Extension spring calculator
Calculate the spring rate, force and shear stress of cylindrical helical extension springs to DIN EN 13906-2. Enter wire diameter, mean coil diameter and number of active coils, specify the initial tension F0 and either deflection or force – the calculator returns spring rate, characteristic line and the stress check against an allowable shear stress with a traffic-light rating, live with every input.
Extension spring calculator (DIN EN 13906-2)
Model: cylindrical helical extension spring made of round wire to DIN EN 13906-2 with initial tension F0, static shear stress check with curvature correction (Bergsträßer or Wahl). No fatigue check; the hooks (end loops) are not evaluated and must be verified separately because of additional bending and notch stresses. A sizing tool, not a substitute for a verified spring design.
Results
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Formulas and fundamentals
The extension spring follows the same basic formulas as the compression spring: the spring rate is R = G·d⁴/(8·Dm³·n) with the shear modulus G, wire diameter d, mean coil diameter Dm and number of active coils n. Deflection s and the force resulting from it are linearly related: F_net = R·s. The difference from the compression spring is the initial tension F0, introduced during coiling, which presses the coils together. The total force at the working point is therefore F = F0 + R·s: only once the external force exceeds F0 does the spring begin to open.
The shear stress in the wire follows from force and geometry as tau = 8·F·Dm/(pi·d³). Because of the wire curvature the stress on the inner side of the coil is higher than this nominal value; this is captured by the stress correction factor k, which depends on the spring index w = Dm/d. The calculator offers the Bergsträßer approximation k = (w+0.5)/(w-0.75) (standard default) and the Wahl approximation k = (4w-1)/(4w-4). The corrected stress tau_k = k·tau governs the strength check.
The static check compares the corrected shear stress tau_k with an allowable shear stress tau_allow. This depends on the material and wire diameter and is usually derived from the minimum tensile strength Rm of the spring wire (e.g. EN 10270). A recommended index range is 4 ≤ w ≤ 20 – outside it the calculator issues a warning. For fluctuating loads a fatigue check is required, and the hooks (end loops) must be verified separately because of additional bending and notch stresses.
Worked example
An extension spring made of spring steel (G = 81,500 N/mm²) has a wire diameter d = 2 mm, mean coil diameter Dm = 20 mm and n = 10 active coils. The spring index is w = Dm/d = 10, the spring rate R = 81,500·2⁴/(8·20³·10) = 2.0375 N/mm.
Without initial tension (F0 = 0) a deflection of s = 20 mm produces the force F = R·s = 40.75 N. The nominal shear stress is tau = 8·40.75·20/(pi·2³) = 259.4 N/mm². With the Bergsträßer correction k = 10.5/9.25 = 1.135 the governing stress is tau_k = 294.5 N/mm².
Against an allowable shear stress of 600 N/mm² the utilisation is about 49 percent – the static check is comfortably satisfied (green). A real extension spring would usually have an initial tension F0 > 0; it shifts the characteristic line upward in parallel so that the spring only opens once F > F0.
Frequently asked questions
What is the initial tension F0 of an extension spring?
When cold-coiling extension springs the coils can be pressed together with a preload. This initial tension F0 must first be overcome before the spring opens. The characteristic line therefore does not start at the origin but at F0: F = F0 + R·s. Compression springs have no counterpart to this because their coils cannot be preloaded together.
How does the calculation differ from the compression spring?
Spring rate and shear stress use the same formulas (R = G·d⁴/(8·Dm³·n) and tau = 8·F·Dm/(pi·d³)). The differences are the initial tension F0 in the force-deflection relation and the absence of solid length and buckling – an extension spring is pulled, not compressed. Instead the hooks (end loops) are the critical feature.
Bergsträßer or Wahl – which correction factor should I use?
Both describe the stress increase on the inner side of the coil and give very similar values for typical spring indices. The Bergsträßer approximation k = (w+0.5)/(w-0.75) is the standard default and preselected here. The Wahl approximation k = (4w-1)/(4w-4) is the Anglo-American alternative. The corrected stress tau_k = k·tau always governs the check.
How large may the allowable shear stress be?
It depends on the material and wire diameter and is derived from the minimum tensile strength Rm of the spring wire (e.g. patented drawn spring steel wire to EN 10270-1). For extension springs the allowable static shear stress is typically somewhat lower than for compression springs because of the additionally loaded hooks. Governing limit values are taken from the material data sheet or the standard.
Why does the spring index w = Dm/d matter?
A small w means a strongly curved wire with high stress concentration on the inner side and difficult manufacture; a very large w leads to unstable, sensitive springs. The proven range is 4 ≤ w ≤ 20, preferably about 7 to 10. Outside this range the calculator points it out with a note.
Is the hook check included?
No. The calculator evaluates the shear stress in the spring body. The hooks (e.g. German loop, hook) experience additional bending and notch stresses and are the most common failure location of extension springs. They must be verified separately. For fluctuating loads a fatigue check for body and hooks is also required.
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