Chapter 5, Problem 75P

To determine

The factor of safety using distortion energy theory.

Answer to Problem 75P

The factor of safety using distortion energy theory for inner radius is $1.92$.

The factor of safety using distortion energy theory for outer radius is $1.95$.

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 inner member.

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

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

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

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

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

Write the expression for radial stress at outer radius for inner member.

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

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

Write the expression for second moment of area.

    $I=\left(\frac{\pi }{64}\right)\left(d^4-d_i^4\right)$                                                                           (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 inner radius.

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

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

Calculate factor of safety for inner radius.

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

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

Write the expression for von Mises stress for outer radius.

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

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

Calculate the factor of safety for outer radius.

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

Here, the factor of safety for outer radius is $n_o$.

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 $45\text{mm}$ for $d_o$ in Equation (III).

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

Substitute $50\text{mm}$ for $d$ in Equation (XV).

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

Substitute $\left(\frac{40\text{mm}}{2}\right)$ for $r_i$, $15.8\text{MPa}$ for $p$, $\left(\frac{45\text{mm}}{2}\right)$ for $r_o$ in Equation (IV).

    $\begin{array}{c}{\left({\sigma }_t\right)}_o=-\left[\frac{{\left(\frac{45\text{mm}}{2}\right)}^2+{\left(\frac{40\text{mm}}{2}\right)}^2}{{\left(\frac{45\text{mm}}{2}\right)}^2-{\left(\frac{40\text{mm}}{2}\right)}^2}\right]\left(15.8\text{MPa}\right)\\ =-\left(8.529\right)\left(15.8\text{MPa}\right)\\ \approx -134.7\text{MPa}\end{array}$

The radial stress for inner radius is zero at inner member.

    ${\left({\sigma }_r\right)}_i=0$

Substitute $\left(\frac{45\text{mm}}{2}\right)$ for $r_o$, $\left(\frac{40\text{mm}}{2}\right)$ for $r_i$, and $15.8\text{MPa}$ for $p$ in Equation (V).

    $\begin{array}{c}{\left({\sigma }_t\right)}_i=-\left(\frac{2{\left(\frac{45\text{mm}}{2}\right)}^2}{{\left(\frac{45\text{mm}}{2}\right)}^2-{\left(\frac{40\text{mm}}{2}\right)}^2}\right)\left(15.8\text{MPa}\right)\\ =-\left(9.5294\right)\left(15.8\text{MPa}\right)\\ \approx -150.6\text{MPa}\end{array}$

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

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

Substitute $50\text{mm}$ for $d$ 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 $M$, $\left(\frac{45\text{mm}}{2}\right)$ for $r_o$ and $\left(181.132\times {10}^3\right){\text{mm}}^4$ for $I$ in Equation (XVI).

    $\begin{array}{c}{\left({\sigma }_x\right)}_o=\pm \left(\frac{\left(675\text{N}\cdot \text{m}\right)\left(\frac{45\text{mm}}{2}\right)}{\left(181.132\times {10}^3\right){\text{mm}}^4}\right)\\ =\pm \left(\frac{\left(675\text{N}\cdot \text{m}\right)\left(22.5\text{mm}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)}{\left(181.132\times {10}^3\right){\text{mm}}^4\left(\frac{{10}^{-12}{\text{m}}^4}{1{\text{mm}}^4}\right)}\right)\\ =\pm \left(83.847\times {10}^6\right)\text{N}/{\text{m}}^2\left(\frac{1\text{MPa}}{{10}^6\text{N}/{\text{m}}^2}\right)\\ \approx \pm \left(83.9\right)\text{MPa}\end{array}$

Substitute $\left(r\right)$ 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 $M$, $\left(\frac{40\text{mm}}{2}\right)$ for $r$ and $\left(181.132\times {10}^3\right){\text{mm}}^4$ for $I$ in Equation (XVII).

    $\begin{array}{c}{\left({\sigma }_x\right)}_i=\pm \left(\frac{\left(675\text{N}\cdot \text{m}\right)\left(\frac{40\text{mm}}{2}\right)}{\left(181.132\times {10}^3\right){\text{mm}}^4}\right)\\ =\pm \left(\frac{\left(675\text{N}\cdot \text{m}\right)\left(20\text{mm}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)}{\left(181.132\times {10}^3\right){\text{mm}}^4\left(\frac{{10}^{-12}{\text{m}}^4}{1{\text{mm}}^4}\right)}\right)\\ =\pm \left(74.531\times {10}^6\right)\text{N}/{\text{m}}^2\left(\frac{1\text{MPa}}{{10}^6\text{N}/{\text{m}}^2}\right)\\ \approx \pm \left(74.5\right)\text{MPa}\end{array}$

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

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

Substitute $r$ for $c$, for outer member 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 $T$, $\left(\frac{40\text{mm}}{2}\right)$ for $r_o$ and $\left(362.26\times {10}^3\right){\text{mm}}^4$ for $J$ in Equation (XIX).

