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Torsion springs calculation

A torsion spring is an elastic mechanical element that stores energy when twisted around its axis, exerting a torque proportional to the angle of rotation. The maximum torque a solid circular shaft of 50 mm diameter in steel with an allowable shear stress of 40.8 MPa can withstand is approximately 1000 N·m / 737.6 lb·ft. The angular deflection of that same shaft subjected to that torque and with a length of 1 m results in 1.2° (0.021 rad), considering a shear modulus of 79 GPa.

Characteristic Description
Physical principle Energy storage by elastic deformation under torsion (shafts) or bending (helical springs).
Constitutive law τ = -κ θ (torque proportional to angle, negative sign indicates opposition).
Stored energy U = ½ κ θ² (for linear behavior).
Typical applications Torsion bars in vehicle suspension, clothespin springs, mousetraps, garage doors.

Torsion springs are classified according to their geometry and the predominant type of stress:

  • Torsion bar (solid or hollow shaft): straight metal or elastomer bar subjected to pure torsion. The stress is shear, maximum at the periphery.
  • Helical torsion spring: wire or strip wound into a helix that works mainly in bending when wound or unwound. Used in clothespins, hinges, and return mechanisms.
  • Spiral spring (clock spring): variant of the helical with concentric flat coils, capable of storing energy for multiple revolutions.
  • Torsion fiber: thin thread (silk, quartz, glass) used in precision instruments (torsion pendulums, galvanometers).

The behavior of a solid or hollow circular cross-section torsion bar, within the elastic limit, is governed by the following expressions (according to data from engineeringtoolbox):

Variable Formula Units
Maximum shear stress on surface τ_max = T·R / J Pa; T in N·m, R in m, J in m⁴
Polar moment of inertia (solid) J = π D⁴ / 32 m⁴; D outer diameter
Polar moment of inertia (hollow) J = π (D⁴ - d⁴) / 32 m⁴; d inner diameter
Maximum torque (solid) T_max = (π/16)·τ_max · D³ N·m
Maximum torque (hollow) T_max = (π/16)·τ_max · (D⁴ - d⁴) / D N·m
Angular deflection (radians) α = L·T / (J·G) rad; L length, G shear modulus
Angular deflection solid (°) α_deg ≈ 584·L·T / (G·D⁴) °; L in m, T in N·m, G in Pa, D in m
Minimum solid diameter D_min = 1,72·(T_max / τ_max)^(1/3) m

For helical torsion springs (according to general spring design principles):

Parameter Expression Units
Spring constant (torque/angle) κ = E·d⁴ / (64·D·N) N·mm/rad; E elastic modulus, d wire ∅, D mean ∅, N active coils
Bending stress in the farthest fiber σ_f = 32·M / (π·d³) Pa; M applied moment
Wahl factor (for curvature correction) K_w = (4C-1)/(4C-4); C = D/d dimensionless

The energy stored in any linear torsion spring is given by U = ½ κ θ², where κ is the torsional stiffness and θ is the torsion angle in radians.

The dimensions of commercial torsion springs are usually standardized according to DIN standards or manufacturer catalogs. Below are typical ranges for helical torsion springs of round wire in AISI 302 stainless steel, based on common offerings:

Parameter Metric range Imperial range
Wire diameter (d) 0.3 – 6 mm 0.012 – 0.236 in
Body outer diameter (De) 3 – 60 mm 0.118 – 2.362 in
Body free length (L0) 5 – 300 mm 0.197 – 11.811 in
Number of coils (N) 2 – 30
Free angle between legs 90° / 120° / 180°
Winding direction Right or left

Solid torsion bars used in vehicle suspension have diameters between 15 mm / 0.59 in and 35 mm / 1.38 in, with effective lengths ranging from 600 mm / 23.6 in to 1500 mm / 59.1 in.

The most common materials for torsion springs and their relevant mechanical properties are listed below.

