Dimensions for a selection of Belleville springs manufactured to DIN 6796 are given in Table 15.6.
Example 15.7
A Belleville spring is required to give a constant force of 200 N ± 10 N over a deflection of ± 0.3 mm. The spring must fit within a 62 mm diameter hole. A carbon spring steel with σ_{uts} = 1700 MPa has been proposed.
Solution
Assume a 60 mm outer diameter to allow some clearance in the hole.
To provide a constant force, an h/t ratio of 1.414 is selected.
The variation of force of ± 5% can be met by choosing an appropriate deflection range to operate in from Fig. 15.33. If the deflection is limited to between 65% and 135% of the flat deflection, then the tolerance on force can be achieved. The nominal force of 200 N will occur in the flat position and the spring will provide a similar force, between 210 N and 190 N, operating on both sides of its centre.
From Eq. (15.54),
$t=\frac{1}{1072}{\left(\frac{{D}_{o}^{2}{F}_{\text{flat}}}{h/t}\right)}^{0.25}=\frac{1}{1072}{\left(\frac{{0.06}^{2}\times 200}{1.414}\right)}^{0.25}=7.88\times {10}^{4}\phantom{\rule{0.5em}{0ex}}\mathrm{m}$
h = 1.414 t = 1.414 × 0.788 = 1.114 mm.
The minimum and maximum deflections are.
δ_{min} = 0.65 h = 0.65 × 1.114 = 0.724 mm.
δ_{max} = 1.35 h = 1.35 × 1.114 = 1.504 mm.
δ_{max} − δ_{min} is greater than the required deflection range of 0.6 mm so the force tolerance can be met.
From Eqs (15.44) and (15.50)–(15.53),
K_{1} = 0.689,
K_{2} = 1.220,
K_{3} = 1.378,
K_{4} = 1.115,
K_{5} = 1.
From Eq. (15.47), σ_{c} = − 840 MPa.
From Eq. (15.48), σ_{ti} = 355 MPa.
From Eq. (15.49), σ_{to} = 658 MPa.
These stresses are well within the capability of a 1700 MPa uts material.
Example 15.8
A Belleville spring is required to give a constant force of 50 N ± 5 N over a deflection of ± 0.2 mm. The spring must fit within a 40 mm diameter hole. A carbon spring steel with σ_{uts} = 1700 MPa has been proposed.
Solution
Assume a 35 mm outer diameter to allow some clearance in the hole.
To provide a constant force, a h/t ratio of 1.414 is selected.
The variation of force of ± 10% can be readily met by choosing an appropriate deflection range to operate in from Fig. 15.33.
If the deflection is limited to between 65% and 135% of the flat deflection, then the tolerance on force can be achieved.
The nominal force of 50 N will occur in the flat position and the spring will provide a similar force, between 55 N and 45 N, operating on both sides of its centre.
From Eq. (15.52),
$t=\frac{1}{1072}{\left(\frac{{D}_{o}^{2}{F}_{\text{flat}}}{h/t}\right)}^{0.25}=\frac{1}{1072}{\left(\frac{{0.035}^{2}\times 50}{1.414}\right)}^{0.25}=4.26\times {10}^{4}\phantom{\rule{0.5em}{0ex}}\mathrm{m}$
h = 1.414 t = 1.414 × 0.426 = 0.602 mm
The minimum and maximum deflections are
δ_{min} = 0.65 h = 0.65 × 0.602 = 0.391 mm
δ_{max} = 1.35 h = 1.35 × 0.602 = 0.812 mm
δ_{max} − δ_{min} is greater than the required deflection range of 0.4 mm so the force tolerance can be met.
From Eqs (15.44) and (15.50)–(15.53),
K_{1} = 0.689
K_{2} = 1.220
K_{3} = 1.378
K_{4} = 1.115
K_{5} = 1
From Eq. (15.47), σ_{c} = − 720 MPa
From Eq. (15.48), σ_{ti} = 305 MPa
From Eq. (15.49), σ_{to} = 564 MPa.
These stresses are well within the capability of a 1700 MPa uts material.
