Mechanical Engineering Exam > Mechanical Engineering Questions > If knuckle joint is to fail by crushing failu...
Start Learning for Free
If knuckle joint is to fail by crushing failure of pin in fork, then determine the diameter of knuckle pin when 50kN axial tensile force act on rods. Given: Max allowable compressive stress=25N/mm², thickness of each eye of fork=25mm.
- a)40mm
- b)50mm
- c)60mm
- d)70mm
Correct answer is option 'A'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
If knuckle joint is to fail by crushing failure of pin in fork, then d...
Explanation: d=P/2aσ = 40mm.
Most Upvoted Answer
If knuckle joint is to fail by crushing failure of pin in fork, then d...
To determine the diameter of the knuckle pin, we need to calculate the maximum allowable compressive stress on the pin.
Given:
Maximum allowable compressive stress = 25 N/mm²
Axial tensile force acting on rods = 50 kN
Since the knuckle joint fails by the crushing failure of the pin in the fork, the maximum compressive stress on the pin will be equal to the tensile stress acting on the rods.
Tensile stress = Force/Area
Converting the axial tensile force to N:
50 kN = 50,000 N
Let's assume the diameter of the knuckle pin as d.
The area of the knuckle pin can be calculated using the formula:
Area = π(d²/4)
Equating the tensile stress and compressive stress:
Force/Area = Maximum allowable compressive stress
50,000 N / (π(d²/4)) = 25 N/mm²
Simplifying the equation:
(50,000 * 4) / (π * d²) = 25
200,000 / (π * d²) = 25
Dividing both sides of the equation by 25:
8,000 / (π * d²) = 1
Rearranging the equation:
d² = 8,000 / (π * 1)
d² = 2,546.48
Taking the square root of both sides:
d ≈ √2,546.48
d ≈ 50.46 mm
Therefore, the diameter of the knuckle pin, when a 50 kN axial tensile force acts on the rods, is approximately 50.46 mm.
Given:
Maximum allowable compressive stress = 25 N/mm²
Axial tensile force acting on rods = 50 kN
Since the knuckle joint fails by the crushing failure of the pin in the fork, the maximum compressive stress on the pin will be equal to the tensile stress acting on the rods.
Tensile stress = Force/Area
Converting the axial tensile force to N:
50 kN = 50,000 N
Let's assume the diameter of the knuckle pin as d.
The area of the knuckle pin can be calculated using the formula:
Area = π(d²/4)
Equating the tensile stress and compressive stress:
Force/Area = Maximum allowable compressive stress
50,000 N / (π(d²/4)) = 25 N/mm²
Simplifying the equation:
(50,000 * 4) / (π * d²) = 25
200,000 / (π * d²) = 25
Dividing both sides of the equation by 25:
8,000 / (π * d²) = 1
Rearranging the equation:
d² = 8,000 / (π * 1)
d² = 2,546.48
Taking the square root of both sides:
d ≈ √2,546.48
d ≈ 50.46 mm
Therefore, the diameter of the knuckle pin, when a 50 kN axial tensile force acts on the rods, is approximately 50.46 mm.
Attention Mechanical Engineering Students!
To make sure you are not studying endlessly, EduRev has designed Mechanical Engineering study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Mechanical Engineering.
|
Explore Courses for Mechanical Engineering exam
|
|
Similar Mechanical Engineering Doubts
Top Courses for Mechanical EngineeringView all
If knuckle joint is to fail by crushing failure of pin in fork, then determine the diameter of knuckle pin when 50kN axial tensile force act on rods. Given: Max allowable compressive stress=25N/mm², thickness of each eye of fork=25mm.a)40mmb)50mmc)60mmd)70mmCorrect answer is option 'A'. Can you explain this answer?
