Wave Spring Worked Example

We begin from first principles:

$$\text{Spring Rate}\left(k\right)=\frac{\text{Load}\left(P\right)}{\text{Deflection}\left(s\right)}$$

And for our purposes here, Load is no different to Force.

$$\frac{\text{Force}\left(P\right)}{\text{Deflection}\left(s\right)}=\frac{E·b·t^3·N^4·D_o}{\left(k\right)·D^3·D_i}$$

The equations used for the calculations of Load, Deflection and Stress are derived from those for a simple beam and have the application of some corrective factors based on empirical measurements and experience.

These equations for loads & stresses, are not precise solutions, but they do provide very useful engineering estimates. As a measure of usefulness, they are more accurate and useful than any of the forecasting models Economists use to explain Recessions, but not nearly as accurate as Cosmologists employ to calculate the Hubble Constant.

Some skepticism should be expected if Forces are provided in Newtons with decimal places.

We can now derive various useful metrics, such as:

1. Required Material Thickness - t, to support a Load: $$t=\sqrt[3]{\frac{\left(k\right)·P·D^3·D_i}{E·s·b·N^4·D_e}}$$
2. Deflection - s, expected from a Load: $$s=\frac{\left(k\right)·P·D^3·D_i}{E·b·t^3·N^4·D_e}$$
3. Induced stress - σ, created by a Load: $$\mathrm{\sigma }=\frac{3·\mathrm{\pi }·P·D}{4·b·t^2·N^2}$$
4. Deflection - s, expected at stress threshold: $$s=\frac{\left(4\right)·\left(k\right)·D^2·\mathrm{\sigma }·D_i}{\left(3\right)·\mathrm{\pi }·E·t·N^2·D_e}$$

Our Task is to select and verify a wave washer which is needed to fit into an 80mm (3.15”) bore and over a 60mm (2.362”) shaft to support a load of between 500 and 550 Newtons (i.e. 112lbs to 124lbs). The Wave Washer Spring must accommodate at least a 1.8 mm (0.071”) deflection. Furthermore the application requires a steady load. The washer will operate in a benign environment where the ambient temperature will never exceed 30°C.

So, here we have a static application. Let us assume a standard single wave washer spring will do the job to start off with. Go to the menu, Our Products → Wave Springs → Single Turn Wave Springs Sizes and see what there is that might cater for our needs.

You should find a 3 Wave Spring of dimension ID: 61.5mm x OD: 79.5mm x t:0.4mm and the Nominal Load stated as 30N at Working Height / H_Load of 3.0mm on Free Height of 6.5mm. The diameter dimensions look OK, deflection is well over the minimum of 1.8mm, but the Design Load is way too low, can we make an easy change to get what we require? It looks like we will have to add a wave or two or change the material thickness.

Start by checking the material thickness t, if we changed it from 0.4mm to 0.6mm or 0.8mm, what would that do for us?

Using, the following equation, see if we get a reasonable material thickness, before considering adding waves:

$$t=\sqrt[3]{\frac{\left(k\right)·P·D^3·D_i}{E·s·b·N^4·D_e}}$$

We Substitute into the equation above, where:

• N = 3, because we have 3 waves
• k = 3.88 because we have N = 3
• P = 525N, mid way between the required Load range of 500 to 550 Newtons
• D = average of 79.5 and 61.5 = 70.5mm
• b = 9.0mm
• s = 2.0mm because we require at least 1.8mm of Working Travel

Giving us:

$$t=\sqrt[3]{\frac{\left(3.88\right)·\left(525\right)·{\left(70.5\right)}^3·\left(61.5\right)}{\left(\mathrm{207\hspace{0.17em}000}\right)·\left(2\right)·\left(9.0\right)·\left(3^4\right)·\left(79.5\right)}}$$

and you should get:

$$t^3=\frac{42.9\times {10}^9}{24.0\times {10}^9}=1.83,\text{so}t=\sqrt[3]{1.83}\approx 1.22\text{mm}$$

We round to a practical strip thickness of t = 1.2 mm. At that thickness the theoretical spring rate is about 248 N/mm, so at the design deflection s = 2.0 mm the theory load is about 496 N (stress about 706 MPa). That is a little below the mid-target of 525 N, but well inside the usual ±15% theory-versus-test band quoted on the intro page, and still inside the 500 to 550 N requirement once manufacturing and test variation are allowed for. (Solving exactly for 525 N at 2.0 mm would give t ≈ 1.22 mm.)

Deflection uses the standard convention s = L₀ − H. With free height L₀ = 6.5 mm and design travel s = 2.0 mm, the working height is H_L = 4.5 mm. The thicker strip still fits the original free-height envelope.

It is a good operating principle to ensure that deflection/travel is maintained between 30% and 70% of Free Height and always make sure that there is adequate clearance between your constraining bore and the loaded Wave Spring. Remember that the diameter of a Wave Spring increases when compressed. The following formula provides its maximum achievable external diameter when compressed to its design/working load.

$${\mathrm{D\text{'}}}_L=\sqrt{D_e^2+\left(0.458\right)·s^2·N^2}$$

At s = 2.0 mm this gives an increased D'_e of about 79.60 mm, so about 0.40 mm total clearance to the 80 mm bore (roughly 0.20 mm per side), which is acceptable.