Question 8.12: For the following thrust bearing (Fig. 8.22), show that the ...
For the following thrust bearing (Fig. 8.22), show that the force on the straight slider in the x direction is the same as that on the guide.
It is given that the velocity profile is
\frac{u}{U}=\left(1-\frac{y}{h}\right)\left[1-3 \frac{y}{h}\left(1-\frac{2}{n+1} \frac{h_{1}}{h}\right)\right]And load
P=\frac{6 \mu U L^{2}}{h_{2}^{2}(n-1)^{2}}\left[\ln n-\frac{2(n-1)}{n+1}\right]Where n=h_{1} / h_{2}
Force on the slider in the x direction is
F_{s} =\int_{0}^{L} \tau_{s} \mathrm{~d} x(1) \frac{\cos \alpha}{\cos \alpha}+\int_{0}^{L}\left(p-p_{0}\right) \frac{\mathrm{d} x}{\cos \alpha}(1) \sin \alpha=\int_{0}^{L} \tau_{s} \mathrm{~d} x+\tan \alpha \int_{0}^{L}\left(p-p_{0}\right) \mathrm{d} x
Now, \tau_{s}=-\mu\left(\frac{\partial u}{\partial y}\right)_{y=h}
And u=U\left(1-\frac{y}{h}\right)\left[1-3 \frac{y}{h}\left(1-\frac{2}{n+1} \cdot \frac{h_{1}}{h}\right)\right] , so we get
\frac{\partial u}{\partial y}=U\left[-\frac{1}{h}-3\left(\frac{1}{h}-\frac{2 y}{h^{2}}\right)\left(1-\frac{2}{n+1} \cdot \frac{h_{1}}{h}\right)\right]
\therefore \tau_{s}=-\mu U\left[-\frac{1}{h}+\frac{3}{h}\left(1-\frac{2}{n+1}, \frac{h_{1}}{h}\right)\right]
=\mu U\left[-\frac{2}{h}+\frac{6}{n+1} \cdot \frac{h_{1}}{h^{2}}\right]
Also, h=h_{1}-\left(h_{1}-h_{2}\right) \frac{x}{L}
Therefore, \mathrm{d} h=-\frac{h_{1}-h_{2}}{L} \mathrm{~d} x
Thus, \int_{0}^{L} \tau_{s} \mathrm{~d} x=\frac{L}{h_{1}-h_{2}} \int_{h_{2}}^{h_{1}} \tau_{s} \mathrm{~d} h
=\frac{\mu L U}{h_{1}-h_{2}} \int_{h_{2}}^{h_{1}}\left(-\frac{2}{h}+\frac{6}{n+1} \cdot \frac{h_{1}}{h^{2}}\right) \mathrm{d} h=\frac{\mu U L}{h_{1}-h_{2}}\left(-2 \ln h-\frac{6}{n+1} \cdot \frac{h_{1}}{h}\right)_{h_{2}}^{h_{1}}
=\frac{\mu U L}{h_{2}(n-1)}\left[-2 \ln n-\frac{6 h_{1}}{n+1}\left(\frac{1}{h_{1}}-\frac{1}{h_{2}}\right)\right]
=\frac{\mu U L}{h_{2}(n-1)}\left[-2 \ln n+\frac{6(n-1)}{n+1}\right]
Also, load P=\int_{0}^{L}\left(p-p_{0}\right) \frac{d x \cos \alpha}{\cos \alpha}
(neglecting contribution of ts to load; a is small)
Therefore, F_{s}=\int_{0}^{L} \tau_{s} \mathrm{~d} x+\tan \alpha(P)
Or F_{s}= \frac{\mu U L}{h_{2}(n-1)}\left(-2 \ln n+\frac{6(n-1)}{n+1}\right)+\left[\frac{h_{1}-h_{2}}{L}\right] \times
\frac{6 \mu U L^{2}}{h_{2}^{2}(n-1)^{2}}\left(\ln n-\frac{2(n-1)}{n+1}\right)
= \frac{\mu U L}{h_{2}(n-1)}\left(4 \ln n-\frac{6(n-1)}{n+1}\right)
F_{G}=\int_{0}^{L} \tau_{G} \mathrm{~d} x
But \tau_{G}=\mu\left(\frac{\partial u}{\partial y}\right)_{y=0}
=\mu U\left[\frac{1}{h}+\frac{3}{h}\left(1-\frac{2}{n+1} \cdot \frac{h_{1}}{h}\right)\right]
The expression is \tau_{G}=\mu U\left[\frac{4}{h}-\frac{6}{n+1} \cdot \frac{h_{1}}{h^{2}}\right]
Therefore, F_{G}=\int_{0}^{L} \tau_{G} \mathrm{~d} x
=\frac{L}{h_{1}-h_{2}} \int_{h_{2}}^{h_{1}} \tau_{G} \mathrm{~d} h (as before)
=\frac{\mu U L}{h_{2}(n-1)} \int_{h_{2}}^{h_{1}}\left(\frac{4}{h}-\frac{6}{n+1} \cdot \frac{h_{1}}{h^{2}}\right) \mathrm{d} h=\frac{\mu U L}{h_{2}(n-1)}\left[4 \ln n+\frac{6 h_{1}}{n+1}\left(\frac{1}{h_{1}}-\frac{1}{h_{2}}\right)\right]
=\frac{\mu U L}{h_{2}(n-1)}\left[4 \ln n-\frac{6(n-1)}{n+1}\right]
which is the same as F_{s} .