Example 15.9
A Belleville spring is required to give a constant force of 10 N ± 1 N over a deflection of ± 0.15 mm. The spring must fit within a 16 mm diameter hole. A carbon spring steel with σ_{uts} = 1700 MPa has been proposed.
Solution
Assume a 14 mm outer diameter to allow some clearance in the hole.
To provide a constant force, a h/t ratio of 1.414 is selected.
The variation of force of ± 10% can be met by choosing an appropriate deflection range to operate in from Fig. 15.33.
If the deflection is limited to between 65% and 135% of the flat deflection, then the tolerance on force can be achieved.
The nominal force of 10 N will occur in the flat position and the spring will provide a similar force, between 11 N and 9 N, operating on both sides of its centre.
From Eq. (15.52),
$t=\frac{1}{1072}{\left(\frac{{D}_{o}^{2}{F}_{\text{flat}}}{h/t}\right)}^{0.25}=\frac{1}{1072}{\left(\frac{{0.014}^{2}\times 10}{1.414}\right)}^{0.25}=1.8\times {10}^{4}\phantom{\rule{0.12em}{0ex}}\mathrm{m}$
h = 1.414 t = 1.414 × 0.18 = 0.255 mm
The minimum and maximum deflections are
δ_{min} = 0.65 h = 0.65 × 0.255 = 0.165 mm
δ_{max} = 1.35 h = 1.35 × 0.255 = 0.344 mm
From Eqs (15.44) and (15.50)–(15.53),
K_{1} = 0.689
K_{2} = 1.220
K_{3} = 1.378
K_{4} = 1.115
K_{5} = 1
From Eq. (15.47), σ_{c} = − 810 MPa
From Eq. (15.48), σ_{ti} = 341 MPa
From Eq. (15.49), σ_{to} = 630 MPa
These stresses are well within the capability of a 1700 MPa uts material.
Table 15.6. Dimensions for a selection of Belleville washer springs manufactured to DIN 6796 from DIN 17222 spring steel
Notation 
D_{i} (mm) 
D_{o} (mm) 
h′ max (mm) 
h′ min (mm) 
t (mm) 
Force (N)^{a}

Test force (N)^{b}

Mass kg/1000 
Core diameter (mm) 
2 
2.2 
5 
0.6 
0.5 
0.4 
628 
700 
0.05 
2 
2.5 
2.7 
6 
0.72 
0.61 
0.5 
946 
1100 
0.09 
2.5 
3 
3.2 
7 
0.85 
0.72 
0.6 
1320 
1500 
0.14 
3 
3.5 
3.7 
8 
1.06 
0.92 
0.8 
2410 
2700 
0.25 
3.5 
4 
4.3 
9 
1.3 
1.12 
1 
3770 
4000 
0.38 
4 
5 
5.3 
11 
1.55 
1.35 
1.2 
5480 
6550 
0.69 
5 
6 
6.4 
14 
2 
1.7 
1.5 
8590 
9250 
1.43 
6 
7 
7.4 
17 
2.3 
2 
1.75 
11,300 
13,600 
2.53 
7 
8 
8.4 
18 
2.6 
2.24 
2 
14,900 
17,000 
3.13 
8 
10 
10.5 
23 
3.2 
2.8 
2.5 
22,100 
27,100 
6.45 
10 
12 
13 
29 
3.95 
3.43 
3 
34,100 
39,500 
12.4 
12 
14 
15 
35 
4.65 
4.04 
3.5 
46,000 
54,000 
21.6 
14 
16 
17 
39 
5.25 
4.58 
4 
59,700 
75,000 
30.4 
16 
18 
19 
42 
5.8 
5.08 
4.5 
74,400 
90,500 
38.9 
18 
20 
21 
45 
6.4 
5.6 
5 
93,200 
117,000 
48.8 
20 
22 
23 
49 
7.05 
6.15 
5.5 
113,700 
145,000 
63.5 
22 
24 
25 
56 
7.75 
6.77 
6 
131,000 
169,000 
92.9 
24 
27 
28 
60 
8.35 
7.3 
6.5 
154,000 
221,000 
113 
27 
30 
31 
70 
9.2 
8 
7 
172,000 
269,000 
170 
30 