Chapter 3.3, Problem 58P

3.57 and 3.58 Two solid steel shafts are fitted with flanges that are then connected by bolts as shown. The bolts are slightly undersized and permit a 1.5° rotation of one flange with respect to the other before the flanges begin to rotate as a single unit. Knowing that G = 77.2 GPa, determine the maximum shearing stress in each shaft when a torque of T of magnitude 500 N·m is applied to the flange indicated.

3.57 The torque T is applied to flange B.

3.58 The torque T is applied to flange C.

To determine

The maximum shearing stress in the shaft AB.

The maximum shearing stress in the shaft CD.

Answer to Problem 58P

The maximum shearing stress in the shaft AB is $\underset{\_}{10.27\text{MPa}}$.

The maximum shearing stress in the shaft CD is $\underset{\_}{48.6\text{MPa}}$.

Explanation of Solution

Given information:

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

The torque applied to the flange C is $500\text{N}\cdot \text{m}$.

The rotation of one flange with respect to the other flange is $1.5°$.

Calculation:

The radius of the shaft AB is $c=15\text{mm}$.

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

$\begin{array}{c}{J}_{AB}=\frac{\pi }{2}{c}^4\\ =\frac{\pi }{2}{\left(15\text{mm}\right)}^4\\ =7.9522\times {10}^4{\text{mm}}^{\text{4}}\\ =7.9522\times {10}^{-8}{\text{m}}^{\text{4}}\end{array}$

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, $7.9522\times {10}^{-8}{\text{m}}^{\text{4}}$ for $J_{AB}$, and 2600 mm for $L_{AB}$ in Equation (1).

$\begin{array}{c}{T}_{AB}=\frac{\left(77.2\text{GPa}\right)\left(7.9522\times {10}^{-8}{\text{m}}^{\text{4}}\right)}{600\text{mm}}{\varphi }_B\\ =\frac{\left(77.2\text{GPa}\times \frac{{10}^9\text{Pa}}{1\text{GPa}}\right)\left(7.9522\times {10}^{-8}{\text{m}}^{\text{4}}\right)}{600\text{mm}\times \frac{{10}^{-3}\text{m}}{1\text{mm}}}{\varphi }_B\\ =10.2318\times {10}^3{\varphi }_B\end{array}$ (2)

The radius of the shaft CD is $c=18\text{mm}$.

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

$\begin{array}{c}{J}_{CD}=\frac{\pi }{2}{c}^4\\ =\frac{\pi }{2}{\left(18\text{mm}\right)}^4\\ =1.64896\times {10}^5{\text{mm}}^{\text{4}}\\ =1.64896\times {10}^{-7}{\text{m}}^{\text{4}}\end{array}$

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

$T_{CD}=\frac{G{J}_{CD}}{L_{CD}}{\varphi }_C$ (3)

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

Substitute $77.2\text{GPa}$ for G, $1.64896\times {10}^{-7}{\text{m}}^{\text{4}}$ for $J_{CD}$, and 900 mm for $L_{CD}$ in Equation (3).

$\begin{array}{c}{T}_{CD}=\frac{\left(77.2\text{GPa}\right)\left(1.64896\times {10}^{-7}{\text{m}}^{\text{4}}\right)}{900\text{mm}}{\varphi }_C\\ =\frac{\left(77.2\text{GPa}\times \frac{{10}^9\text{Pa}}{1\text{GPa}}\right)\left(1.64896\times {10}^{-7}{\text{m}}^{\text{4}}\right)}{900\text{mm}\times \frac{{10}^{-3}\text{m}}{1\text{mm}}}{\varphi }_C\\ =14.1444\times {10}^3{\varphi }_C\end{array}$ (4)

The clearance rotation for flange C is,

$\begin{array}{c}{\varphi }_C\text{'}=1.5°\\ =1.5°\left(\frac{\pi \text{rad}}{180°}\right)\\ =0.026178\text{rad}\end{array}$

Find the torque required to remove the clearance:

Substitute $0.026178\text{rad}$ for ${\varphi }_C\text{'}$ in Equation (2).