Question Description
If knuckle joint is to fail by crushing failure of pin in fork, then determine the diameter of knuckle pin when 50kN axial tensile force act on rods. Given: Max allowable compressive stress=25N/mm², thickness of each eye of fork=25mm.a)40mmb)50mmc)60mmd)70mmCorrect answer is option 'A'. 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 If knuckle joint is to fail by crushing failure of pin in fork, then determine the diameter of knuckle pin when 50kN axial tensile force act on rods. Given: Max allowable compressive stress=25N/mm², thickness of each eye of fork=25mm.a)40mmb)50mmc)60mmd)70mmCorrect answer is option 'A'. 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 If knuckle joint is to fail by crushing failure of pin in fork, then determine the diameter of knuckle pin when 50kN axial tensile force act on rods. Given: Max allowable compressive stress=25N/mm², thickness of each eye of fork=25mm.a)40mmb)50mmc)60mmd)70mmCorrect answer is option 'A'. Can you explain this answer?.
If knuckle joint is to fail by crushing failure of pin in fork, then determine the diameter of knuckle pin when 50kN axial tensile force act on rods. Given: Max allowable compressive stress=25N/mm², thickness of each eye of fork=25mm.a)40mmb)50mmc)60mmd)70mmCorrect answer is option 'A'. 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 If knuckle joint is to fail by crushing failure of pin in fork, then determine the diameter of knuckle pin when 50kN axial tensile force act on rods. Given: Max allowable compressive stress=25N/mm², thickness of each eye of fork=25mm.a)40mmb)50mmc)60mmd)70mmCorrect answer is option 'A'. 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 If knuckle joint is to fail by crushing failure of pin in fork, then determine the diameter of knuckle pin when 50kN axial tensile force act on rods. Given: Max allowable compressive stress=25N/mm², thickness of each eye of fork=25mm.a)40mmb)50mmc)60mmd)70mmCorrect answer is option 'A'. Can you explain this answer?.
Solutions for If knuckle joint is to fail by crushing failure of pin in fork, then determine the diameter of knuckle pin when 50kN axial tensile force act on rods. Given: Max allowable compressive stress=25N/mm², thickness of each eye of fork=25mm.a)40mmb)50mmc)60mmd)70mmCorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Mechanical Engineering.
Download more important topics, notes, lectures and mock test series for Mechanical Engineering Exam by signing up for free.
Here you can find the meaning of If knuckle joint is to fail by crushing failure of pin in fork, then determine the diameter of knuckle pin when 50kN axial tensile force act on rods. Given: Max allowable compressive stress=25N/mm², thickness of each eye of fork=25mm.a)40mmb)50mmc)60mmd)70mmCorrect answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
If knuckle joint is to fail by crushing failure of pin in fork, then determine the diameter of knuckle pin when 50kN axial tensile force act on rods. Given: Max allowable compressive stress=25N/mm², thickness of each eye of fork=25mm.a)40mmb)50mmc)60mmd)70mmCorrect answer is option 'A'. Can you explain this answer?, a detailed solution for If knuckle joint is to fail by crushing failure of pin in fork, then determine the diameter of knuckle pin when 50kN axial tensile force act on rods. Given: Max allowable compressive stress=25N/mm², thickness of each eye of fork=25mm.a)40mmb)50mmc)60mmd)70mmCorrect answer is option 'A'. Can you explain this answer? has been provided alongside types of If knuckle joint is to fail by crushing failure of pin in fork, then determine the diameter of knuckle pin when 50kN axial tensile force act on rods. Given: Max allowable compressive stress=25N/mm², thickness of each eye of fork=25mm.a)40mmb)50mmc)60mmd)70mmCorrect answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice If knuckle joint is to fail by crushing failure of pin in fork, then determine the diameter of knuckle pin when 50kN axial tensile force act on rods. Given: Max allowable compressive stress=25N/mm², thickness of each eye of fork=25mm.a)40mmb)50mmc)60mmd)70mmCorrect answer is option 'A'. Can you explain this answer? tests, examples and also practice Mechanical Engineering tests.
|
|
Explore Courses for Mechanical Engineering exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup