About Helical Compression Spring Calculator
The helical compression spring calculator sizes a round-wire helical spring and predicts how stiff it is and how highly stressed it becomes under load. From the wire diameter d, mean coil diameter D, and number of active coils Na it computes the spring index C = D/d, the spring rate k = G·d⁴/(8·D³·Na), and the deflection δ = F/k for a given axial force F.
It also reports the corrected torsional shear stress τ using the Wahl factor Kw, which accounts for the extra stress on the inside of the coil caused by wire curvature and direct shear. Pick an end-coil treatment (plain, plain-ground, squared, or squared-ground) to set the total coils and solid length, and the tool warns when the spring index falls outside the recommended 4–12 range.
How It Works
- Enter the wire diameter d, mean coil diameter D, number of active coils Na, applied force F, and shear modulus G (default 79,300 MPa for spring steel).
- The calculator finds the spring index C = D/d and the Wahl correction factor Kw = (4C−1)/(4C−4) + 0.615/C, which is largest for tightly wound springs.
- It computes the spring rate k = G·d⁴/(8·D³·Na), the deflection δ = F/k, and the corrected shear stress τ = Kw·8·F·D/(π·d³).
- Choose an end-coil treatment to set total coils Nt and solid length Ls = Nt·d; a warning appears if the spring index is below 4 or above 12.
Worked Example
A spring steel compression spring has wire diameter d = 10 mm, mean coil diameter D = 60 mm, Na = 8 active coils, and shear modulus G = 79,300 MPa, loaded by F = 500 N with squared-and-ground ends. The spring index is C = D/d = 60/10 = 6. The Wahl factor Kw = (4·6−1)/(4·6−4) + 0.615/6 = 23/20 + 0.1025 = 1.15 + 0.1025 = 1.2525. The spring rate k = G·d⁴/(8·D³·Na) = 79,300·10⁴/(8·60³·8) = 793,000,000/13,824,000 = 57.4 N/mm. The deflection δ = F/k = 500/57.4 = 8.72 mm. The corrected shear stress τ = Kw·8·F·D/(π·d³) = 1.2525·8·500·60/(π·1000) = 300,600/3141.6 = 95.7 MPa. Squared-and-ground ends add two inactive coils, so total coils Nt = 8 + 2 = 10 and the solid length Ls = Nt·d = 10·10 = 100 mm. With C = 6 the spring sits comfortably inside the recommended 4–12 index band, so no warning is raised.
Formulas
- Spring index
C = D / d- Wahl curvature correction factor
Kw = (4*C - 1) / (4*C - 4) + 0.615 / C- Spring rate
k = G * d^4 / (8 * D^3 * Na)- Corrected torsional shear stress
tau = Kw * 8 * F * D / (pi * d^3)- Deflection
delta = F / k- Solid length
Ls = Nt * d (Nt = Na + end coils)
Standards & References
- Shigley's Mechanical Engineering Design (helical spring design)
- Wahl curvature correction factor
- Associated Spring / SAE spring design references
Frequently Asked Questions
What is the spring index and what is a good range?
The spring index C = D/d is the ratio of mean coil diameter to wire diameter, and it captures how tightly the spring is wound. A practical range is about 4 to 12. Below 4 the wire is sharply curved, stress concentration is high, and the spring is hard to coil; above 12 the spring is slender and tends to tangle, buckle, or need close-tolerance handling. Indices of roughly 6 to 9 are a common sweet spot.
What does the Wahl factor correct for?
A simple torsion analysis of the wire underestimates the real stress because the inside of each coil is more sharply curved than the outside, and there is an added direct (transverse) shear. The Wahl factor Kw = (4C−1)/(4C−4) + 0.615/C bundles both the curvature effect and the direct-shear effect into one multiplier on the nominal torsional stress, raising the peak stress at the inner fibre. It is largest for low spring indices and approaches 1 as C grows.
Why does the spring rate use the shear modulus G instead of Young’s modulus E?
A helical compression spring deflects mainly because the wire twists in torsion as the coil is loaded, not because it stretches in tension or bends. Torsional behaviour is governed by the shear modulus of rigidity G, so the spring rate k = G·d⁴/(8·D³·Na) uses G (about 79,300 MPa for spring steel) rather than the tensile modulus E. Using E would badly overestimate the stiffness.
How is the solid length related to the end coils?
Solid length is the length when every coil is fully compressed and touching, Ls = Nt·d, where Nt is the total number of coils. The end treatment sets how many inactive coils are added to the active coils Na: plain ends add none, plain-ground add one, and squared or squared-ground ends add two. Squared-and-ground ends give the flattest, most square seating and are common in precision compression springs.