    $\begin{array}{c}{\left({\tau }_{xy}\right)}_i=\left(\frac{\left(900\text{N}\cdot \text{m}\right)\left(\frac{40\text{mm}}{2}\right)}{\left(362.26\times {10}^3\right){\text{mm}}^4}\right)\\ =\left(\frac{\left(900\text{N}\cdot \text{m}\right)\left(\frac{40\text{mm}}{2}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)}{\left(362.26\times {10}^3\right){\text{mm}}^4\left(\frac{{10}^{-12}{\text{m}}^4}{1{\text{mm}}^4}\right)}\right)\\ =\left(49.688\times {10}^6\right)\text{N}/{\text{m}}^2\left(\frac{1\text{MPa}}{{10}^6\text{N}/{\text{m}}^2}\right)\\ \approx 49.7\text{MPa}\end{array}$

Substitute $74.5\text{MPa}$ for ${\sigma }_x$, $-150.6\text{MPa}$ for ${\sigma }_y$ and $49.7\text{MPa}$ for ${\tau }_{xy}$ in Equation (XI).

    $\begin{array}{c}{{\sigma }^{\prime }}_i={\left[\left(\begin{array}{l}{\left(74.5\text{MPa}\right)}^2-\left(\left(74.5\text{MPa}\right)\left(-150\text{MPa}\right)\right)+{\left(-150.6\text{MPa}\right)}^2\\ +3{\left(49.7\text{MPa}\right)}^2\end{array}\right)\right]}^{\frac{1}{2}}\\ ={\left[\left(46815.88{\text{MPa}}^2\right)\right]}^{\frac{1}{2}}\\ \approx 216.37\text{MPa}\end{array}$

Substitute $415\text{MPa}$ for $S_y$ and $216.37\text{MPa}$ for ${{\sigma }^{\prime }}_i$ in Equation (XII).

    $\begin{array}{c}{n}_i=\left(\frac{415\text{MPa}}{216.37\text{MPa}}\right)\\ =1.91801\\ \approx 1.92\end{array}$

Thus, the factor of safety for outer radius is $1.92$.

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

Figure (1)

Substitute $\left(+83.9\text{MPa}\right)$ for ${\sigma }_x$, $-134.7\text{MPa}$ for ${\sigma }_y$, $\left(-15.8\text{MPa}\right)$ for ${\sigma }_z$ and $55.9\text{MPa}$ for ${\tau }_{xy}$ in Equation (XII).

    $\begin{array}{c}{{\sigma }^{\prime }}_o=\left(\frac{1}{\sqrt{2}}\right){\left[\begin{array}{l}{\left(\left(83.9\text{MPa}\right)-\left(-134.7\text{MPa}\right)\right)}^2+{\left(\left(-134.7\text{MPa}\right)-\left(-15.8\text{MPa}\right)\right)}^2\\ +{\left(\left(-15.8\text{MPa}\right)-\left(83.9\text{MPa}\right)\right)}^2+6\left({\left(55.9\text{MPa}\right)}^2\right)\end{array}\right]}^{\frac{1}{2}}\\ =\left(\frac{1}{\sqrt{2}}\right){\left[\left(90612.12\text{MPa}\right)\right]}^{\frac{1}{2}}\\ =212.8522\text{MPa}\\ \approx 212.85\text{MPa}\end{array}$

Substitute $\left(-83.9\text{MPa}\right)$ for ${\sigma }_x$, $-134.7\text{MPa}$ for ${\sigma }_y$, $\left(-15.8\text{MPa}\right)$ for ${\sigma }_z$ and $55.9\text{MPa}$ for ${\tau }_{xy}$ in Equation (XIII).

    $\begin{array}{c}{{\sigma }^{\prime }}_o=\left(\frac{1}{\sqrt{2}}\right){\left[\begin{array}{l}{\left(\left(-83.9\text{MPa}\right)-\left(-134.7\text{MPa}\right)\right)}^2\\ +{\left(\left(-134.7\text{MPa}\right)-\left(-15.8\text{MPa}\right)\right)}^2\\ +{\left(\left(-15.8\text{MPa}\right)-\left(-83.9\text{MPa}\right)\right)}^2\\ +6\left({\left(55.9\text{MPa}\right)}^2\right)\end{array}\right]}^{\frac{1}{2}}\\ =\left(\frac{1}{\sqrt{2}}\right){\left[\left(40104.32\text{MPa}\right)\right]}^{\frac{1}{2}}\\ =141.6056\text{MPa}\\ \approx 141.6\text{MPa}\end{array}$

The value of ${{\sigma }^{\prime }}_o$ is larger for the positive value of ${\sigma }_x$. Thus, the value of ${{\sigma }^{\prime }}_o$ is used as $212.85\text{MPa}$ to calculate the factor of safety.

Substitute $415\text{MPa}$ for $S_y$ and $212.85\text{MPa}$ for ${\sigma }^{\prime }$ in Equation (XIV).

    $\begin{array}{c}{n}_o=\left(\frac{415\text{MPa}}{212.85\text{MPa}}\right)\\ =1.949\\ \approx 1.95\end{array}$

Thus, the factor of safety for outer radius is $1.95$.