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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

Preload F

Final Load F

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:

s/h _{o} |
s[mm] |
F[N] |
s[N/mm ^{2}] |

0.25 | 0.325 | 1524 | 433 |

0.5 | 0.650 | 2701 | 814 |

0.75 | 0.980 | 3659 | 1148 |

1.0 | 1.300 | 4475 | 1421 |

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):

S_{1} = 0.34mm, S_{2} = 0.64mm

σ_{u} = 450 N/mm^{2}

σ_{o}= 804 N/mm^{2}

From the fatigue diagram for group 2 springs figure 19, we obtain σ_{u }450 N/mm^{2} with a maximum stress of σ_{0}= 920 N/mm^{2}. Therefore the spring is fatigue resistant as σ_{o} < σ_{0}.

**Figure 12 - click to enlarge
**

**Example 2: Disc spring with a high h _{o}/t ration**

**Given:
**Guide diameter 30mm

Installed length l

Preload F

Working defl. s

Spring load F

**Required:
**Suitable disc spring dimensions

**Solution:
**Spring inside diameter D

Spring outside diameter D

Because of the very small load range and the small installed length only a spring with a high h

**Selected:
**Disc spring 60 x 30.5 x 1.5mm (figure 13); l

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

K_{1} = 0.688

K_{2} = 1.212

K_{3 }= 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/mm^{2}.

This value lies well under the limit of -1600 N/mm^{2}, 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:

s/h _{o} |
s[mm] |
F[N] |

0.25 | 0.5 | 1338 |

0.5 | 1.0 | 2058 |

0.75 | 1.5 | 2367 |

1.0 | 2.0 | 2469 |

With these 4 points the spring diagram can be drawn.

**Figure 14 - Click to enlarge
**

One can read F_{1} = 2100 N s_{1} = 1.05mm

and for F_{2} = 2400 N s_{2} = 1.61mm

Deflection s_{2} – s_{1}= 0.56mm

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

**This arrangement gives:**

Unloaded length | L_{o}=7.0mm |

Preloaded length | L_{1}=4.90mm |

Preloaded deflection | s_{1}=2.1mm |

Preload | F_{1}=2100N |

Deflection | s_{2}=s_{1}+1.05=3.15mm |

Final load | F_{2}=2390N |

To check the fatigue life we must use the stresses at s_{1} = 1.05 and s_{2} = 1.575mm. Figure 17 shows that point lll is the dominant one, this gives:

s_{1}: σ_{u} = 843 N/mm^{2}

s_{2}: σ_{u} = 1147 N/mm^{2 }

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

H

**Required:
**The spring characteristics and the stresses σ

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 K

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/mm^{2}

As the acceptable value for σ_{OM} is 1600 N/mm^{2}, 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 K_{4 }can be calculated from formula 6: K_{4} = 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.

s/h_{o} |
s[mm] |
F[N] |
[N/mm^{2}] |
[N/mm^{2}] |

0.25 | 1.15 | 66924 | 416 | 389 |

0.5 | 2.3 | 127191 | 890 | 747 |

0.75 | 3.45 | 182838 | 1421 | 1072 |

1.0 | 4.6 | 235503 | 2011 | 1366 |

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} . K_{4} . t’/t

σ_{OM}’ = -1595 N/mm^{2}

^{}

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