Footstep (Pivot) Bearing — Effect of Shaft Radius on Viscous Torque For a vertical shaft running in a footstep bearing with a uniform oil film of thickness t and viscosity mu, if the shaft radius R is doubled (same speed), the torque required to overcome viscous resistance will become:

Introduction:Power loss in bearings often arises from viscous shear in thin lubricant films. In a flat footstep (pivot) bearing supporting a vertical shaft, torque due to viscosity grows rapidly with shaft radius, which is critical for sizing and thermal analysis.

Given Data / Assumptions:

• Flat circular contact area with oil film thickness t (constant).
• Dynamic viscosity mu and angular speed omega are constant.
• No slip at boundaries; laminar shear within the film.

Concept / Approach:The local shear stress is tau = mu * (du/dy) = mu * (omega * r / t). The shear force on an annulus of radius r and width dr is dF = tau * dA = mu * (omega * r / t) * (2 * pi * r * dr). The elemental torque is dT = r * dF.

Step-by-Step Solution:Write dT = r * dF = mu * omega / t * 2 * pi * r^3 * dr.Integrate from r = 0 to R: T = mu * omega / t * 2 * pi * (R^4 / 4).Simplify: T = (pi * mu * omega * R^4) / 2.Thus T ∝ R^4. Doubling R gives T_new / T_old = (2R)^4 / R^4 = 16.

Verification / Alternative check:Dimensional check shows torque scales with viscosity, speed, area, and the fourth power of radius in this configuration—consistent with classical tribology results.

Why Other Options Are Wrong:Linear, quadratic, or cubic scaling (2, 4, 8 times) underestimate the strong dependence of viscous torque on radius for a flat film.

Common Pitfalls:Using average radius approximations without integration; forgetting that film thickness variations or step bearings change exponents.

Final Answer:sixteen times