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The dynamic load capacity of 6306 bearing is 22 kN. The maximum radial load it can sustain to operate at 600 rev/min, for 2000 hours will be equal to
  • a)
    3.16 kN
  • b)
    4.16 kN
  • c)
    6.21 kN
  • d)
    5.29 kN
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
The dynamic load capacity of 6306 bearing is 22 kN. The maximum radia...
Given:dynamic load capacity = 22 x 103 N
Speed(N) = 600 rav/min
Life = 2000 x 60 x 600
72 x 106 rev
Using following relation,
w = 5288.24 N
w = 5.288 kN
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Most Upvoted Answer
The dynamic load capacity of 6306 bearing is 22 kN. The maximum radia...
To determine the maximum radial load that a 6306 bearing can sustain for a given operating condition, we need to consider the dynamic load capacity of the bearing and the operating parameters.

1. Dynamic Load Capacity:
The dynamic load capacity of a bearing refers to the maximum load that the bearing can sustain for a specified number of revolutions without experiencing fatigue failure. In this case, the dynamic load capacity of the 6306 bearing is given as 22 kN.

2. Operating Condition:
The operating condition is defined by the rotational speed (600 rev/min) and the operating time (2000 hours).

3. Calculation of Maximum Radial Load:
To calculate the maximum radial load, we need to consider the bearing life equation. The bearing life equation relates the dynamic load capacity, the equivalent radial load, and the bearing life.

The equivalent radial load is given as:

P = XFr + YFa

Where:
P = Equivalent radial load
Fr = Radial load
Fa = Axial load
X and Y are factors depending on the bearing type and operating conditions.

For a radial load only application, the axial load is assumed to be zero (Fa = 0), and the equivalent radial load simplifies to:

P = Fr

The bearing life equation is given as:

L = (C/P)^p

Where:
L = Bearing life in revolutions
C = Dynamic load capacity
P = Equivalent radial load
p = Exponent for bearing life equation

For deep groove ball bearings like the 6306 bearing, the exponent p is typically taken as 3.

By rearranging the bearing life equation and solving for the equivalent radial load, we get:

P = (C/L)^(1/p)

Substituting the given values, we have:

P = (22 kN / (600 rev/min * 2000 hours))^1/3

Converting the rotational speed to revolutions per hour (600 rev/min * 60 min/hour), we get:

P = (22 kN / (600 rev/hour * 2000 hours))^1/3

Simplifying the equation gives:

P ≈ 5.29 kN

Therefore, the maximum radial load that the 6306 bearing can sustain for this operating condition is approximately 5.29 kN.
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Community Answer
The dynamic load capacity of 6306 bearing is 22 kN. The maximum radia...
Given:dynamic load capacity = 22 x 103 N
Speed(N) = 600 rav/min
Life = 2000 x 60 x 600
72 x 106 rev
Using following relation,
w = 5288.24 N
w = 5.288 kN
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The dynamic load capacity of 6306 bearing is 22 kN. The maximum radial load it can sustain to operate at 600 rev/min, for 2000 hours will be equal toa) 3.16 kNb) 4.16 kNc) 6.21 kNd) 5.29 kNCorrect answer is option 'D'. Can you explain this answer?
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