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A composite spring consists of two close-coiled helical springs connected in series. Each spring has 14 coils at a mean dimeter of 20 mm. The stiffness of the composite spring is 800 N/m. If the wire diameter of one spring is 2.5 mm, find the wire dimeter of the other spring in mm. G = 78 GPa
- a)1.8 mm
- b)2.1 mm
- c)1.3 mm
- d)2.4 mm
Correct answer is option 'A'. Can you explain this answer?
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A composite spring consists of two close-coiled helical springs connec...
Given D =20 mm, n1 = n2 = 14, Keq = 800 N/m d1 = 2.5 mm, d2 = d
Deflection in the spring:
Most Upvoted Answer
A composite spring consists of two close-coiled helical springs connec...
Given:
- Two close-coiled helical springs connected in series
- Each spring has 14 coils at a mean diameter of 20 mm
- Stiffness of composite spring = 800 N/m
- Wire diameter of one spring = 2.5 mm
- Shear modulus (G) = 78 GPa
To find: Wire diameter of the other spring in mm
Approach:
1. Calculate the spring constant (k) of one spring using the given parameters:
- Mean diameter (D) = 20 mm
- Number of coils (n) = 14
- Wire diameter (d) = 2.5 mm
- Shear modulus (G) = 78 GPa
The spring constant of one spring is given by:
k = (Gd⁴n)/(8D³)
k = (78 × 10⁹ × 2.5⁴ × 14)/(8 × 20³)
k = 920.63 N/m
2. Since the two springs are connected in series, the total spring constant (k_total) is given by:
k_total = k₁ + k₂
where k₁ is the spring constant of the spring with wire diameter 2.5 mm and k₂ is the spring constant of the other spring with wire diameter d₂ (to be found).
3. Substitute the values of k_total and k₁ in the above equation and solve for k₂:
800 = 920.63 + k₂
k₂ = 800 - 920.63
k₂ = -120.63 N/m
4. The negative value of k₂ indicates that the other spring is in compression, i.e., it is shorter in length than its free length. Therefore, the wire diameter of the other spring (d₂) must be smaller than 2.5 mm.
5. Calculate the wire diameter (d₂) of the other spring using the equation for spring constant:
k₂ = (Gd₂⁴n)/(8D³)
d₂⁴ = (8k₂D³)/(Gn)
d₂⁴ = (8 × -120.63 × 20³)/(78 × 10⁹ × 14)
d₂⁴ = -0.342
Since the wire diameter cannot be negative, we take the absolute value:
d₂⁴ = 0.342
d₂ = 1.8 mm
Therefore, the wire diameter of the other spring is 1.8 mm. Answer: option A.
- Two close-coiled helical springs connected in series
- Each spring has 14 coils at a mean diameter of 20 mm
- Stiffness of composite spring = 800 N/m
- Wire diameter of one spring = 2.5 mm
- Shear modulus (G) = 78 GPa
To find: Wire diameter of the other spring in mm
Approach:
1. Calculate the spring constant (k) of one spring using the given parameters:
- Mean diameter (D) = 20 mm
- Number of coils (n) = 14
- Wire diameter (d) = 2.5 mm
- Shear modulus (G) = 78 GPa
The spring constant of one spring is given by:
k = (Gd⁴n)/(8D³)
k = (78 × 10⁹ × 2.5⁴ × 14)/(8 × 20³)
k = 920.63 N/m
2. Since the two springs are connected in series, the total spring constant (k_total) is given by:
k_total = k₁ + k₂
where k₁ is the spring constant of the spring with wire diameter 2.5 mm and k₂ is the spring constant of the other spring with wire diameter d₂ (to be found).
3. Substitute the values of k_total and k₁ in the above equation and solve for k₂:
800 = 920.63 + k₂
k₂ = 800 - 920.63
k₂ = -120.63 N/m
4. The negative value of k₂ indicates that the other spring is in compression, i.e., it is shorter in length than its free length. Therefore, the wire diameter of the other spring (d₂) must be smaller than 2.5 mm.
