Chapter 2, Problem 127P

To determine

The force required to maintain the axial movement of the shaft.

Answer to Problem 127P

The force required to maintain the axial movement of the shaft is $69\text{N}$.

Explanation of Solution

Given information:

The diameter of the shaft is $80\text{mm}$, length is $400\text{mm}$, velocity of bearing is $5\text{m}/\text{s}$, the initial clearance between shaft and the bearing is $1.2\text{mm}$, the final clearance between shaft and the bearing is $0.4\text{mm}$ and the dynamic viscosity of the lubricant is $0.10\text{Pa}\cdot \text{s}$.

Write the expression for the shear stress.

  ${\tau }_x=\mu \frac{U}{y\left(x\right)}$   ..... (I)

Here, the shear stress in the horizontal direction is ${\tau }_x$, the dynamic viscosity is $\mu$, the velocity is $U$ and the change in thickness is $y\left(x\right)$.

Write the expression for the variation in clearance.

  $y\left(x\right)=h_1-\frac{x}{L}\left(h_1-h_2\right)$   ..... (II)

Here, the initial clearance is $h_1$, the final clearance is $h_2$, the initial clearance is $h_1$ and the change of position is $x$.

Write the expression for the force required for the axial movement of the shaft.

  $\begin{array}{l}dF=\tau dA\\ dF=\tau \left(\pi D\right)dx\end{array}$   ..... (III)

Here, the force is $F$, the diameter of the shaft is $D$ and the change is position is $dx$.

Integrate Equation (III) under the limit $0$ to $F$ for $dF$ and the limit $0$ to $L$ for $dx$

  ${\displaystyle \underset{0}{\overset{F}{\int }}dF}={\displaystyle \underset{0}{\overset{L}{\int }}{\tau }_x\left(\pi D\right)dx}$   ..... (IV)

Calculation:

Substitute $1.2\text{mm}$ for $h_1$, $0.4\text{mm}$ for $h_2$ and $400\text{mm}$ for $L$ in Equation (II).

  $\begin{array}{c}y\left(x\right)=1.2\text{mm}-\frac{x}{400\text{mm}}\left[\left(1.2\text{mm}-0.4\text{mm}\right)\right]\\ =\left(1.2\text{mm}\times \frac{{10}^{-3}\text{m}}{1\text{mm}}\right)-\left(2\times {10}^{-3}\text{mm}\times \frac{{10}^{-3}\text{m}}{1\text{mm}}\right)\times \left(x\right)\\ =\left(1.2-2x\right)\times {10}^{-3}\text{m}\end{array}$

Substitute $0.10\text{Pa}\cdot \text{s}$ for $\mu$, $5\text{m}/\text{s}$ for $U$ and $\left(1.2-2x\right)\times {10}^{-3}\text{m}$ for $y\left(x\right)$ in Equation (I).

  $\begin{array}{c}{\tau }_x=\left(0.10\text{Pa}\cdot \text{s}\right)\frac{5\text{m}/\text{s}}{\left(1.2-2x\right)\times {10}^{-3}\text{m}}\\ =\frac{0.10\times 5\text{Pa}}{\left(1.2-2x\right)\times {10}^{-3}}\\ =\frac{500\text{Pa}}{\left(1.2-2x\right)}\end{array}$

Substitute $\frac{500\text{Pa}}{\left(1.2-2x\right)}$ for ${\tau }_x$ and $80\text{mm}$ for $D$ in Equation (IV).

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{F}{\int }}dF}={\displaystyle \underset{0}{\overset{L}{\int }}\frac{500\text{Pa}}{\left(1.2-2x\right)}\left(\pi \times 80\text{mm}\right)dx}\\ F-0=500\text{Pa}\left[\pi \times \left(80\text{mm}\times \frac{{10}^{-3}\text{m}}{1\text{mm}}\right)\right]{\displaystyle \underset{0}{\overset{L}{\int }}\frac{1}{\left(1.2-2x\right)}dx}\\ F=500\text{Pa}\times \pi \times 0.08\text{m}\left[\left(\frac{1}{-2}\ln \left(1.2-2L\right)+\left(\frac{1}{2}\right)\ln \left(1.2\right)\right)\right]\end{array}$..... (V)

Substitute $400\text{mm}$ for $L$ in Equation (V).

  $\begin{array}{c}F=500\text{Pa}\times \pi \times 0.08\text{m}\left[-\frac{1}{2}\ln \left(1.2-\left(2\times 400\text{mm}\right)\right)+\frac{1}{2}\ln \left(1.2\text{m}\right)\right]\\ =500\text{Pa}\times \pi \times 0.08\text{m}\left[\frac{-1}{2}\ln \left(1.2-\left(800\text{mm}\times \frac{{10}^{-3}\text{m}}{1\text{mm}}\right)\right)+0.09116\text{m}\right]\\ =69\text{Pa}\cdot {\text{m}}^2\times \left(\frac{1\text{N}/{\text{m}}^2}{1\text{Pa}}\right)\\ =69\text{N}\end{array}$

Conclusion:

The force required to maintain the axial movement of the shaft is $69\text{N}$.