Chapter 3.8, Problem 116P

(a)

To determine

Find the magnitude ${\left({\tau }_{res}\right)}_{\max }$ and location $\left(\rho \right)$ of the maximum residual shearing stress.

Answer to Problem 116P

The magnitude of the maximum residual shearing stress ${\left({\tau }_{res}\right)}_{\max }$ is $\underset{\_}{33.64\text{MPa}}$.

The location of the maximum residual shearing stress $\left(\rho \right)$ is $\underset{\_}{16\text{mm}}$.

Explanation of Solution

Given information:

The radius of the solid shaft $\left(c\right)$ is 16 mm.

The shear stress $\left({\tau }_Y\right)$ is 145 MPa.

The rigidity modulus of steel (G) is 77.2 GPa.

The length of the solid shaft $\left(L\right)$ is 0.6 m.

The angle of twist $\left(\varphi \right)$ increased in shaft is $6°$.

Calculation:

Calculate the maximum shearing strain ${\gamma }_{\max }$ using the relation.

${\gamma }_{\max }=\frac{c\varphi }{L}$

Here, c is radius of solid shaft, $\varphi$ is angle of twist, and L is length of solid shaft.

Substitute 16 mm for c, $6°$ for $\varphi$, and 0.6 m for L.

$\begin{array}{c}{\gamma }_{\max }=\frac{\left(16\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\right)\times \left(\frac{6°}{\left(\frac{180}{\pi }\right)}\right)}{0.6\text{m}}\\ =2.7925\times {10}^{-3}\end{array}$

Calculate the shearing strain $\gamma$ using the relation below.

${\gamma }_Y=\frac{{\tau }_Y}{G}$

Here, $\left({\tau }_Y\right)$ is shearing stress and (G) is rigidity modulus of steel.

Substitute 145 MPa for ${\tau }_Y$ and 77.2 GPa for G.

$\begin{array}{c}{\gamma }_Y=\frac{\left(145\text{MPa}\times \frac{1,000,000\text{Pa}}{1\text{MPa}}\right)}{\left(77.2\text{GPa}\times \frac{1,000,000,000\text{Pa}}{1\text{GPa}}\right)}\\ =1.8782\times {10}^{-3}\end{array}$

Calculate the distribution of shearing strain using the relation.

${\gamma }_Y=\frac{{\rho }_Y}{c}{\gamma }_{\max }$ (1).

Modify the Equation (1).

$\frac{{\rho }_Y}{c}=\frac{{\gamma }_Y}{{\gamma }_{\max }}$

Here, ${\rho }_Y$ is distance from axis of the shaft.

Substitute $1.8782\times {10}^{-3}$ for ${\gamma }_Y$ and $2.7925\times {10}^{-3}$ for ${\gamma }_{\max }$.

$\begin{array}{c}\frac{{\rho }_Y}{c}=\frac{1.8782\times {10}^{-3}}{2.7925\times {10}^{-3}}\\ =0.6726\end{array}$

Find the polar moment of inertia (J) using the relation.

$J=\frac{\pi }{2}{c}^4$

Here, c is radius of solid shaft.

Substitute 16 mm for c.

$\begin{array}{c}J=\frac{\pi }{2}{\left(16\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\right)}^4\\ =102.944\times {10}^{-9}{\text{m}}^{\text{4}}\end{array}$

Calculate the magnitude of torque $T_Y$ using the relation.

$T_Y=\frac{J{\tau }_Y}{c}$

Substitute $102.944\times {10}^{-9}{\text{m}}^{\text{4}}$ for J, 145 MPa for ${\tau }_Y$, and 16 mm for $c$.

$\begin{array}{c}{T}_Y=\frac{102.944\times {10}^{-9}\times \left(145\text{MPa}\times \frac{1,000,000\text{Pa}}{1\text{MPa}}\right)}{\left(16\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\right)}\\ =932.93\text{N}\cdot \text{m}\end{array}$

Calculate the magnitude of torque in unloading $T\text{'}$ using the relation.

$T\text{'}=\frac{4}{3}{T}_Y\left(1-\frac{1}{4}\frac{{\rho }_Y^3}{c^3}\right)$

Substitute $932.93\text{N}\cdot \text{m}$ for $T_Y$ and $0.6726$ for $\frac{{\rho }_Y}{c}$.

$\begin{array}{c}T\text{'}=\frac{4}{3}\times 932.93\times \left(1-\frac{1}{4}{\left(0.6726\right)}^3\right)\\ =1,243.91\times 0.924\\ =1,149.37\text{N}\cdot \text{m}\end{array}$

Calculate the shearing stress in unloading ${\tau }_c^{\text{'}}$ at c using the relation.

${\tau }_c^{\text{'}}=\frac{T\text{'}c}{J}$

Substitute $1149.37\text{N}\cdot \text{m}$ for $T\text{'}$, 16 mm for c, and $102.944\times {10}^{-9}{\text{m}}^{\text{4}}$ for J.

