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If both the mean coil diameter and wire diameter of a helical compression or tension spring be doubled, then the deflection of the spring close coiled under same applied load will
- a)be doubled
- b)be halved
- c)increase four times
- d)get reduced to one – fourth
Correct answer is option 'B'. Can you explain this answer?
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If both the mean coil diameter and wire diameter of a helical compress...
Explanation:
When the mean coil diameter and wire diameter of a helical compression or tension spring are both doubled, the deflection of the spring close coiled under the same applied load will be halved.
Reasoning:
To understand why the deflection is halved, let's consider the formula for the deflection of a helical compression or tension spring:
Deflection (δ) = (8 * F * L * n) / (π * G * d^4)
Where:
- F is the applied load
- L is the length of the spring
- n is the number of active coils
- G is the shear modulus of the spring material
- d is the wire diameter
Effect of doubling the mean coil diameter:
When the mean coil diameter is doubled, the length of the spring (L) also doubles. As a result, the deflection formula becomes:
Deflection (δ) = (8 * F * 2L * n) / (π * G * d^4)
Since the length of the spring (L) is multiplied by 2, the deflection (δ) is also multiplied by 2.
Effect of doubling the wire diameter:
When the wire diameter is doubled, the denominator of the deflection formula becomes 16 times larger. As a result, the deflection formula becomes:
Deflection (δ) = (8 * F * 2L * n) / (π * G * (2d)^4) = (8 * F * 2L * n) / (π * G * 16d^4)
Since the denominator is 16 times larger, the deflection (δ) is halved.
Conclusion:
Combining the effects of doubling the mean coil diameter and doubling the wire diameter, we find that the deflection of the spring is halved. Therefore, the correct answer is option 'B' - the deflection is halved.
When the mean coil diameter and wire diameter of a helical compression or tension spring are both doubled, the deflection of the spring close coiled under the same applied load will be halved.
Reasoning:
To understand why the deflection is halved, let's consider the formula for the deflection of a helical compression or tension spring:
Deflection (δ) = (8 * F * L * n) / (π * G * d^4)
Where:
- F is the applied load
- L is the length of the spring
- n is the number of active coils
- G is the shear modulus of the spring material
- d is the wire diameter
Effect of doubling the mean coil diameter:
When the mean coil diameter is doubled, the length of the spring (L) also doubles. As a result, the deflection formula becomes:
Deflection (δ) = (8 * F * 2L * n) / (π * G * d^4)
Since the length of the spring (L) is multiplied by 2, the deflection (δ) is also multiplied by 2.
Effect of doubling the wire diameter:
When the wire diameter is doubled, the denominator of the deflection formula becomes 16 times larger. As a result, the deflection formula becomes:
Deflection (δ) = (8 * F * 2L * n) / (π * G * (2d)^4) = (8 * F * 2L * n) / (π * G * 16d^4)
Since the denominator is 16 times larger, the deflection (δ) is halved.
Conclusion:
Combining the effects of doubling the mean coil diameter and doubling the wire diameter, we find that the deflection of the spring is halved. Therefore, the correct answer is option 'B' - the deflection is halved.
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If both the mean coil diameter and wire diameter of a helical compression or tension spring be doubled, then the deflection of the spring close coiled under same applied load willa)be doubledb)be halvedc)increase four timesd)get reduced to one fourthCorrect answer is option 'B'. Can you explain this answer?
Question Description
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