Chapter 3, Problem 157RP

Ends A and D of the two solid steel shafts AB and CD are fixed, while ends B and C are connected to gears as shown. Knowing that the allowable shearing stress is 50 MPa in each shaft, determine the largest torque T that can be applied to gear B.

Fig. P3.157

To determine

Find the largest torque (T) that can be applied to gear B.

Answer to Problem 157RP

The largest torque (T) that can be applied to gear B is $\underset{\_}{\text{4}\text{.12}\text{kN}\cdot \text{m}}$.

Explanation of Solution

Given information:

The allowable shearing stress $\left({\tau }_{all}\right)$ is $\text{50MPa}$.

The diameter of the shaft AB $\left(d_{AB}\right)$ is $60\text{mm}$ or $0.060\text{m}$.

The diameter of the shaft CD$\left(d_{CD}\right)$ is $45\text{mm}$ or $0.045\text{m}$.

The radius of the gear B$\left(r_B\right)$ is $\text{100mm}$.

The radius of the gear C $\left(r_C\right)$ is $40\text{mm}$.

The length of the shaft AB $\left(L_{AB}\right)$ is $300\text{mm}$ or $0.3\text{m}$.

The length of the shaft CD $\left(L_{CD}\right)$ is $500\text{mm}$ or $0.5\text{m}$.

Assume that clockwise direction is negative and anticlockwise direction is positive.

Calculation:

Gear B and C:

Calculate the rotation in gear B $\left({\varphi }_B\right)$ using the relation:

$\begin{array}{l}{r}_B{\varphi }_B=r_C{\varphi }_C\\ {\varphi }_B=\frac{r_C}{r_B}{\varphi }_C\end{array}$

Substitute $40\text{mm}$ for $r_C$ and $\text{100mm}$ for $r_B$.

$\begin{array}{c}{\varphi }_B=\frac{40}{100}{\varphi }_C\\ =0.4{\varphi }_C\end{array}$ (1)

Show the free body diagram of the gear C as in Figure 1.

Calculate the force in the shaft CD (F) using the formula:

$\begin{array}{l}{\displaystyle \sum M_C=0}\\ \left(F\times r_C\right)-T_{CD}=0\\ F=\frac{T_{CD}}{r_C}\end{array}$

Show the free body diagram of the gear B as in Figure 2.

Calculate the torque produced in the shaft AB $\left(T_{AB}\right)$ using the formula:

$\begin{array}{l}{\displaystyle \sum M_B=0}\\ -T_{AB}+T-\left(F\times r_B\right)=0\\ T-T_{AB}=F\times r_B\end{array}$

Substitute $\frac{T_{CD}}{r_C}$ for F.

$T-T_{AB}=\frac{T_{CD}}{r_C}\times r_B$

Substitute $40\text{mm}$ for $r_C$ and $\text{100mm}$ for $r_B$.

$\begin{array}{l}T-T_{AB}=\frac{T_{CD}}{40}\times 100\\ T=2.5T_{CD}+T_{AB}\end{array}$ (2)

Shaft AB:

Calculate the radius of the shaft AB $\left(c_{AB}\right)$ using the formula:

$c_{AB}=\frac{d_{AB}}{2}$

Substitute $0.060\text{m}$ for $d_{AB}$.

$\begin{array}{c}{c}_{AB}=\frac{0.060}{2}\\ =0.030\text{m}\end{array}$

Calculate the polar moment of inertia in the shaft AB $\left(J_{AB}\right)$ using the formula:

$J_{AB}=\frac{\pi }{2}{c}^4$

Substitute $0.030\text{m}$ for $c_{AB}$.

$\begin{array}{c}{J}_{AB}=\frac{\pi }{2}{\left(0.030\right)}^4\\ =1.2723\times {10}^{-6}\end{array}$

Calculate the radius of the shaft CD $\left(c_{CD}\right)$ using the formula:

$c_{CD}=\frac{d_{CD}}{2}$

Substitute $0.045\text{m}$ for $d_{CD}$.

$\begin{array}{c}{c}_{CD}=\frac{0.045}{2}\\ =0.0225\text{m}\end{array}$

Calculate the polar moment of inertia in the shaft CD $\left(J_{CD}\right)$ using the formula:

$J_{CD}=\frac{\pi }{2}{c}^4$

Substitute $0.0225\text{m}$ for $c_{CD}$.

$\begin{array}{c}{J}_{CD}=\frac{\pi }{2}{\left(0.0225\right)}^4\\ =4.0258\times {10}^{-7}\end{array}$

Calculate the rotation about gear B $\left({\varphi }_B\right)$ using the formula:

${\varphi }_B={\varphi }_{B/A}=\frac{T_{AB}L}{G{J}_{AB}}$

Substitute $0.3\text{m}$ for L and $1.2723\times {10}^{-6}$ for $J_{AB}$.

