What will be the power lost in friction assuming uniform pressure theory, when a vertical shaft of 100 mm diameter rotating at 150 r.p.m. rests on a flat end foot step bearing? The coefficient of friction is equal to 0.05 and the shaft carries a vertical load of 15 KN.

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 Pressure theory, the effective radius for friction application.

​\({R_{Effe}} = \frac{{2\left( {R_0^3 - R_i^3} \right)}}{{3\left( {R_0^2 - R_i^2} \right)}}\;\;\;\;\; \ldots \left( 1 \right)\)​

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:

R = 50 mm; μ = 0.05; F = 15 kN; N =150 r.p.m. ∴ ω =15.70 rad/s

By using equation (2),

\(\Rightarrow {R_{Effe}} = \frac{{2 \times 100}}{3}\)

⇒ REffe = 33.33 mm

By using equation (3),

∴ P = μ × F × REffe × ω

⇒ P = 0.05 × 15 × 103 × 33.33 × 10-3 × 15.70

⇒ P = 392.7 W