\text{MPa}=\frac{\left(T\right)\left(10\text{mm}\right)}{1.5708\times {10}^{-8}{\text{m}}^4}\\ 75\text{MPa}\times \frac{{10}^6\text{Pa}}{1\text{MPa}}=\frac{\left(T\right)\left(10\text{mm}\times \frac{{10}^{-3}\text{m}}{1\text{mm}}\right)}{1.5708\times {10}^{-8}{\text{m}}^4}\\ T=117.81\text{N}\cdot \text{m}\end{array}$

The radius at gear B is $r_B=60\text{mm}$.

The radius at gear C is $r_C=150\text{mm}$.

The frequency of the shaft CD is,

$\begin{array}{c}{f}_{CD}=\frac{r_B}{r_C}{f}_{AB}\\ =\frac{60\text{mm}}{150\text{mm}}\left(24\text{Hz}\right)\\ =9.6\text{Hz}\end{array}$

Substitute $117.81\text{N}\cdot \text{m}$ for T and $9.6\text{Hz}$ for f in Equation (2).

$\begin{array}{c}P=2\pi \left(9.6\text{Hz}\right)\left(117.81\text{N}\cdot \text{m}\right)\\ =7.11\times {10}^3\text{W}\\ =7.11\text{kW}\end{array}$

For shaft EF:

The polar moment of inertia of shaft EF with radius $c$ is,

$\begin{array}{c}{J}_{EF}=\frac{\pi }{2}{\left(c\right)}^4\\ =\frac{\pi }{2}{\left(14\text{mm}\right)}^4\\ =6.0344\times {10}^4{\text{mm}}^4\\ =6.0344\times {10}^{-8}{\text{m}}^4\end{array}$

Substitute 75 MPa for ${\tau }_{\max }$, 14 mm for $c$, and $6.0344\times {10}^{-8}{\text{m}}^4$ for J in Equation (1).

$\begin{array}{l}75\text{MPa}=\frac{\left(T\right)\left(14\text{mm}\right)}{6.0344\times {10}^{-8}{\text{m}}^4}\\ 75\text{MPa}\times \frac{{10}^6\text{Pa}}{1\text{MPa}}=\frac{\left(T\right)\left(14\text{mm}\times \frac{{10}^{-3}\text{m}}{1\text{mm}}\right)}{6.0344\times {10}^{-8}{\text{m}}^4}\\ T=323.27\text{N}\cdot \text{m}\end{array}$

The radius at gear D is $r_D=60\text{mm}$.

The radius at gear E is $r_E=150\text{mm}$.

The frequency of the shaft EF is,

$\begin{array}{c}{f}_{EF}=\frac{r_D}{r_E}{f}_{CD}\\ =\frac{60\text{mm}}{150\text{mm}}\left(9.6\text{Hz}\right)\\ =3.84\text{Hz}\end{array}$

Substitute $323.27\text{N}\cdot \text{m}$ for T and $3.84\text{Hz}$ for f in Equation (2).

$\begin{array}{c}P=2\pi \left(3.84\text{Hz}\right)\left(323.27\text{N}\cdot \text{m}\right)\\ =7.8\times {10}^3\text{W}\\ =7.8\text{kW}\end{array}$

The maximum allowable power is the smallest value calculated among the shafts AB, CD, and EF.

The allowable power $P_{\text{all}}=7.11\text{kW}$.

Therefore, the maximum power that can be transmitted by the shaft is $\underset{\_}{7.11\text{kW}}$.