Chapter 3.3, Problem 56P

Solve Prob. 3.55, assuming that the shaft AB is replaced by a hollow shaft of the same outer diameter and 25-mm inner diameter.

3.55 Two solid steel shafts (G = 77.2 GPa) are connected to a coupling disk B and to fixed supports at A and C. For the loading shown, determine (a) the reaction at each support, (b) the maximum shearing stress in shaft (c) the maximum shearing stress in shaft BC.

Fig. p3.55

To determine

The reaction at the supports.

Answer to Problem 56P

The reaction at the supports are $\underset{\_}{1,090\text{N}\cdot \text{m}}$ and $\underset{\_}{310\text{N}\cdot \text{m}}$.

Explanation of Solution

Given information:

The modulus of rigidity of solid shafts is $77.2\text{GPa}$.

Inner diameter of the shaft AB is 25 mm.

Calculation:

The outer radius of the shaft AB is $c_2=25\text{mm}$.

The inner radius of the shaft AB is $c_1=12.5\text{mm}$.

The polar moment of inertia of shaft AB of outer radius $c_2$ and inner radius $c_1$ is,

$\begin{array}{c}{J}_{AB}=\frac{\pi }{2}\left(c_2^4-c_1^4\right)\\ =\frac{\pi }{2}\left({\left(25\text{mm}\right)}^4-{\left(12.5\text{mm}\right)}^4\right)\\ =5.7524\times {10}^5{\text{mm}}^{\text{4}}\\ =5.7524\times {10}^{-7}{\text{m}}^{\text{4}}\end{array}$

The torque carried by the shaft AB $\left(T_{AB}\right)$ is expressed as shown below:

$T_{AB}=\frac{G{J}_{AB}}{L_{AB}}{\varphi }_B$ (1)

Here, $G$ is the modulus of rigidity, ${\varphi }_B$ is angle of twist at B, and $L_{AB}$ is the length of the shaft AB.

Substitute $77.2\text{GPa}$ for G, $5.7524\times {10}^{-7}{\text{m}}^{\text{4}}$ for $J_{AB}$, and 200 mm for $L_{AB}$ in Equation (1).

$\begin{array}{c}{T}_{AB}=\frac{\left(77.2\text{GPa}\right)\left(5.7524\times {10}^{-7}{\text{m}}^{\text{4}}\right)}{200\text{mm}}{\varphi }_B\\ =\frac{\left(77.2\text{GPa}\times \frac{{10}^9\text{Pa}}{1\text{GPa}}\right)\left(5.7524\times {10}^{-7}{\text{m}}^{\text{4}}\right)}{200\text{mm}\times \frac{{10}^{-3}\text{m}}{1\text{mm}}}{\varphi }_B\\ =222.043\times {10}^3{\varphi }_B\end{array}$ (2)

The radius of the shaft BC is $c=19\text{mm}$.

The polar moment of inertia of shaft BC of radius $c$ is,

$\begin{array}{c}{J}_{BC}=\frac{\pi }{2}{c}^4\\ =\frac{\pi }{2}{\left(19\text{mm}\right)}^4\\ =2.0471\times {10}^5{\text{mm}}^{\text{4}}\\ =2.0471\times {10}^{-7}{\text{m}}^{\text{4}}\end{array}$

The torque carried by the shaft BC $\left(T_{BC}\right)$ is expressed as shown below:

$T_{BC}=\frac{G{J}_{BC}}{L_{BC}}{\varphi }_B$ (3)

Here, $L_{BC}$ is the length of the shaft BC.

Substitute $77.2\text{GPa}$ for G, $2.0471\times {10}^{-7}{\text{m}}^{\text{4}}$ for $J_{BC}$, and 250 mm for $L_{BC}$ in Equation (3).

$\begin{array}{c}{T}_{BC}=\frac{\left(77.2\text{GPa}\right)\left(2.0471\times {10}^{-7}{\text{m}}^{\text{4}}\right)}{250\text{mm}}{\varphi }_B\\ =\frac{\left(77.2\text{GPa}\times \frac{{10}^9\text{Pa}}{1\text{GPa}}\right)\left(2.0471\times {10}^{-7}{\text{m}}^{\text{4}}\right)}{250\text{mm}\times \frac{{10}^{-3}\text{m}}{1\text{mm}}}{\varphi }_B\\ =63.2144\times {10}^3{\varphi }_B\end{array}$ (4)

The value of total torque in the shaft is $1.4\text{kN}\cdot \text{m}$.

