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A cylindrical shaft is subjected to an alternating stress of 100 MPa. Fatigue strength to sustain 1000 cycles is 490 MPa. If the corrected endurance strength is 70 MPa, estimated shaft life will be
- a)1071 cycles
- b)15000 cycles
- c)281914 cycles
- d)928643 cycles
Correct answer is option 'C'. Can you explain this answer?
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A cylindrical shaft is subjected to an alternating stress of 100 MPa. ...
It is a finite life problem. The line AB is the failure line. Where A {3, log10 ( 0.9σult )} but here it will be A {3, log10 ( 490 )} and B{6, log10 ( σe )} here it is B{6, log10 ( 70)} Therefore F{ log10 N,log10(100)},we have to find N
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A cylindrical shaft is subjected to an alternating stress of 100 MPa. ...
Given data:
Alternating stress, σa = 100 MPa
Fatigue strength for 1000 cycles, σf' = 490 MPa
Corrected endurance strength, σe' = 70 MPa
To estimate the shaft life, we can use the Modified Goodman Diagram approach as follows:
Step 1: Calculate the mean stress, σm
σm = 0 (for fully reversing stress)
Step 2: Calculate the equivalent stress, σeq
σeq = (σa/2) + (σm/2) = σa/2
Step 3: Calculate the factor of safety, Nf
Nf = (σf' / σeq) + (σe' / σa)
Step 4: Calculate the estimated cycles to failure, N
N = 10^(a + b log(Nf))
where a = 5.413 and b = -0.097 (for rotating bending fatigue)
Substituting the given values, we get:
Nf = (490/50) + (70/100) = 10.8
N = 10^(5.413 - 0.097 log(10.8)) = 281914 cycles (approx)
Therefore, the estimated shaft life is 281914 cycles, which is closest to option (C).
Alternating stress, σa = 100 MPa
Fatigue strength for 1000 cycles, σf' = 490 MPa
Corrected endurance strength, σe' = 70 MPa
To estimate the shaft life, we can use the Modified Goodman Diagram approach as follows:
Step 1: Calculate the mean stress, σm
σm = 0 (for fully reversing stress)
Step 2: Calculate the equivalent stress, σeq
σeq = (σa/2) + (σm/2) = σa/2
Step 3: Calculate the factor of safety, Nf
Nf = (σf' / σeq) + (σe' / σa)
Step 4: Calculate the estimated cycles to failure, N
N = 10^(a + b log(Nf))
where a = 5.413 and b = -0.097 (for rotating bending fatigue)
Substituting the given values, we get:
Nf = (490/50) + (70/100) = 10.8
N = 10^(5.413 - 0.097 log(10.8)) = 281914 cycles (approx)
Therefore, the estimated shaft life is 281914 cycles, which is closest to option (C).
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A cylindrical shaft is subjected to an alternating stress of 100 MPa. Fatigue strength to sustain 1000 cycles is 490 MPa. If the corrected endurance strength is 70 MPa, estimated shaft life will bea)1071 cyclesb)15000 cyclesc)281914 cyclesd)928643 cyclesCorrect answer is option 'C'. Can you explain this answer?
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A cylindrical shaft is subjected to an alternating stress of 100 MPa. Fatigue strength to sustain 1000 cycles is 490 MPa. If the corrected endurance strength is 70 MPa, estimated shaft life will bea)1071 cyclesb)15000 cyclesc)281914 cyclesd)928643 cyclesCorrect answer is option 'C'. 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 cylindrical shaft is subjected to an alternating stress of 100 MPa. Fatigue strength to sustain 1000 cycles is 490 MPa. If the corrected endurance strength is 70 MPa, estimated shaft life will bea)1071 cyclesb)15000 cyclesc)281914 cyclesd)928643 cyclesCorrect answer is option 'C'. 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 cylindrical shaft is subjected to an alternating stress of 100 MPa. Fatigue strength to sustain 1000 cycles is 490 MPa. If the corrected endurance strength is 70 MPa, estimated shaft life will bea)1071 cyclesb)15000 cyclesc)281914 cyclesd)928643 cyclesCorrect answer is option 'C'. Can you explain this answer?.
A cylindrical shaft is subjected to an alternating stress of 100 MPa. Fatigue strength to sustain 1000 cycles is 490 MPa. If the corrected endurance strength is 70 MPa, estimated shaft life will bea)1071 cyclesb)15000 cyclesc)281914 cyclesd)928643 cyclesCorrect answer is option 'C'. 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 cylindrical shaft is subjected to an alternating stress of 100 MPa. Fatigue strength to sustain 1000 cycles is 490 MPa. If the corrected endurance strength is 70 MPa, estimated shaft life will bea)1071 cyclesb)15000 cyclesc)281914 cyclesd)928643 cyclesCorrect answer is option 'C'. 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 cylindrical shaft is subjected to an alternating stress of 100 MPa. Fatigue strength to sustain 1000 cycles is 490 MPa. If the corrected endurance strength is 70 MPa, estimated shaft life will bea)1071 cyclesb)15000 cyclesc)281914 cyclesd)928643 cyclesCorrect answer is option 'C'. Can you explain this answer?.
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A cylindrical shaft is subjected to an alternating stress of 100 MPa. Fatigue strength to sustain 1000 cycles is 490 MPa. If the corrected endurance strength is 70 MPa, estimated shaft life will bea)1071 cyclesb)15000 cyclesc)281914 cyclesd)928643 cyclesCorrect answer is option 'C'. Can you explain this answer?, a detailed solution for A cylindrical shaft is subjected to an alternating stress of 100 MPa. Fatigue strength to sustain 1000 cycles is 490 MPa. If the corrected endurance strength is 70 MPa, estimated shaft life will bea)1071 cyclesb)15000 cyclesc)281914 cyclesd)928643 cyclesCorrect answer is option 'C'. Can you explain this answer? has been provided alongside types of A cylindrical shaft is subjected to an alternating stress of 100 MPa. Fatigue strength to sustain 1000 cycles is 490 MPa. If the corrected endurance strength is 70 MPa, estimated shaft life will bea)1071 cyclesb)15000 cyclesc)281914 cyclesd)928643 cyclesCorrect answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
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