calculation examples

The ‘life lines’ also allow the fatigue life to be estimated for various working strokes. In spite of this we show several examples of the calculation and checking of disc springs below.

Example 1: Checking fatigue life of a disc spring

Given:
Spring 45 x 22.4 x 1.75; l_o = 3.05mm (figure 11)
Preload F₁ = 1580N
Final Load F₂ = 2670N
Frequency f = 1000/min

To be determined:
Is the stress within the acceptable range – what is the estimated fatigue life.

Solution:
From the tables of disc springs we can obtain the following data:

With the help of these four points the load and stress relative to the deflection may be drawn.

Figure 11 - click to enlarge
Disc spring 45 x 22.4 x 1.75; l_o = 3.05 mm

The following values may be obtained from the diagram (figure 12): 

S₁ = 0.34mm, S₂ = 0.64mm 
σ_u = 450 N/mm²
σ_o= 804 N/mm² 

From the fatigue diagram for group 2 springs figure 19, we obtain σ_u 450 N/mm² with a maximum stress of σ₀= 920 N/mm². Therefore the spring is fatigue resistant as σ_o < σ₀.         

Figure 12 - click to enlarge
Diagram for spring 45 x 22.4 x 1.75 mm, l_o = 3.05 mm  

Example 2: Disc spring with a high h_o/t ration

Given:
Guide diameter 30mm
Installed length l₁ = 4,9mm
Preload F₁ = 2000 N min.
Working defl. s₂ – s₁ = 1.05mm
Spring load F₂ = 2500 N max.

Required:
Suitable disc spring dimensions 

Solution:
Spring inside diameter D_i = 30.5mm
Spring outside diameter D_e = 60mm (selected because of the favourable D_e/D_i ratio)
Because of the very small load range and the small installed length only a spring with a high h_o/t ratio is suitable. 

Selected:
Disc spring 60 x 30.5 x 1.5mm (figure 13); l_o = 3.5mm h_o/t = 1.333; δ = 1.967 

Calculation:
First the factors are calculated using formula 3, 4 and 5: 

K₁ = 0.688 
K₂ = 1.212 
K₃ = 1.365

Figure 13 - click to enlarge
Disc spring 60 x 30.5 x 1.5 mm

The stress σ_OM can be checked using formula 9: σ_OM = -1048 N/mm².

This value lies well under the limit of -1600 N/mm², the spring will therefore not set. Now the spring loads can be calculated to formula 8a, preferably for the 4 deflections s = 0.25 h_o, s = 0.5 h_o, s = 0.75 h_o and s = h_o.

One obtains the following values:

With these 4 points the spring diagram can be drawn.

Figure 14 - Click to enlarge
Diagram for spring 60 x 30.5 x 1.5 mm, l_o = 3.5 mm

One can read F₁ = 2100 N s₁ = 1.05mm
and for F₂ = 2400 N s₂ = 1.61mm
Deflection s₂ – s₁= 0.56mm 

The deflection of a single spring is not sufficient, therefore two in series must be used.

This arrangement gives:

To check the fatigue life we must use the stresses at s₁ = 1.05 and s₂ = 1.575mm. Figure 17 shows that point lll is the dominant one, this gives:

s₁: σ_u = 843 N/mm² 
s₂: σ_u = 1147 N/mm²

By utilising the fatigue life diagram in figure 19 we can see that the expected life will be in the order of 1,000,000 cycles. 

Example 3: Calculation of a disc spring with contact flats

Given:
Disc spring 200 x 82 x 12mm; l_o = 16.6mm
H_o = 4.6mm; δ = 2.439; h_o/t = 0.383 

Required:
The spring characteristics and the stresses σ_ll and σ_lll. 

Although this spring is to our works standard we show below how the calculations are made and results can be checked in the disc springs dimension tables.

From the formula 3 to 5 we first calculate the constants K₁ to K₃:

K1 = 0.755 
K2 = 1.315 
K3 = 1.541

The static design can be checked by the calculation of σ_OM, the reduced thickness is not considered and we use the values of t and h_o. This gives: 
 
σ_OM = -1579 N/mm²

As the acceptable value for σ_OM is 1600 N/mm², the spring is correct. From figure 6 and considering d and h_o/t the reduction factor t’/t can be obtained: 
 
t’/t = 0.958

Therefore t’ = 11.5mm and h_o’ = 5.1mm. Constant K₄ can be calculated from formula 6: K₄ = 1.0537.

Figure 15 - Click to enlarge
Disc spring 200 x 82 x 12 mm

Figure 16 - click to enlarge
Spring force and stresses for spring 200 x 82 x 12 mm, t’ = 11.5 l0 = 6.6 mm

Now from formula 8b, 11 and 12 the spring force and both stresses can be calculated.

With this spring the greater values of stress are on the inner diameter which should be used. Finally the value of the stress σ_OM for the reduced thickness can be checked:  

σ_OM’ = σ_OM . K₄ . t’/t 
σ_OM’ = -1595 N/mm²