Chapter 5, Problem 74P

Two steel lubes are shrink-filled together where the nominal diameters are 40, 45, and 50 mm. Careful measurement before fitting determined the diametral interference between the tubes to be 0.062 mm. After the fit. the assembly is subjected to a torque of 900 N · m and a bending-moment of 675 N · m. Assuming no slipping between the cylinders, analyze the outer cylinder at the inner and outer radius. Determine the factor of safety using distortion energy with S_v = 415 MPa.

To determine

The factor of safety using distortion energy theory.

Answer to Problem 74P

The factor of safety using distortion energy theory for inner radius is $1.80$.

The factor of safety using distortion energy theory for outer radius is $1.84$.

Explanation of Solution

Write the expression for contact pressure.

$p=\left(\frac{E{\delta }_d}{2d^3}\right)\left[\frac{\left(d_o^2-d^2\right)\left(d^2-d_i^2\right)}{\left(d_o^2-d_i^2\right)}\right]$ (I)

Here, the contact pressure is $p$, the Yong’s modulus is $E$, the diametral interference is ${\delta }_d$, the inner diameter is $d_i$, the outer diameter is $d_o$ and the nominal diameter is $d$.

Write the expression for inner radius.

$r_i=\frac{d_i}{2}$ (II)

Here, the inner radius is $r_i$.

Write the expression for outer radius.

$r_o=\frac{d_o}{2}$ (III)

Here, the outer radius is $r_o$.

Write the expression for tangential stress at outer radius for outer member.

${\left({\sigma }_t\right)}_o=\left[\left(\frac{r_i^{2}p}{\left(r_o^2-r_i^2\right)}\right)\left(2\right)\right]$ (IV)

Here, the tangential stress at outer radius for outer member is ${\left({\sigma }_t\right)}_o$.

Write the expression for tangential stress for inner member.

${\left({\sigma }_t\right)}_i=\left(\frac{r_i^{2}p}{r_o^2-r_i^2}\right)\left(1+\frac{r_o^2}{r_i^2}\right)$ (V)

Here, the tangential stress for inner member is ${\left({\sigma }_t\right)}_i$.

Write the expression for radial stress for inner member.

${\left({\sigma }_r\right)}_i=-p$ (VI)

Here, the radial stress for inner member is ${\left({\sigma }_r\right)}_i$.

Write the expression for second moment of area.

$I=\left(\frac{\pi }{64}\right)\left(d_o^4-d_i^4\right)$ (VII)

Here, the second moment of area is $I$.

Write the expression for stress.

$\left({\sigma }_x\right)=\pm \left(\frac{Mc}{I}\right)$ (VIII)

Here, the stress in $x$ plane is $\left({\sigma }_x\right)$, the bending-moment is $M$, the distance of the member from neutral axis is $c$.

Write the expression for second polar moment of area.

$J=2I$ (IX)

Here, the second polar moment of area is $J$.

Write the expression for shear stress.

$\left({\tau }_{xy}\right)=\left(\frac{Tc}{J}\right)$ (X)

Here, the shear stress is ${\tau }_{xy}$, the torque is $T$.

Write the expression for von Mises stress for outer radius.

${{\sigma }^{\prime }}_o={\left({\sigma }_x^2-{\sigma }_x{\sigma }_y+{\sigma }_y^2+3{\tau }_{xy}^2\right)}^{\frac{1}{2}}$ (XI)

Here, the von Mises stress is ${{\sigma }^{\prime }}_o$, the stress in $y$ plane is ${\sigma }_y$ and shear stress in $xy$ plane is ${\tau }_{xy}$

Calculate factor of safety for outer radius.

$n_o=\left(\frac{S_y}{{{\sigma }^{\prime }}_o}\right)$ (XII)

Here, the factor of safety for outer radius is $n_o$ and the yield strength is $S_y$.

Write the expression for von Mises stress for inner radius.

${{\sigma }^{\prime }}_i=\left(\frac{1}{\sqrt{2}}\right){\left[{\left({\sigma }_x-{\sigma }_y\right)}^2+{\left({\sigma }_y-{\sigma }_z\right)}^2+{\left({\sigma }_z-{\sigma }_x\right)}^2+6{\left({\tau }_{xy}\right)}^2\right]}^{\frac{1}{2}}$ (XIII)

Here, the von Mises stress for inner radius is ${{\sigma }^{\prime }}_i$ and stress in $z$ plane is ${\sigma }_z$.

Calculate the factor of safety for inner radius.

$n_i=\left(\frac{S_y}{{{\sigma }^{\prime }}_i}\right)$ (XIV)

Here, the factor of safety for inner radius is $n_i$.

Write the expression for nominal radius.

$r=\frac{d}{2}$ (XV)

Here, the nominal radius is $r$.

