Question

A Hydrodynamic journal bearing of diameter 250 mm and length 400 mm carries a load of 250 kN. The average bearing pressure is:

• A. $0.25 \text{ N/mm}^2$
• B. $2.5 \text{ N/mm}^2$
• C. $25 \text{ N/mm}^2$
• D. $50 \text{ N/mm}^2$

Hint:

When calculating bearing pressure, always use the Projected Area ($D \times L$), not the surface area ($\pi D L$). Just think of it as the rectangular "shadow" cast by the bearing.

Answer 1

The Correct Option is B

The bearing pressure in a journal bearing is calculated based on the projected area of the bearing. This "projected area" is the equivalent flat surface area that would see the load if the bearing were flat instead of curved. 1. Identifying the Variables: Given data:

• Diameter of bearing ($d$) = 250 mm

• Length of bearing ($L$) = 400 mm

• Load on the bearing ($W$) = 250 kN = $250 \times 10^3$ N

2. Calculating Projected Area: The projected area ($A$) for a journal bearing is the product of its diameter and its length: $$A = d \times L$$ $$A = 250 \text{ mm} \times 400 \text{ mm}$$ $$A = 100,000 \text{ mm}^2 = 10^5 \text{ mm}^2$$

3. Calculating Average Bearing Pressure ($p$): Bearing pressure is the load divided by the projected area: $$p = \frac{W}{A}$$ $$p = \frac{250,000 \text{ N}}{100,000 \text{ mm}^2}$$ $$p = 2.5 \text{ N/mm}^2$$

4. Significance: This average bearing pressure must be kept within specific limits for different materials (like Babbitt or Bronze) to ensure the lubricating oil film can be maintained and to prevent excessive wear or heat generation in the journal.