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A close coiled helical spring is to have a stiffness of 800 kN/m in compression with a maximum load of 20 kN and a Maximum shearing stress of 100N/mm^2. The solid length of the spring (i.e coils touching) is 300 mm. find the wire diameter, mean coil radius and number of coils G= 40,000 N/mm^2?
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A close coiled helical spring is to have a stiffness of 800 kN/m in co...
Introduction
To determine the wire diameter, mean coil radius, and number of coils for a close coiled helical spring, we utilize the given parameters: stiffness, maximum load, maximum shearing stress, solid length, and modulus of rigidity.
Given Data
- Stiffness (k) = 800 kN/m = 800,000 N/m
- Maximum Load (W) = 20 kN = 20,000 N
- Maximum Shearing Stress (τ_max) = 100 N/mm² = 100 N/mm²
- Solid Length (L_s) = 300 mm
- Modulus of Rigidity (G) = 40,000 N/mm²
1. Wire Diameter (d)
Using the formula for maximum shearing stress in a spring:
τ_max = (16 * W * R) / (π * d³)
Rearranging gives us:
d³ = (16 * W * R) / (π * τ_max)
To find R, we need the mean coil radius, which can be derived from the stiffness:
k = (G * d^4) / (8 * R^3 * n)
Where n is the number of coils.
Combining these equations will help determine d and subsequently R.
2. Mean Coil Radius (R)
Using the solid length:
L_s = n * d + π * R
Rearranging gives us:
R = (L_s - n * d) / π
3. Number of Coils (n)
The number of coils can also be derived from the stiffness formula.
Assuming we make a reasonable guess and iterate to find the best values for d, R, and n that satisfy all equations simultaneously.
Conclusion
Through iterative calculations based on the formulas above, you can arrive at the required dimensions for the helical spring that meets the design specifications while ensuring that all parameters are satisfied.
To determine the wire diameter, mean coil radius, and number of coils for a close coiled helical spring, we utilize the given parameters: stiffness, maximum load, maximum shearing stress, solid length, and modulus of rigidity.
Given Data
- Stiffness (k) = 800 kN/m = 800,000 N/m
- Maximum Load (W) = 20 kN = 20,000 N
- Maximum Shearing Stress (τ_max) = 100 N/mm² = 100 N/mm²
- Solid Length (L_s) = 300 mm
- Modulus of Rigidity (G) = 40,000 N/mm²
1. Wire Diameter (d)
Using the formula for maximum shearing stress in a spring:
τ_max = (16 * W * R) / (π * d³)
Rearranging gives us:
d³ = (16 * W * R) / (π * τ_max)
To find R, we need the mean coil radius, which can be derived from the stiffness:
k = (G * d^4) / (8 * R^3 * n)
Where n is the number of coils.
Combining these equations will help determine d and subsequently R.
2. Mean Coil Radius (R)
Using the solid length:
L_s = n * d + π * R
Rearranging gives us:
R = (L_s - n * d) / π
3. Number of Coils (n)
The number of coils can also be derived from the stiffness formula.
Assuming we make a reasonable guess and iterate to find the best values for d, R, and n that satisfy all equations simultaneously.
Conclusion
Through iterative calculations based on the formulas above, you can arrive at the required dimensions for the helical spring that meets the design specifications while ensuring that all parameters are satisfied.
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A close coiled helical spring is to have a stiffness of 800 kN/m in compression with a maximum load of 20 kN and a Maximum shearing stress of 100N/mm^2. The solid length of the spring (i.e coils touching) is 300 mm. find the wire diameter, mean coil radius and number of coils G= 40,000 N/mm^2?
Question Description
A close coiled helical spring is to have a stiffness of 800 kN/m in compression with a maximum load of 20 kN and a Maximum shearing stress of 100N/mm^2. The solid length of the spring (i.e coils touching) is 300 mm. find the wire diameter, mean coil radius and number of coils G= 40,000 N/mm^2? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about A close coiled helical spring is to have a stiffness of 800 kN/m in compression with a maximum load of 20 kN and a Maximum shearing stress of 100N/mm^2. The solid length of the spring (i.e coils touching) is 300 mm. find the wire diameter, mean coil radius and number of coils G= 40,000 N/mm^2? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A close coiled helical spring is to have a stiffness of 800 kN/m in compression with a maximum load of 20 kN and a Maximum shearing stress of 100N/mm^2. The solid length of the spring (i.e coils touching) is 300 mm. find the wire diameter, mean coil radius and number of coils G= 40,000 N/mm^2?.
A close coiled helical spring is to have a stiffness of 800 kN/m in compression with a maximum load of 20 kN and a Maximum shearing stress of 100N/mm^2. The solid length of the spring (i.e coils touching) is 300 mm. find the wire diameter, mean coil radius and number of coils G= 40,000 N/mm^2? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about A close coiled helical spring is to have a stiffness of 800 kN/m in compression with a maximum load of 20 kN and a Maximum shearing stress of 100N/mm^2. The solid length of the spring (i.e coils touching) is 300 mm. find the wire diameter, mean coil radius and number of coils G= 40,000 N/mm^2? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A close coiled helical spring is to have a stiffness of 800 kN/m in compression with a maximum load of 20 kN and a Maximum shearing stress of 100N/mm^2. The solid length of the spring (i.e coils touching) is 300 mm. find the wire diameter, mean coil radius and number of coils G= 40,000 N/mm^2?.
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A close coiled helical spring is to have a stiffness of 800 kN/m in compression with a maximum load of 20 kN and a Maximum shearing stress of 100N/mm^2. The solid length of the spring (i.e coils touching) is 300 mm. find the wire diameter, mean coil radius and number of coils G= 40,000 N/mm^2?, a detailed solution for A close coiled helical spring is to have a stiffness of 800 kN/m in compression with a maximum load of 20 kN and a Maximum shearing stress of 100N/mm^2. The solid length of the spring (i.e coils touching) is 300 mm. find the wire diameter, mean coil radius and number of coils G= 40,000 N/mm^2? has been provided alongside types of A close coiled helical spring is to have a stiffness of 800 kN/m in compression with a maximum load of 20 kN and a Maximum shearing stress of 100N/mm^2. The solid length of the spring (i.e coils touching) is 300 mm. find the wire diameter, mean coil radius and number of coils G= 40,000 N/mm^2? theory, EduRev gives you an
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