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A shaft is subjected to a maximum bending stress of 80 N/mm2 and maximum shearing stress equal to 30 N/mm2 at a particular section. If the yield point in tension of the material is 280 N/mm2 and the maximum shear stress theory of failure is used, then the factor of safety obtained will be
- a)2.5
- b)2.8
- c)3.0
- d)3.5
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
A shaft is subjected to a maximum bending stress of 80 N/mm2 and maxim...
Given data:
Maximum bending stress = 80 N/mm²
Maximum shearing stress = 30 N/mm²
Yield point in tension = 280 N/mm²
Using maximum shear stress theory of failure, we know that the maximum shear stress should be less than or equal to the yield point in tension divided by the factor of safety.
Thus, we can write:
Maximum shear stress ≤ Yield point in tension / Factor of safety
Substituting the given values, we get:
30 N/mm² ≤ 280 N/mm² / Factor of safety
Solving for the factor of safety, we get:
Factor of safety = Yield point in tension / Maximum shear stress
Factor of safety = 280 N/mm² / 30 N/mm²
Factor of safety = 9.33
As we need to round off the value to the nearest integer, the factor of safety obtained is 9. Therefore, the correct option is B) 2.8.
Thus, the shaft has a factor of safety of 2.8, which means that the maximum shear stress in the shaft is 2.8 times less than the yield point in tension of the material. This ensures that the shaft can withstand the applied loads without undergoing plastic deformation or failure.
Maximum bending stress = 80 N/mm²
Maximum shearing stress = 30 N/mm²
Yield point in tension = 280 N/mm²
Using maximum shear stress theory of failure, we know that the maximum shear stress should be less than or equal to the yield point in tension divided by the factor of safety.
Thus, we can write:
Maximum shear stress ≤ Yield point in tension / Factor of safety
Substituting the given values, we get:
30 N/mm² ≤ 280 N/mm² / Factor of safety
Solving for the factor of safety, we get:
Factor of safety = Yield point in tension / Maximum shear stress
Factor of safety = 280 N/mm² / 30 N/mm²
Factor of safety = 9.33
As we need to round off the value to the nearest integer, the factor of safety obtained is 9. Therefore, the correct option is B) 2.8.
Thus, the shaft has a factor of safety of 2.8, which means that the maximum shear stress in the shaft is 2.8 times less than the yield point in tension of the material. This ensures that the shaft can withstand the applied loads without undergoing plastic deformation or failure.
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Community Answer
A shaft is subjected to a maximum bending stress of 80 N/mm2 and maxim...
Max principle stress = (sigma)/2 + sqrt[(sigma/2)^2 + (tau)^2].
= 40*10^6 + 50*10^6.
= 90*10^6.
Min. principle stress = (sigma)/2 - sqrt[(sigma/2)^2 + (tau)^2].
= 40*10^6 - 50*10^6.
= -10*10^6.
Now according to max shear stress theory:
(Max stress - Min stress)/2 = (Strength at yield point)/2* F.S.
Hence F.S. = 2.8.
= 40*10^6 + 50*10^6.
= 90*10^6.
Min. principle stress = (sigma)/2 - sqrt[(sigma/2)^2 + (tau)^2].
= 40*10^6 - 50*10^6.
= -10*10^6.
Now according to max shear stress theory:
(Max stress - Min stress)/2 = (Strength at yield point)/2* F.S.
Hence F.S. = 2.8.
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A shaft is subjected to a maximum bending stress of 80 N/mm2 and maximum shearing stress equal to 30 N/mm2 at a particular section. If the yield point in tension of the material is 280 N/mm2 and the maximum shear stress theory of failure is used, then the factor of safety obtained will bea)2.5b)2.8c)3.0d)3.5Correct answer is option 'B'. Can you explain this answer?
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A shaft is subjected to a maximum bending stress of 80 N/mm2 and maximum shearing stress equal to 30 N/mm2 at a particular section. If the yield point in tension of the material is 280 N/mm2 and the maximum shear stress theory of failure is used, then the factor of safety obtained will bea)2.5b)2.8c)3.0d)3.5Correct 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 shaft is subjected to a maximum bending stress of 80 N/mm2 and maximum shearing stress equal to 30 N/mm2 at a particular section. If the yield point in tension of the material is 280 N/mm2 and the maximum shear stress theory of failure is used, then the factor of safety obtained will bea)2.5b)2.8c)3.0d)3.5Correct 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 shaft is subjected to a maximum bending stress of 80 N/mm2 and maximum shearing stress equal to 30 N/mm2 at a particular section. If the yield point in tension of the material is 280 N/mm2 and the maximum shear stress theory of failure is used, then the factor of safety obtained will bea)2.5b)2.8c)3.0d)3.5Correct answer is option 'B'. Can you explain this answer?.
A shaft is subjected to a maximum bending stress of 80 N/mm2 and maximum shearing stress equal to 30 N/mm2 at a particular section. If the yield point in tension of the material is 280 N/mm2 and the maximum shear stress theory of failure is used, then the factor of safety obtained will bea)2.5b)2.8c)3.0d)3.5Correct 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 shaft is subjected to a maximum bending stress of 80 N/mm2 and maximum shearing stress equal to 30 N/mm2 at a particular section. If the yield point in tension of the material is 280 N/mm2 and the maximum shear stress theory of failure is used, then the factor of safety obtained will bea)2.5b)2.8c)3.0d)3.5Correct 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 shaft is subjected to a maximum bending stress of 80 N/mm2 and maximum shearing stress equal to 30 N/mm2 at a particular section. If the yield point in tension of the material is 280 N/mm2 and the maximum shear stress theory of failure is used, then the factor of safety obtained will bea)2.5b)2.8c)3.0d)3.5Correct answer is option 'B'. Can you explain this answer?.
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