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A closed coil helical spring has 15 coils. If five coils of this spring are removed by cutting, the stiffness of the modified spring will
- a)increase to 2.5 times
- b)increase to 1.5 times
- c)reduce to 0.66 times
- d)remain unaffected
Correct answer is option 'B'. Can you explain this answer?
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A closed coil helical spring has 15 coils. If five coils of this spri...
Problem Analysis:
We are given a closed coil helical spring with 15 coils. We need to determine the change in stiffness of the spring when five coils are removed by cutting.
Solution:
The stiffness of a spring is given by the formula:
Stiffness = (n * G^4 * d^4) / (8 * D^3 * N)
Where:
- n = Number of active coils
- G = Modulus of rigidity
- d = Diameter of the wire
- D = Mean diameter of the coil
- N = Number of total coils
Step 1:
Given data:
- Number of total coils (N) = 15
- Number of active coils (n) = 15 - 5 = 10 (after removing 5 coils)
- All other parameters such as G, d, and D are assumed to remain constant.
Step 2:
We need to calculate the stiffness of the modified spring using the formula mentioned above.
Step 3:
Substituting the given values into the formula, we get:
Stiffness of the modified spring = (10 * G^4 * d^4) / (8 * D^3 * 15)
Step 4:
To compare the stiffness of the modified spring with the original spring, we can express the ratio as:
Stiffness ratio = (Stiffness of the modified spring) / (Stiffness of the original spring)
Step 5:
Since all other parameters such as G, d, and D are assumed to remain constant, we can cancel them in the ratio calculation.
Step 6:
Stiffness ratio = (10 * G^4 * d^4) / (8 * D^3 * 15) / (15 * G^4 * d^4) / (8 * D^3 * 15)
Step 7:
Simplifying the ratio, we get:
Stiffness ratio = 10 / 15 = 2 / 3
Step 8:
The stiffness ratio is equal to 2 / 3, which can also be expressed as 1.5 times.
Conclusion:
Therefore, the stiffness of the modified spring will increase to 1.5 times when five coils are removed by cutting.
We are given a closed coil helical spring with 15 coils. We need to determine the change in stiffness of the spring when five coils are removed by cutting.
Solution:
The stiffness of a spring is given by the formula:
Stiffness = (n * G^4 * d^4) / (8 * D^3 * N)
Where:
- n = Number of active coils
- G = Modulus of rigidity
- d = Diameter of the wire
- D = Mean diameter of the coil
- N = Number of total coils
Step 1:
Given data:
- Number of total coils (N) = 15
- Number of active coils (n) = 15 - 5 = 10 (after removing 5 coils)
- All other parameters such as G, d, and D are assumed to remain constant.
Step 2:
We need to calculate the stiffness of the modified spring using the formula mentioned above.
Step 3:
Substituting the given values into the formula, we get:
Stiffness of the modified spring = (10 * G^4 * d^4) / (8 * D^3 * 15)
Step 4:
To compare the stiffness of the modified spring with the original spring, we can express the ratio as:
Stiffness ratio = (Stiffness of the modified spring) / (Stiffness of the original spring)
Step 5:
Since all other parameters such as G, d, and D are assumed to remain constant, we can cancel them in the ratio calculation.
Step 6:
Stiffness ratio = (10 * G^4 * d^4) / (8 * D^3 * 15) / (15 * G^4 * d^4) / (8 * D^3 * 15)
Step 7:
Simplifying the ratio, we get:
Stiffness ratio = 10 / 15 = 2 / 3
Step 8:
The stiffness ratio is equal to 2 / 3, which can also be expressed as 1.5 times.
Conclusion:
Therefore, the stiffness of the modified spring will increase to 1.5 times when five coils are removed by cutting.
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A closed coil helical spring has 15 coils. If five coils of this spring are removed by cutting, the stiffness of the modified spring willa) increase to 2.5 timesb) increase to 1.5 timesc) reduce to 0.66 timesd) remain unaffectedCorrect answer is option 'B'. Can you explain this answer?
Question Description
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A closed coil helical spring has 15 coils. If five coils of this spring are removed by cutting, the stiffness of the modified spring willa) increase to 2.5 timesb) increase to 1.5 timesc) reduce to 0.66 timesd) remain unaffectedCorrect 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 closed coil helical spring has 15 coils. If five coils of this spring are removed by cutting, the stiffness of the modified spring willa) increase to 2.5 timesb) increase to 1.5 timesc) reduce to 0.66 timesd) remain unaffectedCorrect 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 closed coil helical spring has 15 coils. If five coils of this spring are removed by cutting, the stiffness of the modified spring willa) increase to 2.5 timesb) increase to 1.5 timesc) reduce to 0.66 timesd) remain unaffectedCorrect answer is option 'B'. Can you explain this answer?.
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A closed coil helical spring has 15 coils. If five coils of this spring are removed by cutting, the stiffness of the modified spring willa) increase to 2.5 timesb) increase to 1.5 timesc) reduce to 0.66 timesd) remain unaffectedCorrect answer is option 'B'. Can you explain this answer?, a detailed solution for A closed coil helical spring has 15 coils. If five coils of this spring are removed by cutting, the stiffness of the modified spring willa) increase to 2.5 timesb) increase to 1.5 timesc) reduce to 0.66 timesd) remain unaffectedCorrect answer is option 'B'. Can you explain this answer? has been provided alongside types of A closed coil helical spring has 15 coils. If five coils of this spring are removed by cutting, the stiffness of the modified spring willa) increase to 2.5 timesb) increase to 1.5 timesc) reduce to 0.66 timesd) remain unaffectedCorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
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