Chapter 3.5, Problem 75P

Three shafts and four gears are used to form a gear train that will transmit 7.5 kW from the motor at A to a machine tool at F. (Bearings for the shafts are omitted in the sketch.) Knowing that the frequency of the motor is 30 Hz and that the allowable stress for each shaft is 60 MPa, determine the required diameter of each shaft.

To determine

The required diameter of the shaft AB.

The required diameter of the shaft CD.

The required diameter of the shaft EF.

Answer to Problem 75P

The required diameter of the shaft AB is $\underset{\_}{15.0\text{mm}}$.

The required diameter of the shaft CD is $\underset{\_}{20.4\text{mm}}$.

The required diameter of the shaft EF is $\underset{\_}{27.6\text{mm}}$.

Explanation of Solution

Given information:

The frequency of the motor is 30 Hz.

The power transmitted by the shafts is 7.5 kW.

The allowable shearing stress in each shaft is 60 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)

Here, f is the frequency of the shaft.

For shaft AB:

The frequency of the shaft AB is $f_{AB}=30\text{Hz}$.

Substitute $7.5\text{kW}$ for P and $30\text{Hz}$ for f in Equation (2).

$\begin{array}{l}7.5\text{kW}=2\pi \left(30\text{Hz}\right)\left(T\right)\\ 7.5\text{kW}\left(\frac{{10}^3\text{W}}{1\text{kW}}\right)=2\pi \left(30\text{Hz}\right)\left(T\right)\\ T=39.789\text{N}\cdot \text{m}\end{array}$

The polar moment of inertia of shaft AB with radius $c$ is $J_{AB}=\frac{\pi }{2}{\left(c\right)}^4$.

Substitute 60 MPa for ${\tau }_{\max }$, $39.789\text{N}\cdot \text{m}$ for T, and $\frac{\pi }{2}{\left(c\right)}^4$ for J in Equation (1).

$\begin{array}{l}60\text{MPa}=\frac{\left(39.789\text{N}\cdot \text{m}\right)\left(c\right)}{\frac{\pi }{2}{\left(c\right)}^4}\\ 60\text{MPa}\times \frac{{10}^6\text{Pa}}{1\text{MPa}}=\frac{2\left(39.789\text{N}\cdot \text{m}\right)}{\pi c^3}\\ c^3=4.2217\times {10}^{-7}\\ c=7.50\times {10}^{-3}\text{m}\end{array}$

Diameter of the shaft AB is twice the radius of the shaft AB.

$\begin{array}{c}{d}_{AB}=2\left(7.50\times {10}^{-3}\text{m}\right)\\ =15.0\times {10}^{-3}\text{m}\\ =15.0\times {10}^{-3}\text{m}\times \frac{{10}^3\text{mm}}{1\text{m}}\\ =15.0\text{mm}\end{array}$

Therefore, the required diameter of the shaft AB is $\underset{\_}{15.0\text{mm}}$.

For shaft CD:

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 as follows:

$\begin{array}{c}{f}_{CD}=\frac{r_B}{r_C}{f}_{AB}\\ =\frac{60\text{mm}}{150\text{mm}}\left(30\text{Hz}\right)\\ =12\text{Hz}\end{array}$

Substitute $7.5\text{kW}$ for P and $12\text{Hz}$ for f in Equation (2).

$\begin{array}{l}7.5\text{kW}=2\pi \left(12\text{Hz}\right)\left(T\right)\\ 7.5\text{kW}\left(\frac{{10}^3\text{W}}{1\text{kW}}\right)=2\pi \left(12\text{Hz}\right)\left(T\right)\\ T=99.472\text{N}\cdot \text{m}\end{array}$

The polar moment of inertia of shaft CD with radius $c$ is $J_{CD}=\frac{\pi }{2}{\left(c\right)}^4$.

Substitute 60 MPa for ${\tau }_{\max }$, $99.472\text{N}\cdot \text{m}$ for T, and $\frac{\pi }{2}{\left(c\right)}^4$ for J in Equation (1).

$\begin{array}{l}60\text{MPa}=\frac{\left(99.472\text{N}\cdot \text{m}\right)\left(c\right)}{\frac{\pi }{2}{\left(c\right)}^4}\\ 60\text{MPa}\times \frac{{10}^6\text{Pa}}{1\text{MPa}}=\frac{2\left(99.472\text{N}\cdot \text{m}\right)}{\pi c^3}\\ c^3=1.0554\times {10}^{-6}\\ c=0.01018\text{m}\end{array}$

The diameter of the shaft CD is twice the radius of the shaft CD.

$\begin{array}{c}{d}_{CD}=2\left(0.01018\text{m}\right)\\ =0.02036\text{m}\\ =0.02036\text{m}\times \frac{{10}^3\text{mm}}{1\text{m}}\\ =20.4\text{mm}\end{array}$

Therefore, the required diameter of the shaft CD is $\underset{\_}{20.4\text{mm}}$.

For shaft EF:

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 as follows:

$\begin{array}{c}{f}_{EF}=\frac{r_D}{r_E}{f}_{CD}\\ =\frac{60\text{mm}}{150\text{mm}}\left(12\text{Hz}\right)\\ =4.8\text{Hz}\end{array}$

Substitute $7.5\text{kW}$ for P and $4.8\text{Hz}$ for f in Equation (2).

$\begin{array}{l}7.5\text{kW}=2\pi \left(4.8\text{Hz}\right)\left(T\right)\\ 7.5\text{kW}\left(\frac{{10}^3\text{W}}{1\text{kW}}\right)=2\pi \left(4.8\text{Hz}\right)\left(T\right)\\ T=248.68\text{N}\cdot \text{m}\end{array}$

The polar moment of inertia of shaft EF with radius $c$ is $J_{EF}=\frac{\pi }{2}{\left(c\right)}^4$.

Substitute 60 MPa for ${\tau }_{\max }$, $248.68\text{N}\cdot \text{m}$ for T, and $\frac{\pi }{2}{\left(c\right)}^4$ for J in Equation (1).

$\begin{array}{l}60\text{MPa}=\frac{\left(248.68\text{N}\cdot \text{m}\right)\left(c\right)}{\frac{\pi }{2}{\left(c\right)}^4}\\ 60\text{MPa}\times \frac{{10}^6\text{Pa}}{1\text{MPa}}=\frac{2\left(248.68\text{N}\cdot \text{m}\right)}{\pi c^3}\\ c^3=2.6386\times {10}^{-6}\\ c=0.01382\text{m}\end{array}$

The diameter of the shaft EF is twice the radius of the shaft EF.

$\begin{array}{c}{d}_{EF}=2\left(0.01382\text{m}\right)\\ =0.02764\text{m}\\ =0.02764\text{m}\times \frac{{10}^3\text{mm}}{1\text{m}}\\ \approx 27.6\text{mm}\end{array}$

Therefore, the required diameter of the shaft EF is $\underset{\_}{27.6\text{mm}}$.