5. Calculate the wire diameter (d₂) of the other spring using the equation for spring constant:
k₂ = (Gd₂⁴n)/(8D³)
d₂⁴ = (8k₂D³)/(Gn)
d₂⁴ = (8 × -120.63 × 20³)/(78 × 10⁹ × 14)
d₂⁴ = -0.342
Since the wire diameter cannot be negative, we take the absolute value:
d₂⁴ = 0.342
d₂ = 1.8 mm
Therefore, the wire diameter of the other spring is 1.8 mm. Answer: option A.
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A composite spring consists of two close-coiled helical springs connected in series. Each spring has 14 coils at a mean dimeter of 20 mm. The stiffness of the composite spring is 800 N/m. If the wire diameter of one spring is 2.5 mm, find the wire dimeter of the other spring in mm. G = 78 GPaa)1.8 mmb)2.1mmc)1.3mmd)2.4mmCorrect answer is option 'A'. Can you explain this answer?
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A composite spring consists of two close-coiled helical springs connected in series. Each spring has 14 coils at a mean dimeter of 20 mm. The stiffness of the composite spring is 800 N/m. If the wire diameter of one spring is 2.5 mm, find the wire dimeter of the other spring in mm. G = 78 GPaa)1.8 mmb)2.1mmc)1.3mmd)2.4mmCorrect 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 A composite spring consists of two close-coiled helical springs connected in series. Each spring has 14 coils at a mean dimeter of 20 mm. The stiffness of the composite spring is 800 N/m. If the wire diameter of one spring is 2.5 mm, find the wire dimeter of the other spring in mm. G = 78 GPaa)1.8 mmb)2.1mmc)1.3mmd)2.4mmCorrect 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 A composite spring consists of two close-coiled helical springs connected in series. Each spring has 14 coils at a mean dimeter of 20 mm. The stiffness of the composite spring is 800 N/m. If the wire diameter of one spring is 2.5 mm, find the wire dimeter of the other spring in mm. G = 78 GPaa)1.8 mmb)2.1mmc)1.3mmd)2.4mmCorrect answer is option 'A'. Can you explain this answer?.
A composite spring consists of two close-coiled helical springs connected in series. Each spring has 14 coils at a mean dimeter of 20 mm. The stiffness of the composite spring is 800 N/m. If the wire diameter of one spring is 2.5 mm, find the wire dimeter of the other spring in mm. G = 78 GPaa)1.8 mmb)2.1mmc)1.3mmd)2.4mmCorrect 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 A composite spring consists of two close-coiled helical springs connected in series. Each spring has 14 coils at a mean dimeter of 20 mm. The stiffness of the composite spring is 800 N/m. If the wire diameter of one spring is 2.5 mm, find the wire dimeter of the other spring in mm. G = 78 GPaa)1.8 mmb)2.1mmc)1.3mmd)2.4mmCorrect 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 A composite spring consists of two close-coiled helical springs connected in series. Each spring has 14 coils at a mean dimeter of 20 mm. The stiffness of the composite spring is 800 N/m. If the wire diameter of one spring is 2.5 mm, find the wire dimeter of the other spring in mm. G = 78 GPaa)1.8 mmb)2.1mmc)1.3mmd)2.4mmCorrect answer is option 'A'. Can you explain this answer?.
Solutions for A composite spring consists of two close-coiled helical springs connected in series. Each spring has 14 coils at a mean dimeter of 20 mm. The stiffness of the composite spring is 800 N/m. If the wire diameter of one spring is 2.5 mm, find the wire dimeter of the other spring in mm. G = 78 GPaa)1.8 mmb)2.1mmc)1.3mmd)2.4mmCorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Mechanical Engineering.
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ample number of questions to practice A composite spring consists of two close-coiled helical springs connected in series. Each spring has 14 coils at a mean dimeter of 20 mm. The stiffness of the composite spring is 800 N/m. If the wire diameter of one spring is 2.5 mm, find the wire dimeter of the other spring in mm. G = 78 GPaa)1.8 mmb)2.1mmc)1.3mmd)2.4mmCorrect answer is option 'A'. Can you explain this answer? tests, examples and also practice Mechanical Engineering tests.
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