Material Elastic modulus (E) Shear modulus (G) Tensile strength (Rm) Maximum service temperature
Carbon steel for springs (EN 10270-1) 206 GPa / 29 900 ksi 81.5 GPa / 11 820 ksi 1200 – 2200 MPa / 174 – 319 ksi (depending on ∅) 120 °C / 248 °F
Stainless steel AISI 302/304 193 GPa / 28 000 ksi 70 GPa / 10 150 ksi 1400 – 1800 MPa / 203 – 261 ksi 250 °C / 482 °F
Chrome-silicon steel (ASTM A401) 206 GPa / 29 900 ksi 80 GPa / 11 600 ksi 1400 – 2000 MPa / 203 – 290 ksi 250 °C / 482 °F
Piano wire (ASTM A228) 207 GPa / 30 000 ksi 83 GPa / 12 040 ksi 1600 – 2800 MPa / 232 – 406 ksi 120 °C / 248 °F
Elastomer (natural rubber, hardness 60 ShA) 2 – 10 MPa / 0.29 – 1.45 ksi 0.7 – 3 MPa / 0.10 – 0.43 ksi 15 – 25 MPa / 2.2 – 3.6 ksi 70 °C / 158 °F

Load capacity expresses the maximum allowable torque without exceeding the design stress or causing permanent deformation. For a solid steel torsion bar with τ_adm = 300 MPa / 43.5 ksi, the indicative values are:

Diameter (D) Maximum torque (T_max)
10 mm / 0.394 in 58.9 N·m / 521 lbf·in
15 mm / 0.591 in 198.2 N·m / 1754 lbf·in
20 mm / 0.787 in 471.2 N·m / 4170 lbf·in
25 mm / 0.984 in 920.4 N·m / 8145 lbf·in
30 mm / 1.181 in 1588 N·m / 14 060 lbf·in
35 mm / 1.378 in 2520 N·m / 22 300 lbf·in

For helical springs, the capacity is defined by the torque at maximum deflection (according to working angle). A spring with d = 2 mm / 0.079 in, D = 12 mm / 0.472 in, N = 6 coils and E = 206 GPa, deflected by 90° (1.57 rad), develops a torque of approximately 17.2 N·m / 152 lbf·in, with a corrected bending stress of 1100 MPa.

To choose a suitable torsion spring, the following criteria must be evaluated:

  • Working torque (M) and deflection angle (θ): determine the required torsional stiffness (κ = M / θ).
  • Spring type: torsion bar for large torques with limited radial space; helical spring for moderate torques with limited angular movement; spiral for multiple revolutions.
  • Material: carbon steel for general use, stainless steel if there is risk of corrosion, piano wire for high strengths.
  • Allowable stress: apply safety factor ≥ 1.5 over the elastic limit, and correct for curvature (Wahl) in helical springs.
  • Mounting and space constraints: consider housing diameter, guide shafts or bushings, and attachment legs.
  • Service life: for fatigue (> 10⁵ cycles) limit stress to 60% of the tensile strength.

Mounting torsion springs requires specific precautions:

Aspect Recommendation
Winding direction Select right or left depending on the direction of rotation of the application; reverse if the load is reversed.
Clearance with mandrel or shaft For helical springs, the mandrel diameter should be ≤ 0.9 × spring inner diameter at rest, because when deflected the inner diameter decreases.
Lubrication Apply lithium grease or MoS₂ to the coils for springs working at high frequency or to reduce friction between coils and guides.
End fixation The legs must rest on flat surfaces or bushings with sliding tolerance; avoid stress concentration points using fillet radii.
Mounting on torsion bars Splines or keyways at both ends to transmit torque without slipping; provide sliding fit with lubricated spline.
Corrosion protection Zinc plating, phosphating, or epoxy paint protection in humid environments; in helical springs avoid galvanic contact with the shaft.

The following tables guide the selection of a torsion spring for common applications, considering EN 10270-1 spring steel and mounting on a shaft. Torques are calculated for a 90° deflection.

