We saw in Prob. 10–31E that a slipper-pad bearing can support a large load. If the load were to increase, the gap height would decrease, thereby increasing the pressure in the gap. In this sense, the slipper-pad bearing is “self-adjusting” to varying loads. If the load increases by a factor of 2, calculate how much the gap height decreases. Specifically, calculate the new value of h₀ and the percentage change. Assume that the slope of the upper plate and all other parameters and dimensions stay the same as those in Prob. 10–31E.

Data from Problem 31

A slipper-pad bearing with linearly decreasing gap height (Fig. P10–23) is being designed for an amusement park ride. Its dimensions are h₀ = 1/1000 in

and L = 1.0 in (0.0254 m). The lower plate moves at speed V = 10.0 ft/s (3.048 m/s) relative to the upper plate. The oil is engine oil at 40°C. Both ends of the slipper-pad are exposed to atmospheric pressure, as in Prob. 10–30. 

Calculate the convergence a, and verify that tan α ≅ α for this case.

Calculate the gage pressure halfway along the slipper-pad (at x = 0.5 in). Comment on the magnitude of the gage pressure.

Plot P* as a function of x*, where x* = x/L and P* 5 (P – P_atm)h₀²/μVL. 

Approximately how many pounds (lbf) of weight (load) can this slipper-pad bearing support if it is b = 6.0 in deep (into the page of Fig. P10–23)?

FIGURE P10–23

Transcribed Image Text:

(2.54 x 10-5 m), h 1/2000 in (1.27 x 10-5 m),