$\begin{array}{c}{\tau }_c^{\text{'}}=\frac{1,149.37\times \left(16\times \frac{1\text{m}}{1,000\text{mm}}\right)}{102.944\times {10}^{-9}{\text{m}}^{\text{4}}}\\ =178.64\times {10}^6\text{Pa}\end{array}$

Calculate the shearing stress in unloading $\tau \text{'}$ at $\frac{{\rho }_Y}{c}$ using the relation.

${\tau }_{{\rho }_Y}^{\text{'}}=\frac{T\text{'}c}{J}\frac{{\rho }_Y}{c}$

Substitute $1149.37\text{N}\cdot \text{m}$ for $T\text{'}$, 16 mm for c, $0.6726$ for $\frac{{\rho }_Y}{c}$ and $102.944\times {10}^{-9}{\text{m}}^{\text{4}}$ for J.

$\begin{array}{c}{\tau }_{{\rho }_Y}^{\text{'}}=\frac{1,149.37\times \left(16\times \frac{1\text{m}}{1,000\text{mm}}\right)}{102.944\times {10}^{-9}{\text{m}}^{\text{4}}}\times 0.6726\\ =120.153\times {10}^6\text{Pa}\end{array}$

Calculate the angle of twist in unloading $\varphi \text{'}$ using the relation.

$\varphi \text{'}=\frac{T\text{'}L}{GJ}$

Substitute $1,149.37\text{N}\cdot \text{m}$ for $T\text{'}$, 0.6 m for L, 77.2 GPa for G and $102.944\times {10}^{-9}{\text{m}}^{\text{4}}$ for J.

$\begin{array}{c}\varphi \text{'}=\frac{1,149.37\times 0.6}{\left(77.2\text{GPa}\times \frac{1,000,000,000\text{Pa}}{1\text{GPa}}\right)\times 102.944\times {10}^{-9}}\\ =0.0868\text{rad}\\ \text{=}\left(0.0868\times \frac{180}{\pi }\right)\\ =4.97°\end{array}$

Calculate the residual shearing stress ${\left({\tau }_{res}\right)}_c$ at c using the relation.

${\left({\tau }_{res}\right)}_c={\tau }_Y-{\tau }_c^{\text{'}}$

Substitute 145 MPa for ${\tau }_Y$ and $178.64\times {10}^6\text{Pa}$ for ${\tau }_c^{\text{'}}$.

$\begin{array}{c}{\left({\tau }_{res}\right)}_c=\left(145\text{MPa}\times \frac{1,000,000\text{Pa}}{1\text{MPa}}\right)-178.64\times {10}^6\text{Pa}\\ =-33.64\times {10}^6\text{Pa}\end{array}$

Find the residual shearing stress ${\left({\tau }_{res}\right)}_{{\rho }_Y}$ at c using the relation.

${\left({\tau }_{res}\right)}_{{\rho }_Y}={\tau }_Y-{\tau }_{{\rho }_Y}^{\text{'}}$

Substitute 145 MPa for ${\tau }_Y$ and $120.153\times {10}^6\text{Pa}$ for ${\tau }_{{\rho }_Y}^{\text{'}}$.

$\begin{array}{c}{\left({\tau }_{res}\right)}_{{\rho }_Y}=\left(145\text{MPa}\times \frac{1,000,000\text{Pa}}{1\text{MPa}}\right)-120.153\times {10}^6\text{Pa}\\ =24.847\times {10}^6\text{Pa}\end{array}$

Comparing the above ${\left({\tau }_{res}\right)}_c$ and ${\left({\tau }_{res}\right)}_{{\rho }_Y}$, the residual shearing stress ${\left({\tau }_{res}\right)}_c$ is maximum and located at the center of shaft.

The magnitude of the maximum residual shearing stress $\left({\tau }_{res}\right)$ is $\underset{\_}{33.64\text{MPa}}$.

The location of the maximum residual shearing stress $\left(\rho \right)$ is $\underset{\_}{16\text{mm}}$.

(b)

To determine

Find the permanent angle of twist $\left({\varphi }_{\text{per}}\right)$ in the solid shaft.

Answer to Problem 116P

The permanent angle of twist $\left({\varphi }_{\text{per}}\right)$ is $\underset{\_}{1.03°}$.

Explanation of Solution

Given information:

The angle of twist $\left(\varphi \right)$ increased in shaft $6°$.

Calculation:

Calculate the permanent angle of twist $\left({\varphi }_{per}\right)$ using the relation.

${\varphi }_{per}=\varphi -\varphi \text{'}$

Here, $\varphi \text{'}$ is angle of twist in elastic unloading.

Substitute $6°$ for $\varphi$ and $4.97°$ for $\varphi \text{'}$.

$\begin{array}{c}{\varphi }_{per}=6°-4.97°\\ =1.03°\end{array}$

The permanent angle of twist $\left({\varphi }_{\text{per}}\right)$ is $\underset{\_}{1.03°}$.