$\begin{array}{c}{T}_{CD}^{\text{'}}=14.1444\times {10}^3\left(0.026178\text{rad}\right)\\ =370.27\text{N}\cdot \text{m}\end{array}$

The magnitude of torque applied at the flange is $500\text{N}\cdot \text{m}$.

The torque $T"$ required to cause additional rotation $\varphi "$ is,

$\begin{array}{c}T"=500\text{N}\cdot \text{m}-370.27\text{N}\cdot \text{m}\\ \text{= }129.73\text{N}\cdot \text{m}\end{array}$

The total torque $\left(T"\right)$ is the sum of the torques $T_{AB}^{"}$ and $T_{CD}^{"}$. It is expressed as,

$\begin{array}{l}T"=T_{AB}^{"}+T_{CD}^{"}\\ 129.73\text{N}\cdot \text{m}=10.2318\times {10}^3\varphi "+14.1444\times {10}^3\varphi "\\ 24.3762\times {10}^3\varphi "=129.73\text{N}\cdot \text{m}\\ \varphi "=5.322\times {10}^{-3}\text{rad}\end{array}$

Substitute $5.322\times {10}^{-3}\text{rad}$ for $\varphi "$ in Equation (2).

$\begin{array}{c}{T}_{AB}^{"}=10.2318\times {10}^3\left(5.322\times {10}^{-3}\text{rad}\right)\\ =54.454\text{N}\cdot \text{m}\end{array}$

Substitute $5.322\times {10}^{-3}\text{rad}$ for $\varphi "$ in Equation (4).

$\begin{array}{c}{T}_{CD}^{"}=14.1444\times {10}^3\left(5.322\times {10}^{-3}\text{rad}\right)\\ =75.277\text{N}\cdot \text{m}\end{array}$

Torque in the shaft AB is,

$\begin{array}{c}{T}_{AB}=T_{AB}^{"}\\ =54.454\text{N}\cdot \text{m}\end{array}$

Torque in the shaft CD is,

$\begin{array}{c}{T}_{CD}=T_{CD}^{\text{'}}+T_{CD}^{"}\\ =370.27\text{N}\cdot \text{m}+75.277\text{N}\cdot \text{m}\\ \text{= }445.547\text{N}\cdot \text{m}\end{array}$

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

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

Substitute $54.454\text{N}\cdot \text{m}$ for $T_{AB}$, 15 mm for $c$, and $7.9522\times {10}^{-8}{\text{m}}^{\text{4}}$ for $J_{AB}$ in Equation (5).

$\begin{array}{c}{\tau }_{AB}=\frac{\left(54.454\text{N}\cdot \text{m}\right)\left(15\text{mm}\right)}{7.9522\times {10}^{-8}{\text{m}}^{\text{4}}}\\ =\frac{\left(54.454\text{N}\cdot \text{m}\right)\left(15\text{mm}\times \frac{{10}^{-3}\text{m}}{1\text{mm}}\right)}{7.9522\times {10}^{-8}{\text{m}}^{\text{4}}}\\ =10.271\times {10}^6\text{Pa}\\ \approx 10.27\text{MPa}\end{array}$

Therefore, the maximum shearing stress in the shaft AB is $\underset{\_}{10.27\text{MPa}}$.

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

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

Substitute $445.547\text{N}\cdot \text{m}$ for $T_{CD}$, 18 mm for $c$, and $1.64896\times {10}^{-7}{\text{m}}^{\text{4}}$ for $J_{CD}$ in Equation (6).

$\begin{array}{c}{\tau }_{CD}=\frac{\left(445.547\text{N}\cdot \text{m}\right)\left(18\text{mm}\right)}{1.64896\times {10}^{-7}{\text{m}}^{\text{4}}}\\ =\frac{\left(445.547\text{N}\cdot \text{m}\right)\left(18\text{mm}\times \frac{{10}^{-3}\text{m}}{1\text{mm}}\right)}{1.64896\times {10}^{-7}{\text{m}}^{\text{4}}}\\ =48.636\times {10}^6\text{Pa}\\ \approx 48.6\text{MPa}\end{array}$

Therefore, the maximum shearing stress in the shaft CD is $\underset{\_}{48.6\text{MPa}}$.