Chapter 3.1, Problem 28P

Fig. P3.27 and P3.28

3.28 A torque T = 900 N m is applied to shaft AB of the gear train shown. Knowing that the allowable shearing stress is 80 MPa, determine the required diameter of (a) shaft AB, (b) shaft CD, (c) shaft EF.

To determine

The required diameter of the shaft AB.

Answer to Problem 28P

The required diameter of the shaft AB is $\underset{\_}{38.6\text{mm}}$.

Explanation of Solution

Given information:

The torque applied to the shaft AB is $T=900\text{N}\cdot \text{m}$.

The allowable shear stress is 80 MPa.

Calculation:

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

${\tau }_{\max }=\frac{T_{AB}c}{J}$ (1)

Here, T is the applied internal torque in the shaft AB, J is the polar moment of inertia of the shaft, and c is the radius of the shaft AB.

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

Substitute $\frac{\pi }{2}{c}^4$ for J in Equation (1).

$\begin{array}{c}{\tau }_{\max }=\frac{T_{AB}c}{\frac{\pi }{2}{c}^4}\\ =\frac{2T_{AB}}{\pi c^3}\end{array}$ (2)

The torque in the shaft AB is $T_{AB}=T=900\text{N}\cdot \text{m}$.

Substitute $80\text{MPa}$ for ${\tau }_{\max }$ and $900\text{N}\cdot \text{m}$ for $T_{AB}$ in Equation (2).

$\begin{array}{l}80\text{MPa}=\frac{2\left(900\text{N}\cdot \text{m}\right)}{\pi {\left(c\right)}^3}\\ 80\frac{\text{N}}{{\text{mm}}^{\text{2}}}=\frac{2\left(900\text{N}\cdot \text{m}\times \frac{{10}^3\text{mm}}{1\text{m}}\right)}{\pi {\left(c\right)}^3}\\ c^3=7,161.9724\\ c=19.2757\text{mm}\end{array}$

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

$\begin{array}{c}{d}_{AB}=2\left(19.2757\text{mm}\right)\\ =38.6\text{mm}\end{array}$

Therefore, the required diameter of the shaft AB is $\underset{\_}{38.6\text{mm}}$.

To determine

The required diameter of the shaft CD.

Answer to Problem 28P

The required diameter of the shaft CD is $\underset{\_}{52.3\text{mm}}$.

Explanation of Solution

Given information:

The torque applied to the shaft AB is $T=900\text{N}\cdot \text{m}$.

The allowable shear stress is 80 MPa.

Calculation:

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

${\tau }_{\max }=\frac{T_{CD}c}{J}$ (3)

Here, T is the applied internal torque in the shaft CD, J is the polar moment of inertia of the shaft, and c is the radius of the shaft CD.

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

Substitute $\frac{\pi }{2}{c}^4$ for J in Equation (3).

$\begin{array}{c}{\tau }_{\max }=\frac{T_{CD}c}{\frac{\pi }{2}{c}^4}\\ =\frac{2T_{CD}}{\pi c^3}\end{array}$ (4)

The torque in the shaft CD is expressed as follows:

$\frac{T_{CD}}{r_C}=\frac{T_{AB}}{r_B}$ (5)

Here, $r_C$ is radius at C and $r_B$ is radius at B.

Substitute 75 mm for $r_C$, 30 mm for $r_B$, and T for $T_{AB}$ in Equation (5).

$\begin{array}{c}\frac{T_{CD}}{75\text{mm}}=\frac{T}{30\text{mm}}\\ T_{CD}=2.5T\\ =2.5\left(900\text{N}\cdot \text{m}\right)\\ =2,250\text{N}\cdot \text{m}\end{array}$ (6)

Substitute $80\text{MPa}$ for ${\tau }_{\max }$ and $2,250\text{N}\cdot \text{m}$ for $T_{CD}$ in Equation (4).

$\begin{array}{l}80\text{MPa}=\frac{2\left(2,250\text{N}\cdot \text{m}\right)}{\pi {\left(c\right)}^3}\\ 80\frac{\text{N}}{{\text{mm}}^{\text{2}}}=\frac{2\left(2,250\text{N}\cdot \text{m}\times \frac{{10}^3\text{mm}}{1\text{m}}\right)}{\pi {\left(c\right)}^3}\\ c^3=17,904.9311\\ c=26.161\text{mm}\end{array}$

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

$\begin{array}{c}{d}_{CD}=2\left(26.161\text{mm}\right)\\ =52.3\text{mm}\end{array}$

Therefore, the required diameter of the shaft CD is $\underset{\_}{52.3\text{mm}}$.

To determine

The required diameter of the shaft EF.

Answer to Problem 28P

The required diameter of the shaft EF is $\underset{\_}{75.5\text{mm}}$.

Explanation of Solution

Given information:

The torque applied to the shaft AB is $T=900\text{N}\cdot \text{m}$.

Allowable shear stress is 80 MPa.

Calculation:

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

${\tau }_{\max }=\frac{T_{EF}c}{J}$ (7)

Here, T is the applied internal torque in the shaft EF, J is the polar moment of inertia of the shaft, and c is the radius of the shaft EF.

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

Substitute $\frac{\pi }{2}{c}^4$ for J in Equation (7).

$\begin{array}{c}{\tau }_{\max }=\frac{T_{EF}c}{\frac{\pi }{2}{c}^4}\\ =\frac{2T_{EF}}{\pi c^3}\end{array}$ (8)

The torque in the shaft EF is expressed as follows:

$\frac{T_{EF}}{r_F}=\frac{T_{CD}}{r_D}$ (9)

Here, $r_F$ is radius at F and $r_D$ is radius at D.

Substitute 90 mm for $r_F$, 30 mm for $r_D$, and 2.5T for $T_{CD}$ in Equation (9).

$\begin{array}{c}\frac{T_{EF}}{90\text{mm}}=\frac{2.5T}{30\text{mm}}\\ T_{EF}=7.5T\\ =7.5\left(900\text{N}\cdot \text{m}\right)\\ =6,750\text{N}\cdot \text{m}\end{array}$ (10)

Substitute $80\text{MPa}$ for ${\tau }_{\max }$ and $6,750\text{N}\cdot \text{m}$ for $T_{EF}$ in Equation (8).

$\begin{array}{l}80\text{MPa}=\frac{2\left(6,750\text{N}\cdot \text{m}\right)}{\pi {\left(c\right)}^3}\\ 80\frac{\text{N}}{{\text{mm}}^{\text{2}}}=\frac{2\left(6,750\text{N}\cdot \text{m}\times \frac{{10}^3\text{mm}}{1\text{m}}\right)}{\pi {\left(c\right)}^3}\\ c^3=53,714.7933\\ c=37.731\text{mm}\end{array}$

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

$\begin{array}{c}{d}_{EF}=2\left(37.731\text{mm}\right)\\ =75.5\text{mm}\end{array}$

Therefore, the required diameter of the shaft EF is $\underset{\_}{75.5\text{mm}}$.