$\begin{array}{c}{\varphi }_B={\varphi }_{B/A}=\frac{T_{AB}\times 0.3}{G\times 1.2723\times {10}^{-6}}\\ =235.77\times {10}^3\frac{T_{AB}}{G}\end{array}$

Calculate the rotation about gear C $\left({\varphi }_C\right)$ using the formula:

${\varphi }_C={\varphi }_{C/D}=\frac{T_{CD}L}{G{J}_{CD}}$

Substitute $0.5\text{m}$ for L and $4.0258\times {10}^{-7}$ for $J_{AB}$.

$\begin{array}{c}{\varphi }_C={\varphi }_{C/D}=\frac{T_{CD}\times 0.5}{G\times 4.0258\times {10}^{-7}}\\ =1,241.98\times {10}^3\frac{T_{CD}}{G}\end{array}$

Calculate the torque in CD $\left(T_{CD}\right)$ using the relation:

${\varphi }_B=0.4{\varphi }_C$

Substitute $235.77\times {10}^3\frac{T_{AB}}{G}$ for ${\varphi }_B$ and $1,241.98\times {10}^3\frac{T_{CD}}{G}$ for ${\varphi }_C$.

$\begin{array}{l}235.77\times {10}^3\frac{T_{AB}}{G}=0.4\left(1,241.98\times {10}^3\frac{T_{CD}}{G}\right)\\ 235.77\times {10}^3{T}_{AB}=496.792\times {10}^3{T}_{CD}\\ T_{CD}=\frac{235.77\times {10}^3{T}_{AB}}{496.792\times {10}^3}\end{array}$

$\begin{array}{l}{T}_{CD}=0.4746T_{AB}\\ T_{AB}=\frac{T_{CD}}{0.4746}\end{array}$

Calculate the total torque (T) using the formula:

Consider the equation (2),

$T=2.5T_{CD}+T_{AB}$

Substitute $0.4746T_{AB}$ for $T_{CD}$.

$\begin{array}{c}T=2.5\left(0.4746T_{AB}\right)+T_{AB}\\ =1.1865T_{AB}+T_{AB}\\ =T_{AB}\left(1+1.1865\right)\\ =2.1865T_{AB}\end{array}$ (3)

Substitute $\frac{T_{CD}}{0.4746}$ for $T_{AB}$.

$\begin{array}{c}T=2.1865\left(\frac{T_{CD}}{0.4746}\right)\\ =4.607T_{CD}\end{array}$ (4)

Calculate the torque in the shaft AB $\left(T\right)$ using the formula:

${\tau }_{all}=\frac{T_{AB}c}{J_{AB}}$

Substitute $\text{50MPa}$ for ${\tau }_{all}$, $0.030\text{m}$ for c and $1.2723\times {10}^{-6}$ for $J_{AB}$.

$\begin{array}{l}50\times {10}^6=\frac{T_{AB}\times 0.030}{1.2723\times {10}^{-6}}\\ T_{AB}=\frac{50\times {10}^6\times 1.2723\times {10}^{-6}}{0.030}\\ T_{AB}=2,120.5\text{N}\cdot \text{m}\end{array}$

Substitute $2,120.5\text{N}\cdot \text{m}$ for $T_{AB}$ in Equation (3).

$\begin{array}{c}T=2.1865\left(2,120.5\right)\\ =4.64\text{kN}\cdot \text{m}\end{array}$

Calculate the torque in the shaft CD $\left(T\right)$ using the formula:

${\tau }_{all}=\frac{T_{CD}c}{J_{CD}}$

Substitute $\text{50MPa}$ for ${\tau }_{all}$, $0.0225\text{m}$ for c, and $4.0258\times {10}^{-7}$ for $J_{CD}$.

$\begin{array}{l}50\times {10}^6=\frac{T_{CD}\times 0.0225}{4.0258\times {10}^{-7}}\\ T_{CD}=\frac{50\times {10}^6\times 4.0258\times {10}^{-7}}{0.0225}\\ T_{CD}=894.62\text{N}\cdot \text{m}\end{array}$

Calculate the total torque (T):

Substitute $894.62\text{N}\cdot \text{m}$ for $T_{CD}$ in Equation (4).

$\begin{array}{c}T=4.607\left(894.62\right)\\ =4.12\text{kN}\cdot \text{m}\end{array}$

From the above calculated torque in the shaft AB and CD, take the lesser value.

Thus, the largest torque (T) that can be applied to gear B is $\underset{\_}{\text{4}\text{.12}\text{kN}\cdot \text{m}}$.