Question

The deflection of a spring with 20 active turns for the specified load is 10 mm. If the same spring is cut into two halves and connected parallel and applied the same load, then the deflection would be

• A. \( 10 \text{ mm} \)
• B. \( 20 \text{ mm} \)
• C. \( 5 \text{ mm} \)
• D. \( 2.5 \text{ mm} \)

Hint:

Remember that spring stiffness is inversely proportional to the number of active turns (\( k \propto 1/n \)). When a spring is cut into \( N \) equal parts, each part has \( N \) times the stiffness of the original spring. When springs are connected in parallel, their equivalent stiffness is the sum of individual stiffnesses.

Answer 1

The Correct Option is D

Step 1: Understand the deflection of a helical spring.
The deflection \( \delta \) of a helical spring under an axial load \( W \) is given by the formula:
\[ \delta = \frac{8WD^3n}{Gd^4} \] where:
\( W \) = applied load
\( D \) = mean coil diameter
\( n \) = number of active turns
\( G \) = modulus of rigidity of the spring material
\( d \) = wire diameter
From this formula, we can see that deflection \( \delta \) is directly proportional to the number of active turns \( n \). So, \( \delta \propto n \).
Step 2: Determine the spring constant (stiffness) of the original spring.
Let the original spring have \( n_1 = 20 \) active turns and its deflection be \( \delta_1 = 10 \text{ mm} \) for a load \( W \).
The spring constant \( k \) is defined as \( k = \frac{W}{\delta} \).
So, for the original spring, \( k_1 = \frac{W}{10} \).
From the deflection formula, \( k = \frac{Gd^4}{8D^3n} \). So, \( k \propto \frac{1}{n} \).
Step 3: Analyze the effect of cutting the spring into two halves.
If the original spring with \( n_1 = 20 \) turns is cut into two halves, each half will have:
\( n_2 = \frac{n_1}{2} = \frac{20}{2} = 10 \) active turns. The spring constant of each half-spring (\( k_2 \)) will be: Since \( k \propto \frac{1}{n} \), if \( n \) is halved, \( k \) will be doubled.
So, \( k_2 = 2k_1 \).
Substituting \( k_1 = \frac{W}{10} \), we get \( k_2 = 2 \times \frac{W}{10} = \frac{W}{5} \).
Step 4: Analyze the effect of connecting the two halves in parallel.
When two springs are connected in parallel, their equivalent stiffness \( k_{eq} \) is the sum of their individual stiffnesses. Since both halves are identical, \( k_{eq} = k_2 + k_2 = 2k_2 \). \[ k_{eq} = 2 \times \left(\frac{W}{5}\right) = \frac{2W}{5} \]
Step 5: Calculate the new deflection under the same load.
The same load \( W \) is applied to the parallel combination.
The deflection \( \delta_{eq} \) of the parallel combination is:
\[ \delta_{eq} = \frac{W}{k_{eq}} \] \[ \delta_{eq} = \frac{W}{\frac{2W}{5}} = \frac{5W}{2W} = \frac{5}{2} = 2.5 \text{ mm} \] The final answer is $\boxed{\text{4}}$.