Compression Spring Rate Calculator: Rate, Stress, Solid Height and Buckling
Calculate a helical compression spring's rate k = Gd⁴/(8D³Na) from wire diameter, coil diameter, coils and end type for music wire, hard-drawn, oil-tempered, chrome-silicon, chrome-vanadium, stainless or phosphor bronze, then the deflection and corrected shear stress at your load, wire strength and safety factor, solid height, pitch, buckling, and the coils needed for a target rate. mm or inches, N or lbf.
How it works
k = G d⁴ ÷ (8 D³ Na) · D = OD − d = ID + d · C = D ÷ d · τ = K_B × 8 F D ÷ (π d³), K_B = (4C + 2) ÷ (4C − 3) · Sut = A ÷ dᵐ, Ssy = 0.45 Sut (music, hard-drawn), 0.50 Sut (oil-tempered, Cr-V, Cr-Si), 0.35 Sut (stainless, bronze) · active coils: plain Nt, plain & ground Nt − 1, squared (& ground) Nt − 2 · solid height: d(Nt + 1) plain or squared, d·Nt ground · buckling: stable if L0 < 2.63 D ÷ α
A helical compression spring is a torsion bar wound into a coil. When it is compressed, each active coil twists, so the rate depends on the wire's shear modulus G and the fourth power of its diameter, and falls with the cube of the mean coil diameter and with the number of active coils. End coils that are squared (closed) or ground sit on the seat and do not flex, which is why the active coils are fewer than the total: Nt − 2 for squared ends, Nt − 1 for plain ground ends and Nt for plain ends. The load creates a torque on the wire, and the shear stress is highest on the inside of each coil. The Bergsträsser factor (used in Shigley; within about 1% of the Wahl factor) corrects for that curvature and for direct shear. The allowable stress comes from the wire's tensile strength, which rises as wire gets thinner (Sut = A ÷ dᵐ with Shigley's constants for each ASTM wire), times the static torsional yield fraction for compression springs: 45% for music and hard-drawn wire, 50% for hardened and tempered alloy wire, 35% for stainless and non-ferrous wire. Solid height limits the travel, and a spring that can be closed solid without exceeding the allowable stress is "solid safe". A spring much longer than its diameter can buckle sideways: it is absolutely stable while the free length is below 2.63 D ÷ α, where α depends on how the ends are supported (0.5 between flat parallel plates). The moduli and strength constants are from Shigley Tables 10-4 to 10-6; real wire varies by supplier, so test critical springs.
Worked example
With the defaults: music wire d = 3 mm, OD 25 mm, 8 total coils, squared and ground ends, 60 mm free length, 100 N load. Mean diameter D = 25 − 3 = 22 mm, C = 7.333, Na = 8 − 2 = 6 and G = 81.0 GPa (music wire 1.61–3.18 mm, i.e. 0.064–0.125 in). k = 81,000 × 3⁴ ÷ (8 × 22³ × 6) = 6,561,000 ÷ 511,104 = 12.837 N/mm (73.30 lbf/in). At 100 N it deflects 7.79 mm; K_B = 1.191 so τ = 1.191 × 8 × 100 × 22 ÷ (π × 27) = 247 MPa. The wire strength is Sut = 2211 ÷ 3^0.145 = 1,885 MPa and Ssy = 0.45 × 1,885 = 848 MPa, a safety factor of 3.44. Solid height = 3 × 8 = 24 mm, so there are 36 mm of travel and 462.1 N at solid, where the stress is 1,141 MPa (safety factor 0.74): do not compress it solid. The free length is below the critical 115.7 mm, so it will not buckle between flat plates.
Sources: Shigley's Mechanical Engineering Design, ch. 10 solutions (problems 10-3, 10-6, 10-7: rates, Sut, Ssy, solid length, buckling) — Oakland University ME4300 · Omni Calculator: Spring rate calculator (k = G d⁴ ÷ (8 (OD − d)³ n), inactive end coils) · AmesWeb: Spring steels (ASTM A228, A227, A229, A231, A401 properties and uses)
Frequently asked questions
How do you calculate the rate of a compression spring?
k = G d⁴ ÷ (8 D³ Na), with G the shear modulus, d the wire diameter, D the mean coil diameter (OD − d) and Na the active coils. For example Shigley's music-wire spring with d = 2.5 mm, OD = 31 mm and 13 active coils has k = 81,000 × 2.5⁴ ÷ (8 × 28.5³ × 13) = 1.314 N/mm.
How many coils are active?
Plain ends: all of them (Na = Nt). Plain and ground: Nt − 1. Squared (closed), with or without grinding: Nt − 2. In practice end coils can be partly active, so measured rates are often within ±10% of the calculation.
What is a good spring index?
Between 4 and 12. Below 4 the spring is hard to make and the inner-fibre stress is high; above 12 it tangles, buckles easily and is hard to hold to tolerance.
What stress can a compression spring take?
For static loads with the set not removed, about 45% of the wire's tensile strength for music and hard-drawn wire, 50% for oil-tempered, chrome-vanadium and chrome-silicon, and 35% for stainless and phosphor bronze. Fatigue loading needs a much lower stress.
When will a compression spring buckle?
A steel spring is absolutely stable while its free length is below 2.63 D ÷ α: 5.26 times the mean diameter between flat parallel plates (α = 0.5), 2.63 D with both ends pivoted. Longer springs need a guide rod or bore.
Why is my measured rate different from the calculation?
Common causes are end coils that are partly active, measuring the outside instead of the mean diameter, wire diameter tolerance (rate varies with d⁴, so 2% thinner wire is 8% softer) and a modulus that differs from the handbook value.
Related calculators
Textbook spring-design formulas (Shigley's Mechanical Engineering Design, 10th ed., chapter 10) with idealised assumptions: round wire, uniform pitch, static load at room temperature. Wire strength varies by supplier and batch; use the manufacturer's data and testing, and a qualified engineer's check, for anything safety-related.