Application d (mm / in) Outer diameter (mm / in) No. of coils Torque at 90° (N·m / lbf·in) Free angle (°)
Clothespin 1.0 / 0.039 8 / 0.315 4 0.8 / 7.1 180
Small door hinge 1.5 / 0.059 12 / 0.472 6 2.5 / 22.1 120
Mousetrap 1.8 / 0.071 15 / 0.591 5 4.2 / 37.2 180
Electric access door 2.5 / 0.098 20 / 0.787 8 12.5 / 110.6 90
Timing chain tensioner 3.0 / 0.118 25 / 0.984 10 22.4 / 198.3 120
Sectional door counterweight 5.0 / 0.197 40 / 1.575 15 110.0 / 973.6 270

For steel torsion bars with τ_adm = 400 MPa / 58 000 psi:

Application D (mm / in) Active length (mm / in) Maximum torque (N·m / lbf·in) Maximum angular deflection (°)
Stabilizer bar (light automobile) 18 / 0.709 900 / 35.43 350 / 3098 5.2
Light truck suspension 25 / 0.984 1000 / 39.37 920 / 8145 6.4
Landing gear (light aircraft) 30 / 1.181 800 / 31.50 1600 / 14 160 7.3

What is the difference between a torsion bar and a helical torsion spring?

Section titled “What is the difference between a torsion bar and a helical torsion spring?”

The torsion bar is a solid or hollow element that works in pure shear; a helical spring, although twisted, works essentially in bending of the wire. The torsion bar supports high torques with small angles (typically 5–10°), while the helical allows deflections of up to 360° with moderate torques.

How is the minimum diameter of a torsion bar to transmit 200 N·m with τ_adm = 250 MPa calculated?

Section titled “How is the minimum diameter of a torsion bar to transmit 200 N·m with τ_adm = 250 MPa calculated?”

Applying D_min = 1,72·(T_max / τ_max)^(1/3). For T = 200 N·m and τ = 250 MPa (250×10⁶ Pa) we obtain D_min ≈ 1,72·(200 / 250×10⁶)^(1/3) = 1,72·(8×10⁻⁷)^(1/3) = 1,72·0,00928 m = 0,0160 m, i.e., 16 mm / 0.630 in.

What torsion angle does a solid steel shaft (G = 80 GPa) with D = 20 mm, L = 500 mm and a torque of 50 N·m reach?

Section titled “What torsion angle does a solid steel shaft (G = 80 GPa) with D = 20 mm, L = 500 mm and a torque of 50 N·m reach?”

Using α_deg ≈ 584·L·T / (G·D⁴) = 584·0.5·50 / (80×10⁹·(0.02)⁴) = 14600 / (80×10⁹·1.6×10⁻⁷) = 14600 / 12800 = 1.14°. In radians, 0.0199 rad.

How much energy does a torsion bar with stiffness κ = 5000 N·m/rad store when deflected by 15°?

Section titled “How much energy does a torsion bar with stiffness κ = 5000 N·m/rad store when deflected by 15°?”

First convert 15° to rad: 15 · π/180 = 0.262 rad. Energy U = ½ κ θ² = ½ · 5000 · (0.262)² ≈ 171.6 J / 126.6 ft·lbf.

What spring constant does a helical spring with d = 2.5 mm, D = 15 mm, N = 8 and E = 206 GPa have?

Section titled “What spring constant does a helical spring with d = 2.5 mm, D = 15 mm, N = 8 and E = 206 GPa have?”

κ = E·d⁴ / (64·D·N) = 206×10³ · (2.5⁴) / (64·15·8) ≈ 206×10³·39.06 / (7680) = 8.04×10⁶ / 7680 ≈ 1047 N·mm/rad = 1.047 N·m/rad. Equivalent to 0.0183 N·m/°.

What is the working torque of a clothespin spring with d = 1 mm, D = 8 mm, N = 4, deflected by 180°?

Section titled “What is the working torque of a clothespin spring with d = 1 mm, D = 8 mm, N = 4, deflected by 180°?”

Stiffness κ = 206×10³·1⁴ / (64·8·4) = 206×10³ / 2048 ≈ 100.6 N·mm/rad = 0.1006 N·m/rad. Angle 180° = π rad ≈ 3.14 rad. Torque M = κ·θ = 0.1006·3.14 ≈ 0.316 N·m / 2.80 lbf·in, consistent with the table.