Helical Springs — Effect of Cutting into Two Halves A closely coiled helical spring is cut into two equal halves. How does the stiffness (spring constant) of each resulting spring compare to the original?

Introduction:
Spring stiffness determines load–deflection behavior. Understanding how stiffness changes when altering the number of active coils helps when adapting springs to new applications by cutting or combining coils.

Given Data / Assumptions:

• Closely coiled helical spring with wire diameter d, mean coil diameter D, and N active turns.
• Material shear modulus G unchanged by cutting.
• Ends of the new springs are finished so that the number of active turns halves approximately.

Concept / Approach:
The stiffness k of a closely coiled helical spring is k = (G * d^4) / (8 * D^3 * N). With all other parameters unchanged, k is inversely proportional to the number of active turns N. Cutting the spring into two halves halves N and thus doubles k.

Step-by-Step Solution:
Original stiffness: k_0 = (G * d^4) / (8 * D^3 * N).After cutting: N_new = N / 2.New stiffness: k_new = (G * d^4) / (8 * D^3 * (N / 2)) = 2 * k_0.

Verification / Alternative check:
Series/parallel analogy: two identical springs in series halve stiffness; conversely, halving coil count in a single spring doubles stiffness.

Why Other Options Are Wrong:
same: would require N unchanged, which is not the case.half: contradicts inverse proportionality to N.one-fourth: would occur only if N increased fourfold, not halved.

Common Pitfalls:
Forgetting that wire and mean diameters remain the same after cutting, so only N changes.

Final Answer:
double