A 25 mm diameter shaft is pulled through a cylindrical bearing, see the figure.
ID: 2324466 • Letter: A
Question
A 25 mm diameter shaft is pulled through a cylindrical bearing, see the figure. The lubricant that fills the 0.3 mm gap between the shaft and bearing is an oil having a kinematic viscosity of 8.0 Times 10^-4 m^2/s and a specific gravity of 0.91. The shaft is pulled at a velocity of 3 m/s. Assume the velocity in the gap to be linear, then: Plot the Velocity and Shear Stress profiles between the bearing and the shaft in the boxes below. Determine the shear stress on the surface of the bearing. Determine the force F required to pull the shaft at a velocity of 3 m/s.Explanation / Answer
(a) The plots are linear. For the small clearance between shaft and bearing, the velocity gradient is constant. So the velocity profile is linear. And as the shear stress is directly proportional to velocity gradient, its is also constant in this case.
(b) Shear stress = dynamic viscosity x velocity gradient
= (kinematic viscosity x density of fluid) x (surface velocity / clearance between shaft and bearing)
= (8 x 10-4 x 910) x [3 / (0.3 x 10-3)] Pa [density = specific gravity x 1000]
= 7280 Pa
(c) Force required = Shear stress x surface area
= 7280 x DL [D = diameter of shaft, L = length of the bearing]
= 7280 x x 0.025 x 0.5
= 285.88 N
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