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A 25 mm radius and 300 mm long shaft is supported at the ends by two journal bearings. A load of 2 kN acts on the shaft at a distance of 200 mm from the left bearing. The length to diameter ratio for the bearings is 1.5. The maximum bearings pressure induced is
- a)0.53 MPa
- b)0.35 MPa
- c)5.3 MPa
- d)3.5 MPa
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
A 25 mm radius and 300 mm long shaft is supported at the ends by two ...
From equilibrium conditions,
RA + RB = 2 kN
MA = 0 
RB x 300 = 2 x 200 
l = 1.5d = 1.5 x 50
l = 75 mm
= 0.35 MPa
Most Upvoted Answer
A 25 mm radius and 300 mm long shaft is supported at the ends by two ...
From equilibrium conditions,
RA + RB = 2 kN
MA = 0 
RB x 300 = 2 x 200 
l = 1.5d = 1.5 x 50
l = 75 mm
= 0.35 MPa
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Community Answer
A 25 mm radius and 300 mm long shaft is supported at the ends by two ...
The maximum bearing pressure induced in the journal bearings can be calculated using the equation:
Pmax = (F * L) / (π * d * L / D)
where:
Pmax is the maximum bearing pressure
F is the load acting on the shaft
L is the length of the shaft
d is the diameter of the shaft
L/D is the length to diameter ratio of the bearings
Given:
F = 2 kN = 2000 N
L = 300 mm = 0.3 m
d = 25 mm = 0.025 m
L/D = 1.5
Substituting the given values into the equation, we get:
Pmax = (2000 * 0.3) / (π * 0.025 * 0.3 / 1.5)
Simplifying the equation further:
Pmax = (600) / (π * 0.025 * 0.3 / 1.5)
Pmax = 12000 / (π * 0.025 * 0.3 / 1.5)
Pmax = 12000 / (π * 0.0075)
Pmax ≈ 541,095.9 Pa
Converting the pressure to MPa:
Pmax ≈ 0.541 MPa
Therefore, the maximum bearing pressure induced in the journal bearings is approximately 0.541 MPa. Hence, option B is the correct answer.
Pmax = (F * L) / (π * d * L / D)
where:
Pmax is the maximum bearing pressure
F is the load acting on the shaft
L is the length of the shaft
d is the diameter of the shaft
L/D is the length to diameter ratio of the bearings
Given:
F = 2 kN = 2000 N
L = 300 mm = 0.3 m
d = 25 mm = 0.025 m
L/D = 1.5
Substituting the given values into the equation, we get:
Pmax = (2000 * 0.3) / (π * 0.025 * 0.3 / 1.5)
Simplifying the equation further:
Pmax = (600) / (π * 0.025 * 0.3 / 1.5)
Pmax = 12000 / (π * 0.025 * 0.3 / 1.5)
Pmax = 12000 / (π * 0.0075)
Pmax ≈ 541,095.9 Pa
Converting the pressure to MPa:
Pmax ≈ 0.541 MPa
Therefore, the maximum bearing pressure induced in the journal bearings is approximately 0.541 MPa. Hence, option B is the correct answer.
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A 25 mm radius and 300 mm long shaft is supported at the ends by two journal bearings. A load of 2 kN acts on the shaft at a distance of 200 mm from the left bearing. The length to diameter ratio for the bearings is 1.5. The maximum bearings pressure induced isa) 0.53 MPab) 0.35 MPac) 5.3 MPad) 3.5 MPaCorrect answer is option 'B'. Can you explain this answer?
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A 25 mm radius and 300 mm long shaft is supported at the ends by two journal bearings. A load of 2 kN acts on the shaft at a distance of 200 mm from the left bearing. The length to diameter ratio for the bearings is 1.5. The maximum bearings pressure induced isa) 0.53 MPab) 0.35 MPac) 5.3 MPad) 3.5 MPaCorrect answer is option 'B'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A 25 mm radius and 300 mm long shaft is supported at the ends by two journal bearings. A load of 2 kN acts on the shaft at a distance of 200 mm from the left bearing. The length to diameter ratio for the bearings is 1.5. The maximum bearings pressure induced isa) 0.53 MPab) 0.35 MPac) 5.3 MPad) 3.5 MPaCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 25 mm radius and 300 mm long shaft is supported at the ends by two journal bearings. A load of 2 kN acts on the shaft at a distance of 200 mm from the left bearing. The length to diameter ratio for the bearings is 1.5. The maximum bearings pressure induced isa) 0.53 MPab) 0.35 MPac) 5.3 MPad) 3.5 MPaCorrect answer is option 'B'. Can you explain this answer?.
A 25 mm radius and 300 mm long shaft is supported at the ends by two journal bearings. A load of 2 kN acts on the shaft at a distance of 200 mm from the left bearing. The length to diameter ratio for the bearings is 1.5. The maximum bearings pressure induced isa) 0.53 MPab) 0.35 MPac) 5.3 MPad) 3.5 MPaCorrect answer is option 'B'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A 25 mm radius and 300 mm long shaft is supported at the ends by two journal bearings. A load of 2 kN acts on the shaft at a distance of 200 mm from the left bearing. The length to diameter ratio for the bearings is 1.5. The maximum bearings pressure induced isa) 0.53 MPab) 0.35 MPac) 5.3 MPad) 3.5 MPaCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 25 mm radius and 300 mm long shaft is supported at the ends by two journal bearings. A load of 2 kN acts on the shaft at a distance of 200 mm from the left bearing. The length to diameter ratio for the bearings is 1.5. The maximum bearings pressure induced isa) 0.53 MPab) 0.35 MPac) 5.3 MPad) 3.5 MPaCorrect answer is option 'B'. Can you explain this answer?.
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