Chapter 8.3, Problem 8.96P

To determine

Prove that the magnitude M of the couple required to overcome the frictional resistance of a conical bearing as $M=\frac{2}{3}\frac{{\mu }_{k}P}{\sin \theta }\frac{R_2^3-R_1^3}{R_2^2-R_1^2}$.

Explanation of Solution

Write the equation of the magnitude of the couple as follows;

$M=\frac{2}{3}\frac{{\mu }_{k}P}{\sin \theta }\frac{R_2^3-R_1^3}{R_2^2-R_1^2}$

Here, the coefficient of kinetic friction is ${\mu }_k$, the magnitude of the total axial force is P, the outer radii of the collar is $R_1$, and the inner radii of the collar is $R_2$.

Show the free-body diagram of the conical bearing as in Figure 1.

It has been assumed that the normal force per unit area is inversely proportional to r.

$\Delta A=r\Delta s\Delta \varphi$

Here, the distance between the point to the axis of the shaft is r, the change in distance is $\Delta r$, and the change in azimuthal angle is $\Delta \varphi$.

Here, $\Delta s=\frac{\Delta r}{\sin \theta }$

Find the normal force exerted $\Delta N$ due to the element of area $\Delta A$ using the relation;

$\Delta N=k\Delta A$

Substitute $r\Delta s\Delta \varphi$ for $\Delta A$.

$\Delta N=kr\Delta s\Delta \varphi$

Substitute $\frac{\Delta r}{\sin \theta }$ for $\Delta s$.

$\Delta N=kr\frac{\Delta r}{\sin \theta }\Delta \varphi$ (1)

Find the change in friction force in y-axis;

$\Delta F_y=\Delta N\sin \theta$

Substitute $kr\frac{\Delta r}{\sin \theta }\Delta \varphi$ for $\Delta N$.

$\begin{array}{c}\Delta F_y=kr\frac{\Delta r}{\sin \theta }\Delta \varphi \sin \theta \\ =kr\Delta r\Delta \varphi \end{array}$

Find the value of k using the relation.

$\begin{array}{c}P={\displaystyle \sum \Delta F_y}\\ ={\displaystyle {\int }_0^{2\pi }{\displaystyle {\int }_{R_1}^{R_2}\Delta F_y}}\end{array}$

Substitute $kr\Delta r\Delta \varphi$ for $\Delta F_y$.

$\begin{array}{c}P={\displaystyle {\int }_0^{2\pi }{\displaystyle {\int }_{R_1}^{R_2}kr\Delta r\Delta \varphi }}\\ =k{\left[\frac{r^2}{2}\right]}_{R_1}^{R_2}{\left[\varphi \right]}_0^{2\pi }\\ =k\left[\frac{R_2^2-R_1^2}{2}\right]\left[2\pi \right]\\ =\pi k\left(R_2^2-R_1^2\right)\end{array}$

$k=\frac{P}{\pi \left(R_2^2-R_1^2\right)}$

Substitute $\frac{P}{\pi \left(R_2^2-R_1^2\right)}$ for k in Equation (1).

$\Delta N=\frac{P}{\pi \left(R_2^2-R_1^2\right)}r\frac{\Delta r}{\sin \theta }\Delta \varphi$

Find the change in friction force $\Delta F$ using the relation.

$\Delta F={\mu }_k\Delta N$

Substitute $\frac{P}{\pi \left(R_2^2-R_1^2\right)}r\frac{\Delta r}{\sin \theta }\Delta \varphi$ for $\Delta N$.

$\Delta F={\mu }_k\frac{P}{\pi \left(R_2^2-R_1^2\right)}r\frac{\Delta r}{\sin \theta }\Delta \varphi$

Find the change in moment $\Delta M$ using the relation.

$\Delta M=r\Delta F$

Substitute ${\mu }_k\frac{P}{\pi \left(R_2^2-R_1^2\right)}r\frac{\Delta r}{\sin \theta }\Delta \varphi$ for $\Delta F$.

$\Delta M=r{\mu }_k\frac{P}{\pi \left(R_2^2-R_1^2\right)}r\frac{\Delta r}{\sin \theta }\Delta \varphi$

Integrate the equation to find the couple M.

$M={\displaystyle {\int }_0^{2\pi }{\displaystyle {\int }_{R_1}^{R_2}\Delta M}}$

Substitute $r{\mu }_k\frac{P}{\pi \left(R_2^2-R_1^2\right)}r\frac{\Delta r}{\sin \theta }\Delta \varphi$ for $\Delta M$ and integrate.

$\begin{array}{c}M={\displaystyle {\int }_0^{2\pi }{\displaystyle {\int }_{R_1}^{R_2}r{\mu }_k\frac{P}{\pi \left(R_2^2-R_1^2\right)}r\frac{\Delta r}{\sin \theta }\Delta \varphi }}\\ ={\mu }_k\frac{P}{\pi \sin \theta \left(R_2^2-R_1^2\right)}{\left[\frac{r^3}{3}\right]}_{R_1}^{R_2}{\left[\varphi \right]}_0^{2\pi }\\ ={\mu }_k\frac{P}{\pi \sin \theta \left(R_2^2-R_1^2\right)}\left[\frac{R_2^3-R_1^3}{3}\right]\left[2\pi \right]\\ =\frac{2}{3}\frac{{\mu }_{k}P}{\sin \theta }\frac{\left(R_2^3-R_1^3\right)}{\left(R_2^2-R_1^2\right)}\end{array}$

Hence, it has been proved that the magnitude M of the couple required to overcome the frictional resistance of a conical bearing is $M=\frac{2}{3}\frac{{\mu }_{k}P}{\sin \theta }\frac{\left(R_2^3-R_1^3\right)}{\left(R_2^2-R_1^2\right)}$.