What will be the loss of power in a flat footstep [or flat pivot] bearing due to friction if a load of 20 KN is supported and the shaft is rotating at 10 rad/s? The diameter of the bearing is 16 cm and the coefficient of friction is 0.05. Assume uniform wear. 

Concept:

A footstep bearing is a cylindrical rigid block with a solid foundation. It has a cavity inside it within which the shaft is placed. The footstep bearing supports a vertical shaft. Many times, vertical shafts overhang, thus to support it as an end support the foot-step bearing is used. Instead of radial thrust, footstep bearing works under axial thrust as shown in the figure.

• Here we have to use uniform pressure distribution theory.
• From constant wear theory, the effective radius for friction application.

​\({R_{Effe}} = \frac{{\left( {R_0- R_i} \right)}}{{2}}\;\;\;\;\; \ldots \left( 1 \right)\)​

For footstep bearing,

Ri = 0; Ro = R (Radius of the shaft),

\( \Rightarrow \;{R_{Effe}} = \frac{{2R}}{3}\;\;\;\;\; \ldots \left( 2 \right)\)

∴ Frictional Torque (TEffe) = μ × F × REffe

⇒ Power loss due to friction (P) = TEffe × ω      … (3) 

Calculation:

Given:

 D = 16 cm ∴ R = 8 cm= 80 mm; μ = 0.05; F = 20 kN; ω =10 rad/s

By using equation (2),

\(\Rightarrow {R_{Effe}} = \frac{{ \ 80}}{2}\)

⇒ REffe = 40 mm

By using equation (3),

∴ P = μ × F × REffe × ω

⇒ P = 0.05 × 20 × 103 × 40 × 10-3 × 10

⇒ P = 400 W