Explanation Given information: The diameter of the shaft is 80 mm , length is 400 mm , velocity of bearing is 5 m / s , the initial clearance between shaft and the bearing is 1.2 mm , the final clearance between shaft and the bearing is 0.4 mm and the dynamic viscosity of the lubricant is 0.10 Pa ⋅ s . Write the expression for the shear stress. τ x = μ U y ( x ) ..... (I) Here, the shear stress in the horizontal direction is τ x , the dynamic viscosity is μ , the velocity is U and the change in thickness is y ( x ) . Write the expression for the variation in clearance. y ( x ) = h 1 − x L ( h 1 − h 2 ) ..... (II) Here, the initial clearance is h 1 , the final clearance is h 2 , the initial clearance is h 1 and the change of position is x . Write the expression for the force required for the axial movement of the shaft. d F = τ d A d F = τ ( π D ) d x ..... (III) Here, the force is F , the diameter of the shaft is D and the change is position is d x . Integrate Equation (III) under the limit 0 to F for d F and the limit 0 to L for d x ∫ 0 F d F = ∫ 0 L τ x ( π D ) d x ..... (IV) Calculation: Substitute 1.2 mm for h 1 , 0.4 mm for h 2 and 400 mm for L in Equation (II). y ( x ) = 1.2 mm − x 400 mm [ ( 1.2 mm − 0.4 mm ) ] = ( 1.2 mm × 10 − 3 m 1 mm ) − ( 2 × 10 − 3 mm × 10 − 3 m 1 mm ) × ( x ) = ( 1.2 − 2 x ) × 10 − 3 m Substitute 0.10 Pa ⋅ s for μ , 5 m / s for U and ( 1.2 − 2 x ) × 10 − 3 m for y ( x ) in Equation (I)

4th Edition

ISBN: 2810022150991

Author: CENGEL

Publisher: MCG

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Chapter 2, Problem 120P

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The force required to maintain the axial movement of the shaft.

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Chapter 2, Problem 119EP

Chapter 2, Problem 121P

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Chapter 2 Solutions