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If the wire diameter of a compressive helical spring is increased by 2% the change in
spring stiffness (in %) is __________ (correct to two decimal places)
spring stiffness (in %) is __________ (correct to two decimal places)
Correct answer is '(8.242)'. Can you explain this answer?
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If the wire diameter of a compressive helical spring is increased by 2...
Stiffness of helical spring
Where, d=Spring Wire diameter
D= Mean Coil diameter
n= Number of coils
G= Shear Modulus
d increased by 2% i.e. d’=1.02d
Where, d=Spring Wire diameter
D= Mean Coil diameter
n= Number of coils
G= Shear Modulus
d increased by 2% i.e. d’=1.02d
Most Upvoted Answer
If the wire diameter of a compressive helical spring is increased by 2...
Calculation of Change in Spring Stiffness Due to Increase in Wire Diameter:
Given data:
- Increase in wire diameter = 2%
- Let the original wire diameter be denoted by 'd'
- Let the original spring stiffness be denoted by 'k'
Formula:
- The spring stiffness (k) of a helical spring is given by the formula: k = (G*d^4)/(8*D^3*N)
- Where G = Shear modulus of elasticity of the spring material
- D = Mean diameter of the spring coil
- N = Number of active coils in the spring
Steps:
1. As per the given data, the wire diameter is increased by 2%. Therefore, the new wire diameter (d') can be calculated as:
d' = d + (2/100)*d
d' = d*(1 + 0.02)
d' = 1.02*d
2. Now, the new spring stiffness (k') can be calculated using the same formula as before, but with the new wire diameter value:
k' = (G*d'^4)/(8*D^3*N)
k' = (G*(1.02*d)^4)/(8*D^3*N)
k' = (1.02^4*G*d^4)/(8*D^3*N)
k' = 1.0824*k
3. Finally, the percentage change in spring stiffness can be calculated using the formula:
% change in k = ((k' - k)/k)*100
% change in k = ((1.0824*k - k)/k)*100
% change in k = (0.0824*k)*100
% change in k = 8.24%
Therefore, the change in spring stiffness due to the increase in wire diameter by 2% is 8.24% (correct to two decimal places).
Given data:
- Increase in wire diameter = 2%
- Let the original wire diameter be denoted by 'd'
- Let the original spring stiffness be denoted by 'k'
Formula:
- The spring stiffness (k) of a helical spring is given by the formula: k = (G*d^4)/(8*D^3*N)
- Where G = Shear modulus of elasticity of the spring material
- D = Mean diameter of the spring coil
- N = Number of active coils in the spring
Steps:
1. As per the given data, the wire diameter is increased by 2%. Therefore, the new wire diameter (d') can be calculated as:
d' = d + (2/100)*d
d' = d*(1 + 0.02)
d' = 1.02*d
2. Now, the new spring stiffness (k') can be calculated using the same formula as before, but with the new wire diameter value:
k' = (G*d'^4)/(8*D^3*N)
k' = (G*(1.02*d)^4)/(8*D^3*N)
k' = (1.02^4*G*d^4)/(8*D^3*N)
k' = 1.0824*k
3. Finally, the percentage change in spring stiffness can be calculated using the formula:
% change in k = ((k' - k)/k)*100
% change in k = ((1.0824*k - k)/k)*100
% change in k = (0.0824*k)*100
% change in k = 8.24%
Therefore, the change in spring stiffness due to the increase in wire diameter by 2% is 8.24% (correct to two decimal places).
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If the wire diameter of a compressive helical spring is increased by 2% the change inspring stiffness (in %) is __________ (correct to two decimal places)Correct answer is '(8.242)'. Can you explain this answer?
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If the wire diameter of a compressive helical spring is increased by 2% the change inspring stiffness (in %) is __________ (correct to two decimal places)Correct answer is '(8.242)'. 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 the wire diameter of a compressive helical spring is increased by 2% the change inspring stiffness (in %) is __________ (correct to two decimal places)Correct answer is '(8.242)'. 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 the wire diameter of a compressive helical spring is increased by 2% the change inspring stiffness (in %) is __________ (correct to two decimal places)Correct answer is '(8.242)'. Can you explain this answer?.
If the wire diameter of a compressive helical spring is increased by 2% the change inspring stiffness (in %) is __________ (correct to two decimal places)Correct answer is '(8.242)'. 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 the wire diameter of a compressive helical spring is increased by 2% the change inspring stiffness (in %) is __________ (correct to two decimal places)Correct answer is '(8.242)'. 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 the wire diameter of a compressive helical spring is increased by 2% the change inspring stiffness (in %) is __________ (correct to two decimal places)Correct answer is '(8.242)'. Can you explain this answer?.
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