Conclusion:

Substitute $207\text{GPa}$ for $E$, $0.062\text{mm}$ for ${\delta }_d$, $50\text{mm}$ for $d_o$, $45\text{mm}$ for $d$ and $40\text{mm}$ for $d_i$ in Equation (I).

$\begin{array}{c}p=\left(\frac{\left(207\text{GPa}\right)\left(0.062\text{mm}\right)}{2{\left(45\text{mm}\right)}^3}\right)\left[\frac{\left({\left(50\text{mm}\right)}^2-{\left(45\text{mm}\right)}^2\right)\left({\left(45\text{mm}\right)}^2-{\left(40\text{mm}\right)}^2\right)}{\left({\left(50\text{mm}\right)}^2-{\left(40\text{mm}\right)}^2\right)}\right]\\ =\left(207\text{GPa}\right)\left(\frac{{10}^3\text{MPa}}{1\text{GPa}}\right)\left(7.63069\times {10}^{-5}\right)\\ =15.7955\text{MPa}\\ \approx 15.8\text{MPa}\end{array}$

Substitute $40\text{mm}$ for $d_i$ in Equation (II).

$\begin{array}{c}{r}_i=\frac{d_i}{2}\\ =\frac{40\text{mm}}{2}\end{array}$

Substitute $50\text{mm}$ for $d_o$ in Equation (III).

$\begin{array}{c}{r}_o=\frac{d_o}{2}\\ =\frac{50\text{mm}}{2}\end{array}$

Substitute $45\text{mm}$ for $d$ in Equation (XV).

$\begin{array}{c}r=\frac{d}{2}\\ =\frac{45\text{mm}}{2}\end{array}$

Substitute $\left(\frac{45\text{mm}}{2}\right)$ for $r_i$, $15.8\text{MPa}$ for $p$, $\left(\frac{50\text{mm}}{2}\right)$ for $r_o$ in Equation (IV).

$\begin{array}{c}{\left({\sigma }_t\right)}_o=\left[\left(\frac{{\left(\frac{45\text{mm}}{2}\right)}^2\left(15.8\text{MPa}\right)}{\left({\left(\frac{50\text{mm}}{2}\right)}^2-{\left(\frac{45\text{mm}}{2}\right)}^2\right)}\right)\left(2\right)\right]\\ =\left(\frac{63990\text{MPa}\cdot {\text{mm}}^2}{475{\text{mm}}^2}\right)\\ =134.7157\text{MPa}\\ \approx 134.7\text{MPa}\end{array}$

The radial stress for outer radius is zero. Thus,

${\left({\sigma }_r\right)}_o=0$

Substitute $\left(\frac{45\text{mm}}{2}\right)$ for $r_i$, $\left(\frac{50\text{mm}}{2}\right)$ for $r_o$ and $15.8\text{MPa}$ for $p$ in Equation (V).

$\begin{array}{c}{\left({\sigma }_t\right)}_i=\left(\frac{{\left(\frac{45\text{mm}}{2}\right)}^2\left(15.8\text{MPa}\right)}{\left({\left(\frac{50\text{mm}}{2}\right)}^2-{\left(\frac{45\text{mm}}{2}\right)}^2\right)}\right)\left(1+\frac{{\left(\frac{50\text{mm}}{2}\right)}^2}{{\left(\frac{45\text{mm}}{2}\right)}^2}\right)\\ =\left(15.80\text{MPa}\right)\left(9.5263\right)\\ \approx 150.5\text{MPa}\end{array}$

Substitute $15.8\text{MPa}$ for $p$ in Equation (VI).

${\left({\sigma }_r\right)}_i=-15.8\text{MPa}$

Substitute $50\text{mm}$ for $d_o$ and $40\text{mm}$ for $d_i$ in Equation (VII).

$\begin{array}{c}I=\left(\frac{\pi }{64}\right)\left({\left(50\text{mm}\right)}^4-{\left(40\text{mm}\right)}^4\right)\\ =\left(\frac{\pi }{64}\right)\left(3690000{\text{mm}}^4\right)\\ =\left(181132.4514\right){\text{mm}}^4\\ \approx \left(181.132\times {10}^3\right){\text{mm}}^4\end{array}$

Substitute $\left(r_o\right)$ for $c$, for outer member in Equation (VIII).

${\left({\sigma }_x\right)}_o=\pm \left(\frac{M\left(r_o\right)}{I}\right)$ (XVI)

Here, the stress in outer member is ${\left({\sigma }_x\right)}_o$.

Substitute $675\text{N}\cdot \text{m}$ for $M$, $\left(\frac{50\text{mm}}{2}\right)$ for $r_o$ and $\left(181.132\times {10}^3\right){\text{mm}}^4$ for $I$ in Equation (XVI).

