KEVOS® MAKEOS · Springs
Disc Spring Forces, Stresses and Fatigue Life (Corrected Outside Diameter, Critical Points, Life Equations)
For a stack of disc (Belleville) springs without contact surfaces working between a minimum and a maximum deflection, what forces do they give, what are the stresses at the critical points, which point governs fatigue, how many cycles will they last, and are the dimension ratios sensible?
Disc springs without contact surfaces (Groups 1 and 2, thickness up to 6 mm) by the handbook method: the cone angle tan θ = 2h ÷ (D − d), the corrected outside diameter Da = D − 2t (sin θ + cos θ ÷ 6) for the contact circle, the constants K1, K2 and K3 from δ = D ÷ d, the force F = 4 E s [(h − s ÷ 2)(h − s) t + t³] ÷ [(1 − μ²) K1 Da²] at the minimum and maximum deflection, the compressive stresses at points 0 and 1 and the tensile stresses at points 2 and 3, the critical point by the larger of σmax − 0.5 σmin, the fatigue life N = 10^(a − b (σmax − 0.5 σmin)) for the group, the stress range allowed for the life you need, and the recommended ratios D ÷ d = 1.7 to 2.5, h ÷ t = 0.4 to 1.3 and D ÷ t = 18 to 40.
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How this tool works
Inputs, units and where to find them
| Input | Unit | Typical source |
|---|---|---|
| Outside diameter D | mm | Nominal outside diameter of the disc spring. |
| Inside diameter d | mm | Inside diameter of the disc spring. |
| Thickness t | mm | Thickness of the disc; this tool covers springs without contact surfaces up to 6 mm thick. |
| Cone height h (free height less thickness) | mm | Cone (dish) height of the unloaded spring, the free height less the thickness. |
| Modulus of elasticity E | MPa | Modulus of elasticity of the spring steel. |
| Poisson’s ratio μ | Poisson’s ratio of the spring steel. | |
| Deflections of one disc | ||
| Deflection at minimum load smin | mm | Deflection of one disc at the minimum load, including the preload. |
| Deflection at maximum load smax | mm | Deflection of one disc at the maximum load. |
| Discs in series in the stack | discs | Number of discs stacked in series (alternate directions). |
| Requirements | ||
| Fatigue life you require | cycles | Number of load cycles the stack must survive. |
| Largest compressive stress at point 0 you accept | MPa | Largest compressive stress at point 0 you accept; the handbook gives 1600 N/mm² for most spring steels. |
| Largest tensile stress at points 2 and 3 you accept | MPa | Largest tensile stress at points 2 and 3 you accept; the handbook gives 1300, 1250 and 1200 N/mm² for Groups 1, 2 and 3. |
What each input means
Open an input to see what it is, its units, the worked example and the common mistakes, with a small diagram.
What this means: Outside diameter D
- What it is
- Nominal outside diameter of the disc spring.
- Units and format
- Enter a number in mm: millimetres. The form accepts 0.000001 mm to 1000 mm. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 80 mm as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- Sets the diameter ratio and, through Da, the force.
- Get it right
- Choose it to fit the bore the stack works in, with clearance.
What this means: Inside diameter d
- What it is
- Inside diameter of the disc spring.
- Units and format
- Enter a number in mm: millimetres. The form accepts 0.000001 mm to 1000 mm. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 41 mm as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- Sets the diameter ratio δ = D ÷ d and the constants K1, K2 and K3.
- Get it right
- Choose it to clear the guide rod.
What this means: Thickness t
- What it is
- Thickness of the spring steel the disc is made from.
- Units and format
- Enter a number in mm: millimetres. The form accepts 0.2 mm to 6 mm. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 4 mm as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- Force rises steeply with thickness; it also sets the group for fatigue.
- Get it right
- Below 1.25 mm the spring is Group 1, from 1.25 to 6 mm Group 2.
What this means: Cone height h (free height less thickness)
- What it is
- Cone (dish) height of the unloaded disc.
- Units and format
- Enter a number in mm: millimetres. The form accepts 0.000001 mm to 100 mm. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 2.2 mm as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- Sets the shape of the force curve and how far the disc can deflect.
- Get it right
- The free height of the disc less its thickness.
What this means: Modulus of elasticity E
- What it is
- Modulus of elasticity of the spring steel.
- Units and format
- Enter a number in MPa: megapascals (1 MPa = 1 N/mm²). The form accepts 0.000001 MPa to 1000000 MPa. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 206000 MPa as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- Forces and stresses rise in proportion.
- Get it right
- About 206,000 MPa for spring steels.
What this means: Poisson’s ratio μ
- What it is
- Poisson’s ratio of the spring steel.
- Units and format
- Enter a plain number with no unit. The form accepts 0 to 0.5. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 0.3 as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- Changes forces and stresses slightly.
- Get it right
- Use 0.3 for spring steels.
Deflections of one disc
What this means: Deflection at minimum load smin
- What it is
- Deflection of one disc at the minimum load, including the preload.
- Units and format
- Enter a number in mm: millimetres. The form accepts 0 mm to 100 mm. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 0.9 mm as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- Sets the minimum stresses in the fatigue range.
- Get it right
- Stack deflection at minimum load ÷ discs in series; preload is usually 15 to 20 % of h.
