Chapter 3.8, Problem 107P

(a)

To determine

Find the applied torque and the corresponding angle of twist at the onset of yield.

Answer to Problem 107P

The applied torque at the onset of yield is $\underset{\_}{11.71\text{kN}\cdot \text{m}}$.

The corresponding angle of twist at the onset of yield is $\underset{\_}{3.44°}$.

Explanation of Solution

Given information:

The length of the shaft (L) is 0.9 m.

The shear stress $\left({\tau }_Y\right)$.is 180 MPa

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

The inner diameter of the shaft $\left(d_1\right)$ is 30 mm.

The inner diameter of the shaft $\left(d_2\right)$ is 70 mm.

Calculation:

Find the inner radius $\left(c_1\right)$ of the shaft using the relation:

$c_1=\frac{1}{2}{d}_1$ (1)

Substitute 30 mm for $d_1$ in Equation (1).

$\begin{array}{c}{c}_1=\frac{1}{2}\left(30\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\right)\\ =0.015\text{m}\end{array}$

Find the outer radius $\left(c_2\right)$ of the shaft using the relation:

$c_2=\frac{1}{2}{d}_2$ (2)

Substitute 70 mm for $d_2$ in Equation (2)

$\begin{array}{c}{c}_2=\frac{1}{2}\left(70\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\right)\\ =0.035\text{m}\end{array}$

Find the polar moment of inertia of a shaft (J) using the relation:

$J=\frac{\pi }{2}\left(c_2^4-c_1^4\right)$ (3)

Substitute 0.035 m for $c_2$ and 0.015 m for $c_1$ in Equation (3).

$\begin{array}{c}J=\frac{\pi }{2}\left({0.035}^4-{0.015}^4\right)\\ =2.2777\times {10}^{-6}{\text{m}}^{\text{4}}\end{array}$

At the onset of yield, the stress distribution is the elastic distribution with ${\tau }_{\max }={\tau }_Y$.

Find the applied torque $\left(T_{\text{Y}}\right)$ in the shaft:

$T_Y=\frac{J{\tau }_Y}{c_2}$ (4)

Substitute $2.2777\times {10}^{-6}{\text{m}}^{\text{4}}$ for J, 180 MPa for ${\tau }_Y$, and 0.035 m for $c_2$ in Equation (4).

$\begin{array}{c}{T}_Y=\frac{\left(2.2777\times {10}^{-6}{\text{m}}^4\right)\left(180\text{MPa}\times \frac{{10}^6{\text{N/m}}^2}{1\text{MPa}}\right)}{\left(0.035\text{m}\right)}\\ =11,713.65\text{N}\cdot \text{m}\times \frac{1\text{kN}}{1,000\text{N}}\\ =11.71\text{kN}\cdot \text{m}\end{array}$

Therefore, the applied torque at the onset of yield is $\underset{\_}{11.71\text{kN}\cdot \text{m}}$.

Find the angle of twist $\left(\varphi \right)$ of the shaft:

$\varphi =\frac{TL}{GJ}$ (5)

Substitute $11.71\times {10}^3\text{N}\cdot \text{m}$ for T, 0.9 m for L, $77.2\text{GPa}$ for G, and $2.2777\times {10}^{-6}{\text{m}}^{\text{4}}$ for J in Equation (5).

$\begin{array}{c}\varphi =\frac{\left(11.71\times {10}^3\text{N}\cdot \text{m}\right)\left(0.9\text{m}\right)}{\left(77.2\text{GPa}\times \frac{{10}^9{\text{N/m}}^2}{1\text{GPa}}\right)\left(2.2777\times {10}^{-6}{\text{m}}^{\text{4}}\right)}\\ =59.935\times {10}^{-3}\text{rad}\times \frac{180°}{\pi \text{rad}}\\ =3.434°\end{array}$

Therefore, the corresponding angle of twist at the onset of yield is $\underset{\_}{3.44°}$.

(b)

To determine

Find the applied torque and the corresponding angle of twist of the when the plastic zone is $10\text{mm}$ deep.

Answer to Problem 107P

The applied torque when the plastic zone is 10 mm deep is $\underset{\_}{14.12\text{kN}\cdot \text{m}}$.

The corresponding angle of twist when the plastic zone is 10 mm deep is $\underset{\_}{4.81°}$.

Explanation of Solution

Given information:

The depth of plastic zone (t) is 10 mm.