$\begin{array}{c}{\left({\sigma }_x\right)}_o=\pm \left(\frac{\left(675\text{N}\cdot \text{m}\right)\left(\frac{50\text{mm}}{2}\right)}{\left(181.132\times {10}^3\right){\text{mm}}^4}\right)\\ =\pm \left(\frac{\left(675\text{N}\cdot \text{m}\right)\left(25\text{mm}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)}{\left(181.132\times {10}^3\right){\text{mm}}^4\left(\frac{{10}^{-12}{\text{m}}^4}{1{\text{mm}}^4}\right)}\right)\\ =\pm \left(93.164\times {10}^6\right)\text{N}/{\text{m}}^2\left(\frac{1\text{MPa}}{{10}^6\text{N}/{\text{m}}^2}\right)\\ \approx \pm \left(93.2\right)\text{MPa}\end{array}$

Substitute $\left(r\right)$ for $c$, for inner member in Equation (VIII).

${\left({\sigma }_x\right)}_i=\pm \left(\frac{M\left(r\right)}{I}\right)$ (XVII)

Here, the stress for inner member is ${\left({\sigma }_x\right)}_i$.

Substitute $675\text{N}\cdot \text{m}$ for $M$, $\left(\frac{45\text{mm}}{2}\right)$ for $r$ and $\left(181.132\times {10}^3\right){\text{mm}}^4$ for $I$ in Equation (XVII).

$\begin{array}{c}{\left({\sigma }_x\right)}_i=\pm \left(\frac{\left(675\text{N}\cdot \text{m}\right)\left(\frac{45\text{mm}}{2}\right)}{\left(181.132\times {10}^3\right){\text{mm}}^4}\right)\\ =\pm \left(\frac{\left(675\text{N}\cdot \text{m}\right)\left(22.5\text{mm}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)}{\left(181.132\times {10}^3\right){\text{mm}}^4\left(\frac{{10}^{-12}{\text{m}}^4}{1{\text{mm}}^4}\right)}\right)\\ =\pm \left(83.8476\times {10}^6\right)\text{N}/{\text{m}}^2\left(\frac{1\text{MPa}}{{10}^6\text{N}/{\text{m}}^2}\right)\\ \approx \pm \left(83.9\right)\text{MPa}\end{array}$

Substitute $\left(181.132\times {10}^3\right){\text{mm}}^4$ for $I$ in Equation (IX).

$\begin{array}{c}J=2I\\ =\left(362.26\times {10}^3\right){\text{mm}}^4\end{array}$

Substitute $r_o$ for $c$, for outer member in Equation (X).

${\left({\tau }_{xy}\right)}_o=\left(\frac{T\left(r_o\right)}{J}\right)$ (XVIII)

Here, the shear stress for outer member is ${\left({\tau }_{xy}\right)}_o$.

Substitute $900\text{N}\cdot \text{m}$ for $T$, $\left(\frac{50\text{mm}}{2}\right)$ for $r_o$ and $\left(362.26\times {10}^3\right){\text{mm}}^4$ for $J$ in Equation (XVIII).

$\begin{array}{c}{\left({\tau }_{xy}\right)}_o=\left(\frac{\left(900\text{N}\cdot \text{m}\right)\left(\frac{50\text{mm}}{2}\right)}{\left(362.26\times {10}^3\right){\text{mm}}^4}\right)\\ =\left(\frac{\left(900\text{N}\cdot \text{m}\right)\left(\frac{50\text{mm}}{2}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)}{\left(362.26\times {10}^3\right){\text{mm}}^4\left(\frac{{10}^{-12}{\text{m}}^4}{1{\text{mm}}^4}\right)}\right)\\ =\left(62.11\times {10}^6\right)\text{N}/{\text{m}}^2\left(\frac{1\text{MPa}}{{10}^6\text{N}/{\text{m}}^2}\right)\\ \approx 62.1\text{MPa}\end{array}$

Substitute $r$ for $c$ in Equation (X).

${\left({\tau }_{xy}\right)}_i=\left(\frac{T\left(r\right)}{J}\right)$ (XIX)

Here, the shear stress for inner member is ${\left({\tau }_{xy}\right)}_i$.

Substitute $900\text{N}\cdot \text{m}$ for $T$, $\left(\frac{45\text{mm}}{2}\right)$ for $r_o$ and $\left(362.26\times {10}^3\right){\text{mm}}^4$ for $J$ in Equation (XIX).