What this means: Deflection at maximum load smax
- What it is
- Deflection of one disc at the maximum load.
- Units and format
- Enter a number in mm: millimetres. The form accepts 0.000001 mm to 100 mm. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 1.32 mm as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- Sets the maximum stresses and the force.
- Get it right
- Stack deflection at maximum load ÷ discs in series; it cannot exceed h.
What this means: Discs in series in the stack
- What it is
- Number of discs stacked in series, alternately facing.
- Units and format
- Enter a number in discs: discs. The form accepts 1 discs to 500 discs. Use a whole number. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 6 discs as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- Multiplies the travel, not the force.
- Get it right
- Unloaded stack length = discs × (t + h); it must fit the space.
Requirements
What this means: Fatigue life you require
- What it is
- Number of load cycles the stack must survive.
- Units and format
- Enter a number in cycles: cycles. The form accepts 10000 cycles to 2000000 cycles. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 1000000 cycles as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- Fails the springs when the life is shorter.
- Get it right
- The equations cover 10⁴ to 2 × 10⁶ cycles.
What this means: Largest compressive stress at point 0 you accept
- What it is
- Limit on the compressive stress on the top surface at the mean diameter (point 0).
- Units and format
- Enter a number in MPa: megapascals (1 MPa = 1 N/mm²). The form accepts 0.000001 MPa to 5000 MPa. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 1600 MPa as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- Fails the springs when the stress is higher.
- Get it right
- The handbook gives 1600 N/mm² for most spring steels.
What this means: Largest tensile stress at points 2 and 3 you ac…
- What it is
- Limit on the tensile stress at the bottom inner and outer edges (points 2 and 3).
- Units and format
- Enter a number in MPa: megapascals (1 MPa = 1 N/mm²). The form accepts 0.000001 MPa to 5000 MPa. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 1250 MPa as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- Fails the springs when the stress is higher.
- Get it right
- The handbook gives 1300 (Group 1) and 1250 N/mm² (Group 2).
Method and formulas
- θ = atan (2h ÷ (D − d)); Da = D − 2t (sin θ + cos θ ÷ 6).
- δ = D ÷ d; K1 = (1 ÷ π)((δ − 1) ÷ δ)² ÷ ((δ + 1) ÷ (δ − 1) − 2 ÷ ln δ); K2 = (6 ÷ π)((δ − 1) ÷ ln δ − 1) ÷ ln δ; K3 = (3 ÷ π)(δ − 1) ÷ ln δ.
- k = 4E ÷ ((1 − μ²) K1 Da²); F = k s [(h − s ÷ 2)(h − s) t + t³].
- σ0 = −(3 ÷ π) k t s; σ1 = −k s [K2 (h − s ÷ 2) + K3 t]; σ2 = k s [K3 t − K2 (h − s ÷ 2)]; σ3 = k s [(2K3 − K2)(h − s ÷ 2) + K3 t] ÷ δ.
- Critical point: the larger of σ2max − 0.5 σ2min and σ3max − 0.5 σ3min.
- Group 1 (t < 1.25 mm): N = 10^(10.29085532 − 0.00542096 r); Group 2: N = 10^(10.10734911 − 0.00537616 r); r allowed = (a − log N) ÷ b.
- Example: the critical stress range of 741.07 MPa gives a fatigue life of 1,328,100 cycles.
Limitations
- Disc springs with contact surfaces (all of Group 3 and some of Group 2) need the reduced-thickness method and are not covered.
- The fatigue equations cover limited lives from 10⁴ to 2 × 10⁶ cycles; at 2 × 10⁶ cycles or more the life is taken as unlimited.
- Friction between discs and on the guide is ignored; guide the stack on a hardened rod or in a sleeve.
- Example values are illustrations. Replace them with values for your actual task.
Typical uses
- Check a dynamically loaded disc spring stack for fatigue life
- Find the forces at the preload and working deflections
- Choose a disc spring whose stress range gives the life you need
Who it is for
- Machine designers
- Toolmakers
- Spring and fixture designers
What you get
- Corrected outside diameter Da and constants K1, K2, K3
- Forces at the minimum and maximum deflection
- Stresses at points 0, 1, 2 and 3
- Critical point and fatigue life
- Stress range allowed for the life you need
- Dimension ratio checks and stack lengths
Interpreting the result
- The critical point for fatigue is on the bottom surface, at the inside (point 2) or outside (point 3) edge.
- With h ÷ t above 1.3 the discs in a series stack may deflect unequally; use a single disc where possible.
References
- Machinery’s Handbook, 29th edition (Industrial Press, 2012), printed pp. 354–364. Topic reference; no source tables or recommended settings are reproduced.
Learn the Concepts Behind This Tool
These KEVOS® Library readings explain the method this tool uses, with worked examples.
Useful alongside · Library collection
Specifying springs that work, last and are easy to make and assemble
In Machine Elements and Mechanical Design
What it covers: Springs look simple, yet poorly specified ones sag, break, jam in bores and slow assembly. How to calculate rate and stress, choose materials and ends, and specify springs makers can build.
Useful alongside · KEVOS® original book
Disc spring stacks: large forces in short spaces
What it covers: How disc (Belleville) springs carry large forces in little space: the Almen–László characteristic, series and parallel stacking, stack design for force and travel, and stresses.
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