This document provides specifications for two disc spring designs. It includes dimensions, material properties, calculations of theoretical spring force and stresses for different load cases. Key details include: - Design 1 has an outside diameter of 63mm, inside diameter of 31mm, thickness of 3.5mm, and is made of a material with Young's modulus of 206,000 N/mm^2. - Design 1 can withstand a maximum force of 15,030N over a stroke of 23.1mm when using 22 springs in series. - Design 2 has diameters of 28mm and 14.2mm, thickness of 1mm, and can provide 617.58N of force over a stroke of 0.

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DISC SPRING DESIGN:
Young’s modulus, E = 206000 N/mm2
Poisson’s ratio,  = 0.3
Outside diameter of disc spring, De = 63 mm
Inside diameter of disc spring, Di = 31 mm
Ratio, De/Di = 2.032
 =
i
e
D
D
K1 = (-1)
 x 1
(+1) _ 2 
(-1) ln ()
K1 = 0.701
K2 = 6 x  (-1) _ 1 
ln ()
 x ln ()
K2 = 1.227
K3 = 6 x  (-1) _ 2 
ln ()
 x ln ()
K3 = 1.39
t = 3.5 mm
S = 1.05 mm
ho = 1.4 mm
F = 4 x E x S x t__ x (ho2 + t2 + 0.5S2 –1.5ho. S)
(1-2) x K1 x De2

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F = 1.503x104 N
For S = ho The theoretical force
Spring force
Fc = 4 x E x hox t3
(1-2 ) x K1 x De2
Fc = 1.955x104 N
Stresses
1 = 4 x E x S x (-K2 x (ho – S) – (K3 x t))
(1-2) x K1 x De2 2
1 = -2.03x103 N/mm2
2 = 4 x E x S x (-K2 x (ho – S) + (K3 x t))
(1-2) x K1 x De2 2
2 = 1.296x103 N/mm2
3 = 4 x E x S x ((2k3-K2) x (ho – S) + (K3 x t)) x Di
(1-2) x K1 x De2 2 De
3 = 1.047x103 N/mm2
4 = 4 x E x S x ((2k3-K2) x (ho – S) - (K3 x t)) x Di
(1-2) x K1 x De2 2 De
4 = -1.112x103 N/mm2
Spring are in series
Number of springs, n = 22
Available stroke, Sa = Sxn
Sa = 23.1 mm
Available spring force for Sa, F = 1.503x104 N
Total stroke, St = 6 mm

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Deflection per spring, S1 = St/n
S1 = 0.273 mm
E = 206000 N/mm2
 = 0.3
De = 28 mm
Di = 14.2 mm
De/Di = 1.972
 = De/Di
K1 = (-1)
 x 1
(+1) _ 2 
(-1) ln ()
K1 = 0.689
K2 = 6 x  (-1) _ 1 
ln ()
 x ln ()
K2 = 1.213
K3 = 6 x  (-1) _ 2 
ln ()
 x ln ()
K3 = 1.367
t = 1 mm
ho = 0.8 mm
F = 4 x E x S1 x t__ x (ho2 + t2 + 0.5S12 –1.5ho.S1)
(1-2) x K1 x De2
F = 617.582 N

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For S1 = ho The theoretical force
Spring force
Fc = 4 x E x hox t3
(1-2 ) x K1 x De2
Fc = 1.342x103 N
Stresses
1 = 4 x E x S x (-K2 x (ho – S) – (K3 x t))
(1-2) x K1 x De2 2
1 = -993.705 N/mm2
2 = 4 x E x S x (-K2 x (ho – S) + (K3 x t))
(1-2) x K1 x De2 2
2 = 256.939 N/mm2
3 = 4 x E x S x ((2k3-K2) x (ho – S) + (K3 x t)) x Di
(1-2) x K1 x De2 2 De
3 = 551.22 N/mm2
4 = 4 x E x S x ((2k3-K2) x (ho – S) - (K3 x t)) x Di
(1-2) x K1 x De2 2 De
4 = -319.965 N/mm2

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