$\begin{array}{c}{\left({\tau }_{xy}\right)}_i=\left(\frac{\left(900\text{N}\cdot \text{m}\right)\left(\frac{45\text{mm}}{2}\right)}{\left(362.26\times {10}^3\right){\text{mm}}^4}\right)\\ =\left(\frac{\left(900\text{N}\cdot \text{m}\right)\left(\frac{45\text{mm}}{2}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)}{\left(362.26\times {10}^3\right){\text{mm}}^4\left(\frac{{10}^{-12}{\text{m}}^4}{1{\text{mm}}^4}\right)}\right)\\ =\left(55.899\times {10}^6\right)\text{N}/{\text{m}}^2\left(\frac{1\text{MPa}}{{10}^6\text{N}/{\text{m}}^2}\right)\\ \approx 55.9\text{MPa}\end{array}$

Substitute $-93.2\text{MPa}$ for ${\sigma }_x$, $134.7\text{MPa}$ for ${\sigma }_y$ and $62.11\text{MPa}$ for ${\tau }_{xy}$ in Equation (XI).

$\begin{array}{c}{{\sigma }^{\prime }}_o={\left[\left(\begin{array}{l}{\left(-93.2\text{MPa}\right)}^2-\left(\left(-93.2\text{MPa}\right)\left(134.7\text{MPa}\right)\right)+{\left(134.7\text{MPa}\right)}^2\\ +3{\left(62.11\text{MPa}\right)}^2\end{array}\right)\right]}^{\frac{1}{2}}\\ ={\left[\left(50957.3263{\text{MPa}}^2\right)\right]}^{\frac{1}{2}}\\ \approx 225.73\text{MPa}\end{array}$

Substitute $415\text{MPa}$ for $S_y$ and $225.73\text{MPa}$ for ${{\sigma }^{\prime }}_o$ in Equation (XII).

$\begin{array}{c}{n}_o=\left(\frac{415\text{MPa}}{225.73\text{MPa}}\right)\\ =1.8384\\ \approx 1.84\end{array}$

Thus, the factor of safety for outer radius is $1.84$.

The following diagram shows the 3D stress for outer radius.

Figure (1)

Substitute $\left(+83.9\text{MPa}\right)$ for ${\sigma }_x$, $150.5\text{MPa}$ for ${\sigma }_y$, $\left(-15.8\text{MPa}\right)$ for ${\sigma }_z$ and $55.9\text{MPa}$ for ${\tau }_{xy}$ in Equation (XII).

$\begin{array}{c}{{\sigma }^{\prime }}_i=\left(\frac{1}{\sqrt{2}}\right){\left[\begin{array}{l}{\left(\left(83.9\text{MPa}\right)-\left(150.5\text{MPa}\right)\right)}^2+{\left(\left(150.5\text{MPa}\right)-\left(-15.8\text{MPa}\right)\right)}^2\\ +{\left(\left(-15.8\text{MPa}\right)-\left(83.9\text{MPa}\right)\right)}^2+6\left({\left(55.9\text{MPa}\right)}^2\right)\end{array}\right]}^{\frac{1}{2}}\\ =\left(\frac{1}{\sqrt{2}}\right){\left[\left(60780.2\text{MPa}\right)\right]}^{\frac{1}{2}}\\ =174.3275\text{MPa}\\ \approx \text{174}\text{.33}\text{MPa}\end{array}$

Substitute $\left(-83.9\text{MPa}\right)$ for ${\sigma }_x$, $150.5\text{MPa}$ for ${\sigma }_y$, $\left(-15.8\text{MPa}\right)$ for ${\sigma }_z$ and $55.9\text{MPa}$ for ${\tau }_{xy}$ in Equation (XIII).

$\begin{array}{c}{{\sigma }^{\prime }}_i=\left(\frac{1}{\sqrt{2}}\right){\left[\begin{array}{l}{\left(\left(-83.9\text{MPa}\right)-\left(150.5\text{MPa}\right)\right)}^2+{\left(\left(150.5\text{MPa}\right)-\left(-15.8\text{MPa}\right)\right)}^2\\ +{\left(\left(-15.8\text{MPa}\right)-\left(-83.9\text{MPa}\right)\right)}^2+6\left({\left(55.9\text{MPa}\right)}^2\right)\end{array}\right]}^{\frac{1}{2}}\\ =\left(\frac{1}{\sqrt{2}}\right){\left[\left(105985.52\text{MPa}\right)\right]}^{\frac{1}{2}}\\ =230.20156\text{MPa}\\ \approx 230.2\text{MPa}\end{array}$

The value of ${{\sigma }^{\prime }}_i$ is larger for the negative value of ${\sigma }_x$. Thus, the value of ${{\sigma }^{\prime }}_i$ is used as $230.2\text{MPa}$ to calculate the factor of safety.

Substitute $415\text{MPa}$ for $S_y$ and $230.2\text{MPa}$ for ${\sigma }^{\prime }$ in Equation (XIV).

$\begin{array}{c}{n}_i=\left(\frac{415\text{MPa}}{230.2\text{MPa}}\right)\\ =1.8027\\ \approx 1.80\end{array}$

Thus, the factor of safety for inner radius is $1.80$.