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Sliding Bearings
                          wear rate at sliding interfaces. Combined with Eqs. (12.3) and (12.4), the pv factor is  341
                          calculated by
                                     F       dn
                               pv =    ⋅         ≤ [pv]                                      (12.5)
                                    Bd 60 × 1000
                            When designing a boundary-lubricated bearing, these design criteria must be satis-
                          fied. The allowable values of [p], [v], [pv] of commonly used bearing materials can be
                          foundinTable 12.5later.


                          12.3.2  Hydrodynamically Lubricated Bearings

                          To simplify the analysis, it is assumed that [1–5]:
                          – the lubricant is an incompressible Newtonian fluid with constant viscosity through-
                            out the film;
                          – the lubricant experiences laminar flow, with no slip at boundary surfaces;
                          – the lubricant has no inertial or gravitational forces;
                          – the bushing and journal extend infinitely in the axial direction (z-coordinate), that is,
                            no lubricant flow in the axial direction;
                          – the film is so thin that it experiences negligible pressure variation over its thickness,
                            thus the pressure depends only on coordinate x;
                          – the velocity of any particle of lubricant in the film depends only on coordinates x
                            and y.

                          12.3.2.1  Reynolds Equation
                          Figure 12.8a shows the forces acting on an elemental cube M with dimensions of dx,
                          dy and dz selected from the lubricant film in a journal bearing (Figure 12.8b). The
                          right and left sides of the element are subjected to normal forces due to pressure,
                          while the top and bottom sides of the element are subjected to shear forces due to
                          viscosity and velocity. Establish the force equilibrium equation in the x direction, which
                          gives
                                              (        )               (        )
                               ∑                     p                          
                                  F = pdydz − p +    dx dydz −   dxdz +    +  dy dxdz = 0      (a)
                                   x
                                                     x                        y
                            This reduces to
                                  p      
                                   =                                                           (b)
                                  x    y
                            According to Newton’s law of viscous flow in Eq. (12.1), the shearing stress is simply
                          proportional to the rate of shearing strain, that is,
                                      u
                                  =                                                            (c)
                                      y
                            Substituting Eq. (c) in Eq. (b), and convert it to
                                 2
                                   u  1   p
                                   =                                                           (d)
                                  y 2       x
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