Calculation:

Find the depth of elastic portion $\left({\rho }_Y\right)$:

${\rho }_Y=c_2-t$ (6)

Substitute 0.035 m for $c_2$ and 10 mm for t in Equation (6)

$\begin{array}{c}{\rho }_Y=0.035-\left(10\text{mm}\times \frac{1\text{m}}{1000\text{mm}}\right)\\ =0.035-0.001\\ =0.025\text{m}\end{array}$

Find the polar moment of inertia $\left(J_1\right)$ of elastic portion:

$J_1=\frac{\pi }{2}\left({\rho }_Y^4-c_1^4\right)$ (7)

Substitute 0.025 m for ${\rho }_Y$ and 0.015 m for $c_1$ in Equation (7).

$\begin{array}{c}{J}_1=\frac{\pi }{2}\left({0.025}^4-{0.015}^4\right)\\ =534.07\times {10}^{-9}{\text{m}}^{\text{4}}\end{array}$

Find the torque $\left(T_1\right)$ carried by elastic portion:

$T_1=\frac{J_1{\tau }_Y}{{\rho }_Y}$ (8)

Substitute $534.07\times {10}^{-9}{\text{m}}^{\text{4}}$ for $J_1$, 180 MPa for ${\tau }_1$, and 0.025 m for ${\rho }_Y$ in Equation (8).

$\begin{array}{c}{T}_1=\frac{\left(534.07\times {10}^{-9}{\text{m}}^{\text{4}}\right)\left(180\text{MPa}\times \frac{{10}^6{\text{N/m}}^2}{1\text{MPa}}\right)}{\left(0.025\text{m}\right)}\\ =3.8453\times {10}^3\text{N}\cdot \text{m}\end{array}$

Find the expression of torque $\left(T_2\right)$ carried by plastic portion:

$T_2=2\pi {\displaystyle {\int }_{{\rho }_Y}^{c_2}{\tau }_Y{\rho }^{2}dp}$ (9)

Integrate the Equation (9).

$\begin{array}{c}{T}_2=2\pi {\tau }_Y{\left[\frac{{\rho }^3}{3}\right]}_{{\rho }_Y}^{c_2}\\ =\frac{2}{3}\pi {\tau }_Y\left[c_2^3-{\rho }_Y^3\right]\end{array}$ (10)

Substitute 180 MPa for ${\tau }_Y$, 0.035 m for $c_2$, and 0.025 m for ${\rho }_Y$ in Equation (10).

$\begin{array}{c}{T}_2=\frac{2}{3}\pi \left(180\text{MPa}\times \frac{{10}^6{\text{N/m}}^2}{1\text{MPa}}\right)\left[{\left(0.035\text{m}\right)}^3-{\left(0.025\text{m}\right)}^3\right]\\ =10.273\times {10}^3\text{N}\cdot \text{m}\end{array}$

Find the total applied torque (T):

$T=T_1+T_2$ (11)

Substitute $3.8453\times {10}^3\text{N}\cdot \text{m}$ for $T_1$ and $10.273\times {10}^3\text{N}\text{.m}$ for $T_2$ in Equation (11).

$\begin{array}{c}T=\left(3.8453\times {10}^3\right)+\left(10.273\times {10}^3\right)\\ =14.1183\times {10}^3\text{N}\cdot \text{m}\end{array}$

Therefore, the applied torque when depth of plastic zone is 10 mm is $\underset{\_}{14.1183\times {10}^3\text{N}\cdot \text{m}}$.

Find the angle of twist $\left(\phi \right)$ when the plastic zone is 10 mm

$\phi =\frac{{\tau }_{Y}L}{G{\rho }_Y}$ (12)

Substitute 180 MPa for ${\tau }_Y$, 0.9 m for L, 77.2 GPa for G, and 0.025 m for ${\rho }_Y$ in Equation (12).

$\begin{array}{c}\phi =\frac{\left(180\text{MPa}\times \frac{{10}^6{\text{N/m}}^2}{1\text{MPa}}\right)\left(0.9\text{m}\right)}{\left(77.2\text{GPa}\times \frac{{10}^9{\text{N/m}}^2}{1\text{GPa}}\right)\left(0.025\text{m}\right)}\\ =83.938\times {10}^{-3}\text{rad}\times \frac{180°}{\pi \text{rad}}\\ =4.81°\end{array}$

Therefore, the corresponding angle of twist when plastic zone is 10 mm is $\underset{\_}{4.81°}$.