Helical springs

Why it matters

Helical springs store energy and apply controlled force in valves, clutches, brakes, relays, push-buttons, grippers, suspension and return mechanisms of actuators. In mechatronic devices a spring often sets a preload, a return force or a contact force that a motor or solenoid must overcome, so its stiffness and stress must be right the first time. Spring wire is highly stressed and spring failures are usually fatigue failures.

Key ideas

Geometry. Wire diameter d, mean coil diameter D (outer diameter minus d), spring index C = D/d (practical range about 4 to 12; small C is hard to coil, large C buckles and tangles), number of active coils N, pitch, free length and solid length.

End types and coil counts. Plain, plain and ground, squared (closed), squared and ground. For squared and ground ends - the most common for compression springs - the total coils are Nt = N + 2 and solid length ≈ Nt·d. Other end types change the inactive coils (see your data book).

Stress in the wire. An axial load P on a close-coiled helical spring twists the wire with torque P·D/2, so the wire is mainly in torsion: τ = 8PD/(πd³). Two corrections raise this:

• Direct shear P/(πd²/4) adds on the inner side of the coil - shear stress factor Ks = 1 + 0.5/C.
• Curvature effect - the wire is curved, so the inner fibre is shorter and more highly stressed. The Wahl factor K = (4C − 1)/(4C − 4) + 0.615/C includes both and is used for design: τ = K·8PD/(πd³). The maximum stress is on the inside of the coil.

Deflection and stiffness. Strain energy of torsion gives δ = 8PD³N/(Gd⁴), so the spring rate is k = Gd⁴/(8D³N). Stiffness rises steeply with d (fourth power) and falls with D³ and N. G for spring steel is about 80 GPa (take from your data book).

Combinations. Springs in series (same force, deflections add): 1/k = 1/k₁ + 1/k₂. In parallel or concentric (same deflection, forces add): k = k₁ + k₂. Concentric (nested) springs are wound in opposite hands so the coils do not lock.

Energy stored U = ½·P·δ = ½·k·δ².

Other checks.

• Buckling of compression springs when free length is large relative to D - guide them on a rod or in a tube.
• Surge (resonance): the natural frequency of the spring must be well above the operating frequency (e.g. valve springs, often a factor of about 15-20 is recommended). For a spring between two flat plates, fn = (d/(2π·N·D²))·√(G/(2ρ)).
• Fatigue: springs under fluctuating load are designed with a mean and alternating shear stress (using Ks on the mean and K on the alternating component in many texts) and a fatigue diagram for spring wire.
• Clash allowance: at maximum working load, leave a gap of about 10-15% of the working deflection before solid length.

Formulas

C = D / d

• C: spring index; D: mean coil diameter (m); d: wire diameter (m).

K = (4C − 1)/(4C − 4) + 0.615/C · Ks = 1 + 0.5/C

• K: Wahl stress factor; Ks: direct-shear factor.

τ = K·8·P·D / (π·d³) = K·8·P·C / (π·d²)

• τ: maximum shear stress in the wire (Pa); P: axial load (N).

δ = 8·P·D³·N / (G·d⁴) · k = P/δ = G·d⁴ / (8·D³·N)

• δ: axial deflection (m); N: number of active coils; G: shear modulus (Pa); k: spring rate (N/m).

Series: 1/k = 1/k₁ + 1/k₂ · Parallel: k = k₁ + k₂

U = ½·P·δ

• U: strain energy (J).

Nt = N + 2 (squared and ground) · L_solid ≈ Nt·d

• Nt: total coils; L_solid: solid length (m).

fn = (d / (2π·N·D²))·√(G / (2ρ)) (both ends on flat plates)

• fn: first natural (surge) frequency (Hz); ρ: wire density (kg/m³).

Worked examples

Example 1 (standard). A compression spring has d = 5 mm, D = 40 mm, N = 10 active coils, G = 80 GPa, and carries P = 500 N. Find the maximum shear stress, deflection and stiffness.

1. Spring index: C = D/d = 40/5 = 8.
2. Wahl factor: K = (4 × 8 − 1)/(4 × 8 − 4) + 0.615/8 = 31/28 + 0.0769 = 1.184.
3. Stress: τ = K·8PD/(πd³) = 1.184 × 8 × 500 × 40 / (π × 5³) = 482 MPa.
4. Deflection: δ = 8PD³N/(Gd⁴) = 8 × 500 × 40³ × 10 / (80,000 × 5⁴) = 51.2 mm (with G = 80,000 N/mm²).
5. Stiffness: k = 500/51.2 = 9.77 N/mm.

Example 2 (GATE level). Design a squared-and-ground compression spring to carry 1 kN with a deflection of 25 mm. C = 6, permissible shear stress 400 MPa, G = 80 GPa. Find d, D, N and the solid length.

1. Wahl factor for C = 6: K = 23/20 + 0.615/6 = 1.150 + 0.1025 = 1.2525.
2. From τ = K·8PC/(πd²): d² = 1.2525 × 8 × 1000 × 6 / (π × 400) = 47.84 mm² → d = 6.92 mm → d = 7 mm, D = 6 × 7 = 42 mm.
3. Check stress: τ = 1.2525 × 8 × 1000 × 42 / (π × 7³) = 390.5 MPa < 400 MPa. OK.
4. Active coils: from δ = 8PC³N/(Gd), N = G·d·δ/(8·P·C³) = 80,000 × 7 × 25 / (8 × 1000 × 216) = 8.10 → N = 8 (δ = 24.7 mm, k = 40.5 N/mm; take 8.5 if the full 25 mm is essential).
5. Total coils Nt = 8 + 2 = 10; solid length ≈ 10 × 7 = 70 mm; add the 25 mm working deflection and a clash allowance to get the free length.

Common mistakes

• Using the outer diameter instead of the mean diameter D.
• Forgetting the Wahl factor (underestimates stress by 15-25% at common indices) - or applying it to the deflection formula, where it does not belong.
• Counting total coils instead of active coils in the deflection formula.
• Adding stiffnesses of series springs (that is the parallel rule).
• Mixing G in GPa with dimensions in mm: use G = 80,000 N/mm² with mm, or Pa with m.
• Ignoring buckling and surge for long or fast-cycling springs.

For GATE ME

Expect stiffness and deflection of a helical spring, the effect of changing d, D or N on stiffness (fourth and third powers), springs in series and parallel, the maximum shear stress with a given Wahl factor, and energy stored. Practise ratio questions such as "if the wire diameter is doubled and the coil diameter is unchanged, how does stiffness change?" (16 times).

Quick check

1. C = 5. Find the Wahl factor.
2. Two springs of 10 N/mm and 15 N/mm in series. Equivalent stiffness?
3. Wire diameter halved, everything else the same. Stiffness ratio?
4. Squared-and-ground spring with 12 active coils. Total coils? Answers: 1. K = 19/16 + 0.123 = 1.311. 2. 6 N/mm. 3. 1/16 of the original. 4. 14.