The total torque $\left(T\right)$ is the sum of the torques $T_{AB}$ and $T_{BC}$. It is expressed as follows:

$\begin{array}{l}T=T_{AB}+T_{BC}\\ 1.4\text{kN}\cdot \text{m}=222.043\times {10}^3{\varphi }_B+63.2144\times {10}^3{\varphi }_B\\ 285.2574\times {10}^3{\varphi }_B=1.4\text{kN}\cdot \text{m}\times \frac{{10}^3\text{N}}{1\text{kN}}\\ {\varphi }_B=4.9079\times {10}^{-3}\text{rad}\end{array}$

Substitute $4.9079\times {10}^{-3}\text{rad}$ for ${\varphi }_B$ in Equation (2).

$\begin{array}{c}{T}_{AB}=222.043\times {10}^3\left(4.9079\times {10}^{-3}\text{rad}\right)\\ =1,089.76\text{N}\cdot \text{m}\\ \approx 1,090\text{N}\cdot \text{m}\end{array}$

Substitute $4.9079\times {10}^{-3}\text{rad}$ for ${\varphi }_B$ in Equation (4).

$\begin{array}{c}{T}_{BC}=63.2144\times {10}^3\left(4.9079\times {10}^{-3}\text{rad}\right)\\ =310.25\text{N}\cdot \text{m}\\ \approx 310\text{N}\cdot \text{m}\end{array}$

Therefore, the reaction at the supports are $\underset{\_}{1,090\text{N}\cdot \text{m}}$ and $\underset{\_}{310\text{N}\cdot \text{m}}$.

To determine

The maximum shearing stress in the shaft AB.

Answer to Problem 56P

The maximum shearing stress in the shaft AB is $\underset{\_}{47.4\text{MPa}}$.

Explanation of Solution

Given information:

The modulus of rigidity of solid shafts is $77.2\text{GPa}$.

Calculation:

Refer (a).

The value of torque in the shaft AB is $1,089.76\text{N}\cdot \text{m}$.

The polar moment of inertia of shaft AB is $J_{AB}=5.7524\times {10}^{-7}{\text{m}}^{\text{4}}$.

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

${\tau }_{AB}=\frac{T_{AB}{c}_2}{J_{AB}}$ (5)

Substitute $1,089.76\text{N}\cdot \text{m}$ for $T_{AB}$, 25 mm for $c_2$, and $5.7524\times {10}^{-7}{\text{m}}^{\text{4}}$ for $J_{AB}$ in Equation (5).

$\begin{array}{c}{\tau }_{AB}=\frac{\left(1,089.76\text{N}\cdot \text{m}\right)\left(25\text{mm}\right)}{5.7524\times {10}^{-7}{\text{m}}^{\text{4}}}\\ =\frac{\left(1,089.76\text{N}\cdot \text{m}\right)\left(25\text{mm}\times \frac{{10}^{-3}\text{m}}{1\text{mm}}\right)}{5.7524\times {10}^{-7}{\text{m}}^{\text{4}}}\\ =47.4\times {10}^6\text{Pa}\\ \approx 47.4\text{MPa}\end{array}$

Therefore, the maximum shearing stress in the shaft AB is $\underset{\_}{47.4\text{MPa}}$.

To determine

The maximum shearing stress in the shaft BC.

Answer to Problem 56P

The maximum shearing stress in the shaft BC is $\underset{\_}{28.8\text{MPa}}$.

Explanation of Solution

Given information:

The modulus of rigidity of solid shafts is $77.2\text{GPa}$.

Calculation:

Refer (a).

The value of torque in the shaft BC is $310.25\text{N}\cdot \text{m}$.

The polar moment of inertia of shaft BC of radius $c$ is $J_{BC}=2.0471\times {10}^{-7}{\text{m}}^{\text{4}}$.

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

${\tau }_{BC}=\frac{T_{BC}c}{J_{BC}}$ (6)

Substitute $310.25\text{N}\cdot \text{m}$ for $T_{BC}$, 19 mm for $c$, and $2.0471\times {10}^{-7}{\text{m}}^{\text{4}}$ for $J_{BC}$ in Equation (6).

$\begin{array}{c}{\tau }_{BC}=\frac{\left(310.25\text{N}\cdot \text{m}\right)\left(19\text{mm}\right)}{2.0471\times {10}^{-7}{\text{m}}^{\text{4}}}\\ =\frac{\left(310.25\text{N}\cdot \text{m}\right)\left(19\text{mm}\times \frac{{10}^{-3}\text{m}}{1\text{mm}}\right)}{2.0471\times {10}^{-7}{\text{m}}^{\text{4}}}\\ =28.796\times {10}^6\text{Pa}\\ \approx 28.8\text{MPa}\end{array}$

Therefore, the maximum shearing stress in the shaft BC is $\underset{\_}{28.8\text{MPa}}$.