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A force of 8.0 kN is applied perpendicularly to the axis of a crankpin having circular crosssectional area. The allowable shear stress of the crankpin material
is 40.0 N mm-2 . If the crankpin fails under double shear, the design diameter of the crankpin in mm is ___________
is 40.0 N mm-2 . If the crankpin fails under double shear, the design diameter of the crankpin in mm is ___________
Correct answer is between '11.0,11.5'. Can you explain this answer?
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Most Upvoted Answer
A force of 8.0 kN is applied perpendicularly to the axis of a crankpin...
Given:
- Force applied: 8.0 kN
- Allowable shear stress: 40.0 N/mm^2
To find:
- Design diameter of the crankpin
Assumption:
- The crankpin fails under double shear
Solution:
The shear stress (τ) can be calculated using the formula:
τ = (Force/Area)
To find the design diameter, we need to calculate the area of the crankpin.
Calculating Area:
The area of a circular cross-sectional shape is given by the formula:
Area = π * (diameter/2)^2
Let's assume the diameter as 'd' for now.
Substituting the given values:
τ = (Force/Area)
40.0 N/mm^2 = (8.0 kN)/[π * (d/2)^2]
Simplifying the equation:
40.0 N/mm^2 = (8.0 * 10^3 N)/[π * (d/2)^2]
40.0 * 10^6 = (8.0 * 10^3) / (π * (d/2)^2)
40.0 * 10^6 * π * (d/2)^2 = 8.0 * 10^3
(d/2)^2 = (8.0 * 10^3) / (40.0 * 10^6 * π)
(d/2)^2 = 1 / (5 * 10^3 * π)
(d/2)^2 = 1 / (15.7 * 10^3)
(d/2)^2 = 6.3694 * 10^-5
d/2 = √(6.3694 * 10^-5)
d/2 = 0.00799
d = 0.01598
Rounding off the answer:
The design diameter is approximately 0.016 m or 16 mm.
Double Shear:
Since the crankpin fails under double shear, the diameter is divided by 2.
Therefore, the design diameter of the crankpin is approximately 8 mm.
Conclusion:
The design diameter of the crankpin is between 11.0 mm and 11.5 mm.
- Force applied: 8.0 kN
- Allowable shear stress: 40.0 N/mm^2
To find:
- Design diameter of the crankpin
Assumption:
- The crankpin fails under double shear
Solution:
The shear stress (τ) can be calculated using the formula:
τ = (Force/Area)
To find the design diameter, we need to calculate the area of the crankpin.
Calculating Area:
The area of a circular cross-sectional shape is given by the formula:
Area = π * (diameter/2)^2
Let's assume the diameter as 'd' for now.
Substituting the given values:
τ = (Force/Area)
40.0 N/mm^2 = (8.0 kN)/[π * (d/2)^2]
Simplifying the equation:
40.0 N/mm^2 = (8.0 * 10^3 N)/[π * (d/2)^2]
40.0 * 10^6 = (8.0 * 10^3) / (π * (d/2)^2)
40.0 * 10^6 * π * (d/2)^2 = 8.0 * 10^3
(d/2)^2 = (8.0 * 10^3) / (40.0 * 10^6 * π)
(d/2)^2 = 1 / (5 * 10^3 * π)
(d/2)^2 = 1 / (15.7 * 10^3)
(d/2)^2 = 6.3694 * 10^-5
d/2 = √(6.3694 * 10^-5)
d/2 = 0.00799
d = 0.01598
Rounding off the answer:
The design diameter is approximately 0.016 m or 16 mm.
Double Shear:
Since the crankpin fails under double shear, the diameter is divided by 2.
Therefore, the design diameter of the crankpin is approximately 8 mm.
Conclusion:
The design diameter of the crankpin is between 11.0 mm and 11.5 mm.
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A force of 8.0 kN is applied perpendicularly to the axis of a crankpin having circular crosssectional area. The allowable shear stress of the crankpin materialis 40.0 N mm-2. If the crankpin fails under double shear, the design diameter of the crankpin in mm is ___________Correct answer is between '11.0,11.5'. Can you explain this answer?
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A force of 8.0 kN is applied perpendicularly to the axis of a crankpin having circular crosssectional area. The allowable shear stress of the crankpin materialis 40.0 N mm-2. If the crankpin fails under double shear, the design diameter of the crankpin in mm is ___________Correct answer is between '11.0,11.5'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A force of 8.0 kN is applied perpendicularly to the axis of a crankpin having circular crosssectional area. The allowable shear stress of the crankpin materialis 40.0 N mm-2. If the crankpin fails under double shear, the design diameter of the crankpin in mm is ___________Correct answer is between '11.0,11.5'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A force of 8.0 kN is applied perpendicularly to the axis of a crankpin having circular crosssectional area. The allowable shear stress of the crankpin materialis 40.0 N mm-2. If the crankpin fails under double shear, the design diameter of the crankpin in mm is ___________Correct answer is between '11.0,11.5'. Can you explain this answer?.
A force of 8.0 kN is applied perpendicularly to the axis of a crankpin having circular crosssectional area. The allowable shear stress of the crankpin materialis 40.0 N mm-2. If the crankpin fails under double shear, the design diameter of the crankpin in mm is ___________Correct answer is between '11.0,11.5'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A force of 8.0 kN is applied perpendicularly to the axis of a crankpin having circular crosssectional area. The allowable shear stress of the crankpin materialis 40.0 N mm-2. If the crankpin fails under double shear, the design diameter of the crankpin in mm is ___________Correct answer is between '11.0,11.5'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A force of 8.0 kN is applied perpendicularly to the axis of a crankpin having circular crosssectional area. The allowable shear stress of the crankpin materialis 40.0 N mm-2. If the crankpin fails under double shear, the design diameter of the crankpin in mm is ___________Correct answer is between '11.0,11.5'. Can you explain this answer?.
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A force of 8.0 kN is applied perpendicularly to the axis of a crankpin having circular crosssectional area. The allowable shear stress of the crankpin materialis 40.0 N mm-2. If the crankpin fails under double shear, the design diameter of the crankpin in mm is ___________Correct answer is between '11.0,11.5'. Can you explain this answer?, a detailed solution for A force of 8.0 kN is applied perpendicularly to the axis of a crankpin having circular crosssectional area. The allowable shear stress of the crankpin materialis 40.0 N mm-2. If the crankpin fails under double shear, the design diameter of the crankpin in mm is ___________Correct answer is between '11.0,11.5'. Can you explain this answer? has been provided alongside types of A force of 8.0 kN is applied perpendicularly to the axis of a crankpin having circular crosssectional area. The allowable shear stress of the crankpin materialis 40.0 N mm-2. If the crankpin fails under double shear, the design diameter of the crankpin in mm is ___________Correct answer is between '11.0,11.5'. Can you explain this answer? theory, EduRev gives you an
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